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NVAMediumJEE 2025Arrhenius Equation & Activation Energy

JEE Chemistry 2025 Question with Solution

Consider a complex reaction taking place in three steps with rate constants k1k_1, k2k_2, and k3k_3 respectively. The overall rate constant kk is given by the expression k=k1k3k2k = \sqrt{\frac{k_1 k_3}{k_2}}. If the activation energies of the three steps are 60 kJ mol−160 \, \text{kJ mol}^{-1}, 30 kJ mol−130 \, \text{kJ mol}^{-1}, and 10 kJ mol−110 \, \text{kJ mol}^{-1} respectively, then the overall energy of activation in kJ mol−1\text{kJ mol}^{-1} is _____(Nearest integer).

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: The overall rate constant is

k=k1k3k2k = \sqrt{\frac{k_1 k_3}{k_2}}

and the activation energies are E1=60 kJ mol−1E_1 = 60 \, \text{kJ mol}^{-1}, E2=30 kJ mol−1E_2 = 30 \, \text{kJ mol}^{-1}, and E3=10 kJ mol−1E_3 = 10 \, \text{kJ mol}^{-1}.

Find: The overall activation energy.

Using Arrhenius form for each step,

ki=Aie−Ei/RTk_i = A_i e^{-E_i/RT}

So,

k=k1k3k2=(k1k3k2)1/2k = \sqrt{\frac{k_1 k_3}{k_2}} = \left(\frac{k_1 k_3}{k_2}\right)^{1/2}

Taking logarithm in Arrhenius form, the exponent for the overall rate constant becomes

−ERT=12(−E1RT−E3RT+E2RT)-\frac{E}{RT} = \frac{1}{2}\left(-\frac{E_1}{RT} - \frac{E_3}{RT} + \frac{E_2}{RT}\right)

Hence,

E=E1+E3−E22E = \frac{E_1 + E_3 - E_2}{2}

Substituting the given values,

E=60+10−302=402=20 kJ mol−1E = \frac{60 + 10 - 30}{2} = \frac{40}{2} = 20 \, \text{kJ mol}^{-1}

Therefore, the overall activation energy is 20 kJ mol−120 \, \text{kJ mol}^{-1}.

Using logarithm of the Arrhenius expression

For the relation

k=k1k3k2k = \sqrt{\frac{k_1 k_3}{k_2}}

write it as

k=k11/2k31/2k2−1/2k = k_1^{1/2} k_3^{1/2} k_2^{-1/2}

Now substitute Arrhenius expressions:

k=(A1e−E1/RT)1/2(A3e−E3/RT)1/2(A2e−E2/RT)−1/2k = \left(A_1 e^{-E_1/RT}\right)^{1/2}\left(A_3 e^{-E_3/RT}\right)^{1/2}\left(A_2 e^{-E_2/RT}\right)^{-1/2}

Combining exponential terms,

k=constant×e−(E1+E3−E2)/2RTk = \text{constant} \times e^{-\left(E_1 + E_3 - E_2\right)/2RT}

Comparing with

k=Ae−E/RTk = A e^{-E/RT}

we get

E=E1+E3−E22E = \frac{E_1 + E_3 - E_2}{2}

Thus,

E=60+10−302=20 kJ mol−1E = \frac{60 + 10 - 30}{2} = 20 \, \text{kJ mol}^{-1}

So the numerical value answer is 20.

The solution contains inconsistent intermediate statements such as E=E1+E3−E2E = E_1 + E_3 - E_2 and another incorrect approximation, but its final conclusion is 20, which also matches the correct Arrhenius derivation.

Common mistakes

  • Using E=E1+E3−E2E = E_1 + E_3 - E_2 directly without accounting for the square root. This is wrong because the factor 1/21/2 from x\sqrt{\phantom{x}} must multiply the exponent. Rewrite the given rate constant as a power expression before comparing Arrhenius exponents.

  • Taking square root of activation energies, such as Eoverall=E1E3E2E_{\text{overall}} = \sqrt{\frac{E_1 E_3}{E_2}}. This is wrong because Arrhenius combination occurs through exponents of rate constants, not through direct algebraic operations on the activation energies themselves.

  • Substituting the rate-constant relation numerically without first expressing each kik_i in Arrhenius form. This misses how activation energies combine. Always write ki=Aie−Ei/RTk_i = A_i e^{-E_i/RT} and then compare the exponential term with e−E/RTe^{-E/RT}.

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