Question

When the reaction $\mathrm{A}+\mathrm{B}\iff \mathrm{C}$ is at equilibrium, the concentrations of reactants are as follows: $\left[\mathrm{A}\right]=2\mathrm{mM},\left[\mathrm{B}\right]=3\mathrm{mM},\left[\mathrm{C}\right]=9\mathrm{mM}$. What is the standard free energy change for the reaction?

Short answer

The standard free energy change is ${\mathrm{\Delta G}}^0=-18122.29\mathrm{J}/\mathrm{mol}$.

Step-by-step solution

Given data

The given reaction is,

$\mathrm{A}+\mathrm{B}\iff \mathrm{C}$

${\mathrm{\Delta G}}^0=-18122.29\mathrm{J}/\mathrm{mol}$

A and Bare the substrates and C is the product.

Then,

$\mathrm{A}+\mathrm{B}\iff \mathrm{C}$$\begin{array}{l}\left[\mathrm{A}\right]=0.002\mathrm{M}\\ \left[\mathrm{B}\right]=0.003\mathrm{M}\\ \left[\mathrm{C}\right]=0.009\mathrm{M}\end{array}$$\begin{array}{l}\left[A\right]=0.002M\\ \left[B\right]=0.003M\\ \left[C\right]=0.009M\end{array}$

Calculating the equilibrium constant 

$\begin{array}{l}{K}_{eq}^{\text{'}}=\frac{\left[C\right]}{\left[A\right]\times \left[B\right]}\\ =\frac{\left[0.009M\right]}{\left[0.002M\right]\times \left[0.003M\right]}\\ =1500M\end{array}$

Calculating the standard free energy change 

$\Delta G=\Delta G^0+RT\ln Q$

Where,

$\mathrm{\Delta G}$Change in free energy

$\Delta G^0$Standard free energy change

R gas constant

T Standard temperature

At equilibrium

$\begin{array}{l}\Delta G^0=-RT\ln K_{eq}^{\text{'}}\\ =-2.303\times 8.314J{K}^{-1}mo{l}^{-1}\times 298K\times \ln \left(1500M\right)\\ =-18122.29J/mol\end{array}$

Therefore, the standard free energy change is${\mathrm{\Delta G}}^0=-18122.29\mathrm{J}/\mathrm{mol}$.