91Ó°ÊÓ

The standard enthalpy change upon dissolving one mole of oxygen at 25°C is -11.7 kJ.

The equilibrium constant from the problem 5.85 is given by:

$$\begin{gathered} \ln \left(K\left(T_2\right)\right)=\ln \left(K\left(T_1\right)\right)+\frac{\mathrm{Δ}H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ K\left(T_2\right)=\exp \left(\ln \left(K\left(T_1\right)\right)+\frac{\mathrm{Δ}H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right) \end{gathered}$$

Where,

$K\left(T_1\right)$is the equilibrium constant and is given as $K\left(T_1\right)=\frac{1}{750}$

We need to calculate the equilibrium constant at temperature of T₂ =0°C = 273 K, substitute with the values:

$$\begin{gathered} K(273\mathrm{K})=\exp \left(\ln \left(\frac{1}{750}\right)-\frac{11.7×{10}^3\mathrm{J}}{8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}}\left(\frac{1}{298\mathrm{K}}-\frac{1}{273\mathrm{K}}\right)\right) \\ K(273\mathrm{K})=2.0547×{10}^{-3} \end{gathered}$$

Now we need to calculate the equilibrium constant at temperature of T2 = 100 C = 373 K, substituting the values

$$\begin{gathered} K(373\mathrm{K})=\exp \left(\ln \left(\frac{1}{750}\right)-\frac{11.7×{10}^3\mathrm{J}}{8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}}\left(\frac{1}{298\mathrm{K}}-\frac{1}{373\mathrm{K}}\right)\right) \\ K(373\mathrm{K})=5.159×{10}^{-4} \end{gathered}$$

The equilibrium constant at 0 degree is nearly identical to the equilibrium constant at ambient temperature, but decreases as the temperature rises.