Temperature Dependence, Arrhenius Equation & Collision Theory — IISER Chemistry MCQs with Solutions
Free IISER 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
In the Arrhenius equation $k = A\,e^{-E_a/RT}$, as the temperature approaches infinity, the rate constant $k$ approaches:
A. zero, because the exponential term dominates.
B. $A$, the pre-exponential factor, because $e^{-E_a/RT} \rightarrow 1$. ✓ Correct
C. $e^{E_a}$.
D. $E_a$, the activation energy.
Solution: As $T\rightarrow\infty$, $\dfrac{E_a}{RT}\rightarrow0$ so $e^{-E_a/RT}\rightarrow1$ and $k\rightarrow A$; the frequency factor is the maximum possible rate constant.
Q2 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
On raising the temperature of a reaction, the fraction of molecules possessing energy greater than the activation energy, $e^{-E_a/RT}$:
A. increases, because the exponent $-\dfrac{E_a}{RT}$ becomes less negative as $T$ rises. ✓ Correct
B. first increases and then decreases.
C. decreases, because the molecules move faster.
D. stays constant, since $E_a$ is fixed.
Solution: As $T$ increases, $\dfrac{E_a}{RT}$ decreases, so $e^{-E_a/RT}$ increases and more molecules can cross the energy barrier.
Q3 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A catalyst speeds up a reaction by:
A. providing an alternative path with a lower activation energy, without changing $\Delta H$ of the reaction. ✓ Correct
B. increasing the activation energy of the forward reaction.
C. increasing the value of the equilibrium constant.
D. raising the average kinetic energy of the molecules.
Solution: A catalyst offers a lower-$E_a$ route; it does not alter the thermodynamics ($\Delta H$ or the equilibrium constant $K$), only the speed at which equilibrium is reached.
Q4 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
If the rate of a reaction doubles for every $10\,K$ rise in temperature, then raising the temperature by $20\,K$ increases the rate by a factor of:
A. $4$ ✓ Correct
B. $8$
C. $16$
D. $2$
Solution: Factor $=2^{\Delta T/10}=2^{20/10}=2^2=4$.
Q5 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a reaction, a plot of $\log k$ against $\dfrac{1}{T}$ is a straight line. Its slope equals:
A. $-\dfrac{E_a}{2.303\,R}$ ✓ Correct
B. $+\dfrac{E_a}{2.303\,R}$
C. $\log A$
D. $-\dfrac{E_a}{R}$
Solution: $\log k=\log A-\dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}$, so the slope is $-\dfrac{E_a}{2.303R}$ and the intercept is $\log A$.
Q6 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
According to collision theory, a collision between reactant molecules leads to product only if the molecules:
A. have energy less than the activation energy but collide head-on.
B. collide with energy equal to or greater than the threshold energy AND with the proper orientation. ✓ Correct
C. collide with any energy, provided they are correctly oriented.
D. simply collide, regardless of energy or orientation.
Solution: An effective collision needs both sufficient energy (at least the threshold energy) and the correct orientation; otherwise no reaction occurs.
Q7 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
In the collision-theory expression $k = P\,Z\,e^{-E_a/RT}$, the steric factor $P$ is introduced to account for:
A. the fraction of molecules having energy above the activation energy.
B. the lowering of activation energy by a catalyst.
C. the total number of collisions per unit time.
D. the fraction of energetic collisions in which the molecules are correctly oriented for reaction. ✓ Correct
Solution: The steric (probability) factor $P$ corrects $Z\,e^{-E_a/RT}$ for the orientation requirement; only correctly oriented energetic collisions react.
Q8 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
Which of the following is NOT changed when a catalyst is added to a reaction?
A. The activation energy of the forward reaction.
B. The rate of the reaction.
C. The activation energy of the reverse reaction.
D. The enthalpy change ($\Delta H$) of the reaction. ✓ Correct
Solution: A catalyst lowers both the forward and reverse activation energies and speeds the rate, but $\Delta H$ (a state function) is unchanged.
Q9 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The fraction of molecules with energy $\ge E_a$ is $e^{-E_a/RT}$. For $E_a = 41.57\,kJ\,mol^{-1}$ at $T = 500\,K$, the value of $\dfrac{E_a}{RT}$ is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $10$ ✓ Correct
B. $20$
C. $5$
D. $8.3$
Solution: $\dfrac{E_a}{RT}=\dfrac{41570}{8.314\times500}=\dfrac{41570}{4157}=10$.
Q10 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
According to collision theory, the collision frequency $Z$ of a gas-phase reaction depends on temperature as:
A. $Z$ is independent of $T$
B. $Z \propto \sqrt{T}$ ✓ Correct
C. $Z \propto T^2$
D. $Z \propto T$
Solution: The collision frequency is proportional to the mean molecular speed, which varies as $\sqrt{T}$; the steep rise in rate with $T$ comes mainly from the $e^{-E_a/RT}$ term.
Q11 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The activation energy of a reaction is best described as:
A. the minimum extra energy, above the average energy of the reactants, that colliding molecules must possess to react. ✓ Correct
B. the average kinetic energy of all the reactant molecules.
C. the energy difference between products and reactants.
D. the total energy released when products form.
Solution: $E_a$ is the minimum energy above the average energy of the reactants needed to reach the transition state; threshold energy $=$ average energy $+\,E_a$.
Q12 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A reaction proceeds at rate $r$ at $20\,°C$. If the rate doubles for every $10\,°C$ rise, its rate at $60\,°C$ is:
A. $8\,r$
B. $32\,r$
C. $16\,r$ ✓ Correct
D. $4\,r$
Solution: $\Delta T = 40\,°C$, so factor $=2^{\Delta T/10}=2^4=16$, giving a rate of $16\,r$.
Q13 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction increases four-fold when the temperature is raised from $400\,K$ to $500\,K$. The activation energy is: ($\log 2 = 0.301$, $R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $23.1\,kJ\,mol^{-1}$ ✓ Correct
B. $46.1\,kJ\,mol^{-1}$
C. $11.5\,kJ\,mol^{-1}$
D. $2.3\,kJ\,mol^{-1}$
Solution: $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$; $\log4=0.602=\dfrac{E_a}{19.147}(5\times10^{-4})$, so $E_a=2.31\times10^{4}\,J=23.1\,kJ\,mol^{-1}$.
Q14 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A reaction has $k = 2.0 \times 10^{-3}\,s^{-1}$ at $400\,K$ and activation energy $E_a = 38.3\,kJ\,mol^{-1}$. Its rate constant at $500\,K$ is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $2.0 \times 10^{-2}\,s^{-1}$ ✓ Correct
B. $2.0 \times 10^{-1}\,s^{-1}$
C. $2.0 \times 10^{-3}\,s^{-1}$
D. $4.0 \times 10^{-3}\,s^{-1}$
Solution: $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)=\dfrac{38300}{19.147}(5\times10^{-4})=1.0$, so $k_2=10\times k_1=2.0\times10^{-2}\,s^{-1}$.
Q15 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction, the Arrhenius plot of $\log k$ versus $\dfrac{1}{T}$ is a straight line of slope $-3000\,K$. The activation energy is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $25.0\,kJ\,mol^{-1}$
B. $24.9\,kJ\,mol^{-1}$
C. $114.9\,kJ\,mol^{-1}$
D. $57.4\,kJ\,mol^{-1}$ ✓ Correct
Solution: Slope $=-\dfrac{E_a}{2.303R}\Rightarrow E_a=3000\times2.303\times8.314=5.74\times10^{4}\,J=57.4\,kJ\,mol^{-1}$.
Q16 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction, a plot of $\ln k$ versus $\dfrac{1}{T}$ is a straight line of slope $-6000\,K$. The activation energy is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $50.0\,kJ\,mol^{-1}$
B. $21.7\,kJ\,mol^{-1}$
C. $114.9\,kJ\,mol^{-1}$
D. $49.9\,kJ\,mol^{-1}$ ✓ Correct
Solution: For a $\ln k$ plot the slope $=-\dfrac{E_a}{R}\Rightarrow E_a=6000\times8.314=4.99\times10^{4}\,J=49.9\,kJ\,mol^{-1}$ (no factor $2.303$ for the natural-log plot).
Q17 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
At $300\,K$ a catalyst lowers the activation energy of a reaction by $5.0\,kJ\,mol^{-1}$ (frequency factor unchanged). The rate constant increases by a factor of about: ($R = 8.314\,J\,K^{-1}mol^{-1}$, $e^2 = 7.39$)
A. $2.0$
B. $7.4$ ✓ Correct
C. $5.0$
D. $54.6$
Solution: $\dfrac{k_{cat}}{k}=e^{\Delta E_a/RT}=e^{5000/(8.314\times300)}=e^{2.0}=7.39$.
Q18 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a bimolecular reaction the Arrhenius frequency factor is $A = 1.5 \times 10^{10}\,L\,mol^{-1}s^{-1}$, while collision theory predicts a collision frequency $Z = 3.0 \times 10^{11}\,L\,mol^{-1}s^{-1}$. The steric factor $P$ is:
A. $0.02$
B. $0.5$
C. $0.05$ ✓ Correct
D. $20$
Solution: $P=\dfrac{A}{Z}=\dfrac{1.5\times10^{10}}{3.0\times10^{11}}=0.05$.
Q19 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a bimolecular reaction the collision frequency $Z = 1.0 \times 10^{11}\,L\,mol^{-1}s^{-1}$, the steric factor $P = 0.10$, and $e^{-E_a/RT} = 1.0 \times 10^{-4}$. The rate constant predicted by collision theory is:
A. $1.0 \times 10^{7}\,L\,mol^{-1}s^{-1}$
B. $1.0 \times 10^{6}\,L\,mol^{-1}s^{-1}$ ✓ Correct
C. $1.0 \times 10^{5}\,L\,mol^{-1}s^{-1}$
D. $1.0 \times 10^{11}\,L\,mol^{-1}s^{-1}$
Solution: $k=P\,Z\,e^{-E_a/RT}=0.10\times1.0\times10^{11}\times1.0\times10^{-4}=1.0\times10^{6}\,L\,mol^{-1}s^{-1}$.
Q20 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction at a certain temperature the frequency factor $A = 3.0 \times 10^{11}\,s^{-1}$ and the rate constant $k = 3.0 \times 10^{-3}\,s^{-1}$. The fraction of effective (energetic) collisions, $e^{-E_a/RT}$, is:
A. $1.0 \times 10^{-8}$
B. $9.0 \times 10^{8}$
C. $1.0 \times 10^{14}$
D. $1.0 \times 10^{-14}$ ✓ Correct
Solution: $k=A\,e^{-E_a/RT}\Rightarrow e^{-E_a/RT}=\dfrac{k}{A}=\dfrac{3.0\times10^{-3}}{3.0\times10^{11}}=1.0\times10^{-14}$.
Q21 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A reaction has a temperature coefficient of $3$ (its rate triples for every $10\,K$ rise). By what factor does its rate increase when the temperature is raised by $30\,K$?
A. $27$ ✓ Correct
B. $81$
C. $9$
D. $3$
Solution: Factor $=3^{\Delta T/10}=3^{30/10}=3^3=27$.
Q22 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
Two reactions have the same frequency factor. At $400\,K$, reaction X has $E_a = 50\,kJ\,mol^{-1}$ and reaction Y has $E_a = 60\,kJ\,mol^{-1}$. The ratio $\dfrac{k_X}{k_Y}$ is about: ($R = 8.314$, $e^3 = 20.1$)
A. $0.05$
B. $7.4$
C. $3.0$
D. $20.1$ ✓ Correct
Solution: $\dfrac{k_X}{k_Y}=e^{(E_{a,Y}-E_{a,X})/RT}=e^{10000/(8.314\times400)}=e^{3}=20.1$ (the lower $E_a$ gives the larger $k$).
Q23 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction with temperature coefficient $2$, the rate increases $16$ times when the temperature is raised by $\Delta T$. The value of $\Delta T$ is:
A. $32\,K$
B. $20\,K$
C. $50\,K$
D. $40\,K$ ✓ Correct
Solution: $16=2^{\Delta T/10}=2^4$, so $\dfrac{\Delta T}{10}=4$ and $\Delta T=40\,K$.
Q24 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The fraction of molecules with energy $\ge E_a$ is $f = e^{-E_a/RT}$. For $E_a = 38.3\,kJ\,mol^{-1}$, the ratio $\dfrac{f_{500}}{f_{400}}$ (fractions at $500\,K$ and $400\,K$) is about: ($R = 8.314$, $e^{2.303} = 10$)
A. $10$ ✓ Correct
B. $2.3$
C. $1.0$
D. $100$
Solution: $\dfrac{f_{500}}{f_{400}}=e^{\frac{E_a}{R}\left(\frac{1}{400}-\frac{1}{500}\right)}=e^{\frac{38300}{8.314}(5\times10^{-4})}=e^{2.303}=10$.
Q25 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
At $400\,K$, adding a catalyst makes a reaction proceed $100$ times faster with the same frequency factor. The activation energy is lowered by: ($R = 8.314$, $\log 10 = 1$)
A. $30.6\,kJ\,mol^{-1}$
B. $15.3\,kJ\,mol^{-1}$ ✓ Correct
C. $6.6\,kJ\,mol^{-1}$
D. $7.7\,kJ\,mol^{-1}$
Solution: $\dfrac{k_{cat}}{k}=e^{\Delta E_a/RT}\Rightarrow\Delta E_a=2.303RT\log100=2.303\times8.314\times400\times2=1.53\times10^{4}\,J=15.3\,kJ\,mol^{-1}$.
Q26 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction the collision frequency $Z = 5 \times 10^{10}\,L\,mol^{-1}s^{-1}$ and the energy factor $e^{-E_a/RT} = 2 \times 10^{-5}$. If the experimentally observed rate constant is $k = 5 \times 10^{4}\,L\,mol^{-1}s^{-1}$, the steric factor $P$ is:
A. $5$
B. $0.5$
C. $0.02$
D. $0.05$ ✓ Correct
Solution: $k=P\,Z\,e^{-E_a/RT}\Rightarrow P=\dfrac{k}{Z\,e^{-E_a/RT}}=\dfrac{5\times10^{4}}{5\times10^{10}\times2\times10^{-5}}=\dfrac{5\times10^{4}}{10^{6}}=0.05$.
Q27 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction the forward activation energy is $70\,kJ\,mol^{-1}$ and the reverse activation energy is $90\,kJ\,mol^{-1}$. A catalyst lowers the forward activation energy to $45\,kJ\,mol^{-1}$. The reverse activation energy in the presence of the catalyst is:
A. $90\,kJ\,mol^{-1}$
B. $20\,kJ\,mol^{-1}$
C. $65\,kJ\,mol^{-1}$ ✓ Correct
D. $45\,kJ\,mol^{-1}$
Solution: $\Delta H=E_{a,f}-E_{a,r}=70-90=-20\,kJ$ is unchanged by the catalyst; with new $E_{a,f}=45$, the new $E_{a,r}=45+20=65\,kJ\,mol^{-1}$.
Q28 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction is $1.0 \times 10^{-3}\,s^{-1}$ at $500\,K$ and $1.0 \times 10^{-1}\,s^{-1}$ at $600\,K$. The activation energy is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $49.9\,kJ\,mol^{-1}$
B. $57.4\,kJ\,mol^{-1}$
C. $229.8\,kJ\,mol^{-1}$
D. $114.9\,kJ\,mol^{-1}$ ✓ Correct
Solution: $\log\dfrac{k_2}{k_1}=\log100=2=\dfrac{E_a}{2.303R}\left(\dfrac{1}{500}-\dfrac{1}{600}\right)=\dfrac{E_a}{19.147}(3.33\times10^{-4})$, so $E_a=1.149\times10^{5}\,J=114.9\,kJ\,mol^{-1}$.
Q29 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The collision frequency $Z$ rises only as $\sqrt{T}$, yet reaction rates increase steeply with temperature. The main reason is that:
A. collisions become perfectly elastic at high temperature.
B. the exponential factor $e^{-E_a/RT}$ increases very rapidly with temperature, far outweighing the small $\sqrt{T}$ rise in $Z$. ✓ Correct
C. the activation energy decreases as temperature rises.
D. the steric factor $P$ increases sharply with temperature.
Solution: The dominant temperature dependence is in $e^{-E_a/RT}$, which climbs steeply as $T$ rises; the $\sqrt{T}$ dependence of $Z$ is comparatively minor.
Q30 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
At $500\,K$ a reaction has activation energy $E_a = 95.8\,kJ\,mol^{-1}$. The fraction of collisions that are energetically effective, $e^{-E_a/RT}$, is about: ($R = 8.314$, $\ln x = 2.303\log x$)
A. $1 \times 10^{-23}$
B. $1 \times 10^{-10}$ ✓ Correct
C. $1 \times 10^{10}$
D. $1 \times 10^{-5}$
Solution: $\dfrac{E_a}{RT}=\dfrac{95800}{8.314\times500}=23.05$; $e^{-23.05}=10^{-23.05/2.303}=10^{-10}=1\times10^{-10}$.