JEE Chemistry 2025 Question with Solution

Given: Standard entropies of $X_2$, $Y_2$ and $XY_5$ are $70 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$, $50 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ and $110 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ respectively. Also, $\Delta H = -35 \, \text{kJ mol}^{-1} = -35000 \, \text{J mol}^{-1}$.

Find: The equilibrium temperature $T$.

At equilibrium, Gibbs free energy change is zero:

$$\Delta G = \Delta H - T\Delta S$$

So,

$$0 = \Delta H - T\Delta S$$

First calculate the standard entropy change of reaction:

$$\Delta S = S_{XY_5} - \left( \frac{1}{2}S_{X_2} + \frac{5}{2}S_{Y_2} \right)$$

Substituting the given values:

$$\Delta S = 110 - \left( \frac{1}{2} \times 70 + \frac{5}{2} \times 50 \right)$$ $$\Delta S = 110 - (35 + 125) = 110 - 160 = -50 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$$

Now use

$$T = \frac{\Delta H}{\Delta S}$$

Substitute $\Delta H = -35000 \, \text{J mol}^{-1}$ and $\Delta S = -50 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$:

$$T = \frac{-35000}{-50} = 700 \, \text{K}$$

Therefore, the equilibrium temperature is $700 \, \text{K}$.

Given:

• Reaction:

$$\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5$$

• $\Delta H^\circ = -35 \, \text{kJ mol}^{-1}$
• $S^\circ(X_2) = 70 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$
• $S^\circ(Y_2) = 50 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$
• $S^\circ(XY_5) = 110 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$

Find: Temperature at which the reaction is at equilibrium.

For the reaction,

$$\Delta S^\circ_{\text{rxn}} = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$$

Hence,

$$\Delta S^\circ_{\text{rxn}} = 110 - \left[ \left( \frac{1}{2} \times 70 \right) + \left( \frac{5}{2} \times 50 \right) \right]$$

Evaluate the bracket first:

$$\frac{1}{2} \times 70 = 35$$ $$\frac{5}{2} \times 50 = 125$$

So,

$$\Delta S^\circ_{\text{rxn}} = 110 - (35 + 125)$$ $$\Delta S^\circ_{\text{rxn}} = 110 - 160 = -50 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$$

At equilibrium,

$$\Delta G^\circ = 0$$

Using

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

we get

$$0 = \Delta H^\circ - T\Delta S^\circ$$

Convert enthalpy to joule per mole:

$$\Delta H^\circ = -35 \, \text{kJ mol}^{-1} = -35000 \, \text{J mol}^{-1}$$

Substitute:

$$0 = -35000 - T(-50)$$ $$0 = -35000 + 50T$$ $$50T = 35000$$ $$T = \frac{35000}{50} = 700 \, \text{K}$$

Therefore, the required nearest integer temperature is 700.