Question

For the reaction $\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}$ the initial pressures of gases $\mathit{A}\mathbf{,}\mathit{B}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathit{C}$ are all$\mathbf{0}\mathbf{.}\mathbf{100}\mathit{a}\mathit{t}\mathit{m}$ . Once equilibrium has been established, it is found that .$\mathbf{\left[}\mathit{C}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{040}\mathit{a}\mathit{t}\mathit{m}$ What is $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$for this reaction at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$?

Short answer

The value of ${\text{ΔG}}^{\text{o}}$is $-4.1\text{\hspace{0.17em}kJ/mol}$

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 the standard free energy.

Find the equilibrium pressures

The given reaction is $\text{A(g)}+\text{2B(g)}\rightleftharpoons \text{C(g)}$

The equilibrium pressure of A will be:

$\left[\text{A}\right]=0.160\text{\hspace{0.17em}atm}$

The equilibrium pressure of B will be:

$\left[\text{B}\right]=0.220\text{\hspace{0.17em}atm}$

Now,

$\begin{array}{l}\text{K}=\frac{\left(0.040\right)}{\left(0.160\right){\left(0.220\right)}^2}\\ =5.17\end{array}$

The value ofdata-custom-editor="chemistry" ${\text{ΔG}}^{\text{o}}$ is calculated by ${\text{ΔG}}^{\text{o}}=-\text{RTlnK}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}=-\left(8.314\text{\hspace{0.17em}J/mol.K}\right)\left(298\text{\hspace{0.17em}K}\right)\ln \left(5.17\right)\\ {\text{ΔG}}^{\text{o}}=-4.1\text{\hspace{0.17em}kJ/mol}\end{array}$