Write a relation between $\Delta G$ and $Q$ and define the meaning of each term and answer the following:
$(a)$ Why a reaction proceeds forward when $Q < K$ and no net reaction occurs when $Q = K$.
$(b)$ Explain the effect of increase in pressure in terms of reaction quotient $Q$.
For the reaction: $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_{2}O_{(g)}$

(A) The relation between $\Delta G$ and $Q$ is given by: $\Delta G = \Delta G^{\ominus} + RT \ln Q$
Where:
$\Delta G = \text{Gibbs free energy change of the reaction}$
$\Delta G^{\ominus} = \text{Standard Gibbs free energy change}$
$R = \text{Universal gas constant}$
$T = \text{Absolute temperature in } K$
$Q = \text{Reaction quotient}$
$(a)$ Since $\Delta G^{\ominus} = -RT \ln K$,the equation becomes $\Delta G = RT \ln(Q/K)$.
If $Q < K$,then $\ln(Q/K) < 0$,so $\Delta G < 0$,which means the reaction proceeds in the forward direction.
If $Q = K$,then $\ln(Q/K) = 0$,so $\Delta G = 0$,indicating the system is at equilibrium and no net reaction occurs.
$(b)$ For the reaction $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_{2}O_{(g)}$,the reaction quotient is $Q_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}$.
Increasing the pressure decreases the volume,which increases the molar concentrations. Since there are $4$ moles of gaseous reactants and $2$ moles of gaseous products,the denominator increases more than the numerator. Consequently,$Q_c$ decreases such that $Q_c < K_c$. To re-establish equilibrium,the reaction proceeds in the forward direction.