Question

Given the following standard free energies at \(25^{\circ} \mathrm{C}\), $$ \begin{aligned} \mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \longrightarrow \operatorname{COS}(g)+2 \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-246.5 \mathrm{~kJ} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) & \Delta G^{\circ}=-28.5 \mathrm{~kJ} \end{aligned} $$ find \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction. $$ \mathrm{SO}_{2}(g)+\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{COS}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$

Short answer

Question: What is the standard free energy for the reaction of SO₂(g) + CO(g) + 2H₂(g) → COS(g) + 2H₂O(g) at 25°C, given the following reactions and their standard free energies: 1) SO₂(g) + 3CO(g) → COS(g) + 2CO₂(g) ΔG° = -246.5 kJ 2) CO(g) + H₂O(g) → CO₂(g) + H₂(g) ΔG° = -28.5kJ Answer: The standard free energy for the desired reaction at 25°C is -189.5 kJ.

Step-by-step solution

Identify the desired reaction and given reactions

Recall that the desired reaction is: $$ \mathrm{SO}_{2}(g)+\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{COS}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ The given reactions that we have are: $$ \begin{aligned} \mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \longrightarrow \operatorname{COS}(g)+2 \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-246.5 \mathrm{~kJ} \\ \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) & \Delta G^{\circ}=-28.5 \mathrm{~kJ} \end{aligned} $$

Manipulate the given reactions to match the desired reaction

In order to form the desired reaction, we need to have \(\mathrm{H}_{2} \mathrm{O}(g)\) on the products side instead of the reactants side. To achieve this, we can reverse the second given reaction while changing the sign of its \(\Delta G^{\circ}\): $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \hspace{1cm} \Delta G^{\circ}=28.5 \mathrm{~kJ} $$ Now, we need 2 moles of \(\mathrm{H}_{2}O(g)\) in the products side, so we multiply the new second reaction by 2: $$ 2(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)) \longrightarrow 2(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)) \hspace{1cm} 2\Delta G^{\circ}=2(28.5 \mathrm{~kJ}) = 57 \mathrm{~kJ} $$

Add the manipulated reactions to obtain the desired reaction

Now we can add the first given reaction (unmodified) to the modified second reaction: $$ \begin{aligned} &\mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \longrightarrow \operatorname{COS}(g)+2 \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-246.5 \mathrm{~kJ} \\ + \hspace{0.5cm} &2(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)) \longrightarrow 2(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)) & 2\Delta G^{\circ}=57 \mathrm{~kJ}\\ \hline &\mathrm{SO}_{2}(g)+\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{COS}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta G^{\circ}=\Delta G^{\circ}_{1} + 2\Delta G^{\circ}_{2} \end{aligned} $$

Calculate the standard free energy for the desired reaction

Finally, we can calculate the standard free energy for our desired reaction by adding the values of the previously obtained reactions: $$ \Delta G^{\circ} = \Delta G^{\circ}_{1} + 2\Delta G^{\circ}_{2} = -246.5 \mathrm{~kJ} + 57 \mathrm{~kJ} = -189.5 \mathrm{~kJ} $$ Therefore, the standard free energy for the desired reaction at 25°C is -189.5 kJ.