Problem 167 - A Complete Resource Book in Chemistry for JEE Main

Question

Standard entropy of \(\mathrm{X}_{2}, \mathrm{Y}_{2}\) and \(\mathrm{XY}_{3}\) are 60,40 and 50 \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. For the reaction, \(1 / 2 \mathrm{X}_{2}+3 / 2 \mathrm{Y}_{2} \longrightarrow \mathrm{XY}_{3}, \Delta \mathrm{H}=-30 \mathrm{~kJ}\), to be at equilibrium, the temperature will be (a) \(1250 \mathrm{~K}\) (b) \(500 \mathrm{~K}\) (c) \(750 \mathrm{~K}\) (d) \(1000 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature for equilibrium is 750 K, answer (c).

Step 1: Write the Reaction and Given Data

We have the reaction \( \frac{1}{2} \mathrm{X}_{2} + \frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3} \). The standard entropies given are: \( S^\circ (\mathrm{X}_{2}) = 60 \, \mathrm{J/K\cdot mol} \), \( S^\circ (\mathrm{Y}_{2}) = 40 \, \mathrm{J/K\cdot mol} \), \( S^\circ (\mathrm{XY}_{3}) = 50 \, \mathrm{J/K\cdot mol} \). The change in enthalpy, \( \Delta H = -30 \, \mathrm{kJ} \), and needs to be included in the entropy calculation.

Step 2: Calculate the Change in Entropy (ΔS°)

The standard entropy change for the reaction (\( \Delta S^\circ \)) is calculated as: \[ \Delta S^\circ = S^\circ(\mathrm{XY}_{3}) - \left( \frac{1}{2}S^\circ(\mathrm{X}_{2}) + \frac{3}{2} S^\circ(\mathrm{Y}_{2}) \right) \]Substituting the given values, \( \Delta S^\circ = 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right) = 50 - (30 + 60) = 50 - 90 = -40 \, \mathrm{J/K} \).

Step 3: Use Gibbs Free Energy Equation

For a reaction to be at equilibrium, \( \Delta G^\circ = 0 \). The Gibbs free energy equation is \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \). At equilibrium, substituting \( \Delta G^\circ = 0 \), we rearrange to find \( T \):\[ T = \frac{\Delta H^\circ}{\Delta S^\circ} \].

Step 4: Solve for Temperature T

Substitute the values into the rearranged Gibbs equation: \( \Delta H^\circ = -30 \, \mathrm{kJ/mol} = -30000 \, \mathrm{J/mol} \) and \( \Delta S^\circ = -40 \, \mathrm{J/K \cdot mol} \). Then,\[ T = \frac{-30000}{-40} = 750 \, \text{K} \].

Step 5: Compare with Given Options

The calculated temperature \( 750 \, \text{K} \) matches the option (c). Therefore, the correct answer is (c) \( 750 \, \text{K} \).

Key Concepts

Standard Entropy

Entropy is a measure of the disorder or randomness in a system. In the context of chemistry, standard entropy represents the entropy content of substances at a defined standard state, typically at 298 K and 1 atm pressure. When considering chemical reactions, it's important to understand how entropy changes, as this can influence factors like spontaneity and equilibrium.
In our reaction, we calculate the change in standard entropy, \( \Delta S^\circ \), by applying the formula:

\[ \Delta S^\circ = S^\circ(\mathrm{products}) - S^\circ(\mathrm{reactants}) \]

For \( \mathrm{XY}_{3} \), it is 50 \( \mathrm{J/K\cdot mol} \)

For the reactants \( \frac{1}{2} \mathrm{X}_{2} \) and \( \frac{3}{2} \mathrm{Y}_{2} \) the combined entropy is 90 \( \mathrm{J/K\cdot mol} \)

This results in a \( \Delta S^\circ \) of \(-40 \, \mathrm{J/K} \). A negative entropy change indicates that the product formation, \( \mathrm{XY}_{3} \), is less random compared to the individual reactants.

Enthalpy Change

Enthalpy, represented as \( \Delta H \), is a measure of the total energy of a thermodynamic system. It includes the system's internal energy plus the energy required to create space for it in its environment. When working with chemical reactions, enthalpy changes give us insight into whether a reaction is endothermic or exothermic.
In our chemical equation\[\frac{1}{2} \mathrm{X}_{2} + \frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3} \]we are given a \( \Delta H \) of \(-30 \, \mathrm{kJ/mol} \), indicating that the reaction is exothermic. An exothermic reaction releases energy to its surroundings, typically in the form of heat. This is an essential factor because it affects the temperature required to reach equilibrium. Negative \( \Delta H \) values typically make reactions more spontaneous at lower temperatures, as they release energy which can drive the reaction forward.

Equilibrium Temperature

Equilibrium temperature is a key concept in understanding at which temperature a reaction reaches a balance between reactant and product formation where Gibbs Free Energy \( \Delta G^\circ \) is zero.
The relationship is defined through the Gibbs free energy equation:

\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

At equilibrium, \( \Delta G^\circ = 0 \), allowing us to solve for temperature using:

\[ T = \frac{\Delta H^\circ}{\Delta S^\circ} \]

For the reaction we considered, with \( \Delta H^\circ = -30000 \, \mathrm{J/mol} \) and \( \Delta S^\circ = -40 \, \mathrm{J/K\cdot mol} \), the temperature is calculated as:

\[ T = \frac{-30000}{-40} = 750 \, \text{K} \]

This result indicates the required conditions for the system to achieve equilibrium. This means at 750 K, the rate of the forward and reverse reactions are equal, signifying no net change in concentrations of reactants or products, creating a balanced state.