5.7 Gibbs Energy Change and Equilibrium

We have seen how a knowledge of the sign and magnitude of the free energy change of a chemical reaction allows:

• Prediction of the spontaneity of the chemical reaction.
• Prediction of the useful work that could be extracted from it.

So far we have considered free energy changes in irreversible reactions. Let us now examine the free energy changes in reversible reactions.

‘Reversible’ under strict thermodynamic sense is a special way of carrying out a process such that the system is at all times in perfect equilibrium with its surroundings. When applied to a chemical reaction, the term ‘reversible’ indicates that a given reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. This means that the reactions in both directions should proceed with a decrease in free energy, which seems impossible. It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy.

So, the criterion for equilibrium \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\); is
\(\Delta_r G=0\)
Gibbs energy for a reaction in which all reactants and products are in standard state, \(\Delta_r G^{\ominus}\) is related to the equilibrium constant of the reaction as follows:
\(
0=\Delta_r G^{\ominus}+\mathrm{R} T \ln K
\)
or \(\Delta_r G^{\ominus}=-\mathrm{R} T \ln K\)
or \(\Delta_r G^{\ominus}=-2.303 \mathrm{RT} \log K \dots(6.23)\)
We also know that
\(
\Delta_r G^{\ominus}=\Delta_r H^{\ominus}-T \Delta_r S^{\ominus}=-\mathrm{R} T \ln K \dots(6.24)
\)
For strongly endothermic reactions, the value of \(\Delta_r H^{\ominus}\) may be large and positive. In such a case, the value of \(K\) will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, \(\Delta_r H^{\ominus}\) is large and negative, and \(\Delta_r G^{\ominus}\) is likely to be large and negative too. In such cases, \(K\) will be much larger than 1 . We may expect strongly exothermic reactions to have a large \(K\), and hence can go to near completion. \(\Delta_r G^{\ominus}\) also depends upon \(\Delta_r S^{\ominus}\), if the changes in the entropy of reaction is also taken into account, the value of \(K\) or extent of chemical reaction will also be affected, depending upon whether \(\Delta_r S^{\ominus}\) is positive or negative.

Using equation (6.24),

• It is possible to obtain an estimate of \(\Delta G^{\ominus}\) from the measurement of \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\), and then calculate \(K\) at any temperature for economic yields of the products.
• If \(\mathrm{K}\) is measured directly in the laboratory, the value of \(\Delta G^{\ominus}\) at any other temperature can be calculated. using equation (6.24).

Example 6.12: Calculate \(\Delta_T G^{\ominus}\) for conversion of oxygen to ozone, \(3 / 2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{O}_3(\mathrm{~g})\) at \(298 \mathrm{~K}\). if \(K_p\) for this conversion is \(2.47 \times 10^{-29}\).

Answer: We know \(\Delta_r G^{\ominus}=-2.303 \mathrm{R} T \log K_p\) and \(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)
Therefore, \(\Delta_r G^{\ominus}=\)
\(
\begin{aligned}
– & 2.303\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \\
& \times(298 \mathrm{~K})\left(\log 2.47 \times 10^{-29}\right) \\
= & 163000 \mathrm{~J} \mathrm{~mol}^{-1} \\
= & 163 \mathrm{~kJ} \mathrm{~mol}^{-1} .
\end{aligned}
\)

Example 6.13: Find out the value of the equilibrium constant for the following reaction at \(298 \mathrm{~K}\).
\(
2 \mathrm{NH}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) \leftrightharpoons \mathrm{NH}_2 \mathrm{CONH}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(1)
\)
Standard Gibbs energy change, \(\Delta_r G^{\ominus}\) at the given temperature is \(-13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\).
\(
=\frac{\left(-13.6 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)}{2.303\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K})}=2.38
\)
Hence \(K=\operatorname{antilog} 2.38=2.4 \times 10^2\)

Example 6.14: At \(60^{\circ} \mathrm{C}\), dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Answer: 
\(
\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})
\)
If \(\mathrm{N}_2 \mathrm{O}_4\) is \(50 \%\) dissociated, the mole fraction of both the substances is given by
\(
x_{\mathrm{N}_2 \mathrm{O}_4}=\frac{1-0.5}{1+0.5}: x_{\mathrm{NO}_2}=\frac{2 \times 0.5}{1+0.5}
\)
\(
p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{0.5}{1.5} \times 1 \mathrm{~atm}, p_{\mathrm{NO}_2}=\frac{1}{1.5} \times 1 \mathrm{~atm}
\)
The equilibrium constant \(K_p\) is given by
\(
\begin{aligned}
& K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{1.5}{(1.5)^2(0.5)} \\
& =1.33 \mathrm{~atm} .
\end{aligned}
\)
Since
\(
\Delta_r G^{\ominus}=-\mathrm{R} T \ln K_p
\)
\(
\begin{array}{r}
\Delta_r G^{\ominus}=\left(-8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \times(333 \mathrm{~K}) \times(2.303) \times(0.1239)
\end{array}
\)
\(
=-763.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)