Question

Write a relation between AG 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:
$CO\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons C{H_4}\left( g \right) + {H_2}O\left( g \right)$

Answer 1

We know that the relation between $\Delta G$ and Q is given by the equation, $\Delta G = \Delta {G^0} + RT$ ln Q and also the standard free energy change, i.e., $\Delta {G^0}$ is $- RT$ ln K. So, we can solve (a) part of the question by putting the value of $\Delta {G^0}$ in the equation of the relation between $\Delta G$and Q. For solving the (b) part of the question, we know that on increasing the pressure equilibrium will shift in forward direction it means $Q < K$ and KC will be$\dfrac{{\left[ {C{H_4}} \right]\left[ {{H_2}O} \right]}}{{\left[ {CO} \right]\left[ {H_2^3} \right]}}$.

Complete step by step answer:
The relation between $\Delta G$ and Q is given by,
(a) $\Delta G = \Delta {G^0} + RT$ ln Q
$\Delta G$ = Change in free energy as the reaction proceeds,
$\Delta {G^0}$ = Standard free energy change,
Q = Reaction quotient,
R = Gas constant,
T = Absolute temperature.
Since, $\Delta {G^0} = - RT$ ln K
$\therefore \Delta G = - RT$ ln $K + \;RT$ ln $Q = RT$ ln $\dfrac{Q}{K}$
If $Q < K$, $\Delta G$ will be negative. Reaction will proceed in the forward.
If $Q = K$, $\Delta G = 0$, reaction is in equilibrium and no net reaction is there.
(b) On increasing the pressure equilibrium will shift in forward direction it means $Q < K$.
${K_C} = \dfrac{{\left[ {C{H_4}} \right]\left[ {{H_2}O} \right]}}{{\left[ {CO} \right]\left[ {H_2^3} \right]}}$
On increasing the pressure, the volume is decreased and molar concentration increases. Suppose if the pressure is doubled, then the volume will be halved and hence, molar concentrations will also be doubled.
QC = \dfrac{{\left\\{ {2\left[ {C{H_4}} \right]} \right\\}2\left[ {{H_2}O} \right]}}{{\left\\{ {2\left[ {CO} \right]} \right\\}2\left[ {H_2^3} \right]}} = 1\dfrac{{\left[ {C{H_4}} \right]\left[ {{H_2}O} \right]}}{{4\left[ {CO} \right]\left[ {H_2^3} \right]}} = \dfrac{1}{4}KC
This, ${Q_C}$ is less than ${K_C}$. Hence, to re-establish equilibrium, ${Q_C}$will tend to increase, i.e., equilibrium will shift in the forward direction.

Note:
The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system and the change in the free energy of a system that occurs during a reaction can be measured under any set of conditions and if the data are collected under standard-state conditions, then the result is the standard-state free energy of reaction i.e., $\Delta {G^0}$.