# Question #7b856

Jan 26, 2015

Since this is an equilibrium reaction, the data given point to the formation of more reactants.

The Gibbs free energy change for a system that has a reaction quotient equal to ${Q}_{c}$ is

$\Delta {G}_{\text{rxn}} = \Delta {G}^{\circ} + R T \cdot \ln \left({Q}_{c}\right)$

You know that at equilibrium, $\Delta G = 0$ and $\Delta {G}^{\circ} = - R T \cdot \ln \left({K}_{c}\right)$where ${K}_{c}$ is the equilibrium constant for the reaction.

The above equation can be written as

$\Delta {G}_{\text{rxn}} = - R T \cdot \ln \left({K}_{c}\right) + R T \cdot \ln \left({Q}_{c}\right)$, or

$\Delta {G}_{\text{rxn}} = R T \cdot \ln \left({Q}_{c} / {K}_{c}\right)$

This means that

$\ln \left({Q}_{c} / {K}_{c}\right) = \frac{\Delta {G}_{\text{rxn}}}{R T}$

Since $R$ and $T$ will always be positive, the value on the right of the equation is positive, which means that $\frac{{Q}_{c}}{{K}_{c}}$ is greater than 1, or that ${Q}_{c}$ is greater than ${K}_{c}$ .

This means that the reverse reaction will be spontaneous and the formation of more reactants will be favored.