Question

The standard enthalpy change upon dissolving one mole of oxygen at 25°C is -11.7 kJ. Use this number and the van't Hoff equation (Problem 5.85) to calculate the equilibrium (Henry's law) constant for oxygen in water at 0°C and at 100° C. Discuss the results briefly.

Short answer

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

Step-by-step solution

Given information

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

Explanation

The equilibrium constant from the problem 5.85 is given by:

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

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:

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

Calculations

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

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

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