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 equilibrium temperature is 750 K, so the answer is (c) 750 K.

Step 1: Calculate the Standard Entropy Change (ΔS⁰)

To find the standard entropy change (ΔS⁰) for the reaction, calculate the entropy of products minus the entropy of reactants. The balanced reaction is \( \frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightarrow XY_3 \). Use the given standard entropies: \( S^0(X_2) = 60 \), \( S^0(Y_2) = 40 \), and \( S^0(XY_3) = 50 \) \( \mathrm{JK^{-1}mol^{-1}}\). \[\Delta S^0 = S^0(XY_3) - \left( \frac{1}{2}S^0(X_2) + \frac{3}{2}S^0(Y_2) \right) \]\[\Delta S^0 = 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right) \]\[\Delta S^0 = 50 - (30 + 60) = 50 - 90 = -40 \text{ JK}^{-1} \text{ mol}^{-1} \]

Step 2: Apply Gibbs Free Energy Equation for Equilibrium

For a reaction at equilibrium, the change in Gibbs Free Energy (ΔG) is zero. The equation is:\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]Set \( \Delta G^0 = 0 \) since the reaction is at equilibrium:\[ 0 = -30 \times 10^3 - T \times (-40) \]Convert \( \Delta H^0 \) into Joules: \( -30 \times 10^3 \) J/mol.

Step 3: Solve for the Equilibrium Temperature (T)

Using the equation from Step 2:\[ 0 = -30 \times 10^3 + 40T \]\[ 30,000 = 40T \]\[ T = \frac{30,000}{40} \]\[ T = 750 \text{ K} \]

Key Concepts

Standard Entropy

Standard entropy is a measure of the amount of disorder or randomness in a system. It is denoted by the symbol \( S^0 \) and is expressed in units of \( \mathrm{JK^{-1}mol^{-1}} \). For any given substance, standard entropy can be thought of as the amount of energy spread out at a specific temperature. Each substance has a distinct standard entropy value and these values are typically determined experimentally.
For example, in the original exercise, the standard entropy values are provided as follows:

• \( S^0(\mathrm{X}_2) = 60 \ \mathrm{JK^{-1}mol^{-1}} \)
• \( S^0(\mathrm{Y}_2) = 40 \ \mathrm{JK^{-1}mol^{-1}} \)
• \( S^0(\mathrm{XY}_3) = 50 \ \mathrm{JK^{-1}mol^{-1}} \)

When we compute the entropy change for a reaction, the standard entropies of the products and reactants are compared. A positive entropy change implies increased disorder, while a negative change indicates an increase in order in the reaction system.

Gibbs Free Energy

Gibbs Free Energy, often symbolized as \( \Delta G \), is a thermodynamic quantity used to predict whether a process will proceed spontaneously at constant temperature and pressure. It integrates both enthalpy and entropy to provide a more comprehensive understanding of a reaction's potential.Gibbs Free Energy is calculated using the equation:\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]where:

• \( \Delta H^0 \) is the change in enthalpy (heat content) of the system
• \( T \) is the absolute temperature in Kelvin
• \( \Delta S^0 \) is the change in standard entropy

In the case of equilibrium, as stated in step 2, the Gibbs Free Energy change \( \Delta G^0 = 0 \). This condition aids in determining the temperature at which the reaction reaches equilibrium. Solving for \( T \) at equilibrium involves setting \( \Delta G^0 = 0 \) and plugging into the equation, yielding valuable insights into the balance of energy and order during the reaction.

Reaction Entropy Change

The reaction entropy change \( \Delta S^0 \) quantifies how the disorder within a chemical system alters between reactants and products. Calculating \( \Delta S^0 \) is crucial because it provides insight into the spontaneity and viability of a reaction. It is determined by the formula:\[ \Delta S^0 = S^0(\text{products}) - S^0(\text{reactants}) \]In the given exercise, the values provided for the substances allow for such a computation:\( \Delta S^0 = 50 - \left( \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right) = 50 - 90 = -40 \ \mathrm{JK^{-1}mol^{-1}} \)This result suggests that the reaction leads to a decrease in entropy, or increased order, as the value is negative. This information is valuable for understanding the dynamics of how the system evolves and how these entropic changes, in conjunction with enthalpy change, influence whether the reaction proceeds forward or not.