JEE Main 2023 — Surface Chemistry Question with Solution

Step 1: Identify Activation Energies

• Activation energy for platinum: $E_{a1}=30kJ/mol=30\times 10^{3}J/mol$
• Activation energy for nickel: $E_{a2}=41.4kJ/mol=41.4\times 10^{3}J/mol$

Step 2: Given Constants

• Gas constant: $R=8.3\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$
• Temperature: $T=300\mathrm{K}$
• $\ln 10=2.3$

Step 3: Calculate Ratio of Rate Constants

Using the Arrhenius equation, the ratio of the rate constants can be calculated as:

$\frac{K_2}{K_1}=\exp \left(\frac{E_{a1}-E_{a2}}{RT}\right)$

Step 4: Find Logarithm of the Ratio

Taking the natural logarithm and using $\ln 10=2.3$ to convert to base-10 logarithm:

$\log \frac{K_2}{K_1}=\frac{\ln \left(\frac{K_2}{K_1}\right)}{\ln 10}=\frac{E_{a1}-E_{a2}}{2.3\cdot RT}$

Step 5: Substitute Values

$\log \frac{K_2}{K_1}=\frac{(30\times 10^3-41.4\times 10^3)J/mol}{2.3\times 8.3\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\times 300\mathrm{K}}\approx 1.99$

Step 6: Final Answer

The nearest integer to the calculated value is 2.

So, the logarithm of the ratio of the rates of chemisorption on equal areas of platinum and nickel at 300 K is 2.