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Temperature Dependence, Arrhenius Equation & Collision Theory — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Temperature Dependence, Arrhenius Equation & Collision Theory MCQs with step-by-step solutions (30 questions). Part of Chemical Kinetics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The rate constant of a reaction doubles when the temperature rises from $300\,K$ to $320\,K$. The activation energy is: ($\log 2 = 0.301$, $R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $\approx 27.7\,kJ\,mol^{-1}$ ✓ Correct
B. $\approx 13.8\,kJ\,mol^{-1}$
C. $\approx 55.3\,kJ\,mol^{-1}$
D. $\approx 40\,kJ\,mol^{-1}$
Solution: $\log 2 = \dfrac{E_a}{2.303R}\cdot\dfrac{320-300}{300\times320}$; $E_a = \dfrac{0.301\times19.147}{2.083\times10^{-4}} = 27.7\,kJ\,mol^{-1}$.
Q2 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a reaction at $400\,K$ the fraction of effective collisions is $e^{-E_a/RT} = 2.0 \times 10^{-6}$ and the rate constant is $k = 5.0 \times 10^{4}\,s^{-1}$. The pre-exponential (frequency) factor $A$ is:
A. $2.5 \times 10^{10}\,s^{-1}$ ✓ Correct
B. $4.0 \times 10^{9}\,s^{-1}$
C. $1.0 \times 10^{-1}\,s^{-1}$
D. $2.5 \times 10^{-2}\,s^{-1}$
Solution: $k = A\,e^{-E_a/RT} \Rightarrow A = \dfrac{k}{e^{-E_a/RT}} = \dfrac{5.0\times10^{4}}{2.0\times10^{-6}} = 2.5\times10^{10}\,s^{-1}$.
Q3 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The rate constant of a reaction increases $2.5$-fold when the temperature is raised from $300\,K$ to $310\,K$ (temperature coefficient $= 2.5$). The activation energy is: ($\log 2.5 = 0.398$, $R = 8.314$)
A. $\approx 141\,kJ\,mol^{-1}$
B. $\approx 35.4\,kJ\,mol^{-1}$
C. $\approx 70.9\,kJ\,mol^{-1}$ ✓ Correct
D. $\approx 53.6\,kJ\,mol^{-1}$
Solution: $\log 2.5 = \dfrac{E_a}{2.303R}\cdot\dfrac{10}{300\times310}$; $E_a = \dfrac{0.398\times19.147}{1.075\times10^{-4}} = 70.9\,kJ\,mol^{-1}$.
Q4 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The fraction of molecules with energy $\ge E_a$ is $f = e^{-E_a/RT}$. For $E_a = 50\,kJ\,mol^{-1}$, the ratio $\dfrac{f_{310}}{f_{300}}$ (at $310\,K$ and $300\,K$) is about: ($R = 8.314$)
A. $\approx 1.0$
B. $\approx 1.9$ ✓ Correct
C. $\approx 3.8$
D. $\approx 0.52$
Solution: $\dfrac{f_{310}}{f_{300}} = e^{\frac{E_a}{R}(\frac{1}{300}-\frac{1}{310})} = e^{\frac{50000}{8.314}\times1.075\times10^{-4}} = e^{0.647} = 1.9$.
Q5 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A catalyst lowers the activation energy of a reaction by $8\,kJ\,mol^{-1}$ at $300\,K$ (frequency factor unchanged). The rate constant increases by a factor of about: ($R = 8.314$)
A. $\approx 8$
B. $\approx 25$ ✓ Correct
C. $\approx 2.7$
D. $\approx 3.2$
Solution: Factor $= e^{\Delta E_a/RT} = e^{8000/(8.314\times300)} = e^{3.21} = 24.7 \approx 25$.
Q6 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a bimolecular reaction the measured frequency factor is $A = 6 \times 10^{9}\,L\,mol^{-1}s^{-1}$, while collision theory predicts a collision frequency $Z = 3 \times 10^{11}\,L\,mol^{-1}s^{-1}$. The steric (probability) factor $P$ is:
A. $50$
B. $2 \times 10^{-3}$
C. $0.2$
D. $0.02$ ✓ Correct
Solution: $A = PZ \Rightarrow P = \dfrac{A}{Z} = \dfrac{6\times10^{9}}{3\times10^{11}} = 0.02$.
Q7 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A reaction has $E_a = 100\,kJ\,mol^{-1}$. By what factor does its rate constant increase when the temperature rises from $300\,K$ to $310\,K$? ($R = 8.314$)
A. $\approx 10$
B. $\approx 2.0$
C. $\approx 1.9$
D. $\approx 3.6$ ✓ Correct
Solution: $\log\dfrac{k_2}{k_1} = \dfrac{100000}{2.303\times8.314}\times\dfrac{10}{300\times310} = 0.562$; factor $= 10^{0.562} = 3.6$.
Q8 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a reaction the threshold energy is $E_T = E_a + E_{avg}$, where the average molar energy of the reactants at $300\,K$ is $E_{avg} = \dfrac{3}{2}RT$. If $E_a = 60\,kJ\,mol^{-1}$, the threshold energy is: ($R = 8.314$)
A. $\approx 63.7\,kJ\,mol^{-1}$ ✓ Correct
B. $\approx 62.5\,kJ\,mol^{-1}$
C. $\approx 56.3\,kJ\,mol^{-1}$
D. $\approx 60\,kJ\,mol^{-1}$
Solution: $E_{avg} = \dfrac{3}{2}RT = 1.5\times8.314\times300 = 3.74\,kJ$; $E_T = 60 + 3.74 = 63.7\,kJ\,mol^{-1}$.
Q9 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a reaction the plot of $\log k$ against $\dfrac{1}{T}$ is a straight line of slope $-5000\,K$. The activation energy is: ($R = 8.314$)
A. $\approx 9.6\,kJ\,mol^{-1}$
B. $\approx 191\,kJ\,mol^{-1}$
C. $\approx 41.6\,kJ\,mol^{-1}$
D. $\approx 95.7\,kJ\,mol^{-1}$ ✓ Correct
Solution: For a $\log k$ vs $1/T$ plot, slope $= -\dfrac{E_a}{2.303R}$, so $E_a = 5000\times2.303\times8.314 = 95.7\,kJ\,mol^{-1}$ (the $41.6$ forgets the $2.303$).
Q10 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a reaction the forward activation energy is $80\,kJ\,mol^{-1}$ and the reverse activation energy is $100\,kJ\,mol^{-1}$. A catalyst lowers the forward activation energy to $50\,kJ\,mol^{-1}$. The reverse activation energy in the presence of the catalyst is:
A. $30\,kJ\,mol^{-1}$
B. $50\,kJ\,mol^{-1}$
C. $70\,kJ\,mol^{-1}$ ✓ Correct
D. $100\,kJ\,mol^{-1}$
Solution: $\Delta H = E_{a,f} - E_{a,r} = 80-100 = -20\,kJ$ is unchanged by the catalyst; catalysed reverse $E_a = 50 - (-20) = 70\,kJ\,mol^{-1}$.
Q11 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A reaction has a temperature coefficient of $3$ (its rate constant triples per $10\,K$). By what factor does its rate increase when the temperature is raised by $25\,K$? ($\sqrt{3} = 1.732$)
A. $\approx 9$
B. $\approx 27$
C. $\approx 7.5$
D. $\approx 15.6$ ✓ Correct
Solution: Factor $= 3^{\Delta T/10} = 3^{2.5} = 3^{2}\times3^{0.5} = 9\times1.732 = 15.6$.
Q12 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A reaction has $k = 4.0 \times 10^{-4}\,s^{-1}$ at $300\,K$ and $E_a = 60\,kJ\,mol^{-1}$. Its rate constant at $320\,K$ is: ($R = 8.314$, antilog $0.653 = 4.50$)
A. $\approx 1.2 \times 10^{-3}\,s^{-1}$
B. $\approx 1.8 \times 10^{-3}\,s^{-1}$ ✓ Correct
C. $\approx 8.9 \times 10^{-5}\,s^{-1}$
D. $\approx 4.0 \times 10^{-4}\,s^{-1}$
Solution: $\log\dfrac{k_2}{k_1} = \dfrac{60000}{2.303\times8.314}\times\dfrac{20}{300\times320} = 0.653$; $k_2 = 4.0\times10^{-4}\times4.50 = 1.8\times10^{-3}\,s^{-1}$.
Q13 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a bimolecular gas reaction, collision frequency $Z = 2 \times 10^{11}\,L\,mol^{-1}s^{-1}$, steric factor $P = 0.2$, and $e^{-E_a/RT} = 1 \times 10^{-6}$ at the reaction temperature. The rate constant predicted by collision theory is:
A. $4 \times 10^{4}\,L\,mol^{-1}s^{-1}$ ✓ Correct
B. $1 \times 10^{6}\,L\,mol^{-1}s^{-1}$
C. $4 \times 10^{5}\,L\,mol^{-1}s^{-1}$
D. $2 \times 10^{4}\,L\,mol^{-1}s^{-1}$
Solution: $k = P\,Z\,e^{-E_a/RT} = 0.2\times(2\times10^{11})\times(1\times10^{-6}) = 4\times10^{4}\,L\,mol^{-1}s^{-1}$.
Q14 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
At $500\,K$ a reaction has $E_a = 100\,kJ\,mol^{-1}$. The fraction of collisions that are energetically effective, $e^{-E_a/RT}$, is about: ($R = 8.314$, $\log e = 0.4343$)
A. $\approx 2.4 \times 10^{-11}$
B. $\approx 3.6 \times 10^{-5}$
C. $\approx 1.0 \times 10^{-10}$
D. $\approx 3.6 \times 10^{-11}$ ✓ Correct
Solution: $\dfrac{E_a}{RT} = \dfrac{100000}{8.314\times500} = 24.06$; $\log(e^{-24.06}) = -24.06\times0.4343 = -10.45$, giving $3.6\times10^{-11}$.
Q15 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
Two reactions have the same frequency factor. At $300\,K$, reaction X has $E_a = 60\,kJ\,mol^{-1}$ and reaction Y has $E_a = 75\,kJ\,mol^{-1}$. The ratio $\dfrac{k_X}{k_Y}$ is about: ($R = 8.314$)
A. $\approx 6$
B. $\approx 1.25$
C. $\approx 40$
D. $\approx 410$ ✓ Correct
Solution: $\dfrac{k_X}{k_Y} = e^{(E_{a,Y}-E_{a,X})/RT} = e^{15000/(8.314\times300)} = e^{6.01} = 410$.
Q16 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction is $2.0 \times 10^{-3}\,s^{-1}$ at $300\,K$ and $8.0 \times 10^{-3}\,s^{-1}$ at $320\,K$. Its value at $340\,K$ is: ($\log 2 = 0.301$, $R = 8.314$, antilog $0.531 = 3.40$)
A. $\approx 3.2 \times 10^{-2}\,s^{-1}$
B. $\approx 1.6 \times 10^{-2}\,s^{-1}$
C. $\approx 2.7 \times 10^{-2}\,s^{-1}$ ✓ Correct
D. $\approx 8.0 \times 10^{-3}\,s^{-1}$
Solution: From $300\!\to\!320$: $\log 4 = \dfrac{E_a}{2.303R}(2.083\times10^{-4}) \Rightarrow \dfrac{E_a}{2.303R} = 2890\,K$. Then $\log\dfrac{k_{340}}{k_{320}} = 2890\times1.838\times10^{-4} = 0.531$, so $k_{340} = 8.0\times10^{-3}\times3.40 = 2.7\times10^{-2}\,s^{-1}$.
Q17 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a first-order reaction, a graph of $\ln k$ ($k$ in $s^{-1}$) versus $\dfrac{1}{T}$ is a straight line of slope $-8000\,K$ and intercept $20$. The activation energy and pre-exponential factor are, respectively: ($R = 8.314$, $e^{20} = 4.85 \times 10^{8}$)
A. $8.0\,kJ\,mol^{-1}$ and $4.85 \times 10^{8}\,s^{-1}$
B. $66.5\,kJ\,mol^{-1}$ and $20\,s^{-1}$
C. $153\,kJ\,mol^{-1}$ and $20\,s^{-1}$
D. $66.5\,kJ\,mol^{-1}$ and $4.85 \times 10^{8}\,s^{-1}$ ✓ Correct
Solution: For a $\ln k$ plot, slope $= -\dfrac{E_a}{R} \Rightarrow E_a = 8000\times8.314 = 66.5\,kJ$; intercept $= \ln A \Rightarrow A = e^{20} = 4.85\times10^{8}\,s^{-1}$.
Q18 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
At $500\,K$, adding a catalyst makes a reaction proceed $10^{6}$ times faster with the same frequency factor. The activation energy is lowered by: ($R = 8.314$, $\ln 10 = 2.303$)
A. $\approx 115\,kJ\,mol^{-1}$
B. $\approx 24.9\,kJ\,mol^{-1}$
C. $\approx 6.0\,kJ\,mol^{-1}$
D. $\approx 57.4\,kJ\,mol^{-1}$ ✓ Correct
Solution: $\dfrac{k_{cat}}{k} = e^{\Delta E_a/RT} = 10^{6}$, so $\Delta E_a = RT\ln(10^{6}) = 8.314\times500\times(6\times2.303) = 57.4\,kJ\,mol^{-1}$.
Q19 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For $2HI \rightarrow H_2 + I_2$ at $700\,K$, the experimental Arrhenius factor is $A = 1.6 \times 10^{11}\,L\,mol^{-1}s^{-1}$ while kinetic theory gives a collision frequency $Z = 8 \times 10^{11}\,L\,mol^{-1}s^{-1}$. The steric factor $P$ and its meaning are:
A. $P = 5$; every collision is effective
B. $P = 1.28 \times 10^{23}$; a compound-forming factor
C. $P = 0.2$; the orientation requirement lowers the effective collision rate ✓ Correct
D. $P = 0.2$; the reaction is diffusion-controlled
Solution: $P = \dfrac{A}{Z} = \dfrac{1.6\times10^{11}}{8\times10^{11}} = 0.2$. Since $P<1$, only a fraction of sufficiently energetic collisions have the correct orientation to react.
Q20 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction with $E_a = 80\,kJ\,mol^{-1}$, by what factor does the fraction of molecules able to cross the energy barrier, $e^{-E_a/RT}$, increase when $T$ goes from $300\,K$ to $310\,K$? ($R = 8.314$)
A. $\approx 1.9$
B. $\approx 1.03$
C. $\approx 2.8$ ✓ Correct
D. $\approx 5.6$
Solution: Ratio $= e^{\frac{E_a}{R}(\frac{1}{300}-\frac{1}{310})} = e^{\frac{80000}{8.314}\times1.075\times10^{-4}} = e^{1.035} = 2.8$.
Q21 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A first-order reaction has $k = 1.0 \times 10^{-3}\,s^{-1}$ at $300\,K$ and $E_a = 55.3\,kJ\,mol^{-1}$. When the temperature is raised to $310\,K$, its half-life becomes: ($R = 8.314$, $\log 2 = 0.301$)
A. $\approx 170\,s$
B. $\approx 226\,s$
C. $\approx 693\,s$
D. $\approx 339\,s$ ✓ Correct
Solution: $\log\dfrac{k_{310}}{k_{300}} = \dfrac{55300}{2.303\times8.314}\times\dfrac{10}{300\times310} = 0.311$, so $k_{310} = 2.04\times10^{-3}\,s^{-1}$; $t_{1/2} = \dfrac{0.693}{2.04\times10^{-3}} = 339\,s$.
Q22 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A reaction obeys the Arrhenius equation with $E_a = 52.9\,kJ\,mol^{-1}$. Its temperature coefficient (ratio of rate constants at $308\,K$ and $298\,K$) is about: ($R = 8.314$)
A. $\approx 4.0$
B. $\approx 2.7$
C. $\approx 1.5$
D. $\approx 2.0$ ✓ Correct
Solution: $\log\dfrac{k_{308}}{k_{298}} = \dfrac{52900}{2.303\times8.314}\times\dfrac{10}{298\times308} = 0.301$; ratio $= 10^{0.301} = 2.0$ (why the "doubling per $10\,K$" rule implies $E_a \approx 50\,kJ$).
Q23 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction increases from $1.0 \times 10^{-2}$ to $5.0 \times 10^{-2}\,s^{-1}$ when the temperature is raised from $290\,K$ to $310\,K$. The activation energy is: ($\log 5 = 0.699$, $R = 8.314$)
A. $\approx 120\,kJ\,mol^{-1}$
B. $\approx 60.2\,kJ\,mol^{-1}$ ✓ Correct
C. $\approx 43\,kJ\,mol^{-1}$
D. $\approx 26.1\,kJ\,mol^{-1}$
Solution: $\log 5 = \dfrac{E_a}{2.303R}\cdot\dfrac{20}{290\times310}$; $E_a = \dfrac{0.699\times19.147}{2.225\times10^{-4}} = 60.2\,kJ\,mol^{-1}$.
Q24 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A reaction has $k = 3.0 \times 10^{-2}\,s^{-1}$ at $400\,K$ with $E_a = 76.6\,kJ\,mol^{-1}$. Its pre-exponential factor $A$ is: ($R = 8.314$, $\log 3 = 0.477$, antilog $0.479 = 3.0$)
A. $\approx 3.3 \times 10^{7}\,s^{-1}$
B. $\approx 3.0 \times 10^{8}\,s^{-1}$ ✓ Correct
C. $\approx 1.0 \times 10^{10}\,s^{-1}$
D. $\approx 3.0 \times 10^{-2}\,s^{-1}$
Solution: $\log A = \log k + \dfrac{E_a}{2.303RT} = -1.523 + \dfrac{76600}{2.303\times8.314\times400} = -1.523 + 10.00 = 8.48$; $A = 3.0\times10^{8}\,s^{-1}$.
Q25 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a bimolecular reaction the Arrhenius pre-exponential factor is $A = 2.0 \times 10^{10}\,L\,mol^{-1}s^{-1}$ and the steric factor is $P = 0.04$. The collision frequency $Z$ is:
A. $2.0 \times 10^{10}\,L\,mol^{-1}s^{-1}$
B. $8.0 \times 10^{8}\,L\,mol^{-1}s^{-1}$
C. $5.0 \times 10^{9}\,L\,mol^{-1}s^{-1}$
D. $5.0 \times 10^{11}\,L\,mol^{-1}s^{-1}$ ✓ Correct
Solution: $A = P\,Z \Rightarrow Z = \dfrac{A}{P} = \dfrac{2.0\times10^{10}}{0.04} = 5.0\times10^{11}\,L\,mol^{-1}s^{-1}$.
Q26 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
An exothermic reaction has $\Delta H = -40\,kJ\,mol^{-1}$ and forward activation energy $90\,kJ\,mol^{-1}$. A catalyst lowers the forward $E_a$ by $30\,kJ\,mol^{-1}$. At $300\,K$, the factor by which the reverse rate constant increases due to the catalyst is: ($R = 8.314$, $\ln 10 = 2.303$)
A. $\approx 2.7$
B. $\approx 6.0 \times 10^{6}$
C. $\approx 3.6 \times 10^{2}$
D. $\approx 1.7 \times 10^{5}$ ✓ Correct
Solution: A catalyst lowers the barrier peak, so it lowers both forward and reverse $E_a$ by the same $30\,kJ$. Reverse increase $= e^{30000/(8.314\times300)} = e^{12.03} = 1.7\times10^{5}$ (independent of $\Delta H$).
Q27 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction the fraction of effective collisions is $2.0 \times 10^{-10}$ at $300\,K$ and $8.0 \times 10^{-9}$ at $320\,K$. The activation energy is: ($\log 2 = 0.301$, $R = 8.314$)
A. $\approx 73.6\,kJ\,mol^{-1}$
B. $\approx 147\,kJ\,mol^{-1}$ ✓ Correct
C. $\approx 27.7\,kJ\,mol^{-1}$
D. $\approx 294\,kJ\,mol^{-1}$
Solution: $\dfrac{f_{320}}{f_{300}} = \dfrac{8\times10^{-9}}{2\times10^{-10}} = 40 = e^{\frac{E_a}{R}(\frac{1}{300}-\frac{1}{320})}$; $\log 40 = 1.602 = \dfrac{E_a}{2.303R}(2.083\times10^{-4})$, so $E_a = 147\,kJ\,mol^{-1}$.
Q28 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
Near $300\,K$, a reaction with $E_a = 50\,kJ\,mol^{-1}$ shows what approximate percentage increase in its rate constant per $1\,K$ rise? Use $\dfrac{d\ln k}{dT} = \dfrac{E_a}{RT^2}$. ($R = 8.314$)
A. $\approx 0.7\%$
B. $\approx 2.0\%$
C. $\approx 6.7\%$ ✓ Correct
D. $\approx 20\%$
Solution: $\dfrac{d\ln k}{dT} = \dfrac{E_a}{RT^2} = \dfrac{50000}{8.314\times300^2} = 0.0668\,K^{-1}$, i.e. about $6.7\%$ per kelvin.
Q29 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
Two reactions have Arrhenius parameters: (1) $A_1 = 10^{13}\,s^{-1}$, $E_{a1} = 100\,kJ\,mol^{-1}$; (2) $A_2 = 10^{11}\,s^{-1}$, $E_{a2} = 80\,kJ\,mol^{-1}$. The temperature at which their rate constants are equal is: ($R = 8.314$, $\ln 10 = 2.303$)
A. $\approx 522\,K$ ✓ Correct
B. $\approx 300\,K$
C. $\approx 1044\,K$
D. $\approx 481\,K$
Solution: Set $k_1 = k_2$: $\ln\dfrac{A_1}{A_2} = \dfrac{E_{a1}-E_{a2}}{RT}$, so $RT = \dfrac{20000}{\ln 10^{2}} = \dfrac{20000}{4.605} = 4343$; $T = \dfrac{4343}{8.314} = 522\,K$.
Q30 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a bimolecular reaction at $600\,K$: collision frequency $Z = 2 \times 10^{11}\,L\,mol^{-1}s^{-1}$, steric factor $P = 0.5$, activation energy $E_a = 100\,kJ\,mol^{-1}$. The rate constant is: ($R = 8.314$, $\log e = 0.4343$)
A. $\approx 98\,L\,mol^{-1}s^{-1}$
B. $\approx 1 \times 10^{11}\,L\,mol^{-1}s^{-1}$
C. $\approx 394\,L\,mol^{-1}s^{-1}$
D. $\approx 197\,L\,mol^{-1}s^{-1}$ ✓ Correct
Solution: $\dfrac{E_a}{RT} = \dfrac{100000}{8.314\times600} = 20.05$, so $e^{-20.05} = 1.97\times10^{-9}$; $k = PZe^{-E_a/RT} = 0.5\times2\times10^{11}\times1.97\times10^{-9} = 197\,L\,mol^{-1}s^{-1}$.