Ratio of Chemisorption Rates Using the Arrhenius Equation

For the adsorption of hydrogen on platinum, the activation energy is 30kJ mol^(-1) and for the adsorption of hydrogen on nickel, the activation energy is 41.4kJ mol^(-1). The logarithm of the ratio of the rates of chemisorption on equal areas of the metals at 300K is......... (Nearest integer) Given: ln 10 = 2.3 R = 8.3JK^(-1)mol^(-1)

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Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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The logarithm (base 10) of the ratio of the rates of chemisorption of hydrogen on platinum to nickel at 300 K is 2.

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Step-by-step solution

Given: Activation energy for Pt: $(E_{a})_{1} = 30 {kJ mol}^{- 1} = 30000 {J mol}^{- 1}$ . Activation energy for Ni: $(E_{a})_{2} = 41.4 {kJ mol}^{- 1} = 41400 {J mol}^{- 1}$ . Temperature $T = 300 K$ , $R = 8.3 {J K}^{- 1} {mol}^{- 1}$ , $\ln 10 = 2.303 \approx 2.3$ . Assume equal pre-exponential factors $A$ for equal surface areas.
Find: Find the logarithm of the ratio of the chemisorption rates, $\log_{10} (k_{1} / k_{2})$ (or $\log (r_{Pt} / r_{Ni})$ ).
Strategise: Using the Arrhenius rate equation $k = A e^{- E_{a} / R T}$ , the ratio of rates is $k_{1} / k_{2} = e^{((E_{a})_{2} - (E_{a})_{1}) / (R T)}$ . Taking base-10 logarithm: $\log_{10} (k_{1} / k_{2}) = \frac{(E_{a})_{2} - (E_{a})_{1}}{2.303 R T}$ .
Execute: Substitute values: $\log_{10} (k_{1} / k_{2}) = \frac{(41.4 - 30) \times 10^{3}}{2.3 \times 8.3 \times 300} = \frac{11.4 \times 1000}{5727} = 1.99057 \approx 2$ .
Verify: The metal with lower activation energy (Pt) has a higher rate, yielding a positive log-ratio of approximately 2, corresponding to a rate ratio of $\sim 100$ , which is chemically reasonable for $Δ E_{a} \approx 11.4 kJ/mol$ at room temperature.

Ratio of Chemisorption Rates Using the Arrhenius Equation

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The logarithm (base 10) of the ratio of the rates of chemisorption of hydrogen on platinum to nickel at 300 K is 2.

Why each option works or fails

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