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 \([2008]\) (a) \(1250 \mathrm{~K}\) (b) \(500 \mathrm{~K}\) (c) \(750 \mathrm{~K}\) (d) \(1000 \mathrm{~K}\)

Step-by-Step Solution

Answer

The reaction is at equilibrium at 750 K.

Step 1: Understand the Reaction

The reaction given is \( \frac{1}{2} \mathrm{X}_{2} + \frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3} \). We need to find the temperature at which this reaction is at equilibrium.

Step 2: Write Gibbs Free Energy Equation at Equilibrium

At equilibrium, the change in Gibbs free energy \( \Delta G = 0 \). The equation for Gibbs free energy is given by \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy and \( \Delta S \) is the change in entropy.

Step 3: Calculate Change in Entropy (\(\Delta S\))

The change in entropy for the reaction can be calculated using the standard entropy values: \( \Delta S = S_{\text{products}} - S_{\text{reactants}} = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40) \). Evaluate this expression to obtain \( \Delta S = 50 - 90 = -40 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \).

Step 4: Substitute Values in Gibbs Free Energy Equation

Substitute \( \Delta G = 0 \), \( \Delta H = -30,000 \text{ J/mol} \), and \( \Delta S = -40 \text{ J/K mol} \) into the equation \( \Delta G = \Delta H - T\Delta S \). This simplifies to \( 0 = -30,000 - T(-40) \).

Step 5: Solve for Temperature \(T\)

Rearrange the equation to solve for \(T\): \( T(-40) = -30,000 \) so \( T = \frac{30,000}{40} = 750 \text{ K} \). This is the temperature at which the reaction is at equilibrium.

Key Concepts

Standard Entropy

Standard entropy is a measure of the randomness or disorder within a system. It's usually expressed in units of joules per Kelvin per mole (\( \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \)).
For our reaction, the standard entropy values for \(\mathrm{X}_{2}\), \(\mathrm{Y}_{2}\), and \(\mathrm{XY}_{3}\) are given as 60, 40, and 50 \(\mathrm{JK}^{-1} \mathrm{mol}^{-1}\) respectively.
These values describe the individual entropy contributions of each substance involved in the chemical reaction. The standard entropy helps us calculate the overall change in entropy when reactants are transformed into products.
Understanding standard entropy is crucial because it contributes to determining whether a process is spontaneous and at what temperature the reaction will be at equilibrium.

Equilibrium Temperature

The equilibrium temperature is where a chemical reaction's forward and reverse rates equalize, meaning there's no net change in reactant or product concentrations. To find it, we use the Gibbs Free Energy equation, where \( \Delta G = 0 \) at equilibrium.
Knowing the enthalpy change \( \Delta H \) and entropy change \( \Delta S \), we rearrange the equation \( \Delta G = \Delta H - T \Delta S \) to find the equilibrium temperature \( T = \frac{\Delta H}{\Delta S} \).
For this specific reaction, we calculate an equilibrium temperature of 750 K. This is the temperature at which the energy driving the reaction (enthalpy) is balanced by the dispersal or disorder (entropy) within the system.

Enthalpy Change

Enthalpy change, noted as \( \Delta H \), is the heat absorbed or released during a reaction at constant pressure.
For our given reaction, the enthalpy change is \( -30 \mathrm{kJ} \). This negative value signals that the reaction is exothermic, meaning it releases heat to the surroundings.
Understanding enthalpy change is vital as it indicates how energy changes are involved in making or breaking chemical bonds, affecting reaction spontaneity. It forms one part of the equation to find equilibrium conditions using Gibbs Free Energy principles.

Entropy Change

Entropy change, \( \Delta S \), refers to the difference in disorder between reactants and products. It's calculated from the standard entropies of substances involved in the reaction.
In this exercise, we find \( \Delta S = \text{products' entropy} - \text{reactants' entropy} = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40) = -40 \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).
This negative \( \Delta S \) value suggests that the system's disorder decreases during the reaction, meaning the products are more ordered than the reactants. Knowing \( \Delta S \) helps us assess the balance between energy dispersal and energy release, crucial for determining reaction spontaneity and equilibrium temperature.