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Question 8.2: The equilibrium constant of a particular chemical reaction i...

The equilibrium constant of a particular chemical reaction is doubled when elevating the temperature from 200 K to 300 K. Calculate the reaction enthalpy (considering it as independent of temperature in this range).

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Let us apply this time the definite integral form of the van’t Hoff equation according to (8.78):

  \ln K_{2} – \ln K_{1} =-\frac{\Delta_{\mathrm{r}} H^{\ominus}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) .                       (8.78)

\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta_{\mathrm{r}} H^{\ominus}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) .

Solution of this equation yields \Delta_{\mathrm{r}} H^{\ominus}=3.458 \mathrm{~kJ} / \mathrm{mol}.

Related Answered Questions

Question: 8.1

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 (T/K)^-1 + 20 000 (T/K)^-2 . Calculate the standard reaction enthalpy and entropy at the given temperature. ...

Verified Answer:

Let us start with the form of the van’t Hoff equat...