Chapter 6: Chemical Kinetics
In the early 1970s, two chemists in California — Mario Molina and F. Sherwood Rowland — asked a simple question about a set of very unexciting compounds. Chlorofluorocarbons, then used in refrigerators and spray cans, were prized precisely because they were unreactive. Nothing seemed to attack them. So where did they finally go?
Their answer, published in 1974, was unsettling. Being so stable, these molecules drift slowly upward until they reach the stratosphere, where strong ultraviolet radiation splits them and sets free chlorine atoms. A single chlorine atom then attacks ozone, is regenerated in the very next step, and attacks again — thousands of times over, as sketched in Fig. 6.0. The compounds themselves were harmless. What mattered was how fast, and by what route, they reacted once they got there.
Diagram
Fig. 6.0 – Opening illustration: a CFC molecule rising from ground level to the stratosphere, split by a UV photon to release a Cl atom; beside it a small closed cycle showing Cl + O₃ → ClO + O₂ and ClO + O → Cl + O₂, with the Cl atom labelled “regenerated, not consumed”.
A decade later the Antarctic ozone hole was measured, and in 1987 the Montreal Protocol began phasing these compounds out. In 1995 Molina and Rowland shared the Nobel Prize in Chemistry with Paul Crutzen. The science that made their warning possible was not about which reactions can happen — thermodynamics had already answered that — but about how rapidly they happen, and through which steps. That science is chemical kinetics, and it is the subject of this chapter.
As you work through this chapter, keep these questions in mind — each one will be answered as the ideas behind it are developed.
6.1 Introduction
Three Questions About Every Reaction
Whenever a chemist looks at a chemical reaction, three separate questions have to be answered about it: how far the reaction goes, whether it can happen at all under the given conditions, and how fast it happens. These three characteristics are called the extent of the reaction, its feasibility, and its rate.
You have already met the first two. In Standard XI you learnt that the equilibrium constant of a reaction predicts its extent — that is, how much of the reactants will have turned into products by the time the reaction settles down and stops changing. A large equilibrium constant means the reaction proceeds far towards the products; a small one means very little product is formed.
In Unit 3 of this textbook you learnt that thermodynamic properties such as the change in entropy or the change in enthalpy tell us whether a chemical reaction, written as a chemical equation, will occur under a given set of conditions. This is the question of feasibility: thermodynamics decides whether a reaction is allowed to go forward.
That leaves the third question untouched, and it is a serious gap. Thermodynamics can declare a reaction perfectly feasible and still say nothing whatever about the time it will take. A mixture of hydrogen and oxygen gas is thermodynamically unstable with respect to water, yet the two gases can be kept together in a flask for years with no visible change — until a spark is applied, when the reaction is over in an instant. Feasibility and speed are clearly different questions, and the second needs a branch of chemistry of its own.
The three questions are independent of one another, and it is worth fixing the order in which a chemist asks them, as set out in FC 6.1.
Flowchart
FC 6.1 – The three questions asked about any chemical reaction: balanced chemical equation → Is it feasible under these conditions? (thermodynamics: ΔH, ΔS) → If yes, how far does it go? (equilibrium constant) → How fast does it go, and by what route? (chemical kinetics — this chapter) → answers: rate, rate law, mechanism
Why a Chemist Wants to Know Rates
A chemist wants to know the rates of reactions for more than one reason. Firstly, the study of reaction rates helps us to predict how rapidly a reaction approaches equilibrium. Knowing that a reaction will reach equilibrium is of little practical use if reaching it takes a century; the rate tells us whether the process is worth running at all.
Secondly, rate measurements give information about the mechanism of chemical reactions. The mechanism is the actual route the reaction follows, and it is one of the few things about a reaction that cannot be read off the balanced equation. Rate data is the main experimental evidence a chemist has for working it out.
Reactions Happen in Steps
A number of reactions occur as a sequence of elementary steps, and this sequence of steps constitutes the mechanism of the reaction. In other words, the balanced equation you write is often only a summary — an accounting statement of what went in and what came out. The molecules themselves may have travelled there through two, three or more separate encounters, each one a simple event in its own right.
This idea will be developed carefully later in the chapter. For now, hold on to the picture: a chemical equation tells you the destination, kinetics tells you the road taken and the time on the road.
6.2 Rate of Reactions
The rate of a reaction describes how rapidly the reactants are consumed or the products are formed. It is the chemical equivalent of speed: just as the speed of a car tells you how much distance is covered in a given time, the rate of a reaction tells you how much chemical change occurs in a given time.
The quantity that changes during a reaction is concentration. Reactant concentrations fall as the reaction proceeds, product concentrations rise, and the rate is simply a measure of how quickly this happens.
6.2.1 Average Rate of Chemical Reaction
The average rate of a reaction can be described by knowing the change in concentration of a reactant or a product, divided by the time interval over which the change occurs. Thus,
\[\text{Average rate} = \frac{\text{change in concentration of a species}}{\text{change in time}} = \frac{\Delta c}{\Delta t}\]
Here \(\Delta c\) is the change in concentration, measured in \(\text{mol dm}^{-3}\), and \(\Delta t\) is the time interval over which that change is observed. The Greek letter \(\Delta\) (delta) always means “final value minus initial value”, so its sign carries information — and that sign needs care.
Reading the Minus Sign
Consider the reaction
\[\mathrm{A} \rightarrow \mathrm{B}\]
in which A is consumed and B is produced. Since A is being used up, \([\mathrm{A}]\) falls with time and \(\Delta[\mathrm{A}]\) is a negative quantity. A rate, however, is a measure of how fast something is happening and is always reported as a positive number. A minus sign is therefore placed in front of the reactant term, so that the two negatives cancel:
\[\text{average rate of consumption of A} = -\frac{\Delta[\mathrm{A}]}{\Delta t}\]
For the product B the concentration rises, \(\Delta[\mathrm{B}]\) is already positive, and no correction is needed:
\[\text{Average rate of formation of B} = +\frac{\Delta[\mathrm{B}]}{\Delta t}\]
Because one molecule of A gives exactly one molecule of B here, whatever is lost by A appears as B. The two expressions therefore describe the same single rate:
\[\text{average rate of reaction} = -\frac{\Delta[\mathrm{A}]}{\Delta t} = +\frac{\Delta[\mathrm{B}]}{\Delta t}\]
The dimensions of rate follow directly from this definition: rate is a concentration divided by a time. Since concentration is measured in \(\text{mol dm}^{-3}\) and time in seconds, the dimensions of rate are \(\text{mol dm}^{-3}\ \text{sec}^{-1}\).
6.2.2 Instantaneous Rate of Reaction
An average rate has a serious limitation: it smooths out everything that happens inside the time interval chosen. To determine the instantaneous rate of a reaction, the progress of the reaction is followed by measuring the concentrations of reactant or product for different time intervals.
The changes in concentration are relatively fast in the beginning, and later become slow. This is why an average taken over a long interval is misleading — the reaction was not going at that speed throughout. The concentration of a reactant or of a product plotted against time is shown in Fig. 6.1 (a) and 6.1 (b), and you can see the curvature that carries this information.
Diagram
Fig. 6.1 – Determination of instantaneous rate. (a) Reactant concentration vs time: a curve falling steeply at first and flattening later, with a tangent drawn to the curve at time t₁; label the axes “reactant concentration” and “time”, mark t₁ on the time axis. (b) Product concentration vs time: a curve rising steeply at first and levelling off later, with a tangent drawn at the same time t₁; label the axes “product concentration” and “time”.
A tangent drawn to the curve at time \(t_1\) gives the rate of the reaction at that moment. The slope thus obtained gives the instantaneous rate of the reaction at time \(t_1\). A tangent is simply the straight line that just touches the curve without cutting across it, and its steepness matches the steepness of the curve at exactly that point — which is why its slope is the rate at that instant and not an average over a stretch of time.
The instantaneous rate \(dc/dt\) is represented by replacing \(\Delta c/\Delta t\) by the derivative \(dc/dt\) in the expression of average rate. The switch from \(\Delta\) to \(d\) is a switch from “change over a measurable interval” to “change over an interval so small that the curve has no time to bend”. In chemical kinetics we are concerned with instantaneous rates.
For the reaction \(\mathrm{A} \rightarrow \mathrm{B}\),
\[\text{Rate of consumption of A at any time } t = -\frac{d[\mathrm{A}]}{dt}\]
\[\text{Rate of formation of B at any time } t = \frac{d[\mathrm{B}]}{dt}\]
\[\text{Rate of reaction at time } t = -\frac{d[\mathrm{A}]}{dt} = \frac{d[\mathrm{B}]}{dt}\]
When Stoichiometry Is Not One-to-One
For the reaction involving one mole of A and B each, the rate of consumption of A equals the rate of formation of B. This is not true for the reactions involving different stoichiometries. Consider, for example, a reaction:
\[\mathrm{A} + 3\,\mathrm{B} \rightarrow 2\,\mathrm{C}\]
When one mole of A and three moles of B are consumed, two moles of C are formed. The stoichiometric coefficients of the three species are different. Thus the rate of consumption of B is three times the rate of consumption of A. Likewise the rate of formation of C is twice the rate of consumption of A. We write,
\[-\frac{d[\mathrm{B}]}{dt} = -3\,\frac{d[\mathrm{A}]}{dt} \qquad \text{and} \qquad \frac{d[\mathrm{C}]}{dt} = -2\,\frac{d[\mathrm{A}]}{dt}\]
This creates an awkward situation. Three different species are changing at three different speeds, so “the rate of the reaction” would have three different values depending on which species you happened to watch. The cure is to divide each term by the stoichiometric coefficient of that species. Doing so brings all three to a common value. With this,
\[-\frac{d[\mathrm{A}]}{dt} = -\frac{1}{3}\,\frac{d[\mathrm{B}]}{dt} = \frac{1}{2}\,\frac{d[\mathrm{C}]}{dt}\]
or
\[\text{rate of reaction} = -\frac{d[\mathrm{A}]}{dt} = -\frac{1}{3}\,\frac{d[\mathrm{B}]}{dt} = \frac{1}{2}\,\frac{d[\mathrm{C}]}{dt}\]
Now a single number describes the reaction, whichever species is monitored in the laboratory.
The General Rate Expression
In general, for the reaction
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D},\]
\[\text{rate} = -\frac{1}{a}\,\frac{d[\mathrm{A}]}{dt} = -\frac{1}{b}\,\frac{d[\mathrm{B}]}{dt} = \frac{1}{c}\,\frac{d[\mathrm{C}]}{dt} = \frac{1}{d}\,\frac{d[\mathrm{D}]}{dt}\]
This one line is worth memorising, because writing it down correctly is the first step in a large family of examination questions. Reactants take the minus sign, products take the plus sign, and every term is divided by its own coefficient from the balanced equation.
6.3 Rate of Reaction and Reactant Concentration
The rate of a reaction at a given temperature, for a given time instant, depends on the concentration of the reactant. Such a rate–concentration relation is called the rate law.
This dependence is easy to picture. Reactant molecules must meet before they can react, and the more crowded the solution or the vessel, the more often those meetings occur. As a reaction proceeds and the reactants are used up, the crowd thins, the meetings become rarer, and the reaction slows down — which is exactly the flattening you saw in the concentration–time curves of the previous section.
6.3.1 Rate Law
Consider the general reaction,
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D} \qquad \ldots (6.1)\]
The rate of reaction at a given time is proportional to the molar concentration of the reactants at that time, raised to simple powers. Thus
\[\text{Rate of reaction} \propto [\mathrm{A}]^x[\mathrm{B}]^y\]
or
\[\text{rate} = k\,[\mathrm{A}]^x[\mathrm{B}]^y \qquad \ldots (6.2)\]
where \(k\), the proportionality constant, is called the rate constant. The rate constant is independent of concentration and varies with temperature. For unit concentrations of A and B, \(k\) is equal to the rate of the reaction — a useful way to remember what it means physically. Equation (6.2) is called the differential rate law.
The powers \(x\) and \(y\) of the concentration terms A and B in the rate law are not necessarily equal to the stoichiometric coefficients (\(a\) and \(b\)) appearing in Eq. (6.1). Thus \(x\) and \(y\) may be simple whole numbers, zero or a fraction. You have to realise that \(x\) and \(y\) are experimentally determined.
This point deserves emphasis, because it is where most mistakes are made. The rate law in Eq. (6.2) is determined experimentally, and it expresses the rate of a chemical reaction in terms of the molar concentrations of the reactants. It is not predicted from the stoichiometries of the reactants. You cannot look at a balanced equation and write down its rate law; you must be told the experimental result, or be given data from which to work it out.
What the Exponents Tell Us
The exponents \(x\) and \(y\) appearing in the rate law tell us how a change in concentration affects the rate of the reaction. Four cases are worth examining.
- For \(x = y = 1\), Eq. (6.2) gives
\[\text{rate} = k[\mathrm{A}][\mathrm{B}]\]
The equation implies that the rate of the reaction depends linearly on the concentrations of A and B. If either the concentration of A or of B is doubled, the rate would be doubled.
- For \(x = 2\) and \(y = 1\), Eq. (6.2) then leads to
\[\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]\]
If the concentration of A is doubled keeping that of B constant, the rate of reaction will increase by a factor of 4. The reason is arithmetic: the doubled concentration is squared, and \(2^2 = 4\).
If \(x = 0\), the rate is independent of the concentration of A. Since any quantity raised to the power zero equals one, the term \([\mathrm{A}]^0\) simply drops out of the rate law. Adding more A then makes no difference at all to how fast the reaction goes.
If \(x < 0\), the rate decreases as \([\mathrm{A}]\) increases. A negative exponent puts the concentration term in the denominator, so the substance actually slows its own reaction — an unusual but real situation.
The factor is always 2 raised to the power of the exponent, and this single observation is enough to answer a large number of examination questions on rate laws.
The three common possibilities are also easy to recognise from a graph of rate against concentration, as drawn in Fig. 6.12.
Diagram
Fig. 6.12 – Three plots of rate (y-axis) against reactant concentration [A] (x-axis) on the same set of axes: a horizontal line (rate independent of [A]), a straight line through the origin (rate ∝ [A]), and an upward-curving parabola through the origin (rate ∝ [A]²); label each curve only as “curve I”, “curve II”, “curve III”.
6.3.2 Writing the Rate Law
Since the rate law cannot be deduced from the equation, writing one down is simply a matter of translating an experimental statement into symbols. Consider the reaction,
\[2\,\mathrm{H_2O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]
If the rate of the reaction is proportional to the concentration of \(\mathrm{H_2O_2}\), the rate law is given by
\[\text{rate} = k[\mathrm{H_2O_2}]\]
Notice that the coefficient 2 in the balanced equation has left no trace in the rate law. The experiment says “proportional to the first power of \([\mathrm{H_2O_2}]\)”, and that is what is written.
Two phrases occur again and again in such statements, and each translates directly. “Proportional to the square of the concentration” means an exponent of 2. “Independent of the concentration” means an exponent of 0, so that species does not appear in the rate law at all.
6.3.3 Order of the Reaction
For the reaction,
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\]
if the rate of the reaction is given as
\[\text{rate} = k[\mathrm{A}]^x[\mathrm{B}]^y,\]
then the sum \(x + y\) gives the overall order of the reaction.
The individual exponents also have names of their own: \(x\) is the order with respect to A, and \(y\) is the order with respect to B. An examination question may ask for either the individual orders or the overall order, so read the wording carefully. For example:
- For the reaction,
\[2\,\mathrm{H_2O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]
the experimentally determined rate law is rate \(= k[\mathrm{H_2O_2}]\). The reaction is of first order.
- If the experimentally determined rate law for the reaction
\[\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightarrow 2\,\mathrm{HI}\text{(g)}\]
is rate \(= k[\mathrm{H_2}][\mathrm{I_2}]\), the reaction is of first order in \(\mathrm{H_2}\) and \(\mathrm{I_2}\) each, and hence overall of second order. Here \(1 + 1 = 2\).
Key Points About the Order of Reaction
The order of a chemical reaction is experimentally determined.
The order can be an integer or a fraction. Look at the reaction,
\[\mathrm{CH_3CHO}\text{(g)} \rightarrow \mathrm{CH_4}\text{(g)} + \mathrm{CO}\text{(g)}.\]
The rate law for the reaction was found to be rate \(= k[\mathrm{CH_3CHO}]^{3/2}\). Here the order of the reaction is 3/2. A fractional order is a strong hint that the reaction proceeds through a complicated multi-step route, since no single collision can involve one and a half molecules.
- The order of the reaction can be zero for:
\[\mathrm{NO_2}\text{(g)} + \mathrm{CO}\text{(g)} \rightarrow \mathrm{NO}\text{(g)} + \mathrm{CO_2}\text{(g)}\]
The rate expression for this is: rate \(= k[\mathrm{NO_2}]^2\). This shows that the order of reaction with respect to \(\mathrm{NO_2}\) is 2 and with CO is zero, or the rate is independent of the concentration of CO. The overall order of reaction is 2. Note that a species can be a genuine reactant, consumed in the reaction, and still have zero order — being needed and controlling the speed are two different things.
- Only a few reactions of third order are known. Reactions with orders higher than three are rare. This is because a third-order rate law would require three particles to come together, and a simultaneous four-body encounter is so improbable that it effectively never happens.
| [X] mol \(\text{dm}^{-3}\) | [Y] mol \(\text{dm}^{-3}\) | Initial rate mol \(\text{dm}^{-3}\text{s}^{-1}\) | |
|---|---|---|---|
| (i) | 0.10 | 0.10 | \(2.0 \times 10^{-4}\) |
| (ii) | 0.20 | 0.10 | \(8.0 \times 10^{-4}\) |
| (iii) | 0.20 | 0.30 | \(8.0 \times 10^{-4}\) |
Determine the order with respect to X and to Y, write the rate law, state the overall order, and calculate the rate constant with its correct unit.
6.4 Molecularity of Elementary Reactions
At the end of the previous section a question was left hanging: if the powers in a rate law do not come from the balanced equation, where do they come from? The answer lies inside the reaction, in the steps the molecules actually go through.
Complex reactions are those which constitute a series of elementary reactions. The balanced equation you write for such a reaction is the sum of those steps, and, like any sum, it hides the individual terms that produced it.
6.4.1 Elementary Reaction
Consider,
\[\mathrm{O_3}\text{(g)} \rightarrow \mathrm{O_2}\text{(g)} + \mathrm{O}\text{(g)}\]
\[\mathrm{C_2H_5I}\text{(g)} \rightarrow \mathrm{C_2H_4}\text{(g)} + \mathrm{HI}\text{(g)}\]
These reactions occur in a single step and cannot be broken down further into simpler reactions. These are elementary reactions.
An elementary reaction is therefore not a summary of anything — it is the event itself, exactly as the molecules experience it. This is what makes elementary reactions special, and it is the reason the whole of this section is built on them.
6.4.2 Molecularity of Reaction
The molecularity refers to how many reactant molecules are involved in the reaction. In the above reactions there is only one reactant molecule. These are unimolecular reactions, or their molecularity is one. Such a reaction is simply a molecule falling apart on its own, without needing to meet anything else.
\[\mathrm{O_3}\text{(g)} + \mathrm{O}\text{(g)} \rightarrow 2\,\mathrm{O_2}\text{(g)}\]
\[2\,\mathrm{NO_2}\text{(g)} \rightarrow 2\,\mathrm{NO}\text{(g)} + \mathrm{O_2}\text{(g)}\]
The elementary reactions involving two reactant molecules are bimolecular reactions, or they have molecularity as two. Here the reaction cannot begin until two particles actually collide. Note the second example carefully: the two colliding particles need not be different — two molecules of \(\mathrm{NO_2}\) striking each other is still a bimolecular event.
Because molecularity counts particles, it can only be a whole number, and it can only be counted for a single step. Asking for the molecularity of an overall complex reaction is meaningless — there is no one collision to count.
The rarity of termolecular steps is not an accident. Getting two particles to meet is easy; getting a third to arrive at the same point at the same instant is enormously less likely. This is the physical reason behind the observation made in the previous section that reactions of order higher than three are practically unknown.
6.4.3 Order and Molecularity of Elementary Reactions
The rate law for the elementary reaction
\[2\,\mathrm{NO_2}\text{(g)} \rightarrow 2\,\mathrm{NO}\text{(g)} + \mathrm{O_2}\text{(g)}\]
is found to be rate \(= k[\mathrm{NO_2}]^2\). The reaction is second order and bimolecular. The order of reaction is 2 and its molecularity is also 2.
For the elementary reaction,
\[\mathrm{C_2H_5I}\text{(g)} \rightarrow \mathrm{C_2H_4}\text{(g)} + \mathrm{HI}\text{(g)}\]
\[\text{rate} = k[\mathrm{C_2H_5I}]\]
It is unimolecular and first order. However, the order and molecularity of the reaction may or may not be the same.
This is the key to the whole matter: for an elementary reaction, order equals molecularity. For a complex reaction, the overall order is an experimental result that need not match anything in the balanced equation.
6.4.4 Rate Determining Step
A number of chemical reactions are complex. They take place as a series of elementary steps. One of these steps is slower than the others. The slowest step is the rate determining step.
The slowest step determines the rate of the overall reaction. The reason is a matter of common sense: no matter how quickly the other steps can run, they cannot process material faster than the slow step delivers it to them.
Consider,
\[2\,\mathrm{NO_2Cl}\text{(g)} \rightarrow 2\,\mathrm{NO_2}\text{(g)} + \mathrm{Cl_2}\text{(g)}.\]
The reaction takes place in two steps:
\(\mathrm{NO_2Cl}\text{(g)} \xrightarrow{\ k_1\ } \mathrm{NO_2}\text{(g)} + \mathrm{Cl}\text{(g)} \quad\) (slow)
\(\mathrm{NO_2Cl}\text{(g)} + \mathrm{Cl}\text{(g)} \xrightarrow{\ k_2\ } \mathrm{NO_2}\text{(g)} + \mathrm{Cl_2}\text{(g)} \quad\) (fast)
Adding the two steps and cancelling whatever appears on both sides gives
\[\text{Overall } 2\,\mathrm{NO_2Cl}\text{(g)} \rightarrow 2\,\mathrm{NO_2}\text{(g)} + \mathrm{Cl_2}\text{(g)}\]
The two steps and their outcome are set out in FC 6.2.
Flowchart
FC 6.2 – Two-step decomposition of nitryl chloride: NO₂Cl → NO₂ + Cl (slow, rate determining, k₁) → the Cl atom produced is passed to step 2 → Cl + NO₂Cl → NO₂ + Cl₂ (fast, k₂) → add the two steps and cancel Cl → overall 2NO₂Cl → 2NO₂ + Cl₂; side branch: rate law taken from the slow step only → rate = k[NO₂Cl], first order
The first step being slower than the second, it is the rate determining step.
The rate law is
\[\text{rate} = k[\mathrm{NO_2Cl}]\]
This also represents the rate law of the overall reaction. The reaction thus is of the first order. Because the slow step is an elementary reaction, its rate law can be written straight from its own equation — and only one molecule of \(\mathrm{NO_2Cl}\) appears in it.
Reaction Intermediate
In the above reaction Cl is formed in the first step and consumed in the second. Such a species represents the reaction intermediate. The concentration of a reaction intermediate does not appear in the rate law.
An intermediate is easy to spot in a mechanism: look for the species that turns up on the product side of one step and on the reactant side of another. It is never present in the bottle at the start and never left over at the end, which is exactly why it vanishes when the steps are added up.
Distinction Between Order and Molecularity of a Reaction
Table 6.1 sets the two ideas side by side. They are easily confused because both are small whole numbers attached to a reaction, but they belong to different worlds — one to the laboratory, one to the mechanism.
Table 6.1: Distinction between order and molecularity of a reaction
| Order | Molecularity |
|---|---|
| 1. It is experimentally determined property. | i. It is theoretical entity. |
| 2. It is the sum of powers of the concentration terms of reactants those appear in the rate equation. | ii. It is the number of reactant molecules taking part in an elementary reaction. |
| 3. It may be an integer, fraction or zero. | iii. It is integer. |
\(\mathrm{O_3}\text{(g)} \rightarrow \mathrm{O_2}\text{(g)} + \mathrm{O}\text{(g)} \quad\) (slow)
\(\mathrm{O}\text{(g)} + \mathrm{O_3}\text{(g)} \rightarrow 2\,\mathrm{O_2}\text{(g)} \quad\) (fast)
Write the overall balanced reaction, state the molecularity of each step, write the predicted rate law and the overall order, and identify the reaction intermediate.
6.5 Integrated Rate Law
We introduced the differential rate law earlier. It describes how the rate of a reaction depends on the concentration of the reactants in terms of derivatives.
Useful as it is, the differential rate law answers only one question: how fast is the reaction right now, at this concentration. It does not tell a chemist what he usually wants to know — how much reactant will be left after half an hour, or how long the reaction must be left running to reach a required conversion.
The differential rate laws are converted into integrated rate laws. These tell us the concentrations of reactants for different times. Integration is the mathematical operation that adds up a rate over a stretch of time to give a total change, and that is exactly the step needed here.
6.5.1 Integrated Rate Law for the First Order Reactions in Solution
Consider the first order reaction,
\[\mathrm{A} \rightarrow \text{product} \qquad \ldots (6.3)\]
The differential rate law is given by
\[\text{rate} = -\frac{d[\mathrm{A}]}{dt} = k\,[\mathrm{A}] \qquad \ldots (6.4)\]
where \([\mathrm{A}]\) is the concentration of the reactant at time \(t\).
Rearranging Eq. (6.4),
\[\frac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\,dt \qquad \ldots (6.5)\]
The rearrangement collects every term containing concentration on the left and every term containing time on the right, so that each side can be integrated separately.
Let \([\mathrm{A}]_0\) be the initial concentration of the reactant A at time \(t = 0\). Suppose \([\mathrm{A}]_t\) is the concentration of A at time \(= t\).
The equation (6.5) is integrated between the limits \([\mathrm{A}] = [\mathrm{A}]_0\) at \(t = 0\) and \([\mathrm{A}] = [\mathrm{A}]_t\) at \(t = t\)
\[\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t} \frac{d[\mathrm{A}]}{[\mathrm{A}]} = -k \int_{0}^{t} dt\]
On integration,
\[\Big[\ln[\mathrm{A}]\Big]_{[\mathrm{A}]_0}^{[\mathrm{A}]_t} = -k\,(t)\Big|_{0}^{t}\]
Substitution of limits gives
\[\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -k\,t\]
or
\[\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt \qquad \ldots (6.6)\]
using the rule that the difference of two logarithms equals the logarithm of their ratio. Turning the ratio upside down changes the sign, giving
\[k = \frac{1}{t}\ln\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\]
Converting \(\ln\) to \(\log_{10}\), we write
\[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t} \qquad \ldots (6.7)\]
Eq. (6.7) gives the integrated rate law for the first order reactions.
Other Forms of the First Order Rate Law
The rate law can be written in the following forms.
- Eq. (6.6) is
\[\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt\]
By taking antilog of both sides we get
\[\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = e^{-kt} \quad \text{or} \quad [\mathrm{A}]_t = [\mathrm{A}]_0\,e^{-kt} \qquad \ldots (6.8)\]
This exponential form is the one to reach for when the concentration at a given time is wanted directly. It also shows the character of first order decay: the concentration never reaches exactly zero, but falls by the same fraction in every equal interval of time.
- Let ‘\(a\)’ \(\text{mol dm}^{-3}\) be the initial concentration of A at \(t = 0\).
Let \(x\) \(\text{mol dm}^{-3}\) be the concentration of A that decreases (reacts) during time \(t\). The concentration of A that remains unreacted at time \(t\) would be \((a - x)\) \(\text{mol/dm}^3\).
Substitution of \([\mathrm{A}]_0\) and \([\mathrm{A}]_t = (a - x)\)
\[k = \frac{2.303}{t}\log_{10}\frac{a}{(a-x)} \qquad \ldots (6.9)\]
This third form is the most convenient when a question is phrased in terms of percentages, because you may then take \(a = 100\) and let \(x\) be the percentage decomposed.
Equations (6.7), (6.8) and (6.9) represent the integrated rate law of first order reactions.
6.5.2 Units of Rate Constant for the First Order Reaction
The integrated rate law is
\[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\]
Because \(\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\) is a unitless quantity, the dimensions of \(k\) will be \((\text{time})^{-1}\). The two concentrations are divided by one another, so their units cancel before the logarithm is taken — a logarithm can only be taken of a pure number.
The units of \(k\) will be \(\text{s}^{-1}\), \(\text{min}^{-1}\) or \((\text{hour})^{-1}\).
Notice what this implies. For a first order reaction the unit of \(k\) carries no concentration term at all, which is why an examiner can hand you nothing but the unit \(\text{s}^{-1}\) and expect you to identify the reaction as first order.
6.5.3 Half Life of the First Order Reactions (\(t_{1/2}\))
Radioactive processes follow the first order kinetics. The half life of a reaction is the time required for the reactant concentration to fall to one half of its initial value.
6.5.4 Half Life and Rate Constant of the First Order Reaction
The integrated rate law for the first order reaction is
\[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\]
where \([\mathrm{A}]_0\) is the initial concentration of reactant at \(t = 0\). It falls to \([\mathrm{A}]_t\) at time \(t\) after the start of the reaction. The time required for \([\mathrm{A}]_0\) to become \([\mathrm{A}]_0/2\) is denoted as \(t_{1/2}\)
or
\[[\mathrm{A}]_t = [\mathrm{A}]_0/2 \qquad \text{at } t = t_{1/2}\]
Putting this condition in the integrated rate law we write
\[k = \frac{2.303}{t_{1/2}}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_0/2}\]
\[= \frac{2.303}{t_{1/2}}\log_{10} 2\]
\[= \frac{2.303}{t_{1/2}} \times 0.3010\]
\[k = \frac{0.693}{t_{1/2}}\]
\[t_{1/2} = \frac{0.693}{k} \qquad \ldots (6.10)\]
Watch what happened in the very first line: the initial concentration cancelled out, leaving only the number 2 inside the logarithm. Eq. (6.10) shows that the half life of a first order reaction is independent of the initial reactant concentration. This is shown in Fig. 6.2 as a plot of \([\mathrm{A}]_t\) versus \(t\).
Diagram
Fig. 6.2 – Half life period of a first order reaction: a smooth decay curve of [A]t (y-axis) against time (x-axis); mark [A]₀ on the y-axis, then [A]₀/2, [A]₀/4 and [A]₀/8, dropping horizontal and vertical guide lines to the time axis to show three successive equal intervals each labelled t₁/₂.
Fig. 6.2 makes the point visually: each time the concentration halves, the clock has advanced by the same interval, whether the curve is high up or far down.
6.5.5 Graphical Representation of the First Order Reactions
A straight-line graph is the most reliable way of confirming an order experimentally, because the eye can judge straightness far better than it can judge the shape of a curve. Three different straight lines can be obtained for a first order reaction, depending on which equation is plotted.
- The differential rate law for the first order reaction \(\mathrm{A} \rightarrow \mathrm{P}\) is
\[\underset{y}{\text{rate}} = -\frac{d[\mathrm{A}]}{dt} = \underset{m}{k}\ \underset{x}{[\mathrm{A}]_t} + \underset{c}{0}\]
The equation is of the form \(y = mx + c\). A plot of rate versus concentration \([\mathrm{A}]_t\) is a straight line passing through the origin. This is shown in Fig. 6.3. The slope of the straight line \(= k\).
Diagram
Fig. 6.3 – Variation of rate with [A]: a straight line rising from the origin, x-axis labelled “initial concentration”, y-axis labelled “rate”; mark the slope as equal to k.
- From Eq. (6.7) the integrated rate law is
\[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\]
On rearrangement, the equation becomes
\[\frac{kt}{2.303} = \log_{10}[\mathrm{A}]_0 - \log_{10}[\mathrm{A}]_t\]
Hence,
\[\underset{y}{\log_{10}[\mathrm{A}]_t} = \underset{m}{-\frac{k}{2.303}}\ \underset{x}{t} + \underset{c}{\log_{10}[\mathrm{A}]_0}\]
The equation is of the straight line. A graph of \(\log_{10}[\mathrm{A}]_t\) versus \(t\) yields a straight line with slope \(-k/2.303\) and y-axis intercept as \(\log_{10}[\mathrm{A}]_0\). This is shown in Fig. 6.4.
Diagram
Fig. 6.4 – A plot showing log₁₀[A]t vs time t: a straight line sloping downwards from left to right, y-axis labelled log₁₀[A]t and x-axis labelled “time t”; mark the intercept on the y-axis as log₁₀[A]₀ and the slope as −k/2.303.
- Eq. (6.7) gives
\[\underset{y}{\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}} = \underset{m}{\frac{k}{2.303}}\ \underset{x}{t}\]
The equation has a straight line form \(y = mx\). Hence, the graph of \(\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\) versus \(t\) is a straight line passing through the origin as shown in Fig. 6.5.
Diagram
Fig. 6.5 – A plot of log₁₀([A]₀/[A]t) vs time t: a straight line rising from the origin, y-axis labelled log₁₀([A]₀/[A]t), x-axis labelled t; mark the slope as k/2.303.
There is no constant term in this last equation, which is why this line, unlike the one in Fig. 6.4, must pass through the origin.
6.5.6 Examples of First Order Reactions
Some examples of reactions of first order are:
\[2\,\mathrm{H_2O_2}\text{(l)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}, \qquad \text{rate} = k[\mathrm{H_2O_2}]\]
\[2\,\mathrm{N_2O_5}\text{(g)} \rightarrow 4\,\mathrm{NO_2}\text{(g)} + \mathrm{O_2}\text{(g)}, \qquad \text{rate} = k\,[\mathrm{N_2O_5}]\]
In both cases the coefficient 2 in the equation has again left no mark on the rate law, a reminder of how thoroughly order and stoichiometry are separate matters.
6.5.7 Integrated Rate Law for Gas Phase Reactions
For a gas-phase reaction it is far easier to fit a pressure gauge to a sealed vessel than to withdraw and analyse samples. Fortunately, at constant temperature and volume the pressure of a gas is directly proportional to its concentration, as \(P = (n/V)RT\) shows, so pressures may be used in place of concentrations throughout.
For the gas phase reaction,
\[\mathrm{A}\text{(g)} \rightarrow \mathrm{B}\text{(g)} + \mathrm{C}\text{(g)}\]
let the initial pressure of A be \(P_i\), that decreases by \(x\) within time \(t\).
Pressure of reactant A at time \(t\)
\[P_A = P_i - x \qquad \ldots (6.11)\]
The pressures of the products B and C at time \(t\) are
\[P_B = P_C = x\]
Each molecule of A that disappears produces one molecule of B and one of C, so the pressure lost by A appears twice over on the product side. The total pressure at time \(t\) is then
\[P = P_i - x + x + x = P_i + x\]
Hence,
\[x = P - P_i \qquad \ldots (6.12)\]
This is the crucial link: the rise in total pressure measured on the gauge is exactly the fall in the pressure of A.
The pressure of A, \(P_A\), at time \(t\) is obtained by substitution of Eq. (6.12) into Eq. (6.11). Thus
\[P_A = P_i - (P - P_i) = P_i - P + P_i = 2P_i - P\]
The integrated rate law turns out to be
\[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\]
The concentration is now expressed in terms of pressures.
Thus, \([\mathrm{A}]_0 = P_i\) and \([\mathrm{A}]_t = P_A = 2P_i - P\)
Substitution gives in the above
\[k = \frac{2.303}{t}\log_{10}\frac{P_i}{2P_i - P} \qquad \ldots (6.13)\]
\(P\) is the total pressure of the reaction mixture at time \(t\).
6.5.8 Zero Order Reactions
The rate of a zero order reaction is independent of the reactant concentration. Such a reaction proceeds at a fixed speed, from beginning to end, no matter how much reactant is present — behaviour that seems to contradict everything said so far, and which therefore needs a special explanation. That explanation is given by the examples at the end of this subsection.
Integrated Rate Law for Zero Order Reactions
For a zero order reaction,
\[\mathrm{A} \rightarrow \mathrm{P}\]
the differential rate law is given by
\[\text{rate} = -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^0 = k \qquad \ldots (6.14)\]
since any quantity raised to the power zero is 1. Rearrangement of Eq. (6.14) gives
\[d[\mathrm{A}] = -k\,dt\]
Integration between the limits \([\mathrm{A}] = [\mathrm{A}]_0\) at \(t = 0\) and \([\mathrm{A}] = [\mathrm{A}]_t\) at \(t = t\) gives
\[\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t} d[\mathrm{A}] = -k\int_{0}^{t} dt\]
\[\text{or } [\mathrm{A}]_t - [\mathrm{A}]_0 = -kt\]
Hence,
\[k\,t = [\mathrm{A}]_0 - [\mathrm{A}]_t \qquad \ldots (6.15)\]
No logarithm appears anywhere in this result. The concentration term stayed outside the integral because it was not there at all, and that single difference is what separates zero order behaviour from first order behaviour.
Units of Rate Constant of Zero Order Reactions
\[k = \frac{[\mathrm{A}]_0 - [\mathrm{A}]_t}{t} = \frac{\text{mol L}^{-1}}{t} = \text{mol dm}^{-3}\ \text{t}^{-1}\]
The units of the rate constant of a zero order reaction are the same as that of the rate. This follows directly from Eq. (6.14), where the rate and the rate constant are simply equal to one another.
Half Life of Zero Order Reactions
The rate constant of a zero order reaction is given by eqn (6.15)
\[k = \frac{[\mathrm{A}]_0 - [\mathrm{A}]_t}{t}\]
Using the conditions \(t = t_{1/2}\), \([\mathrm{A}]_t = [\mathrm{A}]_{1/2}\), Eq. (6.15) becomes
\[k = \frac{[\mathrm{A}]_0 - [\mathrm{A}]_{1/2}}{t_{1/2}} = \frac{[\mathrm{A}]_0}{2\,t_{1/2}}\]
Half the initial concentration has been subtracted from the whole, leaving half of it in the numerator. Hence,
\[t_{1/2} = \frac{[\mathrm{A}]_0}{2k} \qquad \ldots 6.16\]
The half life of a zero order reaction is proportional to the initial concentration of the reactant.
Here, unlike in the first order case, \([\mathrm{A}]_0\) refused to cancel. Start with twice as much reactant and the reaction — running at a fixed speed — needs twice as long to get through half of it.
Graphical Representation of Zero Order Reactions
The rate law in Eq. (6.15) gives
\[\underset{y}{[\mathrm{A}]_t} = \underset{m}{-k}\ \underset{x}{t} + \underset{c}{[\mathrm{A}]_0} \qquad \ldots 6.17\]
which is a straight line given by \(y = mx + c\).
A plot of \([\mathrm{A}]_t\) versus \(t\) is a straight line as shown in Fig. 6.6.
Diagram
Fig. 6.6 – [A]t vs t for a zero order reaction: a straight line falling steadily from left to right, y-axis labelled [A]t and x-axis labelled “time”; mark the intercept on the y-axis as [A]₀ and the slope as −k.
The slope of the straight line is \(-k\) and its intercept on the y-axis is \([\mathrm{A}]_0\).
The \(t_{1/2}\) of a zero order reaction is directly proportional to the initial concentration.
Compare Fig. 6.6 with Fig. 6.2: the zero order reaction falls along a straight line and would in principle reach zero concentration at a definite moment, while the first order curve merely approaches the axis and never quite arrives.
Given nothing but a set of concentration–time readings, this table is enough to identify the order, and FC 6.3 sets out the procedure.
Flowchart
FC 6.3 – Identifying the order from concentration–time data: measure [A] at several times → plot [A]t against t → if this plot is a straight line → zero order, slope = −k → if it is curved → plot log₁₀[A]t against t → if this plot is a straight line → first order, slope = −k/2.303 → in either case read k from the slope → cross-check using the half life: constant t₁/₂ confirms first order, t₁/₂ proportional to [A]₀ confirms zero order
Examples of Zero Order Reactions
Here follow some examples.
Decomposition of \(\mathrm{NH_3}\) on platinum metal
\[2\,\mathrm{NH_3}\text{(g)} \rightarrow \mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)}\]
The metal surface gets completely covered by a layer of \(\mathrm{NH_3}\) molecules. The number of \(\mathrm{NH_3}\) molecules attached on the platinum surface is small compared to the total ammonia. A large number of the \(\mathrm{NH_3}\) molecules tend to remain as gas, which do not react. The molecules present on the metal surface only react. The rate of the reaction is thus independent of the total concentration of \(\mathrm{NH_3}\) and remains constant.
The surface, in other words, is the scarce resource. Once it is fully occupied, extra ammonia in the vessel has nowhere to go and cannot contribute to the reaction.
- Decomposition of nitrous oxide in the presence of a Pt catalyst.
\[2\,\mathrm{N_2O}\text{(g)} \xrightarrow{\ \mathrm{Pt}\ } 2\,\mathrm{N_2}\text{(g)} + \mathrm{O_2}\text{(g)}\]
- The catalytic decomposition of \(\mathrm{PH_3}\) on hot tungsten at high pressure.
All three examples share the same feature — a solid surface of fixed area doing the work — which is why zero order kinetics is so characteristic of reactions on catalysts.
6.5.9 Pseudo-First Order Reactions
Certain reactions which are expected to be of higher order follow the first order kinetics. Consider the hydrolysis of methyl acetate.
\[\mathrm{CH_3COOCH_3}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightarrow \mathrm{CH_3COOH}\text{(aq)} + \mathrm{CH_3OH}\text{(aq)}\]
The rate law is
\[\text{rate} = k'\,[\mathrm{CH_3COOCH_3}]\,[\mathrm{H_2O}]\]
The reaction is expected to follow the second order kinetics; however, it obeys the first order.
The reason is that the solvent water is present in such large excess that the change in its concentration is negligible compared to the initial one, or its concentration remains constant. A little arithmetic makes the point: a litre of water contains roughly 55 moles, so consuming even a tenth of a mole of it during the reaction alters \([\mathrm{H_2O}]\) by well under one per cent — a change no experiment would notice.
Thus \([\mathrm{H_2O}] = \text{constant} = k''\). The rate law becomes
\[\text{rate} = k'\,[\mathrm{CH_3COOCH_3}]\,k''\]
\[= k\,[\mathrm{CH_3COOCH_3}]\]
where \(k = k'k''\).
The reaction is thus of first order.
The prefix pseudo means false or apparent, and it is well chosen: the reaction only wears the disguise of a first order reaction. Change the conditions so that water is no longer in excess, and the true second order character would reappear.
6.6 Collision Theory of Bimolecular Reactions
So far the rate law has been treated as something to be measured. The obvious next question is why a reaction should have any particular rate at all — what is going on among the molecules that makes one reaction finish in a flash and another crawl. Collision theory is the simplest picture that answers this, and it was built for bimolecular reactions, where two particles must meet before anything can happen.
6.6.1 Collision Between Reactant Molecules
Chemical reactions occur as a result of collisions between the reactant species. It may be expected that the rate of the reaction is equal to the rate of collision.
If that expectation were correct, kinetics would be a very short subject: count the collisions and you would have the rate. But the arithmetic does not agree. For the gas-phase reactions the number of collisions is far more, and typically many powers of tens compared to the observed rate.
In other words, molecules collide vastly more often than they react. Something is filtering the collisions, and the rest of this section identifies what.
6.6.2 Activation
For the reaction to occur the colliding reactant molecules must possess the minimum kinetic energy. This minimum kinetic energy is the activation energy. The reaction would occur only if colliding molecules possess kinetic energies equal to or greater than the activation energy.
Here is the first filter. A gentle collision merely bounces the molecules apart, because bonds in the reactants must be stretched and broken before new ones can form, and breaking a bond costs energy. That energy has to come from the motion of the colliding pair itself.
6.6.3 Orientation of Reactant Molecules
The requirement for successful collision described above is sufficient for reactions involving simple molecules (or ions), however not for those involving complex molecules.
Besides the above considerations the colliding molecules must have proper orientation. The molecules need to be so oriented relative to each other that the reacting groups approach closely.
This is the second filter, and it is a purely geometrical one. A molecule is not a featureless ball; the bond that has to break sits at a particular place on it, and a collision that lands anywhere else achieves nothing.
Consider,
\[\mathrm{A} + \mathrm{C\!-\!B} \rightarrow \mathrm{A\!-\!B} + \mathrm{C}\]
The two possibilities are drawn in Fig. 6.13.
Diagram
Fig. 6.13 – Orientation of colliding molecules for the reaction A + C–B → A–B + C. (i) Unsuccessful collision: sphere A approaching the C end of the C–B molecule, with a second frame showing A and C–B separating unchanged; label “improper orientation — no reaction”. (ii) Successful collision: sphere A approaching the B end of the C–B molecule, with a second frame showing A–B formed and C released; label “proper orientation — reaction occurs”.
The collision of A with C approaching toward A would not lead to reaction. No reaction will take place. The reactant molecules would collide and separate owing to the improper orientation of C–B. A has struck the wrong end of the molecule: the atom it needs to bond with, B, is on the far side and out of reach.
The reaction is successful as a result of proper orientation of C–B. A fraction of such collisions bring forth conversion of reactants to products. Here A arrives at the B end, close enough for the new A–B bond to begin forming as the old C–B bond loosens.
6.6.4 Potential Energy Barrier
Consider again the reaction
\[\mathrm{A} + \mathrm{C\!-\!B} \rightarrow \mathrm{A\!-\!B} + \mathrm{C}\]
During the course of collision, the new bond A–B develops. At the same time the bond C–B breaks. A configuration in which all the three atoms are weakly connected together is called the activated complex.
\[\mathrm{A} + \mathrm{C\!-\!B} \rightarrow \mathrm{A}\cdots\mathrm{B}\cdots\mathrm{C} \rightarrow \mathrm{A\!-\!B} + \mathrm{C}\]
The activated complex is not a product and cannot be bottled. It exists for an instant at the top of the barrier and then falls one way or the other — forward to products, or back to reactants.
To attain the configuration \(\mathrm{A}\cdots\mathrm{B}\cdots\mathrm{C}\), atoms need to gain energy, which comes from the kinetic energy of the colliding molecules. Notice the exchange taking place: the motion of the approaching particles is converted, at the instant of impact, into the potential energy of a strained arrangement of atoms. This is why slow-moving molecules cannot react — they have nothing to pay with.
The energy barrier between reactants and products is shown in Fig. 6.7. The reactant molecules need to climb up and overcome this before they get converted to products.
Diagram
Fig. 6.7 – Potential energy barrier: a curve of energy (y-axis) against reaction progress, starting at the reactant level, rising to a rounded peak and falling to a lower product level; mark the vertical distance from reactants to peak as Ea and the vertical distance from reactants to products as ΔH, and label the peak “activated complex”.
The height of the barrier is called activation energy (\(E_a\)). Thus the reactant molecules transform to products only if they possess energy equal to or greater than such activation energy.
Reading the Energy Profile
A diagram such as Fig. 6.7 repays careful reading, because a whole family of questions is built on it. The horizontal axis is not time but the progress of the reaction — the journey from reactants, through the activated complex, to products. The vertical axis is potential energy.
Three distances on the diagram matter. Measured upwards from the reactant level to the peak is \(E_a\) for the forward reaction. Measured upwards from the product level to the same peak is \(E_a\) for the reverse reaction. And measured from the reactant level to the product level is \(\Delta H\), the enthalpy change, which is negative when the products lie lower, as they do in Fig. 6.7.
The three are connected by
\[E_a(\text{forward}) - E_a(\text{reverse}) = \Delta H\]
For an exothermic reaction the products sit below the reactants, so the climb back up is steeper than the climb forward, and the reverse activation energy is the larger of the two. For an endothermic reaction the opposite holds.
How Many Collisions Actually Succeed
A fraction of molecules those possess energy greater than \(E_a\) is given by
\[f = e^{-E_a/RT}\]
where \(R\) is the molar gas constant, \(8.314\ \text{J K}^{-1}\text{mol}^{-1}\), and \(T\) is the temperature in kelvin. Read the expression as a competition between two energies: \(E_a\), the barrier that must be cleared, and \(RT\), a measure of the energy the molecules actually have at that temperature. When the barrier is much larger than \(RT\), the exponent is a large negative number and \(f\) becomes vanishingly small.
For a gaseous reaction at 298 K, \(E_a = 75\ \text{kJ/mol}\). The fraction of successful collisions is given by
\[f = e^{-E_a/RT} = e^{-75000/8.314 \times 298} = 7 \times 10^{-14}\]
or only 7 collisions in \(10^{14}\) collisions are sufficiently energetic to lead to the reaction.
As a result only a few collisions lead to products. The number of successful collisions is further reduced by the orientation requirement already discussed.
Two conditions are worth fixing firmly in mind. All collisions of reactant molecules do not lead to a chemical reaction. The colliding molecules need to possess certain energy which is greater than the activation energy \(E_a\), and proper orientation.
6.7 Temperature Dependence of Reaction Rates
It has been observed that the rates of most of the chemical reactions usually increase with temperature. In everyday life we see that fuels such as oil and coal are inert at room temperature but burn rapidly at higher temperatures. Many foods spoil rapidly at room temperature and last longer in a freezer.
The question is where this effect enters the mathematics of kinetics. A rate depends on only two things — the concentrations, and the rate constant. The concentrations change only a little with temperature. The rate constant shows a strong dependence on the temperature.
So the whole of the temperature effect is carried by \(k\), and finding how \(k\) varies with \(T\) is the business of this section.
6.7.1 Arrhenius Equation
Arrhenius suggested that the rate of a reaction varies with temperature as
\[k = A\,e^{-E_a/RT} \qquad \ldots 6.18\]
where \(k\) is the rate constant, \(E_a\) is the activation energy, \(R\) the molar gas constant, \(T\) the temperature in kelvin, and \(A\) is the pre-exponential factor. Eq. (6.18) is called the Arrhenius equation.
The equation is worth reading as a sentence rather than as a formula. The exponential term \(e^{-E_a/RT}\) is exactly the fraction of molecules able to clear the barrier, which you met at the end of the previous section. The factor \(A\) accounts for how often the molecules meet at all, and for whether they are correctly aimed when they do. Multiply the two together — how many encounters, and what fraction of them succeed — and you have the rate constant.
The pre-exponential factor \(A\) and the rate constant have the same unit in the case of first order reactions. Besides, \(A\) is found to be related to the frequency of collisions.
6.7.2 Graphical Determination of Activation Energy
Taking logarithm of both sides of eqn (6.18) we obtain
\[\ln k = -\frac{E_a}{RT} + \ln A \qquad \ldots (6.19)\]
Taking logarithms is the standard way of dealing with an exponential, because it brings the exponent down to the level of ordinary algebra where a straight line can be recognised.
Converting the natural base to base 10 we write
\[\underset{y}{\log_{10}k} = \underset{m}{-\frac{E_a}{2.303\,R}}\ \underset{x}{\frac{1}{T}} + \underset{c}{\log_{10}A} \qquad \ldots (6.20)\]
This equation is of the form of a straight line \(y = mx + c\).
Note carefully what is plotted on the horizontal axis: not the temperature itself, but its reciprocal \(1/T\). Only in this form does the relation become a straight line.
The Arrhenius plot of \(\log_{10}k\) versus \(1/T\) giving a straight line is shown in Fig. (6.8). A slope of the line is \(-E_a/2.303R\) with its intercept being \(\log_{10}A\).
Diagram
Fig. 6.8 – Variation of log₁₀k with 1/T: a straight line sloping downwards from left to right, y-axis labelled log₁₀k and x-axis labelled 1/T; mark the slope as −Ea/2.303R and the intercept on the y-axis as log₁₀A.
From a slope of the line the activation energy can be determined. Since the slope is negative and equals \(-E_a/2.303R\), measuring the slope of the line and multiplying its magnitude by \(2.303R\) returns \(E_a\) directly. This is the standard experimental method: measure \(k\) at several temperatures, plot, and read the barrier height off the graph.
Eq. (6.18) shows that with an increase of temperature, \(-E_a/RT\) and in turn the rate of reaction would increase.
6.7.3 Determination of Activation Energy
Drawing a full graph requires rate constants at many temperatures. In practice two temperatures are often enough, and the working below turns the Arrhenius equation into a form that uses just two measurements.
For two different temperatures \(T_1\) and \(T_2\)
\[\log_{10}k_1 = \log_{10}A - \frac{E_a}{2.303\,R\,T_1} \qquad \ldots (6.21)\]
\[\log_{10}k_2 = \log_{10}A - \frac{E_a}{2.303\,R\,T_2} \qquad \ldots (6.22)\]
where \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\) respectively. Subtracting Eq. (6.21) from Eq. (6.22),
\[\log_{10}k_2 - \log_{10}k_1 = -\frac{E_a}{2.303\,R}\,\frac{1}{T_2} + \frac{E_a}{2.303\,R}\,\frac{1}{T_1}\]
The unknown term \(\log_{10}A\) has cancelled between the two equations, which is the entire purpose of subtracting them. Hence,
\[\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]
\[= \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \qquad \ldots (6.23)\]
The last step simply combines the two reciprocals over a common denominator. Equation (6.23) is the workhorse of this topic: given any four of the five quantities \(k_1\), \(k_2\), \(T_1\), \(T_2\) and \(E_a\), it delivers the fifth.
6.7.4 Graphical Description of Effect of Temperature
It has been realized that the average kinetic energy of molecules is proportional to temperature. The collision theory suggested that a bimolecular reaction occurs only if the reacting molecules have sufficient kinetic energies (at least \(E_a\)) and proper orientation when they collide.
At a given temperature, the fraction of molecules with their kinetic energy equal to or greater than \(E_a\) may lead to the product. With an increase of temperature the fraction of molecules having their energies equal to or greater than (\(\geq E_a\)) would increase. The rate of the reaction thus would increase.
The molecules in a sample do not all move at the same speed. At any instant some are slow, some are fast, and most lie somewhere in between — which is why a distribution has to be drawn rather than a single value. This is depicted by plotting a fraction of molecules with given kinetic energy versus kinetic energy for two different temperatures \(T_1\) and \(T_2\) (\(T_2\) being \(> T_1\)) in Fig. 6.9.
Diagram
Fig. 6.9 – Comparison of fraction of molecules activated at T₁ and T₂: two distribution curves on the same axes, x-axis “kinetic energy” and y-axis “fraction of molecules with given kinetic energy”; the T₁ curve is taller with its peak further left, the T₂ curve is lower and broader with its peak shifted right; mark a vertical line at Ea well to the right of both peaks and shade the two areas beyond it, labelling them (a) for T₁ and (b) for T₂.
The area under the curve is proportional to the number of molecules with those values of kinetic energy. The total area is the same at \(T_1\) and \(T_2\), because heating the sample does not create or destroy any molecules — it only redistributes their energies.
The areas (a) and (b) represent the fraction of molecules with kinetic energy exceeding \(E_a\) at \(T_1\) and \(T_2\) respectively (since \(T_2 > T_1\)). This indicates that the fraction of molecules possessing energies larger than \(E_a\) increases with temperature. The rate of reaction increases accordingly.
Look at where the line marking \(E_a\) has been drawn — far out to the right, well beyond the peak of either curve. This is the crucial detail. Because the reacting molecules come only from the thin tail of the distribution, a small shift of the whole curve to the right adds a large proportion to that tail, even though the peak itself has barely moved.
6.8 Effect of a Catalyst on the Rate of Reaction
The previous section showed one way of speeding up a reaction: raise the temperature, and more molecules will carry enough energy to cross the barrier. There is a second way, and it is often the better one — leave the temperature alone and lower the barrier instead.
A catalyst is a substance added to the reactants that increases the rate of the reaction without itself being consumed in the reaction.
The words “without itself being consumed” are the heart of the definition. A catalyst does take part — it must, or it could do nothing — but it is returned unchanged at the end, which is why a small quantity can process an unlimited amount of reactant.
Consider
\[2\,\mathrm{KClO_3}\text{(s)} \xrightarrow{\ \mathrm{MnO_2}\ } 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}\]
Here \(\mathrm{MnO_2}\) is the catalyst. It has been observed that the decomposition rate increases with the addition of the catalyst. Notice that \(\mathrm{MnO_2}\) appears above the arrow rather than among the reactants, which is the standard way of writing a catalyst: it is present in the flask but does not enter the balanced equation, and none of it is used up.
An Alternative Pathway
A catalyst provides an alternative pathway associated with lower activation energy.
This is the essential idea, and it is worth stating carefully. The catalyst does not push the molecules over the existing barrier, and it does not give them extra energy. It offers them a different route to the same destination — one that runs through a lower pass.
Fig. 6.10 compares the potential energy barriers for the catalysed and uncatalysed reactions.
Diagram
Fig. 6.10 – Potential energy barriers for catalysed and uncatalysed reactions: two curves drawn on the same axes from the same reactant level to the same product level, with energy on the y-axis and “reaction coordinate” on the x-axis; the outer curve has a tall peak labelled (Ea)₁ for the uncatalysed reaction and the inner curve a lower peak labelled (Ea)₂ for the catalysed reaction.
The barrier for the uncatalysed reaction \((E_a)_1\) is larger than that for the same reaction in the presence of a catalyst \((E_a)_2\).
Look carefully at where the two curves begin and end in Fig. 6.10. They start at the same reactant level and finish at the same product level; only the height of the peak between them differs. The catalyst has therefore changed nothing about the reactants or the products, and \(\Delta H\) for the reaction is exactly what it was before. What has changed is the difficulty of getting from one to the other.
Consider the decomposition of \(\mathrm{H_2O_2}\) in aqueous solution catalysed by \(\mathrm{I^-}\) ions.
\[2\,\mathrm{H_2O_2}\text{(l)} \xrightarrow{\ \mathrm{I^-}\ } 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]
At room temperature the rate of reaction is slower in the absence of the catalyst, with its activation energy being \(76\ \text{kJ mol}^{-1}\). In the presence of the iodide ion catalyst \(\mathrm{I^-}\) the reaction is faster, since the activation energy decreases to \(57\ \text{kJ mol}^{-1}\).
A drop of \(19\ \text{kJ mol}^{-1}\) may not look dramatic written down, but the Arrhenius equation turns it into something spectacular, as the next worked example shows.
Why More Molecules Can React
Fig. 6.11 shows a plot of the fraction of molecules as a function of energy.
Diagram
Fig. 6.11 – Comparison of fraction of molecules for catalysed and uncatalysed reactions: a single energy-distribution curve, x-axis “energy” and y-axis “fraction of molecules, dn/N”; draw two vertical lines on the right-hand side, the nearer one labelled (Ea)₂ “in the presence of catalyst” and the further one labelled (Ea)₁ “before use of catalyst”, and shade the two areas beyond each line to show that the catalysed area is the larger.
A catalyst lowers the threshold energy. Consequently more molecules acquire the minimum amount of energy and tend to cross the energy barrier. The fraction of activated molecules is greater for the catalysed reaction. The rate of the catalysed reaction is thus larger than the reaction with no catalyst.
Compare this figure with Fig. 6.9 of the previous section and the contrast becomes clear. Raising the temperature left the barrier where it was and moved the distribution curve towards it; adding a catalyst leaves the distribution curve exactly where it is and moves the barrier towards the molecules. Either way, the shaded area beyond the line grows — and that shaded area is the reaction rate.
6.9 Summary
Every chemical reaction raises three separate questions: how far it goes (extent, answered by the equilibrium constant), whether it can occur (feasibility, answered by thermodynamics), and how fast it goes (rate). Chemical kinetics is the branch of chemistry dealing with the rates of chemical reactions and the factors affecting them.
Rates are studied for two reasons: they tell us how rapidly a reaction approaches equilibrium, and they give information about the mechanism — the sequence of elementary steps by which the reaction actually occurs.
The average rate of a reaction is the change in concentration of a species divided by the time interval, \(\text{average rate} = \Delta c/\Delta t\). A minus sign is attached to every reactant term because reactant concentration falls with time. The units of rate are \(\text{mol dm}^{-3}\ \text{s}^{-1}\).
The instantaneous rate is the rate at a particular moment, obtained from the slope of the tangent to the concentration–time curve at that point. It is written with derivatives, \(-d[\mathrm{A}]/dt\), and it is the instantaneous rate that kinetics is concerned with.
When the stoichiometric coefficients differ, each term is divided by its own coefficient so that a single number describes the reaction. For \(a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\), \[\text{rate} = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{b}\frac{d[\mathrm{B}]}{dt} = \frac{1}{c}\frac{d[\mathrm{C}]}{dt} = \frac{1}{d}\frac{d[\mathrm{D}]}{dt}\]
The rate law relates rate to reactant concentration: \(\text{rate} = k[\mathrm{A}]^x[\mathrm{B}]^y\), called the differential rate law. The rate constant \(k\) is independent of concentration, varies with temperature, and equals the rate when all concentrations are unity.
The exponents \(x\) and \(y\) are determined experimentally and are not necessarily the stoichiometric coefficients. They may be whole numbers, zero, fractional or negative. The rate law can never be written from a balanced overall equation.
The overall order of a reaction is the sum of the powers of the concentration terms in the rate law. Order may be an integer, a fraction (as in \(\text{rate} = k[\mathrm{CH_3CHO}]^{3/2}\)) or zero (as for CO in \(\mathrm{NO_2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO_2}\)). Third order reactions are few and higher orders are rare.
Elementary reactions occur in a single step and cannot be broken down further. The molecularity of an elementary reaction is the number of reactant molecules taking part in it — one for unimolecular, two for bimolecular. For an elementary reaction alone, order and molecularity are equal.
Order is an experimentally determined property and may be an integer, fraction or zero; molecularity is a theoretical entity, is defined only for an elementary step, and is always an integer.
Complex reactions are a series of elementary steps. The slowest step is the rate determining step and it fixes the rate of the overall reaction. A species produced in one step and consumed in a later one is a reaction intermediate; its concentration does not appear in the rate law.
Integrated rate laws convert the differential rate law into an equation giving concentration as a function of time. For a first order reaction: \[k = \frac{2.303}{t}\log_{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}, \qquad [\mathrm{A}]_t = [\mathrm{A}]_0e^{-kt}, \qquad k = \frac{2.303}{t}\log_{10}\frac{a}{(a-x)}\]
Since the logarithmic term is unitless, the rate constant of a first order reaction has the dimensions of \((\text{time})^{-1}\), with units \(\text{s}^{-1}\), \(\text{min}^{-1}\) or \(\text{hour}^{-1}\).
The half life \(t_{1/2}\) is the time in which the reactant concentration falls to half its initial value. For a first order reaction \(t_{1/2} = 0.693/k\), which is independent of the initial concentration. Radioactive processes follow first order kinetics.
A first order reaction gives three straight lines: rate against \([\mathrm{A}]_t\) (through the origin, slope \(k\)); \(\log_{10}[\mathrm{A}]_t\) against \(t\) (slope \(-k/2.303\), intercept \(\log_{10}[\mathrm{A}]_0\)); and \(\log_{10}([\mathrm{A}]_0/[\mathrm{A}]_t)\) against \(t\) (through the origin, slope \(k/2.303\)).
Decomposition of \(\mathrm{H_2O_2}\) and of \(\mathrm{N_2O_5}\) are examples of first order reactions.
For a first order gas-phase reaction \(\mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)}\) at constant volume, pressures replace concentrations. With \(P_A = 2P_i - P\), \[k = \frac{2.303}{t}\log_{10}\frac{P_i}{2P_i - P}\] where \(P_i\) is the initial pressure of A and \(P\) the total pressure at time \(t\).
For a zero order reaction the rate is independent of reactant concentration: \(\text{rate} = k[\mathrm{A}]^0 = k\), and the integrated form is \(kt = [\mathrm{A}]_0 - [\mathrm{A}]_t\). The rate constant has the same units as the rate, \(\text{mol dm}^{-3}\text{t}^{-1}\).
The half life of a zero order reaction, \(t_{1/2} = [\mathrm{A}]_0/2k\), is directly proportional to the initial concentration. A plot of \([\mathrm{A}]_t\) against \(t\) is a straight line of slope \(-k\) and intercept \([\mathrm{A}]_0\).
Zero order behaviour appears when a metal surface is saturated: the decomposition of \(\mathrm{NH_3}\) on platinum, of \(\mathrm{N_2O}\) on a Pt catalyst, and of \(\mathrm{PH_3}\) on hot tungsten at high pressure. Only the molecules on the surface react, so the total concentration makes no difference.
Pseudo-first order reactions are of higher order but obey first order kinetics, because one reactant is present in such large excess that its concentration stays effectively constant. In the hydrolysis of methyl acetate, \(\text{rate} = k'[\mathrm{CH_3COOCH_3}][\mathrm{H_2O}]\) becomes \(\text{rate} = k[\mathrm{CH_3COOCH_3}]\) with \(k = k'k''\).
Collision theory explains why molecules collide far more often than they react. A collision leads to reaction only if the colliding molecules possess at least the activation energy, and only if they are properly oriented so that the reacting groups approach closely.
At the moment of a successful collision the atoms pass through the activated complex, a configuration in which the old bond is not yet broken and the new one not yet formed. The height of the energy barrier between reactants and products is the activation energy \(E_a\), and the fraction of molecules able to cross it is \(f = e^{-E_a/RT}\). At 298 K with \(E_a = 75\ \text{kJ mol}^{-1}\) this is only about 7 collisions in \(10^{14}\).
Rate constants rise strongly with temperature, following the Arrhenius equation \[k = A\,e^{-E_a/RT}\] where \(A\) is the pre-exponential factor, related to the frequency of collisions.
Taking logarithms gives the straight-line form \[\log_{10}k = -\frac{E_a}{2.303\,R}\cdot\frac{1}{T} + \log_{10}A\] so a plot of \(\log_{10}k\) against \(1/T\) has slope \(-E_a/2.303R\) and intercept \(\log_{10}A\), from which \(E_a\) can be determined.
From rate constants at two temperatures, \[\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1T_2}\right)\]
Raising the temperature widens the energy distribution of the molecules. The total area under the curve is unchanged, but the area beyond \(E_a\) grows, so a larger fraction of molecules can react and the rate increases.
A catalyst increases the rate of a reaction without being consumed, by providing an alternative pathway of lower activation energy — as \(\mathrm{MnO_2}\) does for the decomposition of \(\mathrm{KClO_3}\). For the decomposition of \(\mathrm{H_2O_2}\), \(\mathrm{I^-}\) ions lower \(E_a\) from \(76\ \text{kJ mol}^{-1}\) to \(57\ \text{kJ mol}^{-1}\).
Because the catalyst lowers the threshold energy, a greater fraction of molecules can cross the barrier and the catalysed reaction is faster than the uncatalysed one at the same temperature.
The chapter as a whole moves from measuring a rate to explaining it, and the route is summarised in FC 6.4.
Flowchart
FC 6.4 – The logic of chemical kinetics: measure concentration against time → obtain average and instantaneous rates → fit the experimental rate law, rate = k[A]ˣ[B]ʸ → read off the order → branch A: integrate the rate law → first order (log form, t½ = 0.693/k) or zero order (linear form, t½ = [A]₀/2k) → predict concentration at any time; branch B: interpret the rate law → elementary steps, molecularity, rate determining step, intermediate → collision theory: energy ≥ Ea plus correct orientation → the two ways to raise the rate → raise T (Arrhenius, shifts the energy distribution) or add a catalyst (lowers Ea, alternative pathway)
6.10 Objective Questions
- The rate law for the reaction \(a\mathrm{A} + b\mathrm{B} \rightarrow \mathrm{P}\) is rate \(= k[\mathrm{A}][\mathrm{B}]\). The rate of reaction doubles if
- concentrations of A and B are both doubled
- [A] is doubled and [B] is kept constant
- [B] is doubled and [A] is halved
- [A] is kept constant and [B] is halved
- The order of the reaction for which the units of rate constant are \(\text{mol dm}^{-3}\ \text{s}^{-1}\) is
- 1
- 3
- 0
- 2
- The rate constant for the reaction \(2\,\mathrm{N_2O_5}\text{(g)} \rightarrow 2\,\mathrm{N_2O_4}\text{(g)} + \mathrm{O_2}\text{(g)}\) is \(4.98 \times 10^{-4}\ \text{s}^{-1}\). The order of reaction is
- 2
- 1
- 0
- 3
- Time required for 90% completion of a certain first order reaction is \(t\). The time required for 99.9% completion will be
- \(t\)
- \(2t\)
- \(t/2\)
- \(3t\)
- Slope of the graph \(\ln[\mathrm{A}]_t\) versus \(t\) for a first order reaction is
- \(-k\)
- \(k\)
- \(k/2.303\)
- \(-k/2.303\)
- What is the half life of a first order reaction if the time required to decrease the concentration of reactant from 0.8 M to 0.2 M is 12 h?
- 12 h
- 3 h
- 1.5 h
- 6 h
- The reaction, \(3\,\mathrm{ClO^-} \rightarrow \mathrm{ClO_3^{-}} + 2\,\mathrm{Cl^-}\) occurs in two steps,
- \(2\,\mathrm{ClO^-} \rightarrow \mathrm{ClO_2^{-}}\)
- \(\mathrm{ClO_2^{-}} + \mathrm{ClO^-} \rightarrow \mathrm{ClO_3^{-}} + \mathrm{Cl^-}\) The reaction intermediate is
- \(\mathrm{Cl^-}\)
- \(\mathrm{ClO_2^{-}}\)
- \(\mathrm{ClO_3^{-}}\)
- \(\mathrm{ClO^-}\)
- The elementary reaction \(\mathrm{O_3}\text{(g)} + \mathrm{O}\text{(g)} \rightarrow 2\,\mathrm{O_2}\text{(g)}\) is
- unimolecular and second order
- bimolecular and first order
- bimolecular and second order
- unimolecular and first order
- Rate law for the reaction, \(2\,\mathrm{NO} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{NOCl}\) is rate \(= k[\mathrm{NO}]^2[\mathrm{Cl_2}]\). Thus \(k\) would increase with
- increase of temperature
- increase of concentration of NO
- increase of concentration of \(\mathrm{Cl_2}\)
- increase of concentrations of both \(\mathrm{Cl_2}\) and NO
- For an endothermic reaction, \(\mathrm{X} \rightleftharpoons \mathrm{Y}\). If \(E_f\) is the activation energy of the forward reaction and \(E_r\) that for the reverse reaction, which of the following is correct?
- \(E_f = E_r\)
- \(E_f < E_r\)
- \(E_f > E_r\)
- \(\Delta H = E_f - E_r\) is negative
- The activation energies for two reactions are \(E_a\) and \(E'_a\) with \(E_a > E'_a\). If the temperature of the reacting system increases from \(T_1\) to \(T_2\), predict which of the following is correct?
- \(\dfrac{k'_1}{k_1} = \dfrac{k'_2}{k_2}\)
- \(\dfrac{k'_1}{k_1} > \dfrac{k'_2}{k_2}\)
- \(\dfrac{k'_1}{k_1} < \dfrac{k'_2}{k_2}\)
- \(\dfrac{k'_1}{k_1} < 2\,\dfrac{k'_2}{k_2}\) (\(k\) values are rate constants at the lower temperature and \(k'\) values at the higher temperature.)
6.11 Check Your Concepts
Assertion: The half life of a first order reaction does not depend on the initial concentration of the reactant. Reason: For a first order reaction the rate is directly proportional to the concentration of the reactant. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Explain why the exponents in a rate law cannot be obtained from the balanced equation of an overall reaction, but can be written directly from the equation of an elementary step.
The rate law of a certain reaction is found to be rate \(= k[\mathrm{A}]^{3/2}\). What does a fractional order tell you about the way the reaction occurs? Could an elementary reaction ever have such an order? Give your reason.
A zero order reaction and a first order reaction are each allowed to run for a very long time. Explain why the reactant in the zero order reaction is completely consumed at a definite moment, while the reactant in the first order reaction is not.
Two reactions have exactly the same value of \(\Delta H\) but very different rates at the same temperature. Explain how this is possible, referring to the potential energy profile of a reaction.
In the reaction \(\mathrm{NO_2}\text{(g)} + \mathrm{CO}\text{(g)} \rightarrow \mathrm{NO}\text{(g)} + \mathrm{CO_2}\text{(g)}\), carbon monoxide is genuinely consumed and yet the order with respect to CO is zero. Explain how a reactant that is used up can have no influence on the rate.
Both a reaction intermediate and a catalyst take part in a reaction, and neither appears in the overall balanced equation. Distinguish clearly between the two, stating where each appears in the mechanism and whether each appears in the rate law.
Assertion: A catalyst does not increase the amount of product obtained when a reversible reaction reaches equilibrium. Reason: A catalyst lowers the activation energy of the forward and the reverse reaction to the same extent. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Match the graph in Column I with the information it provides in Column II.
| Column I | Column II |
|---|---|
| a. \([\mathrm{A}]_t\) against \(t\), a straight line of slope \(-k\) | i. first order, from the differential rate law |
| b. \(\log_{10}[\mathrm{A}]_t\) against \(t\), a straight line of slope \(-k/2.303\) | ii. determination of activation energy |
| c. rate against \([\mathrm{A}]\), a straight line through the origin | iii. zero order |
| d. \(\log_{10}k\) against \(1/T\), a straight line of slope \(-E_a/2.303R\) | iv. first order, from the integrated rate law |
Molecules in a gas collide enormously more often than they react. Name the two requirements that a collision must satisfy in order to produce products, and state which of the two is affected by the size and complexity of the reacting molecules.
A rise of 10 K raises the average kinetic energy of the molecules by only a few per cent, yet it may double the rate of a reaction. Explain this apparent mismatch using the energy distribution curve.
Two reactions have the same pre-exponential factor, but the activation energy of the second is twice that of the first. State which reaction is faster at a given temperature and explain your answer using the Arrhenius equation.
Arrange the following three reactions in increasing order of the fraction of molecules able to react at a fixed temperature, and give the reason for your order: \(E_a = 40\ \text{kJ mol}^{-1}\), \(E_a = 80\ \text{kJ mol}^{-1}\), \(E_a = 120\ \text{kJ mol}^{-1}\).
Explain why the molecularity of a reaction can never be zero or fractional, whereas the order of a reaction may be either.
The decomposition of ammonia on a platinum surface is a zero order reaction. Explain why, and predict what would happen to the order if the pressure of ammonia above the surface were made extremely small.
A first order reaction is repeated with twice the initial concentration of reactant, everything else being unchanged. State, with reasons, what happens to (a) the initial rate, (b) the half life, and (c) the rate constant. Answer the same three parts for a zero order reaction.
The rate law for the hydrolysis of methyl acetate in water contains the concentrations of two species, yet the reaction is described as first order. Explain the term used for such a reaction and state what the experimentally measured rate constant actually represents.
Assertion: For an exothermic reaction, the activation energy of the reverse reaction is greater than that of the forward reaction. Reason: In an exothermic reaction the products lie at a lower energy than the reactants. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Why must the temperature be expressed in kelvin and the activation energy in \(\text{J mol}^{-1}\) when using the Arrhenius equation with \(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\)?
A complex reaction proceeds by two elementary steps. The rate law of the overall reaction is found to contain only the concentration of the reactant appearing in the first step. What does this tell you about the relative speeds of the two steps, and what would the rate law contain if the second step were the slower one instead?
Explain why a catalyst is written above the arrow of a chemical equation rather than among the reactants, and state what would be observed about the mass of the catalyst at the end of the reaction.
Sketch, on the same axes, the potential energy profile of an uncatalysed reaction and of the same reaction in the presence of a catalyst. Mark \((E_a)_1\), \((E_a)_2\) and \(\Delta H\), and state which of these three quantities the catalyst has changed.
For the reaction, \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}\), what is the relationship among \(\dfrac{d[\mathrm{N_2}]}{dt}\), \(\dfrac{d[\mathrm{H_2}]}{dt}\) and \(\dfrac{d[\mathrm{NH_3}]}{dt}\)?
For the reaction, \(\mathrm{CH_3Br}\text{(aq)} + \mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{CH_3OH}\text{(aq)} + \mathrm{Br^-}\text{(aq)}\), the rate law is rate \(= k[\mathrm{CH_3Br}][\mathrm{OH^-}]\).
- How does the reaction rate change if \([\mathrm{OH^-}]\) is decreased by a factor of 5?
- What is the change in rate if the concentrations of both reactants are doubled?
What is the relationship between the coefficients of reactants in a balanced equation for an overall reaction and the exponents in the rate law? In what case are the coefficients the exponents?
Why do all collisions between reactant molecules not lead to a chemical reaction?
What is the activation energy of a reaction?
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Write the Arrhenius equation and explain the terms involved in it.
What is the rate determining step?
Write the relationships between rate constant and half life of first order and zeroth order reactions.
How do the half lives of first order and zero order reactions change with the initial concentration of reactants?
How is the instantaneous rate of reaction determined?
Distinguish between order and molecularity of a reaction.
A reaction takes place in two steps,
\(\mathrm{NO}\text{(g)} + \mathrm{Cl_2}\text{(g)} \rightarrow \mathrm{NOCl_2}\text{(g)}\)
\(\mathrm{NOCl_2}\text{(g)} + \mathrm{NO}\text{(g)} \rightarrow 2\,\mathrm{NOCl}\text{(g)}\)
- Write the overall reaction.
- Identify the reaction intermediate.
- What is the molecularity of each step?
Obtain the relationship between the rate constant and half life of a first order reaction.
How will you represent a zeroth order reaction graphically?
What are pseudo-first order reactions? Give one example and explain why it is pseudo-first order.
How does a catalyst increase the rate of reaction? Explain with the help of a potential energy diagram for catalysed and uncatalysed reactions.
Explain with the help of the Arrhenius equation how the rate of reaction changes with (a) temperature and (b) activation energy.
Derive the integrated rate law for a first order reaction.
How will you represent first order reactions graphically?
Derive the integrated rate law for the first order reaction, \(\mathrm{A}\text{(g)} \rightarrow \mathrm{B}\text{(g)} + \mathrm{C}\text{(g)}\), in terms of pressure.
What is a zeroth order reaction? Derive its integrated rate law. What are the units of the rate constant?
How will you determine activation energy: (a) graphically using the Arrhenius equation (b) from rate constants at two different temperatures?
Explain graphically the effect of temperature on the rate of reaction.
If you wish to determine the reaction order and rate constant for the reaction, \(2\,\mathrm{AB_2} \rightarrow \mathrm{A_2} + 2\,\mathrm{B_2}\): a) What data would you collect? b) How would you use these data to determine whether the reaction is zeroth or first order?
6.12 Practice With Numericals
For the reaction \[4\,\mathrm{NH_3}\text{(g)} + 3\,\mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{N_2}\text{(g)} + 6\,\mathrm{H_2O}\text{(g)}\] nitrogen is formed at a rate of \(0.020\ \text{mol dm}^{-3}\text{s}^{-1}\). Write the general rate expression for the reaction, and calculate the rate of the reaction, the rates of consumption of \(\mathrm{NH_3}\) and \(\mathrm{O_2}\), and the rate of formation of \(\mathrm{H_2O}\).
For the reaction \(\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}\) the following initial-rate data were obtained at a constant temperature.
| [A]/mol \(\text{dm}^{-3}\) | [B]/mol \(\text{dm}^{-3}\) | Initial rate/mol \(\text{dm}^{-3}\text{s}^{-1}\) | |
|---|---|---|---|
| (i) | 0.10 | 0.10 | \(3.0 \times 10^{-4}\) |
| (ii) | 0.20 | 0.10 | \(6.0 \times 10^{-4}\) |
| (iii) | 0.10 | 0.20 | \(1.2 \times 10^{-3}\) |
Determine the order with respect to each reactant, write the rate law, state the overall order and calculate the rate constant with its unit.
State the order of each of the following reactions from the unit of its rate constant: (a) \(5.0 \times 10^{-2}\ \text{s}^{-1}\), (b) \(2.0 \times 10^{-3}\ \text{mol dm}^{-3}\text{s}^{-1}\), (c) \(1.4 \times 10^{-2}\ \text{mol}^{-1}\text{dm}^{3}\text{s}^{-1}\).
The rate law for a reaction is rate \(= k[\mathrm{A}]^2[\mathrm{B}]\), with \(k = 2.0 \times 10^{-2}\ \text{mol}^{-2}\text{dm}^{6}\text{s}^{-1}\). Calculate the rate of the reaction when \([\mathrm{A}] = 0.50\ \text{mol dm}^{-3}\) and \([\mathrm{B}] = 0.20\ \text{mol dm}^{-3}\), and state by what factor the rate would change if \([\mathrm{A}]\) alone were trebled.
The rate constant of a first order reaction is \(2.3 \times 10^{-3}\ \text{s}^{-1}\). Calculate its half life, and the time required for the concentration of the reactant to fall from \(0.50\ \text{mol dm}^{-3}\) to \(0.125\ \text{mol dm}^{-3}\).
A first order reaction is 20% complete in 15 minutes. Calculate the rate constant of the reaction and the time required for the reaction to be 60% complete.
The concentration of the reactant in a certain reaction falls from \(0.80\ \text{mol dm}^{-3}\) to \(0.40\ \text{mol dm}^{-3}\) in 10 minutes, and from \(0.40\ \text{mol dm}^{-3}\) to \(0.20\ \text{mol dm}^{-3}\) in the next 10 minutes. Deduce the order of the reaction, giving your reasoning, and calculate its rate constant.
The half life of a first order reaction is 30 minutes. Calculate the rate constant of the reaction and the percentage of reactant decomposed after 90 minutes.
A zero order reaction has a rate constant of \(1.5 \times 10^{-2}\ \text{mol dm}^{-3}\text{s}^{-1}\) and an initial reactant concentration of \(0.60\ \text{mol dm}^{-3}\). Calculate the concentration of the reactant after 20 s, the time at which the concentration falls to \(0.15\ \text{mol dm}^{-3}\), and the half life of the reaction.
For the first order gas phase reaction \(\mathrm{A}\text{(g)} \rightarrow \mathrm{B}\text{(g)} + \mathrm{C}\text{(g)}\) carried out at constant volume, the initial pressure of A is 0.60 bar and the total pressure of the mixture after 300 s is 0.75 bar. Calculate the pressure of A remaining at 300 s, the rate constant of the reaction, and its half life.
The rate constant of a reaction increases four times when the temperature is raised from 300 K to 320 K. Calculate the activation energy of the reaction. (\(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\))
The activation energy of a reaction is \(120\ \text{kJ mol}^{-1}\) and its rate constant at 400 K is \(5.0 \times 10^{-4}\ \text{s}^{-1}\). Calculate the rate constant at 450 K. (\(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\))
For a certain first order reaction the Arrhenius plot of \(\log_{10}k\) against \(1/T\) is a straight line of slope \(-6.5 \times 10^{3}\ \text{K}\) with an intercept of 12.0 on the \(\log_{10}k\) axis. Calculate the activation energy of the reaction in \(\text{kJ mol}^{-1}\) and the pre-exponential factor with its unit. (\(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\))
Calculate the fraction of molecules possessing energy equal to or greater than an activation energy of \(80\ \text{kJ mol}^{-1}\) at 300 K and at 350 K. Hence calculate the factor by which this fraction increases over the 50 K rise. (\(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\))
A catalyst lowers the activation energy of a reaction from \(90\ \text{kJ mol}^{-1}\) to \(60\ \text{kJ mol}^{-1}\) without altering the pre-exponential factor. Calculate the factor by which the rate constant increases at 298 K. (\(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\))
The half life of a first order reaction is 6.93 minutes. Calculate its rate constant, and the fraction of the reactant remaining unreacted after 20.79 minutes.
For the reaction \(2\mathrm{A} + \mathrm{B} \rightarrow\) products, find the rate law from the following data.
| [A]/M | [B]/M | rate/M \(\text{s}^{-1}\) |
|---|---|---|
| 0.3 | 0.05 | 0.15 |
| 0.6 | 0.05 | 0.30 |
| 0.6 | 0.2 | 1.20 |
In a first order reaction, the concentration of reactant decreases from \(20\ \text{mmol dm}^{-3}\) to \(8\ \text{mmol dm}^{-3}\) in 38 minutes. What is the half life of the reaction?
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react?
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at \(25\ ^\circ\text{C}\) is \(3.7 \times 10^{-5}\ \text{s}^{-1}\). What is the rate constant at \(30\ ^\circ\text{C}\)? (\(R = 8.314\ \text{J/K mol}\))
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K?
The rate constant of a reaction at \(500\ ^\circ\text{C}\) is \(1.6 \times 10^{3}\ \text{M}^{-1}\text{s}^{-1}\). What is the frequency factor of the reaction if its activation energy is 56 kJ/mol?
Show that the time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life.
The rate constant for a first order reaction is given by \(\log_{10}k = 14.34 - \dfrac{1.25 \times 10^{4}}{T}\). Calculate the activation energy of the reaction.
What fraction of molecules in a gas at 300 K collide with an energy equal to an activation energy of 50 kJ/mol?
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