For the adsorption of hydrogen on platinum, the activation energy is 30 kJ mol-1 and for the adsorption of hydrogen on nickel, the activation energy is 41.4 kJ mol-1 . The logarithm of the ratio of the rates of chemisorption on equal areas of the metals at 300 K is ……… (Nearest integer) Given: ln 10 = 2. 3 R = 8.3 JK-1 mol-1
Correct Answer: 2
Solution and Explanation
Given:
- Activation energy for hydrogen adsorption on platinum (\( E_a^{Pt} \)): 30 kJ/mol = \( 30 \times 1000 = 30000 \, \text{J/mol} \)
- Activation energy for hydrogen adsorption on nickel (\( E_a^{Ni} \)): 41.4 kJ/mol = \( 41.4 \times 1000 = 41400 \, \text{J/mol} \)
- Temperature (\( T \)): 300 K
- Gas constant (\( R \)): 8.3 J/K/mol
- \( \ln 10 = 2.3 \)
Formula:
The rate of chemisorption is proportional to \( e^{-E_a / RT} \). Therefore, the ratio of rates is:
\[ \frac{r_{Pt}}{r_{Ni}} = \frac{e^{-E_a^{Pt} / RT}}{e^{-E_a^{Ni} / RT}} = e^{\frac{E_a^{Ni} - E_a^{Pt}}{RT}} \]
Step-by-Step Calculation:
- Difference in activation energies: \[ E_a^{Ni} - E_a^{Pt} = 41400 - 30000 = 11400 \, \text{J/mol} \]
- Substitute into the exponent: \[ \frac{E_a^{Ni} - E_a^{Pt}}{RT} = \frac{11400}{8.3 \times 300} \]
- Calculate: \[ \frac{11400}{2490} \approx 4.58 \]
- Ratio of rates (logarithmic form): \[ \ln \left( \frac{r_{Pt}}{r_{Ni}} \right) = 4.58 \]
- Convert to base-10 logarithm: \[ \log \left( \frac{r_{Pt}}{r_{Ni}} \right) = \frac{\ln \left( \frac{r_{Pt}}{r_{Ni}} \right)}{\ln 10} = \frac{4.58}{2.3} \approx 2 \]
Conclusion:
The logarithm of the ratio of chemisorption rates is approximately 2.
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