Question

Consider this reaction: I₂(g)+Cl₂(g)⇌2ICl(g) K=6.91×103 at 460 K

Calculate ΔrG for the reaction at 460 K under each of the following conditions:

A) Standard Conditions

B) at Equilibrium

C) PI2 = 0.325 bar PCl2 = 0.387 bar PICl = 2.55 bar

Answer 1

The reaction is shown below.

\(\mathrm{I}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ICl}(\mathrm{g})\)

Equilibrium constant of the reaction \(\mathrm{K}\) is \(6.91 \times 10^{3} .\)

a)

Calculate the value of \(\Delta \mathrm{G}\) for the reaction by using the following formula.

\(\Delta \mathrm{G}^{\circ}=-\mathrm{RTlnK}\)

Here, \(\mathrm{R}\) is gas constant, and \(\mathrm{T}\) is temperature

Substituite \(8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\) for \(\mathrm{R}, 460 \mathrm{~K}\) for \(\mathrm{T},\) and \(6.91 \times 10^{3}\) for \(\mathrm{K}\).

\(\Delta \mathrm{G}^{\circ}=-8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K} \times(460 \mathrm{~K}) \times \ln \left(6.91 \times 10^{3}\right)\)

$$ \begin{aligned} \Delta \mathrm{G}^{\circ} &=-8.314 \times 460 \times 8.840 \mathrm{~J} / \mathrm{mol} \\ &=-33808 \mathrm{~J} / \mathrm{mol}=-33.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Therefore, \(\Delta \mathrm{G}^{\circ}\) of the reaction at \(460 \mathrm{~K}\) is \(-33.8 \mathrm{~kJ} / \mathrm{mol}\)

b)

At equilibrium, the Gibbs free energy change \(\Delta G^{\circ}\) will be zero. Since, at equilibrium no reaction proceeds in either direction forward or backward.

Therefore, at equilibrium Gibbs free energy change is \(0 \mathrm{~kJ} / \mathrm{mol}\).

c)

Calculate \(\Delta \mathrm{G}_{\mathrm{rxn}}\) by using the following formula.

\(\Delta G_{r x n}=\Delta G^{0}+R T \ln Q\)

The value for \(\Delta G^{\circ}\) is \(-33.8 \mathrm{~kJ} / \mathrm{mol}\).

Here, \(Q\) is reaction quotient, \(\mathrm{Q}=\mathrm{Kp}\)

Write the equilibrium quotient for the reaction as shown below. \(\mathrm{Q}=\frac{P_{\mathrm{IC} 1}^{2}}{P_{\mathrm{I}_{2}} P_{\mathrm{Cl}_{2}}}\)

Substitute the values.

$$ \mathrm{Q}=\frac{(2.55 \mathrm{bar})^{2}}{0.325 \mathrm{bar} \times 0.387 \mathrm{bar}}=51.699 $$

Now calculate \(\Delta G_{\mathrm{rxn}}\) by substituting \(-33.8 \mathrm{~kJ} / \mathrm{mol}\) for \(\Delta \mathrm{G}^{\circ}, 8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\) for \(\mathrm{R}\)

\(460 \mathrm{~K}\) for \(\mathrm{T},\) and 51.699 for \(\mathrm{Q}\)

$$ \begin{aligned} \Delta G_{\mathrm{rxn}} &=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \mathrm{Q} \\ \Delta \mathrm{G}_{\mathrm{rxn}} &=-33.8 \mathrm{~kJ} / \mathrm{mol}+8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 460 \mathrm{~K} \times \ln (51.699) \\ &=-33.8 \mathrm{~kJ} / \mathrm{mol}+8.314 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \times 460 \times 3.945 \\ &=-33.8 \mathrm{~kJ} / \mathrm{mol}+8.314 \times 460 \times 3.945 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\ &=-18.71 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Therefore, \(\Delta \mathrm{G}_{\mathrm{rxn}}\) for the given reaction is \(-18.71 \mathrm{~kJ} / \mathrm{mol}\)