Question

Consider the following reaction at $\mathbf{298}\mathit{K}$

$\mathbf{2}\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{S}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

An equilibrium mixture contains ${\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{a}\mathit{n}\mathit{d}\mathit{S}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$ at partial pressure of respectively. Using data from appendix 4 , determine the equilibrium partial pressure of $\mathit{S}{\mathit{O}}_{\mathbf{2}}$in the mixture. Will this reaction be most favoured at a high or a low temperature assuming standard conditions?

Short answer

The equilibrium pressure of${\text{SO}}_{\text{2}}$ is very negligible that is ${\text{P}}_{{\text{SO}}_{\text{2}}}=1.0\times {10}^{-12}\text{atm}$

Step-by-step solution

Definition of equilibrium pressure

The changes ΔH and ΔS in enthalpy and entropy as a function of temperature is defined as equilibrium pressure.

Calculation of  ΔGo

$\begin{array}{l}{\text{ΔG}}_{\text{R}}^{\text{0}}=\left[2\Delta {\text{G}}_{\text{f}}^{\text{0}}{\text{(SO}}_{\text{3}}\text{(g)}\right)]-[2\Delta {\text{G}}_{\text{f}}^{\text{0}}{\text{(SO}}_{\text{2}}\text{(g))}+1\Delta {\text{G}}_{\text{f}}^{\text{0}}{\text{(O}}_{\text{2}}\text{(g))}]\\ {\text{ΔG}}_{\text{R}}^{\text{0}}=-742+600\\ {\text{ΔG}}_{\text{R}}^{\text{0}}=-142\text{kJ/mol}\end{array}$

Pressure of ${\text{SO}}_{\text{2}}$

$\begin{array}{l}{\text{ΔG}}_{\text{R}}^{\text{0}}=-\text{RTln}\frac{{\left[{\text{P}}_{{\text{SO}}_{\text{3}}}\right]}^{\text{2}}}{{\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\left[{\text{P}}_{{\text{O}}_{\text{2}}}\right]}\\ \frac{{\left[{\text{P}}_{{\text{SO}}_{\text{3}}}\right]}^{\text{2}}}{{\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\left[{\text{P}}_{{\text{O}}_{\text{2}}}\right]}=7.75\times {10}^{24}\\ {\text{P}}_{{\text{SO}}_{\text{2}}}=1.0\times {10}^{-12}\text{atm}\end{array}$

The equilibrium pressure of ${\text{SO}}_{\text{2}}$ is very negligible.