At a constant surface coverage, the pressure needed for a given amount of adsorption was measured at several temperatures. A plot of $lo g_{10} p$ against $1/ T$ is linear with slope $- 1950 K$ . Using $2.303$ and $R = 8.314 J mo l^{- 1} K^{- 1}$ , and the relation $lo g_{10} p = \frac{Δ H}{2.303 RT} + constant$ at constant coverage, the enthalpy of adsorption $Δ H$ is:

Answer: C

At constant coverage

ln p = Δ H / (RT) + C

, so

lo g_{10} p

has slope

Δ H / (2.303 R)

. Then

Δ H = - 1950 \times 2.303 \times 8.314 = - 37337 J mo l^{- 1} = - 37.3 kJ mo l^{- 1}

, negative for adsorption. Omitting

2.303

gives

- 16.2