Question 8.1: The equilibrium constant of a particular chemical reaction i...
The equilibrium constant of a particular chemical reaction in the vicinity of T = 502.3 K can be calculated using the following equation:
\ln K = 1-1 000 (\frac{T}{K} )^{-1} +20 000 (\frac{T}{K} )^{-2}.Calculate the standard reaction enthalpy and entropy at the given temperature.
Let us start with the form of the van’t Hoff equation according to (8.73):
\left(\frac{\partial \ln K_{a}}{\partial T}\right )_{P} = \frac{\Delta _{r}H^{\ominus }}{RT^{2} }.As we can see, by calculating the derivative of ln K with respect to the temperature from the given formula, we readily obtain the reaction enthalpy (7.652 kJ mol^{-1}). The reaction entropy can be calculated using the following relation:
\Delta _{r}S^{\ominus } = \frac{\Delta _{r}H^{\ominus }-\Delta _{r}G^{\ominus }}{T}.The missing \Delta _{r}G^{\ominus } can be calculated using the relation (8.19):
-\Delta _{r}G^{\ominus } = RT\ln K_{a}The result is \Delta _{r}S^{\ominus } = 7.655 J/ (mol K).
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Question: 8.2
Verified Answer:
Let us apply this time the definite integral form ...