Question

For the process $\frac{3}{2}A \rightarrow B$, at 298 K, $\Delta G^\circ$ is $163\ \text{kJ mol}^{-1}$. The composition of the reaction mixture is $[B]=1$ and $[A]=10000$. Predict the direction of the reaction and the relation between reaction quotient $(Q)$ and the equilibrium constant $(K)$.}

• A. forward direction because $Q > K$
• B. reverse direction because $Q > K$
• C. forward direction because $Q < K$
• D. reverse direction because $Q < K$
• E. it is at equilibrium as $Q = K$

Hint:

Always compare $Q$ with $K$:
• $Q < K$ → forward reaction
• $Q > K$ → backward reaction
• Large $+\Delta G^\circ$ → very small $K$

Answer 1

The Correct Option is B

Concept:
The spontaneity and direction of a chemical reaction are determined using Gibbs free energy and the relation between reaction quotient $(Q)$ and equilibrium constant $(K)$: \[ \Delta G = \Delta G^\circ + RT \ln Q \] Also, \[ \Delta G^\circ = -RT \ln K \] From these relations:
• If $Q < K$, then $\Delta G < 0$ → reaction proceeds in forward direction
• If $Q > K$, then $\Delta G > 0$ → reaction proceeds in reverse direction
• If $Q = K$, system is at equilibrium

Step 1: Write expression for reaction quotient $Q$.
For the reaction: \[ \frac{3}{2}A \rightarrow B \] \[ Q = \frac{[B]}{[A]^{3/2}} \] Substitute given values: \[ Q = \frac{1}{(10000)^{3/2}} \]

Step 2: Simplify $Q$.
\[ 10000 = 10^4 \Rightarrow (10^4)^{3/2} = 10^{6} \] \[ Q = \frac{1}{10^6} = 10^{-6} \]

Step 3: Determine equilibrium constant $K$ using $\Delta G^\circ$.
\[ \Delta G^\circ = -RT \ln K \] Given $\Delta G^\circ = +163\ \text{kJ mol}^{-1}$ (positive), therefore: \[ \ln K < 0 \Rightarrow K < 1 \] Thus, equilibrium strongly favors reactants and $K$ is very small.

Step 4: Compare $Q$ and $K$.
We found: \[ Q = 10^{-6} \] Since $K$ is much smaller than 1 (due to large positive $\Delta G^\circ$), we get: \[ Q > K \]

Step 5: Determine direction of reaction.
Since: \[ Q > K \Rightarrow \Delta G > 0 \] The reaction will proceed in the reverse direction to attain equilibrium.