At a fixed surface coverage, a gas needs a pressure of 10 kPa at 300 K and 40 kPa at 320 K. Use ln(p_2/p_1)=(Delta H)/(R)((1)/(T_2)-(1)/(T_1)) with R=8.314 J mol^-1K^-1 and ln4=1.386. Calculate Delta H of adsorption in kJ mol^-1 (nearest integer, include the sign).

JEE Advanced levelNumerical answerDifficulty 3 of 3sc043

At a fixed surface coverage, a gas needs a pressure of $10 kPa$ at $300 K$ and $40 kPa$ at $320 K$ . Use $ln (p_{2} / p_{1}) = \frac{Δ H}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ with $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 4 = 1.386$ . Calculate $Δ H$ of adsorption in $kJ mo l^{- 1}$ (nearest integer, include the sign).

Numerical answer type, answer in kJ/mol, tolerance 1.

1/ T_{2} - 1/ T_{1} = 1/320 - 1/300 = - 2.0833 \times 1 0^{- 4} K^{- 1}

Δ H = 8.314 \times 1.386/ (- 2.0833 \times 1 0^{- 4}) = - 55323 J mo l^{- 1} = - 55.3 kJ mo l^{- 1}

, i.e. about

- 55 kJ mo l^{- 1}

(negative, exothermic).