Question

Calculate$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$ for ${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$at $\mathbf{600}\mathbf{.}\mathit{K}$using the following data

${\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{a}\mathit{t}\mathbf{600}\mathbf{.}\mathit{K}$

$\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{37}}\mathit{a}\mathit{t}\mathbf{600}\mathbf{.}\mathit{K}$

Short answer

The value of${\text{ΔG}}^{\text{o}}$ is$140.89\text{\hspace{0.17em}K}$

Step-by-step solution

Definition of standard free energy

The amount of energy released in the conversion of reactants to products under standard conditions is defined as standard free energy

Calculation of  ΔGo

At ,$600.\text{\hspace{0.17em}K}$

${\text{H}}_{\text{2}}\text{(g)}+{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)\hspace{0.17em}K}=2.3\times {10}^6\mathrm{...}\left(1\right)$

${\text{2H}}_{\text{2}}\text{(g)}+{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2{\text{H}}_{\text{2}}\text{O(g)\hspace{0.17em}K}=1.8\times {10}^{37}\mathrm{...}\left(2\right)$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}=\text{RTlnK}\\ {\text{ΔG}}^{\text{o}}=-{\text{73072.18J\hspace{0.17em}mol}}^{-1}\\ {\text{ΔG}}^{\text{2o}}=-{\text{427922.09\hspace{0.17em}J\hspace{0.17em}mol}}^{-1}\\ {\text{ΔG}}^{\text{3o}}=-{\text{213961.045\hspace{0.17em}J\hspace{0.17em}mol}}^{-1}\end{array}$

To obtain the required equation, reaction$\left(2\right)$ is multiplied by $12$coefficient

${\text{H}}_{\text{2}}\text{g}+{\text{12O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{H}}_{\text{2}}\text{O(g)}\mathrm{...}\text{(3)}$

To obtain the required equation, reaction $\left(3\right)$is subtracted from reaction$\left(1\right)$

${\text{ΔG}}^{\text{o}}=(-73072.18\text{\hspace{0.17em}J/mol})-(-213961.045\text{\hspace{0.17em}J/mol})$

${\text{ΔG}}^{\text{o}}=140.89\text{\hspace{0.17em}k}$