Showing the 12 most recent of 15 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A first order reaction is $50 \%$ complete in 30 minutes at 300 K and in 10 minutes at 320 K . The activation energy of the reaction ( $E_a$ ) is: $\left[R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 2=0.3010 ; \log 3=0.4771\right]$ (A) $75.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (B) $43.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (C) $23.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (D) $52.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

   ›Reveal solutionSolution

   For a first‑order reaction, the half‑life is inversely proportional to the rate constant. Using the two half‑lives at different temperatures in the Arrhenius equation gives the activation energy. The result is approximately 43.8 kJ mol⁻¹, which corresponds to option (B).

   Concept & Intuition

   For a first‑order reaction, the half‑life $t_{1/2}$ is related to the rate constant $k$ by

   $t_{1/2} = \frac{\ln 2}{k}.$

   Thus, $k \propto 1/t_{1/2}$. When temperature changes, the rate constant changes according to the Arrhenius equation:

   $\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).$

   Since $k \propto 1/t_{1/2}$, we can replace the ratio of rate constants by the inverse ratio of half‑lives. This lets us find $E_a$ directly from the given half‑life data.

   Step‑by‑step solution

   1. Write the half‑life relation for a first‑order reaction

   $t_{1/2} = \frac{\ln 2}{k} \quad \Rightarrow \quad k = \frac{\ln 2}{t_{1/2}}.$

   So at temperature $T_1 = 300\ \text{K}$, $t_{1/2}^{(1)} = 30\ \text{min}$; at $T_2 = 320\ \text{K}$, $t_{1/2}^{(2)} = 10\ \text{min}$.

   2. Form the ratio of rate constants

   $\frac{k_2}{k_1} = \frac{\ln 2 / t_{1/2}^{(2)}}{\ln 2 / t_{1/2}^{(1)}} = \frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \frac{30}{10} = 3.$

   3. Apply the Arrhenius equation in logarithmic form

   $\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).$

   Substitute $\frac{k_2}{k_1} = 3$ and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$:

   $\ln 3 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right).$

   4. Compute the temperature difference term

   $\frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800}.$

   So

   $\ln 3 = \frac{E_a}{8.314} \times \frac{1}{4800}.$

   5. Solve for $E_a$

   $E_a = 8.314 \times 4800 \times \ln 3.$

   Use $\ln 3 = 2.3026 \times \log_{10} 3$ (since $\ln x = 2.3026 \log_{10} x$). Given $\log 3 = 0.4771$, …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.The rate constants for two different reactions, $\mathrm{k}_1$ and $\mathrm{k}_2$ are $10^{16} \cdot \mathrm{e}^{-2000 / \mathrm{T}}$ and $10^{15} \cdot \mathrm{e}^{-1000 / \mathrm{T}}$ respectively. The temperature at which $k_1=k_2$ is (A) $1000 K$ (B) $\frac{2000}{2.303} K$ (C) $\frac{1000}{2.303} K$ (D) $2000 K$

   ›Reveal solutionSolution

   The key idea is to set the two Arrhenius‑type rate constants equal and solve for temperature. The result is $T = \frac{1000}{2.303}$ K, which corresponds to option (C).

   We are given two rate constants of the Arrhenius form:

   $k_1 = 10^{16} e^{-2000/T}$ and $k_2 = 10^{15} e^{-1000/T}$.

   The problem asks for the temperature at which they are equal.

   The natural approach is to set $k_1 = k_2$ and solve for $T$. Because the expressions involve exponentials, we will use logarithms to bring the exponents down.

   1. Set the two expressions equal

   $10^{16} e^{-2000/T} = 10^{15} e^{-1000/T}$

   2. Divide both sides by $10^{15}$ to simplify the pre‑exponential factors:

   $\frac{10^{16}}{10^{15}} e^{-2000/T} = e^{-1000/T}$

   $10 \, e^{-2000/T} = e^{-1000/T}$

   3. Take the natural logarithm of both sides

   $\ln(10) + \ln\!\left(e^{-2000/T}\right) = \ln\!\left(e^{-1000/T}\right)$

   Since $\ln(e^u) = u$, this becomes:

   $\ln(10) - \frac{2000}{T} = -\frac{1000}{T}$

   4. Collect the terms with $T$ Add $\frac{2000}{T}$ to both sides:

   $\ln(10) = \frac{2000}{T} - \frac{1000}{T}$

   $\ln(10) = \frac{1000}{T}$

   5. Solve for $T$

   $T = \frac{1000}{\ln(10)}$

   Now recall that $\ln(10) \approx 2.302585$, so we can write:

   $T = \frac{1000}{2.303} \ \text{K}$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.The hydrogenation of Ethyne is carried out at 600 K . The same reaction when carried out in presence of a catalyst maintaining the same rate constant, the temperature required is only 400 K . If the catalyst lowers the Activation energy of the reaction by $20 \mathrm{~kJ} / \mathrm{mol}$, what is the value of $\mathrm{E}_{\mathrm{a}}$ ? (A) $60 \mathrm{~kJ} / \mathrm{mol}$ (B) $100 \mathrm{~kJ} / \mathrm{mol}$ (C) $50 \mathrm{~kJ} / \mathrm{mol}$ (D) $80 \mathrm{~kJ} / \mathrm{mol}$

   ›Reveal solutionSolution

   The key idea is that the catalyst keeps the rate constant the same at a lower temperature by reducing the activation energy. Using the Arrhenius equation in logarithmic form, we equate the rate constants at the two different conditions and solve for the original activation energy, which comes out to be $60\ \mathrm{kJ/mol}$.

   Concept and Intuition

   The Arrhenius equation tells us that the rate constant $k$ depends on both the activation energy $E_a$ and the temperature $T$:

   $k = A e^{-E_a/(RT)}$

   If a catalyst lowers $E_a$ by $\Delta E$, the reaction can proceed at the same rate (same $k$) at a lower temperature. Here, the uncatalyzed reaction at 600 K has the same rate constant as the catalyzed reaction at 400 K, with the catalyst reducing $E_a$ by 20 kJ/mol. We set the two Arrhenius expressions equal and solve for the original $E_a$.

   Step-by-step solution

   1. Write the Arrhenius equation for both cases. For the uncatalyzed reaction at $T_1 = 600\ \mathrm{K}$:

   $k = A e^{-E_a/(R T_1)}$

   For the catalyzed reaction at $T_2 = 400\ \mathrm{K}$, the activation energy is $E_a - 20$ (in kJ/mol):

   $k = A e^{-(E_a - 20)/(R T_2)}$

   The pre-exponential factor $A$ is assumed unchanged by the catalyst.

   2. Equate the two expressions for $k$. Since the rate constant is the same:

   $A e^{-E_a/(R T_1)} = A e^{-(E_a - 20)/(R T_2)}$

   Cancel $A$ (non-zero) and take natural logs:

   $-\frac{E_a}{R T_1} = -\frac{E_a - 20}{R T_2}$

   Multiply both sides by $-R$:

   $\frac{E_a}{T_1} = \frac{E_a - 20}{T_2}$

   3. Substitute the temperatures (in Kelvin). $$$$

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.(i) and

   (ii) are 2 chemical reactions carried out at TK. (i). $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{W}$ with $\mathrm{k}_1$ as rate constant. (ii). $\mathrm{X} \rightarrow \mathrm{Z}+\mathrm{W}$ with $k_2$ as rate constant. The Activation energy for reaction

   (ii) is 3 times that of reaction (i). What is the expression for $k_1$ ? (A) $k_1=-2 k_2 e^{2 E a_1 / R T}$ (B) $k_1=k_2 e^{2 E a_1 / R T}$ (C) $k_1=2 k_2 e^{E a_1 / R T}$ (D) $k_1=k_2 / 2 e^{-E a_1 / R T}$

   ›Reveal solutionSolution

   The problem gives two parallel reactions from the same reactant X, with activation energies related by $E_{a2} = 3E_{a1}$. Using the Arrhenius equation and the fact that the pre-exponential factors are equal (same reactant, similar reactions), we find $k_1 = k_2 e^{2E_{a1}/RT}$, which corresponds to option (B).

   The key idea here is the Arrhenius equation, which relates the rate constant $k$ to the activation energy $E_a$ and temperature $T$:

   $k = A e^{-E_a / RT}$

   where $A$ is the pre-exponential factor (frequency factor). For two reactions starting from the same reactant and producing similar products, it is reasonable to assume the pre-exponential factors are equal: $A_1 = A_2$. The problem gives $E_{a2} = 3E_{a1}$. We can then write expressions for $k_1$ and $k_2$ and eliminate the common $A$ to relate them.

   1. Write the Arrhenius equations for both reactions For reaction (i): $k_1 = A_1 e^{-E_{a1}/RT}$ For reaction (ii): $k_2 = A_2 e^{-E_{a2}/RT}$ Given $E_{a2} = 3E_{a1}$ and assuming $A_1 = A_2 = A$, we have:

   $k_1 = A e^{-E_{a1}/RT}, \quad k_2 = A e^{-3E_{a1}/RT}$

   2. Divide the two equations to eliminate $A$

   $\frac{k_1}{k_2} = \frac{A e^{-E_{a1}/RT}}{A e^{-3E_{a1}/RT}} = e^{(-E_{a1} + 3E_{a1})/RT} = e^{2E_{a1}/RT}$

   3. Solve for $k_1$ Multiply both sides by $k_2$: $k_1 = k_2 \, e^{2E_{a1}/RT}$ …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.When the temperature of a reaction $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ is increased from 300 K to 310 K the rate constant increases by $12 \%$. What is the Activation energy of the reaction? (A) $163.9 \mathrm{~kJ} / \mathrm{mol}$ (B) $85.69 \mathrm{~kJ} / \mathrm{mol}$ (C) $192.17 \mathrm{~kJ} / \mathrm{mol}$ (D) $8.76 \mathrm{~kJ} / \mathrm{mol}$

   ›Reveal solutionSolution

   From the Arrhenius equation with $k_2/k_1=1.12$ between $300\ \text{K}$ and $310\ \text{K}$, $E_a\approx 8.76\ \text{kJ/mol}$ — option (D).

   A $12\%$ increase means $\dfrac{k_2}{k_1}=1.12$. The two-temperature Arrhenius form:

   $\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right).$

   $\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{300}-\frac{1}{310}=\frac{10}{93000}=1.075\times10^{-4}\ \text{K}^{-1},$

   $\ln(1.12)=0.1133.$

   Solving for $E_a$ (with $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$): …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two statements, One Assertion (A) and the other Reason (R) are given. Choose the right option. Assertion: The rate constant $(\mathrm{k})$ for a chemical reaction gets nearly doubled for a $10^0$ rise in temperature. Reason: The number of bimolecular collisions between reactant molecules increase with increase in temperature (A) A is correct but R is wrong. (B) Both A and R are correct but R is not the correct explanation of A . (C) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$. (D) A is wrong but R is correct.

   ›Reveal solutionSolution

   The assertion that the rate constant roughly doubles for a 10 °C rise is empirically true for many reactions near room temperature, but the reason given — increased bimolecular collision frequency — is not the correct explanation; the real cause is the exponential increase in the fraction of molecules with energy above the activation barrier.

   The key concept here is the Arrhenius equation, which separates the effect of temperature on reaction rates into two parts: the collision frequency and the fraction of collisions that are energetic enough to overcome the activation energy. While both factors increase with temperature, the dominant effect for a typical 10 °C rise is the exponential increase in the fraction of high-energy molecules, not the modest increase in collision frequency.

   1. Understanding the Assertion (A) The statement that the rate constant $k$ nearly doubles for a 10 °C rise is a well-known rule of thumb for many reactions near room temperature. This comes from the Arrhenius equation:

   $k = A e^{-E_a/(RT)}$

   where $A$ is the pre-exponential factor (related to collision frequency and orientation), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature.

   For a typical activation energy of about 50 kJ/mol, a 10 °C rise from 300 K to 310 K gives:

   $\frac{k_{310}}{k_{300}} = \exp\left[-\frac{E_a}{R}\left(\frac{1}{310} - \frac{1}{300}\right)\right] \approx \exp\left( \frac{50\,000}{8.314} \times \frac{10}{300 \times 310} \right) \approx e^{0.65} \approx 1.92$

   So the assertion is correct for many common reactions.

   2. Understanding the Reason (R) The reason states that the number of bimolecular collisions between reactant molecules increases with temperature. This is true — from kinetic molecular theory, the collision frequency is proportional to $\sqrt{T}$. For a 10 °C rise from 300 K to 310 K, the increase is:

   $\frac{\sqrt{310}}{\sqrt{300}} \approx 1.016$

   That’s only about a 1.6% increase, far from the ~100% increase in the rate constant. So while the reason is factually correct, it is not sufficient to explain the near-doubling of $k$.

   3. Why the reason fails as an explanation …

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7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two chemical reactions of the same order have equal Frequency factor value. Their Activation energies differ by $26.8 \mathrm{~kJ} / \mathrm{mol}$. At 300 K if $k_2=x k_1$ find the value of $x$. (A) $4.631 \times 10^4$ (B) $1.143 \times 10^3$ (C) $2.286 \times 10^3$ (D) 4.665

   ›Reveal solutionSolution

   The ratio of rate constants for two reactions with equal frequency factors depends only on the difference in activation energies via the Arrhenius equation. Using $\Delta E_a = 26.8\ \mathrm{kJ/mol}$ at $T=300\ \mathrm{K}$, we find $x = e^{\Delta E_a/(RT)} \approx 4.631 \times 10^4$, so option (A) is correct.

   The key idea is the Arrhenius equation:

   $k = A e^{-E_a/(RT)}$

   When two reactions have the same frequency factor $A$, the ratio of their rate constants depends only on the difference in activation energies. This lets us directly compute $x = k_2/k_1$ without needing absolute values.

   1. Write the Arrhenius expressions For reaction 1: $k_1 = A e^{-E_{a1}/(RT)}$ For reaction 2: $k_2 = A e^{-E_{a2}/(RT)}$ Since $A$ is equal, the ratio is:

   $\frac{k_2}{k_1} = e^{-(E_{a2} - E_{a1})/(RT)}$

   2. Interpret the given difference The activation energies differ by $26.8\ \mathrm{kJ/mol}$. We are not told which is larger, but the problem states $k_2 = x k_1$. If $E_{a2} < E_{a1}$, then $k_2 > k_1$ and $x > 1$. Given the options, $x$ is large, so we take $E_{a1} - E_{a2} = 26.8\ \mathrm{kJ/mol}$ (i.e., reaction 2 has the lower activation energy). Thus:

   $\frac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/(RT)} = e^{\Delta E_a/(RT)}$

   3. Plug in values with consistent units $\Delta E_a = 26.8\ \mathrm{kJ/mol} = 26800\ \mathrm{J/mol}$ $R = 8.314\ \mathrm{J/(mol \cdot K)}$ $T = 300\ \mathrm{K}$ So: $$$$

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8. KCET 2024Set B-21 markMCQ
   Q.Which one of the following does not represent Arrhenius equation? (A) $\log k = \log A - \dfrac{E_a}{2.303RT}$ (B) $k = A e^{-E_a/RT}$ (C) $\ln k = -\dfrac{E_a}{RT} + \ln A$ (D) $k = A e^{E_a/RT}$

   ›Reveal solutionSolution

   Three options are algebraic rearrangements of $k = Ae^{-E_a/RT}$; the odd one out has lost the minus sign in the exponent.

   Step 1 — The Arrhenius equation.

   $k = A\,e^{-E_a/RT}$

   The exponential factor $e^{-E_a/RT}$ is the Boltzmann fraction — the fraction of molecular collisions with energy at least $E_a$. A fraction must be $\le 1$, which forces the exponent to be negative.

   Step 2 — Check each option.

   (B) $k = Ae^{-E_a/RT}$ — this is the Arrhenius equation. ✓ (represents it)

   (C) Take $\ln$ of (B):

   $\ln k = \ln A + \ln\!\left(e^{-E_a/RT}\right) = \ln A - \frac{E_a}{RT} = -\frac{E_a}{RT} + \ln A. \;✓$

   (A) Convert (C) to base-10 using $\ln y = 2.303\log y$:

   $2.303\log k = 2.303 \log A - \frac{E_a}{RT} \;\Rightarrow\; \log k = \log A - \frac{E_a}{2.303\,RT}. \;✓$

   (D) $k = Ae^{+E_a/RT}$ — the sign is flipped. ✗ …

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.In the presence of a catalyst at a given temperature of $27^{\circ} \mathrm{C}$, the Activation energy of a specific reaction is reduced by $100 \mathrm{~J} / \mathrm{mol}$. What is the ratio between the rate constants for the catalysed $(\mathrm{k}_2)$ and uncatalysed $(\mathrm{k}_1)$ reactions? (A) $1.04 \times 10^{-2}$ (B) $2.32 \times 10^{-2}$ (C) 1.97 (D) 1.04

   ›Reveal solutionSolution

   Lowering $E_a$ by 100 J/mol at 300 K raises the rate constant by a factor $e^{100/RT}=e^{0.0401}\approx1.04$.

   From the Arrhenius equation, at the same temperature:

   $\frac{k_2}{k_1}=\frac{A e^{-E_{a2}/RT}}{A e^{-E_{a1}/RT}}=e^{(E_{a1}-E_{a2})/RT}=e^{\Delta E_a/RT}$ …

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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.The Activation energy for the reaction $A \rightarrow B+C$, at a temperature $\mathrm{TK}$ was $0.04606 \mathrm{~RT} \mathrm{~J} / \mathrm{mol}$. What is the ratio of Arrhenius factor to the Rate constant for this reaction? (A) 1.585 (B) $3.2 \times 10^{-2}$ (C) $1.047 \times 10^{-2}$ (D) 1.047

    ›Reveal solutionSolution

    $A/k=e^{E_a/RT}=e^{0.04606}\approx1.047$.

    Arrhenius equation: $k=Ae^{-E_a/RT}$, so

    $\frac{A}{k}=e^{E_a/RT}$

    Here $E_a=0.04606\,RT$, hence $E_a/RT=0.04606$. …

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11. COMEDK 2024Set 2024-M1 markMCQ
    Q.The rate constant for the reaction $\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$ at $500 \mathrm{~K}$ is given as $0.004 \mathrm{~s}^{-1}$. At what temperature will the rate constant become $0.014 \mathrm{~s}^{-1}$ ? $\mathrm{E}_{\mathrm{a}}$ for the reaction is $18.231 \mathrm{~kJ}$. (A) 950 K (B) 597 K (C) 700 K (D) 800 K

    ›Reveal solutionSolution

    Apply the two-temperature Arrhenius equation with $k_1=0.004\,\text{s}^{-1}$ at $500\,\text{K}$ and $k_2=0.014\,\text{s}^{-1}$; solving gives $T_2\approx700\,\text{K}$.

    $\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

    With $E_a=18231\,\text{J mol}^{-1}$, $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$:

    $\ln\frac{0.014}{0.004}=\ln(3.5)=1.253,\qquad \frac{E_a}{R}=\frac{18231}{8.314}=2193.$ …

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12. KCET 2023Set D-21 markMCQ
    Q.At $500$ K, for a reversible reaction $\mathrm{A}_2(g) + \mathrm{B}_2(g) \rightleftharpoons 2\mathrm{AB}(g)$ in a closed container, $K_c = 2 \times 10^{-5}$. In the presence of catalyst, the equilibrium is attaining $10$ times faster. The equilibrium constant $K_c$ in the presence of catalyst at the same temperature is (A) $2 \times 10^{-4}$ (B) $2 \times 10^{-6}$ (C) $2 \times 10^{-10}$ (D) $2 \times 10^{-5}$

    ›Reveal solutionSolution

    A catalyst does not change the equilibrium constant — it only speeds up the rate at which equilibrium is reached. So $K_c$ remains $2 \times 10^{-5}$ at the same temperature. The correct option is (D).

    The key idea here is simple but often misunderstood: a catalyst affects the rate of a reaction, not the position of equilibrium. Let’s see why.

    1. What a catalyst does

       A catalyst provides an alternative reaction pathway with a lower activation energy. This means both the forward and reverse reactions are accelerated equally. Because it speeds up both directions by the same factor, the ratio of the forward rate constant to the reverse rate constant — which is exactly $K_c$ — stays unchanged.

    2. Equilibrium constant depends only on temperature

       For a given reaction, $K_c$ is a function of temperature alone. It is determined by the standard Gibbs free energy change:

    $\Delta G^\circ = -RT \ln K_c$

    A catalyst does not alter $\Delta G^\circ$; it only lowers the activation barrier. So at the same temperature, $K_c$ is fixed.

    3. The “10 times faster” detail …

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