Q.The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate $E_a$.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In the Arrhenius equation, $\exp\left(-\dfrac{E_a}{RT}\right)$, is equal to (A) Frequency factor (B) Fraction of molecules that have energy higher than $E_a$ (C) Rate of the reaction (D) Rate constant of the reaction

   ›Reveal solutionSolution

   This tests the meaning of each term in the Arrhenius equation; the exponential term represents the fraction of molecules with energy exceeding the activation energy.

   Concept and Intuition

   The Arrhenius equation, $k = A e^{-E_a/RT}$, splits the rate constant into two physically distinct factors: $A$ (the frequency/pre-exponential factor, related to collision frequency and orientation) and $e^{-E_a/RT}$ (the fraction of molecular collisions that have enough energy to cross the activation energy barrier, derived from the Maxwell-Boltzmann energy distribution). Only molecules with energy $\geq E_a$ can react upon collision; the exponential term quantifies what fraction of the population meets this threshold at a given temperature.

   Step-by-Step Solution

   1. Write the Arrhenius equation: $k = A\,e^{-E_a/RT}$.
   2. Recognize $A$ is the frequency factor — related to the rate of collisions and their proper orientation, independent of energy considerations.
   3. Recognize $e^{-E_a/RT}$ arises from integrating the Maxwell-Boltzmann distribution above the energy threshold $E_a$ — it represents the fraction of molecules whose energy exceeds $E_a$ at temperature $T$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.At 300°C, decomposition of azomethane follows first order kinetics. Rate constant of this reaction at this temperature is $2.5 \times 10^{-4} s^{-1}$. If the activation energy of the reaction is 42 kcal mol$^{-1}$, what is the temperature (in K) at which the half-life of the reaction is 138.6 seconds? ($R = 2$ cal $K^{-1}mol^{-1}$, $\log 20 = 1.30$) (A) 725 (B) 425 (C) 525 (D) 625

   ›Reveal solutionSolution

   This tests the Arrhenius equation relating rate constants at two temperatures; solving gives $T_2 \approx 625\,K$.

   Concept and Intuition

   A faster half-life means a larger rate constant. The Arrhenius equation connects the ratio of two rate constants at two temperatures to the activation energy — a bigger $E_a$ needs a bigger temperature jump to boost $k$ by the same factor.

   Step-by-Step Solution

   1. $T_1 = 300°C = 573\,K$, $k_1 = 2.5\times10^{-4}\,s^{-1}$.
   2. Desired half-life $t_{1/2}=138.6\,s$ for a first-order reaction: $k_2 = \dfrac{0.693}{138.6} = 5\times10^{-3}\,s^{-1}$.
   3. Ratio: $k_2/k_1 = \dfrac{5\times10^{-3}}{2.5\times10^{-4}} = 20$.
   4. Arrhenius (two-temperature form): $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$.
   5. $\dfrac{E_a}{2.303R} = \dfrac{42000}{2.303\times2} = 9118.5$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The rate constant of a first order reaction at 600 K is $1.6\times10^{-5}s^{-1}$. If its activation energy is 198.87 $kJ\ mol^{-1}$, what is the rate constant (in $s^{-1}$) at 700 K? ($R = 8.3\ J\ mol^{-1}K^{-1}$) (antilog(0.4771) = 3.0) (A) $4.8\times10^{-4}$ (B) $4.8\times10^{-3}$ (C) $4.8\times10^{-2}$ (D) $3.2\times10^{-4}$

   ›Reveal solutionSolution

   The Arrhenius equation converts rate constants at two temperatures via activation energy; the given antilog hint (antilog(0.4771)=3.0) is a shortcut for the final power-of-ten step, giving $k_2=4.8\times10^{-3}\ s^{-1}$.

   Concept and Intuition

   The two-point form of the Arrhenius equation lets us find how much a rate constant changes when temperature changes, purely from the activation energy — a higher $E_a$ makes the rate far more sensitive to temperature. The problem hands us the exact antilog we'll need at the end, which is a strong signal the intended computation lands exactly on $\log(k_2/k_1)=2.4771$.

   Step-by-Step Solution

   1. Write the two-temperature Arrhenius relation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$.
   2. Compute $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{600}-\dfrac{1}{700} = \dfrac{700-600}{600\times700} = \dfrac{100}{420000} = \dfrac{1}{4200}$.
   3. Compute $\dfrac{E_a}{2.303R} = \dfrac{198870}{2.303\times8.3} = \dfrac{198870}{19.1149} \approx 10403.9$.
   4. Multiply: $10403.9 \times \dfrac{1}{4200} \approx 2.4771$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The k value of the reaction A $\to$ products is $5 \times 10^3\ s^{-1}$ at 300 $K$. Its activation energy is $50\ kJ\,mol^{-1}$. At $T(K)$, its value becomes $1.0 \times 10^4\ s^{-1}$. What is the value of T (in K)? ($R = 8.3\ J\,mol^{-1}K^{-1}$), ($\log 2 = 0.3$) (A) $397$ (B) $311$ (C) $286$ (D) $345$

   ›Reveal solutionSolution

   This tests the two-point form of the Arrhenius equation to find an unknown temperature given a rate constant ratio. The answer works out to T ≈ 311 K.

   Concept and Intuition

   The Arrhenius equation $k = A e^{-E_a/RT}$ can be converted, for two temperatures, into a linear relation between $\log(k_2/k_1)$ and $(1/T_1 - 1/T_2)$, letting us solve for an unknown temperature when the rate constants and activation energy are known.

   Step-by-Step Solution

   1. Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$.
   2. Given $k_1 = 5\times10^3\ s^{-1}$ at $T_1 = 300\ K$, and $k_2 = 1.0\times10^4\ s^{-1}$ at $T_2 = T$ (to find). Ratio $k_2/k_1 = \dfrac{1.0\times10^4}{5\times10^3} = 2$.
   3. Given $\log 2 = 0.3$, so LHS $= 0.3$.
   4. Compute $\dfrac{E_a}{2.303R} = \dfrac{50000}{2.303 \times 8.3} = \dfrac{50000}{19.115} \approx 2615.7$.
   5. So $0.3 = 2615.7\left(\dfrac{1}{300} - \dfrac{1}{T}\right)$, giving $\dfrac{1}{300} - \dfrac{1}{T} = \dfrac{0.3}{2615.7} \approx 1.1468\times10^{-4}$. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.At 600 K, the time taken for the completion of 10% of a first order reaction is same as that of its 20% completion at 610 K. What is the value of ratio of rate constants $\left(\dfrac{k_{610}}{k_{600}}\right)$? ($\log(1.111) = 0.0457$; $\log(1.25) = 0.0969$) (A) 1.211 (B) 2.118 (C) 2.511 (D) 2.711

   ›Reveal solutionSolution

   Same reaction time for 10% completion at 600K and 20% completion at 610K lets the time cancel, leaving $k_{610}/k_{600}=\log(1.25)/\log(1.111)\approx2.12$.

   Concept and Intuition

   For a first-order reaction, $k=\dfrac{1}{t}\ln\dfrac{1}{1-x}$ where $x$ is the fraction reacted. If the same time $t$ produces different completions at two temperatures, that common $t$ cancels out when taking the ratio of the two rate constants, leaving a ratio of two logarithmic (completion) factors only.

   Step-by-Step Solution

   1. At 600 K, 10% complete in time $t$: $k_{600} = \dfrac{1}{t}\ln\dfrac{1}{0.90} = \dfrac{2.303}{t}\log(1.111)$.
   2. At 610 K, 20% complete in the same time $t$: $k_{610} = \dfrac{1}{t}\ln\dfrac{1}{0.80} = \dfrac{2.303}{t}\log(1.25)$.
   3. Ratio: $\dfrac{k_{610}}{k_{600}} = \dfrac{\log(1.25)}{\log(1.111)} = \dfrac{0.0969}{0.0457} \approx 2.12$. …

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6. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Given below are two statements Statement I: Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values Statement II: In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than $E_a$ Correct answer is (A) Statements I and II both are correct (B) Statements I and II both are not correct (C) Statement I is correct but statement II is not correct (D) Statement I is not correct but statement II is correct

   ›Reveal solutionSolution

   Statement I is correct because reaction orders are experimentally determined and can be zero, positive integers, or fractions. Statement II is incorrect because the frequency factor (pre-exponential factor) is not a fraction of molecules with energy above $E_a$; that fraction is given by the exponential term $e^{-E_a/RT}$. Therefore, the correct option is (C).

   The Arrhenius equation is the backbone of chemical kinetics when it comes to temperature dependence. It is written as:

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

   Here, $k$ is the rate constant, $A$ is the frequency factor (also called the pre-exponential factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The key intuition:

   • $A$ represents how often molecules collide in the correct orientation (a frequency, not a fraction).
   • The exponential term $e^{-E_a/RT}$ is the fraction of molecules that have energy equal to or greater than $E_a$. Mixing these two up is a classic mistake.

   Now, let’s evaluate each statement carefully.

   1. Statement I: "Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values"

      • The order of a reaction is defined as the sum of the exponents of concentration terms in the experimentally determined rate law.
      • It is not derived from the stoichiometric coefficients (that would be molecularity, which applies only to elementary steps).
      • Orders can indeed be zero (e.g., decomposition of ammonia on a platinum surface), positive integers (most common), or positive fractions (e.g., the reaction $H_2 + Br_2 \rightarrow 2HBr$ has order 1.5).
      • Negative orders are also possible, but the statement only claims zero, positive integers, or positive fractions — all of which are valid.
      • Conclusion: Statement I is correct.

   2. Statement II: "In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than $E_a$"

      • Let’s revisit the Arrhenius equation: $k = A e^{-E_a/RT}$. …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.The rate constant of a first order reaction at 400 and 500 K is respectively $2\times10^{-5}\ \mathrm{s^{-1}}$ and $4\times10^{-3}\ \mathrm{s^{-1}}$. What is the approximate activation energy (in $\mathrm{kJ\,mol^{-1}}$)? ($R = 8.3\ \mathrm{J\,mol^{-1}K^{-1}}$; $\log 2 = 0.3$) (A) 880 (B) 88 (C) 38.2 (D) 8.8

   ›Reveal solutionSolution

   Plugging the two rate constants and temperatures into the two-point Arrhenius equation gives an activation energy of about 88 kJ/mol.

   Concept and Intuition

   The Arrhenius equation's two-temperature form lets you find activation energy directly from two rate constants at two known temperatures, without knowing the pre-exponential factor.

   Step-by-Step Solution

   1. Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$.
   2. $\dfrac{k_2}{k_1} = \dfrac{4\times10^{-3}}{2\times10^{-5}} = 200$; $\log 200 = \log 2 + \log 100 = 0.3 + 2 = 2.3$.
   3. $\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{400} - \dfrac{1}{500} = \dfrac{500-400}{400\times500} = \dfrac{100}{200000} = 5\times10^{-4}\ \mathrm{K^{-1}}$.
   4. $2.3 = \dfrac{E_a}{2.303 \times 8.3} \times 5\times10^{-4}$.
   5. $2.303 \times 8.3 = 19.115$. …

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8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The rate constant of a first order reaction is $3.46 \times 10^{-2}\ s^{-1}$ at 298K. What is the rate constant of the reaction at 350 K if its activation energy is $50.1\ kJ\ mol^{-1}$? ($R = 8.314\ J\ K^{-1}mol^{-1}$) (log 2 = 0.3010) (A) $0.592\ s^{-1}$ (B) $0.692\ s^{-1}$ (C) $0.792\ s^{-1}$ (D) $0.892\ s^{-1}$

   ›Reveal solutionSolution

   This applies the two-temperature Arrhenius equation to find the rate constant at a higher temperature given the activation energy; the result is closest to $0.692\ s^{-1}$.

   Concept and Intuition

   The Arrhenius equation, $k = Ae^{-E_a/RT}$, shows that rate constants increase sharply with temperature for reactions with a sizeable activation energy. Taking the equation at two temperatures and dividing eliminates the pre-exponential factor $A$, giving a convenient logarithmic two-point formula for finding a rate constant at a new temperature.

   Step-by-Step Solution

   1. Write the two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$.
   2. Compute $\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{298} - \dfrac{1}{350} = \dfrac{350-298}{298\times350} = \dfrac{52}{104300} \approx 4.986\times10^{-4}\ \text{K}^{-1}$.
   3. Compute $\dfrac{E_a}{2.303R} = \dfrac{50100}{2.303\times8.314} = \dfrac{50100}{19.147} \approx 2616.6$.
   4. Multiply: $\log\dfrac{k_2}{k_1} \approx 2616.6 \times 4.986\times10^{-4} \approx 1.305$.
   5. So $\dfrac{k_2}{k_1} = 10^{1.305} = 10 \times 10^{0.305}$. Using $\log 2 = 0.3010$, $10^{0.305}$ is just slightly above 2, i.e. $\approx 2.02$, giving $k_2/k_1 \approx 20.2$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The rate constant of a first order reaction was doubled when the temperature was increased from 300 to 310 K. What is its approximate activation energy (in $\mathrm{kJ\,mol^{-1}}$)? ($R = 8.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$; $\log 2 = 0.3$) (A) 5.33 (B) 533.3 (C) 53333 (D) 53.33

   ›Reveal solutionSolution

   Plugging the rate-doubling data into the two-point Arrhenius equation gives $E_a \approx 53.33\ \mathrm{kJ\,mol^{-1}}$ — option (D).

   Concept and Intuition

   The Arrhenius equation, $k = Ae^{-E_a/RT}$, links a reaction's rate constant to temperature through its activation energy. When we know the rate constant at two temperatures, we can eliminate the pre-exponential factor $A$ and solve directly for $E_a$ using the two-temperature form of the equation — this is the standard way "activation energy from a rate-doubling" problems are solved.

   Step-by-Step Solution

   1. Two-point Arrhenius form: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$.
   2. Here $k_2/k_1 = 2$ (rate doubled), so $\log 2 = 0.3$ (given).
   3. $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{300}-\dfrac{1}{310} = \dfrac{310-300}{300\times310} = \dfrac{10}{93000} = 1.0753\times10^{-4}\ \mathrm{K^{-1}}$.
   4. $0.3 = \dfrac{E_a}{2.303\times 8.3}\times 1.0753\times10^{-4}$.
   5. $2.303\times 8.3 = 19.115$. …

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10. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.The rate constants of a first order reaction at 300 K is $k_1$ and 400 K is $k_2$. What is the value of $\ln\dfrac{k_2}{k_1}$ if activation energy of reaction is $41.5\ \mathrm{kJ\,mol^{-1}}$? ($R = 8.3\ \mathrm{J\,K^{-1}\,mol^{-1}}$) (A) 1.809 (B) 4.166 (C) 2.083 (D) 3.618

    ›Reveal solutionSolution

    Plugging $E_a$, $R$, $T_1=300$ K and $T_2=400$ K into the two-point Arrhenius equation gives $\ln(k_2/k_1)\approx 4.166$.

    Concept and Intuition

    The Arrhenius equation $k = A\,e^{-E_a/RT}$ shows that rate constants at two temperatures are related by $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$, since the pre-exponential factor $A$ cancels out. A higher activation energy or a larger temperature gap gives a bigger ratio $k_2/k_1$.

    Step-by-Step Solution

    1. Convert $E_a$ to joules: $41.5\ \mathrm{kJ/mol} = 41500\ \mathrm{J/mol}$.
    2. Compute $\dfrac{E_a}{R} = \dfrac{41500}{8.3} = 5000$.
    3. Compute $\dfrac1{T_1}-\dfrac1{T_2} = \dfrac1{300}-\dfrac1{400} = \dfrac{4-3}{1200} = \dfrac1{1200}$. …

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11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The rate constant, k for a first order reaction, $\mathrm{C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)}$ is x $\mathrm{s^{-1}}$ at 600 K and 4x $\mathrm{s^{-1}}$ at 700 K. The energy of activation of the reaction (in kJ $\mathrm{mol^{-1}}$) is (log 4 = 0.6, R = 8.3 J $\mathrm{K^{-1}}$ $\mathrm{mol^{-1}}$) (A) 48.16 (B) 58.16 (C) 38.16 (D) 28.16

    ›Reveal solutionSolution

    Plugging the given rate-constant ratio and temperatures into the two-point Arrhenius equation gives an activation energy of 48.16 kJ/mol.

    Concept and Intuition

    The Arrhenius equation relates the rate constant at two different temperatures to the activation energy: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$. Given rate constants at two temperatures, we can solve directly for $E_a$.

    Step-by-Step Solution

    1. $k_1 = x$ at $T_1=600$ K, $k_2=4x$ at $T_2=700$ K, so $k_2/k_1 = 4$.
    2. $\ln4 = 2.303\log4 = 2.303\times0.6 = 1.3818$.
    3. $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac1{600}-\dfrac1{700} = \dfrac{700-600}{600\times700}=\dfrac{100}{420000}=2.381\times10^{-4}\ \text{K}^{-1}$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.The rate constant of a reaction at 500K and 700K are 0.02 s$^{-1}$ and 0.2 s$^{-1}$ respectively. The activation energy of the reaction (in kJ mol$^{-1}$) is ($R=8.3$ J K$^{-1}$ mol$^{-1}$) (A) 66.90 (B) 33.45 (C) 22.30 (D) 44.45

    ›Reveal solutionSolution

    Applying the two-point Arrhenius equation gives activation energy $\approx 33.45$ kJ/mol.

    Concept and Intuition

    The Arrhenius equation $k = A e^{-E_a/RT}$ links rate constant to temperature through the activation energy. Taking the ratio of rate constants at two temperatures eliminates the pre-exponential factor $A$, letting us solve directly for $E_a$.

    Step-by-Step Solution

    1. Write $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$.
    2. $\dfrac{k_2}{k_1} = \dfrac{0.2}{0.02}=10$, so $\ln 10 = 2.303$.
    3. $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{500}-\dfrac{1}{700} = \dfrac{700-500}{500\times700} = \dfrac{200}{350000} = 5.714\times10^{-4}\ \text{K}^{-1}$. …

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