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.

Question

The equilibrium constant of a particular chemical reaction in the vicinity of T = 502.3 K can be calculated using the following equation:

[latex]\ln K = 1-1 000 (\frac{T}{K} )^{-1} +20 000 (\frac{T}{K} )^{-2}.[/latex]

Calculate the standard reaction enthalpy and entropy at the given temperature.

Accepted Answer

Let us start with the form of the van’t Hoff equation according to (8.73):

[latex]\left(\frac{\partial \ln K_{a}}{\partial T}\right )_{P} = \frac{\Delta _{r}H^{\ominus }}{RT^{2} }.[/latex]

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 [latex]mol^{-1}[/latex]). The reaction entropy can be calculated using the following relation:

[latex]\Delta _{r}S^{\ominus } = \frac{\Delta _{r}H^{\ominus }-\Delta _{r}G^{\ominus }}{T}.[/latex]

The missing [latex]\Delta _{r}G^{\ominus }[/latex] can be calculated using the relation (8.19):

[latex]-\Delta _{r}G^{\ominus } = RT\ln K_{a}[/latex]

The result is [latex]\Delta _{r}S^{\ominus }[/latex] = 7.655 J/ (mol K).

Question 8.1: The equilibrium constant of a particular chemical reaction i...

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