"Chemical Kinetics" — Practice

Question 1

For a reaction, doubling $\left[	ext{A}ight]$ quadruples the rate while doubling $\left[	ext{B}ight]$ leaves it unchanged. Write the rate law, give the overall order, and state the units of $k$.

Accepted Answer

Doubling $\left[	ext{A}ight]$ multiplies the rate by four, so the order in $	ext{A}$ is $2$.

Doubling $\left[	ext{B}ight]$ changes nothing, so the order in $	ext{B}$ is $0$.

Rate $=k\left[	ext{A}ight]^{2}$, overall order $2$.

Units: $	ext{mol}^{-1}	ext{L}\,	ext{s}^{-1}$, that is L mol$^{-1}$ s$^{-1}$.

Question 2

A first-order reaction is $20\%$ complete in $10$ minutes. Find the rate constant, the half-life, and the time for $90\%$ completion.

Accepted Answer

$k=\dfrac{2.303}{10}\log\dfrac{100}{80}=\dfrac{2.303	imes0.0969}{10}=0.0223$ min$^{-1}$

$t_{1/2}=\dfrac{0.693}{0.0223}=31.1$ min

$t_{90}=\dfrac{2.303}{0.0223}\log10=\dfrac{2.303}{0.0223}=103$ min

As a check, $90\%$ completion is $\log10$ against $\log2$ for the half-life, a ratio of $3.32$ — and $31.1	imes3.32=103$ min.

Question 3

The half-life of a reaction is $50$ minutes at an initial concentration of $0.1$ M and $100$ minutes at $0.05$ M. Determine the order.

Accepted Answer

$t_{1/2}\propto\left[	ext{A}ight]_0^{1-n}$

Halving the concentration doubled the half-life, so $t_{1/2}\propto\left[	ext{A}ight]_0^{-1}$.

$1-n=-1\qquad\Rightarrow\qquad n=2$

The reaction is **second order**, and no graph or rate constant was needed to establish it.

Question 4

A zero-order reaction has $k=2	imes10^{-3}$ mol L$^{-1}$ s$^{-1}$ and an initial concentration of $0.5$ M. Find the half-life and the time for complete consumption.

Accepted Answer

$t_{1/2}=\dfrac{\left[	ext{A}ight]_0}{2k}=\dfrac{0.5}{2	imes2	imes10^{-3}}=125$ s

Complete consumption: $\left[	ext{A}ight]=0$ when $t=\dfrac{\left[	ext{A}ight]_0}{k}=\dfrac{0.5}{2	imes10^{-3}}=250$ s

Exactly twice the half-life, which is characteristic of zero order: the reaction genuinely finishes rather than approaching completion asymptotically.

Question 5

The rate constant of a reaction triples when the temperature rises from $290$ K to $300$ K. Find the activation energy.

Accepted Answer

$\ln3=\dfrac{E_a}{8.314}\left(\dfrac{1}{290}-\dfrac{1}{300}ight)=\dfrac{E_a}{8.314}	imes1.149	imes10^{-4}$

$1.0986=\dfrac{E_a}{8.314}	imes1.149	imes10^{-4}$

$E_a=\dfrac{1.0986	imes8.314}{1.149	imes10^{-4}}=7.95	imes10^{4}$ J mol$^{-1}$

About $79$ kJ mol$^{-1}$, noticeably higher than the roughly $54$ kJ mol$^{-1}$ that a mere doubling would imply.

Question 6

Two reactions have activation energies of $40$ and $80$ kJ mol$^{-1}$. Compare the factor by which each speeds up when the temperature rises from $300$ K to $310$ K.

Accepted Answer

For $40$ kJ mol$^{-1}$: $\ln\dfrac{k_2}{k_1}=\dfrac{40000}{8.314}	imes1.075	imes10^{-4}=0.517$, so the factor is $1.68$.

For $80$ kJ mol$^{-1}$: $\ln\dfrac{k_2}{k_1}=1.035$, so the factor is $2.81$.

The reaction with the **higher** activation energy is far more temperature-sensitive.

This is precisely why heating a mixture of competing reactions favours whichever route has the larger barrier.

Question 7

Estimate the fraction of molecules with energy exceeding $100$ kJ mol$^{-1}$ at $300$ K and at $600$ K, and comment.

Accepted Answer

At $300$ K: $\dfrac{E_a}{RT}=\dfrac{100000}{8.314	imes300}=40.1$, giving $f=e^{-40.1}=4	imes10^{-18}$.

At $600$ K: $\dfrac{E_a}{RT}=20.05$, giving $f=e^{-20.05}=2	imes10^{-9}$.

Doubling the temperature multiplies the reactive fraction by about $5	imes10^{8}$.

The average molecular energy has only doubled, but the tail beyond a high threshold has grown by nine orders of magnitude.

Question 8

A reaction $2	ext{A}+	ext{B}ightarrow	ext{P}$ proceeds by $	ext{A}+	ext{B}ightarrow	ext{I}$ (fast equilibrium) followed by $	ext{I}+	ext{A}ightarrow	ext{P}$ (slow). Derive the rate law.

Accepted Answer

Rate $=k_2\left[	ext{I}ight]\left[	ext{A}ight]$

The fast equilibrium gives $K=\dfrac{\left[	ext{I}ight]}{\left[	ext{A}ight]\left[	ext{B}ight]}$, so $\left[	ext{I}ight]=K\left[	ext{A}ight]\left[	ext{B}ight]$.

Rate $=k_2K\left[	ext{A}ight]^{2}\left[	ext{B}ight]$

Second order in $	ext{A}$ and first in $	ext{B}$, third order overall — which here does match the stoichiometry, though only by coincidence.

Question 9

For the decomposition of ozone, the mechanism is $	ext{O}_3ightleftharpoons	ext{O}_2+	ext{O}$ (fast equilibrium) followed by $	ext{O}+	ext{O}_3ightarrow2	ext{O}_2$ (slow). Derive the rate law and explain the observed inhibition by oxygen.

Accepted Answer

Rate $=k_2\left[	ext{O}ight]\left[	ext{O}_3ight]$

From the equilibrium, $K=\dfrac{\left[	ext{O}_2ight]\left[	ext{O}ight]}{\left[	ext{O}_3ight]}$, so $\left[	ext{O}ight]=\dfrac{K\left[	ext{O}_3ight]}{\left[	ext{O}_2ight]}$.

Rate $=\dfrac{k_2K\left[	ext{O}_3ight]^{2}}{\left[	ext{O}_2ight]}$

The rate is **inversely** proportional to oxygen, so accumulating product slows the reaction — a negative order, which no stoichiometric reading could ever produce.

Question 10

Explain why the acid-catalysed hydrolysis of an ester in dilute aqueous solution appears first order, and how the true order could be established.

Accepted Answer

Water is present at about $55$ M and is barely depleted, so its concentration is effectively constant and is absorbed into the observed rate constant.

The acid is a catalyst and is not consumed at all.

Only the ester concentration changes appreciably, so the reaction appears first order with $k'=k\left[	ext{H}_2	ext{O}ight]$.

The true order in water could be established by running the reaction in a non-aqueous solvent with controlled small amounts of water added.

Question 11

A catalyst raises the rate of a reaction by a factor of $10^{6}$ at $300$ K. Find the reduction in activation energy it produces.

Accepted Answer

$\ln\left(10^{6}ight)=\dfrac{\Delta E_a}{8.314	imes300}$

$13.82=\dfrac{\Delta E_a}{2494}$

$\Delta E_a=3.45	imes10^{4}$ J mol$^{-1}$, about $34.5$ kJ mol$^{-1}$

A millionfold acceleration from a barrier reduction smaller than a typical hydrogen bond energy — which is why enzymes achieve such enormous rate enhancements from modest binding effects.

Question 12

A substance decomposes by two parallel first-order paths whose activation energies are $60$ and $90$ kJ mol$^{-1}$, with equal pre-exponential factors. Find the product ratio at $300$ K and at $500$ K.

Accepted Answer

$\dfrac{k_1}{k_2}=e^{\left(90000-60000ight)/RT}$

At $300$ K: $e^{30000/2494}=e^{12.03}=1.7	imes10^{5}$

At $500$ K: $e^{30000/4157}=e^{7.22}=1.4	imes10^{3}$

The low-barrier product dominates overwhelmingly at low temperature, but its advantage falls by more than a hundredfold on heating — which is the basis of thermodynamic against kinetic control in synthesis.

Practice — "Chemical Kinetics"

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