$N_{2}$ gas adsorbed on charcoal to the extent of 0.92 $c m^{3} / g r a m$ at a pressure of 4.8 atm and temperature of 190 K. But at 250 K the same amount of adsorption is achieved only when the pressure was increased to 32 atm. Find out the magnitude of Molar Enthalpy of Adsorption of $N_{2}$ on charcoal? Give your answer in kJ/mol

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answer is 12.3.

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Detailed Solution

$\log \frac{p_{1}}{p_{2}} = \frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] \log \frac{4.8}{32} = \frac{Δ H}{2.303 R} [\frac{1}{190} - \frac{1}{250}] \log \frac{3}{20} = \frac{Δ H}{2.303 R} [\frac{60}{190 \times 250}] - 0.82 = \frac{Δ H}{2.303 R} [\frac{60}{190 \times 250}] Δ H = - 12.4 k J$

Magnitude of $Δ H = 12.4 k J / m o l$

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