Question

Adsorption of a toxic gas on 1.0 g activated charcoal is 0.75 cm$^3$ both at 25 atm, 140 K and at 30.0 atm, 280 K. The isosteric enthalpy for adsorption of the gas in kJ mol$^{-1}$ (rounded off to two decimal places) is ________. (Given: $R = 8.314$ J K$^{-1}$ mol$^{-1}$)

Hint:

For physisorption, $\Delta H_\text{ads}$ is typically between 10–40 kJ mol$^{-1}$ and decreases with increasing temperature.

Answer 1

Correct Answer: -5.81

Step 1: Apply Clausius–Clapeyron equation.
At constant coverage, \[ \ln \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_\text{ads}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Given $P_1 = 25$, $P_2 = 30$, $T_1 = 140$, $T_2 = 280$.
Step 2: Substitute values.
\[ \ln \left(\frac{30}{25}\right) = \frac{\Delta H_\text{ads}}{8.314}\left(\frac{1}{140} - \frac{1}{280}\right) \] \[ 0.182 = \frac{\Delta H_\text{ads}}{8.314}(0.00357) \] \[ \Delta H_\text{ads} = \frac{0.182 \times 8.314}{0.00357} = 424.6 \, \text{J mol}^{-1} = 12.0 \, \text{kJ mol}^{-1} \]
Step 3: Conclusion.
The isosteric enthalpy of adsorption is 12.0 kJ mol$^{-1}$.