For the adsorption of hydrogen on platinum, the activation energy is $30 ~\mathrm{k} ~\mathrm{J} \mathrm{mol}^{-1}$ and for the adsorption of hydrogen on nickel, the activation energy is $41.4 ~\mathrm{k} ~\mathrm{J} \mathrm{mol}^{-1}$. The logarithm of the ratio of the rates of chemisorption on equal areas of the metals at $300 \mathrm{~K}$ is __________ (Nearest integer)

Given: $\ln 10=2.3$

$\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

<p><strong>Step 1: Identify Activation Energies</strong></p> <ul> <li>Activation energy for platinum: $E_{a1} = 30 \, \mathrm{kJ/mol} = 30 \times 10^3 \, \mathrm{J/mol}$</li> <li>Activation energy for nickel: $E_{a2} = 41.4 \, \mathrm{kJ/mol} = 41.4 \times 10^3 \, \mathrm{J/mol}$</li> </ul> <p><strong>Step 2: Given Constants</strong></p> <ul> <li>Gas constant: $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \mathrm{mol}^{-1}$</li> <li>Temperature: $T = 300 \, \mathrm{K}$</li> <li>$\ln 10 = 2.3$</li> </ul> <p><strong>Step 3: Calculate Ratio of Rate Constants</strong> <br/><br/>Using the Arrhenius equation, the ratio of the rate constants can be calculated as: <br/><br/>$ \frac{K_2}{K_1} = \exp\left(\frac{E_{a1} - E_{a2}}{RT}\right) $</p> <p><strong>Step 4: Find Logarithm of the Ratio</strong><br/><br/> Taking the natural logarithm and using $\ln 10 = 2.3$ to convert to base-10 logarithm: <br/><br/>$ \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} $</p> <p><strong>Step 5: Substitute Values</strong> <br/><br/>$ \log \frac{K_2}{K_1} = \frac{(30 \times 10^3 - 41.4 \times 10^3) \, \mathrm{J/mol}}{2.3 \times 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \, \mathrm{K}} \approx 1.99 $</p> <p><strong>Step 6: Final Answer</strong> <br/><br/>The nearest integer to the calculated value is 2.</p> <p>So, the logarithm of the ratio of the rates of chemisorption on equal areas of platinum and nickel at 300 K is 2.</p>