Question

12) The change in enthalpy (ΔHorxn) for a reaction is -25.4 kJ/mol . The equilibrium constant for the reaction is 1.4×103 at 298 K. What is the equilibrium constant for the reaction at 602 K ?

13) A reaction has an equilibrium constant of 7.9×103 at 298 K. At 761 K , the equilibrium constant is 0.30. Find ΔHorxn for the reaction.

Answer 1

The relationship between equilibrium constants at different temperatures, enthaply, and temperatures can be expressed as

$$\text{ln}\frac{K_2}{K_1}=\frac{\Delta H}{R}\bigg(\frac{1}{T_1}-\frac{1}{T_2}\bigg),\\ \space\\ K_2=K_1e^{\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})}=\\ \space\\ =1400\cdot2.72^{\frac{-25400}{8.314}(\frac{1}{298}-\frac{1}{602})}=7.90.$$

Likewise, we can calculate the change in enthalpy:

$$\Delta H=\frac{R\text{ ln}\frac{K_2}{K_1}}{\frac{1}{T_1}-\frac{1}{T_2}}=-41.5\cdot10^3\text{ kJ/mol}.$$