"Chemical Kinetics" — Practice

Question 1

For a reaction between A and B, tripling $[A]$ at fixed $[B]$ triples the rate, while doubling $[B]$ at fixed $[A]$ quadruples it. Write the rate law, give the overall order and the units of $k$.

Accepted Answer

Tripling A triples the rate, so the order in A is 1. Doubling B quadruples the rate, so the order in B is 2. The rate law is $	ext{rate} = k[A][B]^2$ and the overall order is 3. The units follow from $\mathrm{mol^{1-n}L^{n-1}s^{-1}}$ with $n = 3$, giving $\mathrm{L^2\,mol^{-2}\,s^{-1}}$. Note that the exponents were read from experiment, and nothing in the balanced equation was needed or used.

Question 2

State four differences between order and molecularity.

Accepted Answer

Order is determined experimentally while molecularity is deduced from a proposed mechanism. Order applies to any reaction, whereas molecularity is defined only for an elementary step. Order may be zero, fractional or negative, while molecularity is always a positive whole number. Molecularity can never exceed three, since simultaneous four-body collisions are vanishingly rare, whereas order has no such upper limit. The two coincide only for a single-step reaction.

Question 3

The reaction $\mathrm{2NO_2 + F_2 \rightarrow 2NO_2F}$ is found to be first order in each reactant. What does this reveal about the mechanism?

Accepted Answer

The balanced equation would suggest second order in $\mathrm{NO_2}$, so the measured law rules out a single elementary step. A rate law first order in each reactant means the rate-determining step involves one molecule of each, so the slow step is one $\mathrm{NO_2}$ meeting one $\mathrm{F_2}$. The second $\mathrm{NO_2}$ is consumed in a subsequent fast step, which does not appear in the rate law because it never limits anything. The general principle is that the rate law reflects the slow step and everything preceding it.

Question 4

A first-order reaction is 25 per cent complete in 40 minutes. Calculate its rate constant and its half-life.

Accepted Answer

If 25 per cent has reacted, 75 per cent remains. Using $k = (2.303/t)\log([A]_0/[A])$, we have $k = (2.303/40)\log(100/75)$. The logarithm is 0.1249, so $k = 0.05758 	imes 0.1249 = 7.19 	imes 10^{-3}\ \mathrm{min^{-1}}$. The half-life is $0.693/k = 96.4$ minutes. Note that the half-life exceeds the time given, which is consistent: less than half the material has yet reacted.

Question 5

A reaction's concentration falls from 1.00 M to 0.50 M in 15 minutes and from 0.50 M to 0.25 M in a further 15 minutes. Identify the order and find the rate constant.

Accepted Answer

The two successive half-lives are equal at 15 minutes each, and a half-life independent of concentration is the signature of first order. Had the reaction been zero order the second half-life would have been 7.5 minutes, half the first; had it been second order the second half-life would have been 30 minutes, double the first. The rate constant is $k = 0.693/15 = 0.0462\ \mathrm{min^{-1}}$.

Question 6

A zero-order reaction has $k = 0.020\ \mathrm{mol\ L^{-1}\ s^{-1}}$ and starts at 0.80 M. Find the half-life and the time for complete reaction.

Accepted Answer

For zero order the half-life is $[A]_0/2k = 0.80/0.040 = 20$ seconds. Complete reaction occurs when the concentration reaches zero, at $t = [A]_0/k = 0.80/0.020 = 40$ seconds. Zero order is the only case where a reaction genuinely finishes at a definite time rather than approaching completion asymptotically, and its successive half-lives halve, being 20 s then 10 s then 5 s.

Question 7

A rate constant is quoted as $4.5 	imes 10^{-3}\ \mathrm{L\ mol^{-1}\ s^{-1}}$. What is the order, and what does the concentration dependence look like?

Accepted Answer

Units of L mol$^{-1}$ s$^{-1}$ correspond to $n = 2$ in $\mathrm{mol^{1-n}L^{n-1}s^{-1}}$, so the reaction is second order overall. The rate therefore quadruples when a single reactant's concentration is doubled, or doubles when either of two reactants is doubled if the order is one in each. The half-life is inversely proportional to initial concentration, so a more dilute solution takes proportionally longer to reach the halfway point and successive half-lives double.

Question 8

The rate of a reaction triples when the temperature is raised from 300 K to 310 K. Calculate the activation energy.

Accepted Answer

Using $\log(k_2/k_1) = (E_a/2.303R)\left[(T_2-T_1)/(T_1T_2)ight]$, we have $\log 3 = 0.4771$, $T_2 - T_1 = 10$ and $T_1T_2 = 93000$. So $E_a = 0.4771 	imes 2.303 	imes 8.314 	imes 9300 = 8.50 	imes 10^4\ \mathrm{J\ mol^{-1}}$, about 85.0 kJ mol$^{-1}$. A larger activation energy than one giving a mere doubling is expected, since a bigger barrier makes a reaction more temperature sensitive.

Question 9

Explain why a 10 K rise near room temperature can double a reaction rate when it raises the mean molecular speed by under 2 per cent.

Accepted Answer

Because the rate depends on the fraction of molecules with energy above the activation barrier, not on the average energy. That fraction sits far out in the tail of the Maxwell-Boltzmann distribution, where the curve falls steeply, so a small rightward shift of the whole distribution produces a large proportional increase in the area beyond the barrier. The dependence is exponential through $e^{-E_a/RT}$. Note also that the effect shrinks at higher temperature, since the Arrhenius bracket contains the change in $1/T$ rather than in $T$.

Question 10

A catalyst lowers the activation energy of a reaction from 100 kJ mol$^{-1}$ to 80 kJ mol$^{-1}$ at 300 K. By what factor does the rate increase?

Accepted Answer

The ratio of rate constants is $e^{\Delta E_a/RT}$, assuming the frequency factor is unchanged. The difference is 20000 J mol$^{-1}$ and $RT = 8.314 	imes 300 = 2494$, so the exponent is 8.02 and the factor is $e^{8.02} \approx 3.0 	imes 10^{3}$. A one-fifth reduction in the barrier speeds the reaction three thousandfold, because the barrier sits in an exponent. Enzymes do far more: catalase removes 67 kJ mol$^{-1}$ and gains a factor near $5 	imes 10^{11}$.

Question 11

Explain why the hydrolysis of ethyl acetate in dilute aqueous solution is described as pseudo-first order.

Accepted Answer

The true rate law is second order overall, first order in the ester and first order in water. But in dilute solution water is also the solvent, present at about 55 M, so consuming even 0.1 mol L$^{-1}$ of ester changes the water concentration by under 0.2 per cent. Water's concentration is therefore effectively constant and is absorbed into the rate constant, leaving a rate that depends only on the ester. The measured constant is the true one multiplied by the water concentration, and the units betray this: $k_{obs}$ is in s$^{-1}$ while the true $k_2$ is in L mol$^{-1}$ s$^{-1}$.

Question 12

Why does a catalyst not change the equilibrium constant of a reaction?

Accepted Answer

Because it provides an alternative pathway with a lower activation energy, and both the forward and reverse reactions travel that same pathway. The barrier is lowered by the same amount in each direction, so both rate constants rise by the same factor and the ratio at which they balance, which is the equilibrium constant, is unchanged. The initial and final states are untouched, so $\Delta H$ and $\Delta G$ are also unchanged. A catalyst shortens the time to reach equilibrium and never alters where it lies.

Practice — "Chemical Kinetics"

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