Temperature Dependence, Arrhenius Equation & Collision Theory — NEET Chemistry MCQs with Solutions
Free NEET 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 $k = A\,e^{-E_a/RT}$, the factor $A$ represents:
A. the frequency (pre-exponential) factor. ✓ Correct
B. the temperature coefficient of the reaction.
C. the fraction of molecules that collide per second.
D. the activation energy of the reaction.
Solution: $A$ is the pre-exponential (frequency) factor, related to the total collision frequency and the orientation requirement.
Q2 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
If the rate of a reaction roughly doubles for every $10\,°C$ rise, the rate increases by a factor of about ______ when the temperature is raised by $30\,°C$.
A. $6$
B. $8$ ✓ Correct
C. $3$
D. $30$
Solution: Factor $= 2^{\Delta T/10} = 2^{30/10} = 2^3 = 8$.
Q3 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
A catalyst increases the rate of a reaction mainly by:
A. increasing the enthalpy change $\Delta H$ of the reaction.
B. providing an alternative path with a lower activation energy. ✓ Correct
C. increasing the value of the equilibrium constant.
D. increasing the activation energy of the reaction.
Solution: A catalyst offers a new pathway of lower $E_a$, so a larger fraction of molecules can cross the barrier; $\Delta H$ and $K$ are unchanged.
Q4 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
As the temperature is raised, the fraction of molecules having energy greater than the activation energy, $e^{-E_a/RT}$:
A. increases ✓ Correct
B. becomes zero
C. remains constant
D. decreases
Solution: As $T$ increases, $E_a/RT$ decreases, so $e^{-E_a/RT}$ increases; more molecules cross the energy barrier and the rate rises.
Q5 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
For a plot of $\log k$ against $\dfrac{1}{T}$ (Arrhenius plot), the slope is equal to:
A. $\log A$
B. $+\dfrac{E_a}{2.303R}$
C. $-\dfrac{E_a}{2.303R}$ ✓ Correct
D. $-\dfrac{E_a}{R}$
Solution: $\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}$, so the slope is $-\dfrac{E_a}{2.303R}$.
Q6 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
According to collision theory, a collision leads to a reaction only if the colliding molecules have:
A. energy less than the activation energy.
B. the correct orientation but no minimum energy requirement.
C. energy equal to or greater than the threshold energy and the proper orientation. ✓ Correct
D. any amount of energy and any orientation.
Solution: Only collisions that are both sufficiently energetic (≥ threshold energy) and properly oriented are effective.
Q7 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
In the collision-theory expression $k = P\,Z\,e^{-E_a/RT}$, the steric (probability) factor $P$ accounts for:
A. the fraction of molecules having the minimum energy.
B. the increase in collision frequency with temperature.
C. the fraction of collisions in which the molecules are properly oriented. ✓ Correct
D. the total number of collisions per unit time.
Solution: $P$ corrects the collision-theory rate for the requirement of correct orientation; $Z$ gives collision frequency and $e^{-E_a/RT}$ the energy requirement.
Q8 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
Which of the following is NOT changed by adding a catalyst to a reaction?
A. The rate of the forward reaction.
B. The enthalpy change $\Delta H$ of the reaction. ✓ Correct
C. The rate of the reverse reaction.
D. The activation energy of the reaction.
Solution: A catalyst lowers $E_a$ and speeds both forward and reverse reactions, but the thermodynamic $\Delta H$ (and $K$) stay the same.
Q9 — Temperature Dependence, Arrhenius Equation & Collision Theory · easy · numerical
The temperature coefficient of a reaction is defined as the ratio of its rate constants at:
A. two temperatures differing by $1\,K$.
B. two temperatures differing by $10\,K$, i.e. $k_{308}/k_{298}$. ✓ Correct
C. two temperatures differing by $100\,K$.
D. $0\,K$ and $273\,K$.
Solution: The temperature coefficient is $\dfrac{k_{T+10}}{k_{T}}$, usually taken as $\dfrac{k_{308}}{k_{298}}$, and is about 2–3 for most reactions.
Q10 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction doubles when the temperature is raised from $300\,K$ to $310\,K$. The activation energy is: ($\log 2 = 0.301$, $R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $107\,kJ\,mol^{-1}$
B. $5.76\,kJ\,mol^{-1}$
C. $53.6\,kJ\,mol^{-1}$ ✓ Correct
D. $26.8\,kJ\,mol^{-1}$
Solution: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$; $0.301 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{10}{300\times310}$, giving $E_a \approx 53.6\,kJ\,mol^{-1}$.
Q11 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction increases threefold when the temperature rises from $300\,K$ to $310\,K$. The activation energy is: ($\log 3 = 0.4771$, $R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $\approx 170\,kJ\,mol^{-1}$
B. $\approx 85\,kJ\,mol^{-1}$ ✓ Correct
C. $\approx 53.6\,kJ\,mol^{-1}$
D. $\approx 47.7\,kJ\,mol^{-1}$
Solution: $0.4771 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{10}{300\times310}$; solving gives $E_a \approx 8.5\times10^{4}\,J\,mol^{-1} = 85\,kJ\,mol^{-1}$.
Q12 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The Arrhenius plot of $\ln k$ versus $\dfrac{1}{T}$ for a reaction is a straight line of slope $-1.0 \times 10^{4}\,K$. The activation energy is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $83.14\,kJ\,mol^{-1}$ ✓ Correct
B. $19.15\,kJ\,mol^{-1}$
C. $831.4\,kJ\,mol^{-1}$
D. $8.314\,kJ\,mol^{-1}$
Solution: For $\ln k$ vs $1/T$, slope $= -\dfrac{E_a}{R}$, so $E_a = -\text{slope}\times R = 1.0\times10^{4}\times8.314 = 83140\,J\,mol^{-1} = 83.14\,kJ\,mol^{-1}$.
Q13 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The fraction of molecules with energy greater than $E_a$ is $e^{-E_a/RT}$. For $E_a = 50\,kJ\,mol^{-1}$ at $T = 500\,K$, the value of $\dfrac{E_a}{RT}$ is about: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $6.0$
B. $24.0$
C. $1.2$
D. $12.0$ ✓ Correct
Solution: $\dfrac{E_a}{RT} = \dfrac{50000}{8.314\times500} = \dfrac{50000}{4157} \approx 12.0$.
Q14 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction with temperature coefficient $2$, by what factor does the rate increase when the temperature is raised from $300\,K$ to $340\,K$?
A. $4$
B. $16$ ✓ Correct
C. $32$
D. $8$
Solution: Factor $= 2^{\Delta T/10} = 2^{40/10} = 2^4 = 16$.
Q15 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate of a reaction (temperature coefficient $= 2$) increases 32 times when the temperature is raised by $\Delta T\,°C$. The value of $\Delta T$ is:
A. $50\,°C$ ✓ Correct
B. $32\,°C$
C. $25\,°C$
D. $40\,°C$
Solution: $2^{\Delta T/10} = 32 = 2^5 \Rightarrow \dfrac{\Delta T}{10} = 5 \Rightarrow \Delta T = 50\,°C$.
Q16 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A catalyst lowers the activation energy of a reaction from $60\,kJ\,mol^{-1}$ to $50\,kJ\,mol^{-1}$ at $500\,K$. Assuming the same frequency factor, the ratio $\dfrac{k_{catalysed}}{k_{uncatalysed}}$ is about: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $\approx 0.09$
B. $\approx 11$ ✓ Correct
C. $\approx 2.4$
D. $\approx 120$
Solution: $\dfrac{k_{cat}}{k_{uncat}} = e^{(E_a - E_a^{\prime})/RT} = e^{10000/(8.314\times500)} = e^{2.41} \approx 11$.
Q17 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction the experimentally observed rate constant is much smaller than that predicted by the collision-frequency term $Z\,e^{-E_a/RT}$. This is best explained by:
A. a steric factor $P$ less than 1, reflecting the need for correct molecular orientation. ✓ Correct
B. the collision frequency $Z$ being independent of temperature.
C. the activation energy being negative.
D. the reaction having zero activation energy.
Solution: Not every energetic collision is effective; the orientation requirement makes $P < 1$, so the true $k = P\,Z\,e^{-E_a/RT}$ is less than the simple collision-theory value.
Q18 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction is $1 \times 10^{-4}\,s^{-1}$ at $300\,K$ and $1 \times 10^{-3}\,s^{-1}$ at $350\,K$. The activation energy is about: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $80.4\,kJ\,mol^{-1}$
B. $4.02\,kJ\,mol^{-1}$
C. $19.1\,kJ\,mol^{-1}$
D. $40.2\,kJ\,mol^{-1}$ ✓ Correct
Solution: $\log\dfrac{k_2}{k_1} = \log 10 = 1 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{350-300}{300\times350}$, giving $E_a \approx 40.2\,kJ\,mol^{-1}$.
Q19 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
At $300\,K$ a reaction has frequency factor $A = 4 \times 10^{10}\,s^{-1}$ and rate constant $k = 4 \times 10^{-2}\,s^{-1}$. The value of $e^{-E_a/RT}$ (the fraction of effective collisions) is:
A. $10^{12}$
B. $10^{-8}$
C. $10^{-12}$ ✓ Correct
D. $10^{-14}$
Solution: Since $k = A\,e^{-E_a/RT}$, $e^{-E_a/RT} = \dfrac{k}{A} = \dfrac{4\times10^{-2}}{4\times10^{10}} = 10^{-12}$.
Q20 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
If $e^{-E_a/RT} = 10^{-12}$ at $300\,K$, the activation energy of the reaction is: ($R = 8.314\,J\,K^{-1}mol^{-1}$, $\ln x = 2.303\log x$)
A. $\approx 33.2\,kJ\,mol^{-1}$
B. $\approx 27.6\,kJ\,mol^{-1}$
C. $\approx 5.75\,kJ\,mol^{-1}$
D. $\approx 68.9\,kJ\,mol^{-1}$ ✓ Correct
Solution: $\dfrac{E_a}{RT} = 2.303\times12 = 27.64$, so $E_a = 27.64\times8.314\times300 \approx 68.9\times10^{3}\,J\,mol^{-1} = 68.9\,kJ\,mol^{-1}$.
Q21 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For the Arrhenius plot of $\ln k$ against $\dfrac{1}{T}$, the intercept (value of $\ln k$ at $\dfrac{1}{T} = 0$) equals:
A. $A$
B. $-\dfrac{E_a}{R}$
C. $\ln A$ ✓ Correct
D. $\log A$
Solution: $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$; at $1/T = 0$ the intercept is $\ln A$.
Q22 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A catalyst is added to a reversible reaction that is at equilibrium. Which statement is correct?
A. It shifts the equilibrium towards the products.
B. It changes the standard free-energy change $\Delta G^{\circ}$ of the reaction.
C. It speeds up the forward and reverse reactions equally, so equilibrium is reached faster but $K_{eq}$ is unchanged. ✓ Correct
D. It increases the equilibrium constant $K_{eq}$.
Solution: A catalyst lowers $E_a$ for both directions equally; it changes only how fast equilibrium is attained, not its position or $K_{eq}$.
Q23 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
According to collision theory the collision frequency $Z$ varies only as $\sqrt{T}$, yet reaction rates rise sharply with temperature. The main reason is that:
A. the steric factor $P$ increases strongly with temperature.
B. the activation energy decreases as temperature rises.
C. the collision frequency $Z$ itself increases extremely rapidly with $T$.
D. the exponential factor $e^{-E_a/RT}$ increases rapidly with temperature. ✓ Correct
Solution: A $10\,K$ rise changes $Z$ (∝ $\sqrt{T}$) only slightly, but sharply increases $e^{-E_a/RT}$, which dominates the temperature dependence of the rate.
Q24 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction is $1 \times 10^{-3}\,s^{-1}$ at $300\,K$. If its activation energy is $40.2\,kJ\,mol^{-1}$, its value at $350\,K$ is: ($R = 8.314\,J\,K^{-1}mol^{-1}$)
A. $1 \times 10^{-2}\,s^{-1}$ ✓ Correct
B. $1 \times 10^{-4}\,s^{-1}$
C. $5 \times 10^{-3}\,s^{-1}$
D. $2 \times 10^{-3}\,s^{-1}$
Solution: $\log\dfrac{k_2}{k_1} = \dfrac{40200}{2.303\times8.314}\times\dfrac{50}{300\times350} \approx 1$, so $k_2 = 10\,k_1 = 1\times10^{-2}\,s^{-1}$.
Q25 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
A reaction proceeds at rate $r$ at $25\,°C$. Assuming the rate doubles for every $10\,°C$ rise, its rate at $55\,°C$ is:
A. $8r$ ✓ Correct
B. $6r$
C. $16r$
D. $3r$
Solution: $\Delta T = 55 - 25 = 30\,°C$, so rate $= r\times2^{30/10} = r\times2^3 = 8r$.
Q26 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
Two reactant molecules collide with energy greater than the activation energy, yet no reaction occurs. The most likely reason is that:
A. the collision frequency was too high.
B. the activation energy had become negative.
C. the temperature was above the threshold value.
D. the molecules were not properly oriented at the instant of collision. ✓ Correct
Solution: Effective collisions require both sufficient energy and correct orientation; an energetic but wrongly oriented collision is ineffective (captured by the steric factor $P$).
Q27 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a hypothetical reaction the activation energy $E_a = 0$. According to the Arrhenius equation, its rate constant:
A. increases exponentially with temperature.
B. is zero at all temperatures.
C. is independent of temperature and equal to $A$. ✓ Correct
D. decreases as temperature increases.
Solution: With $E_a = 0$, $e^{-E_a/RT} = e^0 = 1$, so $k = A$ regardless of temperature.
Q28 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
Two reactions have the same frequency factor $A$. Reaction X has $E_a = 40\,kJ\,mol^{-1}$ and reaction Y has $E_a = 80\,kJ\,mol^{-1}$ at the same temperature. Which is faster and why?
A. Y is faster, because a higher activation energy gives a larger rate constant.
B. X is faster, because a lower activation energy gives a larger $e^{-E_a/RT}$. ✓ Correct
C. Both have the same rate, because $A$ is the same.
D. Y is faster, because $E_a$ does not affect the rate constant.
Solution: A smaller $E_a$ makes $e^{-E_a/RT}$ larger, so with equal $A$ reaction X has the larger rate constant and is faster.
Q29 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
For a reaction the collision frequency $Z = 10^{11}$ and the energy factor $e^{-E_a/RT} = 10^{-5}$ (consistent units). If the observed rate constant is $k = 10^{5}$, the steric factor $P$ is:
A. $0.01$
B. $10$
C. $0.1$ ✓ Correct
D. $1$
Solution: $k = P\,Z\,e^{-E_a/RT} \Rightarrow P = \dfrac{k}{Z\,e^{-E_a/RT}} = \dfrac{10^{5}}{10^{11}\times10^{-5}} = \dfrac{10^{5}}{10^{6}} = 0.1$.
Q30 — Temperature Dependence, Arrhenius Equation & Collision Theory · medium · numerical
The rate constant of a reaction increases four-fold 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. $\approx 13.8\,kJ\,mol^{-1}$
B. $\approx 27.6\,kJ\,mol^{-1}$
C. $\approx 55.3\,kJ\,mol^{-1}$ ✓ Correct
D. $\approx 110.6\,kJ\,mol^{-1}$
Solution: $\log 4 = 0.602 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{20}{300\times320}$, giving $E_a \approx 55.3\,kJ\,mol^{-1}$.