3. Chemical Kinetics

What you will learn

• Define normalised, average and instantaneous rates.
• Find empirical orders from controlled data.
• Separate order from elementary-step molecularity.
• Apply zero- and first-order integrated laws and half-lives.
• Use Arrhenius comparisons with absolute temperature.
• Explain catalysis and the roles of energy and orientation in collisions.

Big idea: a possible reaction still needs a possible route.

Reaction speed connects observations in a container with molecular events. A concentration–time trace is not just a picture: its slope tells how quickly composition changes, and its shape tests a proposed rate law.

This chapter develops that connection in stages. Begin with a measured rate, identify a concentration dependence, integrate a suitable model, then ask why temperature and catalysts change its behaviour. Keep the assumptions visible at every stage.

Nineteenth-century measurements of sugar reactions helped establish quantitative time-dependent chemistry. Later work on mass action and Arrhenius temperature dependence supplied compact mathematical models. Their continuing value comes from comparing predictions with measurements, not from assuming every reaction has the same mechanism.

Thermodynamics asks whether a reaction is favoured. Kinetics asks how quickly the observable composition changes.

A reaction can be energetically favourable and still proceed very slowly because reaching a transition state requires a difficult molecular rearrangement. A catalyst can offer a faster route without changing the initial and final equilibrium states. Never infer speed from the sign of Gibbs energy alone.

For $aA+bB\rightarrow cC+dD$, the reaction rate is defined using stoichiometrically normalised concentration changes at constant volume.

$$v=-\frac1a\frac{d[A]}{dt}=-\frac1b\frac{d[B]}{dt}=\frac1c\frac{d[C]}{dt}=\frac1d\frac{d[D]}{dt}$$

Reactant concentrations decrease, so their derivatives are negative. The minus signs make the forward rate positive. Product appearance rates need not equal individual reactant disappearance rates; the stoichiometric coefficients connect them.

$$\mathrm{N_{2}} + 3 \mathrm{H_{2}} \longrightarrow 2 \mathrm{NH_{3}}$$

Hydrogen disappears at $0.0150\,\mathrm{mol\,L^{-1}\,s^{-1}}$ during the written ammonia reaction. Find the normalised reaction rate and ammonia appearance rate.

Divide the hydrogen disappearance rate by its coefficient. Multiply the resulting reaction rate by the product coefficient.

Result: $v=0.005\,\mathrm{mol\,L^{-1}\,s^{-1}}$ and $d[\mathrm{NH_3}]/dt=0.01\,\mathrm{mol\,L^{-1}\,s^{-1}}$.

An average rate uses a finite time interval: the slope of a secant on a concentration–time curve. An instantaneous rate uses the tangent at one time. A curved graph therefore has different average rates over different intervals. The initial rate is the instantaneous rate at the start of an experiment.

A reactant concentration falls from $1.20$ to $0.900\,\mathrm{mol\,L^{-1}}$ over $50.0\,\mathrm s$. Find its average disappearance rate.

Subtract the final concentration from the initial one and divide by the elapsed time.

Result: $0.006\,\mathrm{mol\,L^{-1}\,s^{-1}}$. This is the reactant disappearance rate; use a stoichiometric divisor if the question asks for a normalised reaction rate.

Rate units combine a concentration unit with inverse time. A question may use seconds, minutes or hours; keep the time unit consistent with the reported rate constant. A gas reaction can be followed through pressure when temperature, volume and stoichiometry supply a valid pressure–concentration relation.

1. What geometric feature gives an instantaneous rate on a concentration–time graph?
2. Does a favourable equilibrium imply a fast reaction?

Answers

1. The tangent slope, with the sign and coefficient appropriate to the species.
2. No. A large kinetic barrier can make a favourable reaction slow.

A balanced equation tells you what is conserved. Experiments tell you how the rate responds to composition.

Concentration or gas pressure changes how often relevant particles meet. Temperature changes molecular energy distributions and many rate constants. Catalysts alter available pathways. For a heterogeneous reaction, exposed surface area can affect the observed rate. Light can matter when photon absorption begins a reaction.

An empirical rate law may be written $v=k[A]^x[B]^y$. The exponents are found from measured rates under controlled conditions. They are not generally the coefficients of the overall balanced reaction. A proposed mechanism must be consistent with the measured law.

Use the initial-rate data to determine the exponents and rate constant.

From I to II only A doubles, while rate increases fourfold. Thus the A exponent is two. From I to III only B doubles, while rate doubles. Thus the B exponent is one.

Substitute one complete trial into $k=v/([A]^2[B])$.

Result: $v=k[A]^2[B]$, total order $=3$, and $k=2\,\mathrm{L^2\,mol^{-2}\,s^{-1}}$.

The total order is the sum of the empirical exponents. Orders can be zero, fractional or negative in suitable measured rate laws. They describe a dependence under specified conditions, not a count of molecules in an imagined overall collision.

$$[k]=(\text{concentration})^{1-n}(\text{time})^{-1}$$

For a zero-order law, the rate constant has rate units. For first order it has inverse-time units. For second order it has inverse-concentration, inverse-time units. Write the units from the measured law instead of memorising a table without understanding it.

Molecularity counts the reacting species in one elementary step. It is a positive integer for the elementary step being described. An overall multi-step equation does not have a molecularity that can be read simply by adding its coefficients. An elementary step can have a rate law linked to its molecularity, but an overall empirical law needs kinetic evidence.

In a pseudo-first-order experiment, one reactant is present in such large excess that its concentration changes negligibly. Its nearly constant contribution is absorbed into an observed rate constant. This does not change the underlying stoichiometry or automatically make the true overall order one.

1. Rate doubles when A doubles while all other variables stay fixed. What exponent is consistent with this comparison?
2. Can the order of a measured overall reaction be fractional?

Answers

1. One.
2. Yes. Molecularity of a specified elementary step, however, is an integer.

An integrated rate equation predicts composition after time has passed. Its graph gives a practical way to test the proposed order.

For disappearance of A with $-d[A]/dt=k$, integrate to obtain $[A]_t=[A]_0-kt$. A plot of concentration against time is straight, with slope $-k$, only while the zero-order model remains valid.

A zero-order model has $[A]_0=0.600\,\mathrm M$ and $k=0.00500\,\mathrm{M\,s^{-1}}$. Find concentration after $80.0\,\mathrm s$.

Use $[A]_t=[A]_0-kt$. Check that the prediction has not crossed below zero.

Result: $[A]_t=0.2\,\mathrm M$. The model must not be extended to negative concentration.

For $-d[A]/dt=k[A]$, separation and integration give $\ln([A]_0/[A]_t)=kt$. Equivalent forms are useful for different tasks.

$$[A]_t=[A]_0e^{-kt},\qquad k=\frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]_t}$$

A straight line of natural logarithm of concentration against time has slope minus the first-order rate constant. The amount may be concentration, or another measured quantity proportional to unreacted A under fixed conditions. Do not insert total gas pressure if product gases also contribute.

For a gas reaction $A\rightarrow B+C$ at fixed temperature and volume, the initial pressure of pure A is $0.800\,\mathrm{bar}$. After $120\,\mathrm s$, total pressure is $1.200\,\mathrm{bar}$. Find a first-order rate constant.

Let the pressure-equivalent amount reacted be $u$. Then $p_A=p_0-u$ while $p_{\rm total}=p_0+u$. Hence $p_A=2p_0-p_{\rm total}$.

Only the remaining A pressure belongs in the concentration ratio.

Result: $p_A=0.4\,\mathrm{bar}$ and $k=0.0057762\,\mathrm{s^{-1}}$.

A half-life is the time for the reactant amount to fall to half its starting value for the interval being considered. Substitute that fraction into the integrated law instead of treating every reaction as first order.

$$t_{1/2}^{\rm zero}=\frac{[A]_0}{2k},\qquad t_{1/2}^{\rm first}=\frac{\ln2}{k}$$

The zero-order example above has half-life $60\,\mathrm s$. A first-order reaction whose half-life is $90.0\,\mathrm s$ has $k=0.00770164\,\mathrm{s^{-1}}$. Its successive half-lives are equal, so after two half-lives one-quarter remains.

Pseudo-first-order hydrolysis illustrates an important approximation. If solvent water is in large excess, its activity is effectively constant. A measured first-order dependence on the dilute reactant can then describe the experiment even if water participates in the chemical step.

1. A first-order reactant goes through three half-lives. What fraction remains?
2. Which graph is linear for a first-order disappearance?

Answers

1. $0.125$.
2. Natural logarithm of reactant concentration against time.

Temperature changes the population of molecules able to reach a transition state. That is why a modest temperature change can produce a large rate change.

The activation energy describes the temperature sensitivity in an Arrhenius model. It is connected with the energetic barrier along a pathway, not with the net reaction enthalpy. An exothermic reaction can still require activation.

$$k=Ae^{-E_a/(RT)},\qquad\ln k=\ln A-\frac{E_a}{R}\frac1T$$

Temperature must be absolute. The pre-exponential factor includes encounter and orientation effects in this simple picture. When the model applies over the selected range, a plot of natural logarithm of rate constant against inverse temperature has slope minus activation energy divided by the gas constant.

$$\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac1{T_1}-\frac1{T_2}\right)$$

A rate constant increases from $0.00300\,\mathrm{s^{-1}}$ at $310\,\mathrm K$ to $0.0120\,\mathrm{s^{-1}}$ at $330\,\mathrm K$. Find the Arrhenius activation energy.

The ratio is greater than one, and $(1/T_1-1/T_2)$ is positive. Their combination gives a positive activation energy.

Use $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ before converting joules to kilojoules.

Result: $E_a=58953.71\,\mathrm{J\,mol^{-1}}$, or $58.954\,\mathrm{kJ\,mol^{-1}}$.

A catalyst provides an alternative sequence of elementary steps. Its lower effective barrier can increase the rate without changing the energies of the same initial reactants and final products. The reverse reaction also has access to the alternative pathway. Therefore a catalyst changes how quickly equilibrium is reached, not the equilibrium constant at the same temperature.

A reaction-coordinate diagram uses progress along a pathway on the horizontal axis, not elapsed time. A peak represents a transition state in the simplified picture. The reactant-to-peak energy difference is the forward barrier; the reactant-to-product difference is the reaction energy change. A multi-step pathway can have several peaks.

1. Does a catalyst change the equilibrium constant for the same overall reaction at a fixed temperature?
2. What is the slope of ln k against 1/T in the Arrhenius model?

Answers

1. No.
2. $-E_a/R$.

A collision is an opportunity to react, not a guarantee of reaction.

Collision theory starts with encounters between particles. A useful encounter needs enough energy along the relevant reaction motion and a suitable molecular orientation. A collision that merely exchanges kinetic energy can leave both molecules chemically unchanged.

For a simple bimolecular gas model, a useful symbolic expression is $k=PZ_{AB}e^{-E_a/(RT)}$. Here $Z_{AB}$ is a collision-frequency factor and $P$ is an orientation or steric factor. The exact units depend on the convention used for the collision factor.

The exponential factor estimates the temperature-sensitive fraction associated with crossing an energy requirement. The steric factor recognises that a reactive site must face the correct partner. Large molecules can collide frequently while only a small set of orientations leads to products.

Two proposed pathways have similar encounter frequency and activation energy, but one requires a very specific end-on approach. Which additional factor should be examined?

Energy alone does not determine the probability of success. An approach that places the reactive atoms in the wrong orientation may not form the required transition state.

Result: The steric or orientation factor; the more restrictive pathway can have a smaller effective pre-exponential contribution.

This model is most direct for simple gas-phase elementary reactions. Solvent motion, diffusion, complex mechanisms and quantum effects can require richer models. Use it to explain why encounter frequency, energy and orientation all matter, rather than to claim that every measured rate law comes from one collision.

A strong exam explanation separates three changes. Raising concentration changes encounter opportunities. Raising temperature changes the energy distribution and often the encounter frequency. Adding a catalyst changes the pathway. These changes are not interchangeable, even when each can make a reaction faster.

1. Name two requirements beyond mere contact for a successful elementary collision.
2. Is a reaction-coordinate axis a time axis?

Answers

1. Sufficient relevant energy and a suitable orientation.
2. No. It tracks progression along a molecular pathway.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

3.1 Distinguish thermodynamic feasibility from reaction speed.

3.2 For $2A\rightarrow3B$, A disappears at $0.020\,\mathrm{M\,s^{-1}}$. Find reaction rate and B appearance rate.

3.3 Concentration falls from $0.850$ to $0.610\,\mathrm M$ in $40.0\,\mathrm s$. Find average disappearance rate.

3.4 Explain why a tangent rate and an interval-average rate can differ.

3.5 Why is a rate law normally established experimentally?

3.6 A concentration is doubled and rate rises eightfold with other variables fixed. Find its exponent.

3.7 B is doubled but the rate does not change. What measured order in B is consistent with this test?

3.8 For $v=k[A]^{1/2}[B]$, find total order.

3.9 Find the units of k for an overall third-order law with rate in $\mathrm{mol\,L^{-1}\,s^{-1}}$.

3.10 Why is “fractional molecularity” inappropriate for a specified elementary step?

3.11 A zero-order law has $[A]_0=0.450\,\mathrm M$ and $k=0.00300\,\mathrm{M\,s^{-1}}$. Find concentration after $60.0\,\mathrm s$.

3.12 Find the zero-order half-life for the preceding conditions.

3.13 Why must the zero-order linear law stop being extrapolated after reactant exhaustion?

3.14 A first-order reaction has $k=0.0120\,\mathrm{s^{-1}}$. Find its half-life.

3.15 For that first-order reaction, what fraction remains after $100\,\mathrm s$?

3.16 A first-order reactant falls to one-fifth of its initial amount in $200\,\mathrm s$. Find k.

3.17 Which graph distinguishes a zero-order law most directly?

3.18 Which graph distinguishes a first-order law most directly?

3.19 What fraction remains after four first-order half-lives?

3.20 What creates pseudo-first-order behaviour?

3.21 At fixed volume and temperature, pure A begins at $0.600\,\mathrm{bar}$ in $A\rightarrow B+C$. Later total pressure is $0.900\,\mathrm{bar}$. Find remaining A pressure.

3.22 Why is total pressure unsuitable as the unreacted-A concentration in that gas example?

3.23 Find the slope of an Arrhenius ln k versus 1/T line for $E_a=45000\,\mathrm{J\,mol^{-1}}$ and $R=8.314$.

3.24 If activation energy is positive, what does the Arrhenius model predict on warming?

3.25 For $E_a=50000\,\mathrm{J\,mol^{-1}}$, find $k(320\,\mathrm K)/k(300\,\mathrm K)$ using $R=8.314$.

3.26 What changes in a catalyst’s reaction-coordinate diagram, and what remains fixed?

3.27 Why does a catalyst speed approach to equilibrium from either side?

3.28 Give one reason why frequent collisions can still produce a slow reaction.

3.29 A measured reaction order disagrees with the exponents predicted by a proposed single elementary step. What should be reconsidered?

3.30 Outline a reliable method for assigning order from kinetic data.

Answers

3.1 Feasibility concerns the direction of Gibbs-energy change; speed concerns the kinetic route and barrier.

3.2 $v=0.01\,\mathrm{M\,s^{-1}}$ and B appears at $0.03\,\mathrm{M\,s^{-1}}$.

3.3 $0.006\,\mathrm{M\,s^{-1}}$.

3.4 A curved concentration–time graph changes slope, so a tangent and a secant usually describe different slopes.

3.5 The overall stoichiometry does not uniquely specify the elementary pathway or its concentration dependence.

3.6 The exponent is $3$.

3.7 Zero order in B over the tested conditions.

3.8 $1.5$.

3.9 $\mathrm{L^2\,mol^{-2}\,s^{-1}}$.

3.10 Molecularity counts discrete reacting species in that elementary event.

3.11 $0.27\,\mathrm M$.

3.12 $75\,\mathrm s$.

3.13 It would predict negative concentration, which is physically impossible.

3.14 $57.762\,\mathrm s$.

3.15 $0.30119$.

3.16 $0.0080472\,\mathrm{s^{-1}}$.

3.17 Concentration against time is linear with a constant negative slope during its valid range.

3.18 Natural logarithm of concentration against time is linear.

3.19 $0.0625$.

3.20 A participating reactant is kept in such large excess that its activity or concentration is effectively constant.

3.21 $0.3\,\mathrm{bar}$.

3.22 Products contribute to the measured total pressure, while the rate-law amount must refer to remaining A.

3.23 $-5412.56\,\mathrm K$.

3.24 The exponential suppression becomes smaller and the rate constant increases, if the other model parameters remain applicable.

3.25 $3.5005$.

3.26 The available path and its barriers change. The energies of the same initial reactants and final products, and hence their net energy difference, remain fixed.

3.27 The alternative pathway is available in both directions; it does not selectively change the equilibrium condition.

3.28 Only a small fraction may have suitable energy and orientation for productive rearrangement.

3.29 The proposed mechanism, the elementary-step assumption or the conditions used to infer the rate law.

3.30 Hold other variables fixed when comparing initial rates, test appropriate integrated plots, state the range of validity and check the derived rate-constant units.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

What changes in a catalyst’s reaction-coordinate diagram, and what remains fixed?

The available path and its barriers change. The energies of the same initial reactants and final products, and hence their net energy difference, remain fixed.

Why does a catalyst speed approach to equilibrium from either side?

The alternative pathway is available in both directions; it does not selectively change the equilibrium condition.

Give one reason why frequent collisions can still produce a slow reaction.

Only a small fraction may have suitable energy and orientation for productive rearrangement.

A measured reaction order disagrees with the exponents predicted by a proposed single elementary step. What should be reconsidered?

The proposed mechanism, the elementary-step assumption or the conditions used to infer the rate law.

Outline a reliable method for assigning order from kinetic data.

Hold other variables fixed when comparing initial rates, test appropriate integrated plots, state the range of validity and check the derived rate-constant units.