Q.Write a relation between ΔG and Q and define the meaning of each term and answer the following :

Concept: Equilibrium Constant and Reaction Quotient

The fundamental relation connecting Gibbs free energy change, reaction quotient, and equilibrium constant is:

where $\Delta G$ is the free energy change under given conditions, $\Delta G^\circ$ is the standard free energy change, $R$ is the gas constant, $T$ is temperature (K), and $Q$ is the reaction quotient (ratio of product to reactant activities at any instant). At equilibrium, $\Delta G = 0$ and $Q = K$, giving $\Delta G^\circ = -RT \ln K$.

1. Direction of reaction: When $Q < K$, we have $\ln Q < \ln K$, so $\Delta G = RT(\ln Q - \ln K) < 0$. A negative $\Delta G$ means the forward reaction is spontaneous—the system must form more products to reach equilibrium. When $Q = K$, substitution yields $\Delta G = 0$; the system is at equilibrium and no net change occurs.
2. Effect of pressure increase: …

The Fundamental Relation

The connection between spontaneity and composition is captured by

where:

• $\Delta G$ is the Gibbs free energy change under the current, non-standard conditions — it tells us whether the reaction will proceed forward ($\Delta G < 0$), backward ($\Delta G > 0$), or sit at equilibrium ($\Delta G = 0$).
• $\Delta G^\circ$ is the standard Gibbs free energy change, measured when all species are at unit activity (1 bar for gases, 1 M for solutions). It is a fixed number for a given reaction at a given temperature.
• $R$ is the universal gas constant ($8.314 \, \text{J mol}^{-1} \text{K}^{-1}$).
• $T$ is the absolute temperature in kelvin.
• $Q$ is the reaction quotient, the ratio of product activities to reactant activities raised to their stoichiometric coefficients, evaluated at any moment:

At equilibrium, $\Delta G = 0$ and $Q$ becomes the equilibrium constant $K$, so

Substituting this back into the first equation gives an alternative form:

This version makes the role of $Q$ and $K$ transparent.

(a) Why $Q < K$ drives the reaction forward, and $Q = K$ means equilibrium

The sign of $\Delta G$ determines direction. From $\Delta G = RT \ln \frac{Q}{K}$:

1. When $Q < K$: The ratio $\frac{Q}{K} < 1$, so $\ln \frac{Q}{K} < 0$, which makes $\Delta G < 0$. A negative free energy change means the forward reaction is spontaneous — the system can lower its free energy by converting reactants into products. The reaction proceeds forward until $Q$ rises to equal $K$.

2. When $Q = K$: Now $\frac{Q}{K} = 1$, so $\ln 1 = 0$ and $\Delta G = 0$. The system is at equilibrium. The forward and reverse rates are equal, and there is no net change in composition. The free energy is at a minimum with respect to the reaction coordinate; any shift in either direction would increase $G$.

3. When $Q > K$: The ratio $\frac{Q}{K} > 1$, so $\ln \frac{Q}{K} > 0$ and $\Delta G > 0$. The reverse reaction is spontaneous; the system will convert products back into reactants until $Q$ falls to $K$.

(b) Effect of increasing pressure on the reaction $\text{CO (g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$

First, count moles of gas on each side:

• Reactants: $1 + 3 = 4$ moles of gas
• Products: $1 + 1 = 2$ moles of gas

The forward reaction reduces the total number of gas molecules.

What happens when we increase pressure?

For an ideal gas, partial pressure is proportional to molar concentration at constant temperature. The reaction quotient in terms of partial pressures is

When we suddenly compress the system (increase total pressure), every partial pressure scales up by the same factor, say $\alpha > 1$. The new quotient becomes …