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NEET-UG and JEE Main Chemistry · Class 12

Arrhenius equation and activation energy — NEET & JEE Questions

Arrhenius equation and activation energy is a topic in Chemical Kinetics, a Class 12 Chemistry chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 47 published NEET-UG and JEE Main questions on Arrhenius equation and activation energy, each with the marking scheme of the real paper.

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Arrhenius equation and activation energy MCQs with solutions

10 of the 47 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    The Arrhenius equation is commonly written as

    • A. k=Ae−Ea/(RT)k=Ae^{-E_a/(RT)}
    • B. k=AeEaRTk=Ae^{E_aRT}
    • C. k=Ea/(RT)k=E_a/(RT)
    • D. k=ARTk=ART
  2. Question 2 · difficulty L2 · understanding

    Decomposition of a hydrocarbon follows the equation k=(5.5×1011 s−1)e−28000 K T\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}. The activation energy of reaction is ____\_\_\_\_ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}. (Nearest Integer) Given : R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

  3. Question 3 · difficulty L2 · understanding

    For a reaction, given below is the graph of ln⁡k\ln k vs 1T{1 \over T}. The activation energy for the reaction is equal to ____________ cal mol−1\mathrm{cal} \,\mathrm{mol}^{-1}. (nearest integer) (Given : R=2 cal K−1  mol−1\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1} )

  4. Question 4 · difficulty L2 · understanding

    The equation k = (6.5 ×\times 10 12 s −-1 )e −-26000K/T is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol −-1 . [nearest integer] (Given : R = 8.314 J K −-1 mol −-1 )

  5. Question 5 · difficulty L2 · understanding

    The rate constant for a first order reaction is given by the following equation: ln⁡k=33.24−2.0×104 KT\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T} The activation energy for the reaction is given by ____________ kJ mol −-1 . (In nearest integer) (Given : R = 8.3 J K −-1 mol −-1 )

  6. Question 6 · difficulty L2 · understanding

    If the activation energy of a reaction is 80.9 kJ mol −-1 , the fraction of molecules at 700 K, having enough energy to react to form products is e −-x . The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K −-1 mol −-1 ]

  7. Question 7 · difficulty L2 · understanding

    Consider the given figure and choose the correct option :

    • A. Activation energy of both forward and backward reaction is E1+E2E_1+E_2 and reactant is more stable than product.
    • B. Activation energy of forward reaction is E1+E2E_1+E_2 and product is more stable than reactant.
    • C. Activation energy of backward reaction is E1\mathrm{E}_1 and product is more stable than reactant.
    • D. Activation energy of forward reaction is E1+E2E_1+E_2 and product is less stable than reactant.
  8. Question 8 · difficulty L2 · understanding

    For a first-order reaction A →\to P, the temperature (T) dependent rate constant (k) was found to follow the equation log⁡k=−(2000)1T+6.0\log k = - (2000){1 \over T} + 6.0. The pre-exponential factor A and activation energy EaE_a, respectively, are

    • A. 1.0×106 s−11.0\times10^6~\mathrm{s^{-1}} and 9.2 kJ mol−1^{-1}
    • B. 6.0 s−16.0~\mathrm{s^{-1}} and 16.6 kJ mol−1^{-1}
    • C. 1.0×106 s−11.0\times10^6~\mathrm{s^{-1}} and 16.6 kJ mol−1^{-1}
    • D. 1.0×106 s−11.0\times10^6~\mathrm{s^{-1}} and 38.3 kJ mol−1^{-1}
  9. Question 9 · difficulty L3 · understanding

    For reaction A → P, rate constant k=1.5×103 s−1k = 1.5 \times 10^3 \ \mathrm{s}^{-1} at 27∘C27^{\circ}\mathrm{C} If activation energy for the above reaction is 60 kJ mol−160\ \mathrm{kJ}\ \mathrm{mol}^{-1}, then the temperature (in ∘C^{\circ}\mathrm{C}) at which rate constant, k=4.5×103 s−1k = 4.5 \times 10^3\ \mathrm{s}^{-1} is ______. (Nearest integer) Given : log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48, R=8.3 J K−1 mol−1R = 8.3\ \mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}, ln⁡10=2.3\ln 10 = 2.3

  10. Question 10 · difficulty L3 · understanding

    Consider A→k1 B\mathrm{A} \xrightarrow{\mathrm{k}_1} \mathrm{~B} and C→k2D\mathrm{C} \xrightarrow{\mathrm{k}_2} \mathrm{D} are two reactions. If the rate constant (k1)\left(\mathrm{k}_1\right) of the A⟶B\mathrm{A} \longrightarrow \mathrm{B} reaction can be expressed by the following equation log⁡10k=14.34−1.5×104 T/K\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}} and activation energy of C⟶DC \longrightarrow D reaction (Ea2)\left(E a_2\right) is 15\frac{1}{5} th of the A⟶BA \longrightarrow B reaction (Ea1)\left(E a_1\right), then the value of (Ea2)\left(E a_2\right) is ____\_\_\_\_ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}. (Nearest Integer)

Common mistakes in Arrhenius equation and activation energy

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Assumes higher temperature increases the activation-energy barrier.
  • Forgets the negative sign or inverts the $E_a/R$ ratio.
  • Assumes activation energy affects only equilibrium and not the rate constant.

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