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MCQMediumJEE 2026Arrhenius Equation & Activation Energy

JEE Chemistry 2026 Question with Solution

At 27∘C27^\circ C, in presence of a catalyst, activation energy of a reaction is lowered by 10 kJmol−110\,kJ mol^{-1}. The logarithm of the ratio k(catalysed)k(uncatalysed)\dfrac{k(catalysed)}{k(uncatalysed)} is _____.

(Consider that the frequency factor for both the reactions is same)

  • A

    0.17410.1741

  • B

    1.7411.741

  • C

    3.4823.482

  • D

    17.4117.41

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: At 27∘C27^\circ C, the activation energy is lowered by 10 kJmol−110\,kJ mol^{-1} and the frequency factor is same for both reactions.

Find: The logarithm of the ratio k(catalysed)k(uncatalysed)\dfrac{k(catalysed)}{k(uncatalysed)}.

Using Arrhenius equation,

k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}

Since the frequency factor AA is same for both reactions,

ln⁡(kcku)=Eu−EcRT\ln\left(\frac{k_c}{k_u}\right) = \frac{E_u - E_c}{RT}

Substituting the given values,

Eu−Ec=10 kJmol−1=104 Jmol−1E_u - E_c = 10\,kJ mol^{-1} = 10^4\,J mol^{-1} T=27∘C=300 KT = 27^\circ C = 300\,K R=8.314 Jmol−1K−1R = 8.314\,J mol^{-1}K^{-1}

Therefore,

ln⁡(kcku)=1048.314×300≈4.01\ln\left(\frac{k_c}{k_u}\right) = \frac{10^4}{8.314 \times 300} \approx 4.01

Converting natural logarithm to common logarithm,

log⁡(kcku)=4.012.303≈1.741\log\left(\frac{k_c}{k_u}\right) = \frac{4.01}{2.303} \approx 1.741

The solution then states that since the question asks logarithm of ratio multiplied by 1010,

=17.41= 17.41

Hence, the correct option according to the provided solution is D. Note that this conclusion follows the source solution text, although the intermediate calculation gives 1.7411.741 for the common logarithm.

Source Hint and Interpretation

Hint from source: A small decrease in activation energy leads to a very large increase in reaction rate.

This means the ratio k(catalysed)k(uncatalysed)\dfrac{k(catalysed)}{k(uncatalysed)} increases exponentially because the exponential term in Arrhenius equation depends on activation energy.

Using the source working,

ln⁡(kcku)≈4.01\ln\left(\frac{k_c}{k_u}\right) \approx 4.01

and

log⁡(kcku)≈1.741\log\left(\frac{k_c}{k_u}\right) \approx 1.741

However, the provided solution explicitly concludes option (4)\text{(4)} and marks D as correct.

Common mistakes

  • Using the Arrhenius equation separately for the two reactions and forgetting to take the ratio. This is inefficient and can cause cancellation errors. Since the frequency factor is same, directly write the logarithm of the ratio and use the activation-energy difference.

  • Not converting 10 kJmol−110\,kJ mol^{-1} into Jmol−1J mol^{-1}. The gas constant R=8.314R = 8.314 is in SI units, so activation energy must be written as 104 Jmol−110^4\,J mol^{-1}.

  • Confusing ln⁡\ln with log⁡\log. The Arrhenius form first gives a natural logarithm. To convert to common logarithm, divide by 2.3032.303.

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