Question

Calculate or the reactionat when the equilibrium concentrations are. Is the reaction exergonic or endergonic under standard conditions?

Short answer

The standard free energy change for the reaction is ${\mathrm{\Delta G}}^{0\text{'}}=5.703{\mathrm{kJmol}}^{-1}$.

Step-by-step solution

Given data

${\mathrm{\Delta G}}^{0\text{'}}=5.703\mathrm{kJmo}{l}^{-1}\mathrm{A}+\mathrm{B}\iff \mathrm{C}+\mathrm{D}$

Concentration of the reactants is $\left[\mathrm{A}\right]=10\mathrm{\mu M};\left[\mathrm{B}\right]=15\mathrm{\mu M};$

Concentration of the products is $\left[\mathrm{C}\right]=3\mathrm{\mu M};\left[\mathrm{B}\right]=5\mathrm{\mu M};$

Temperature, T=298 K

Calculating the standard free energy change 

$\begin{array}{l}{\mathrm{\Delta G}}^{0\text{'}}=-{\mathrm{RTlnK}}_{\mathrm{eq}}\\ =-\mathrm{RTln}\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}\end{array}$

On substituting the standard free energy change values

$\begin{array}{l}\Delta G^{0\text{'}}=-8.314\times 298\times \ln \frac{3\times 5}{10\times 15}\\ =-2477.57\times \left(\ln 1-\ln 10\right)\\ =-2477.57\times -2.302\\ =5703.37kJmo{l}^{-1}\\ \Delta G^{0\text{'}}=5.703kJmo{l}^{-1}\end{array}$

Therefore, the standard free energy change for the reaction is$\begin{array}{l}{\mathrm{\Delta G}}^{0\text{'}}=5.703{\mathrm{kJmol}}^{-1}\end{array}$.

Hence, the reaction is nonspontaneous or endergonic at standard conditions.