Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

The enthalpy and entropy change for the reaction, \(\mathrm{Br}_{2}(1)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrCl}(\mathrm{g})\) are \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(105 \mathrm{~J} \mathrm{~mol}^{-1}\) respectively. The temperature at which the reaction will be in equilibrium is (a) 450 (b) 300 (c) \(285.7\) (d) 273

Step-by-Step Solution

Answer

The equilibrium temperature is 285.7 K, which is option (c).

Step 1: Understanding the Problem

We are given the enthalpy change \( \Delta H = 30 \text{ kJ/mol} \) and the entropy change \( \Delta S = 105 \text{ J/mol K} \) for a reaction. We need to find the temperature at which the reaction is at equilibrium.

Step 2: Equilibrium Condition

For a reaction to be at equilibrium, the change in Gibbs free energy \( \Delta G \) is zero. The equation relating these quantities is: \[ \Delta G = \Delta H - T \Delta S = 0 \] where \( T \) is the temperature in Kelvin.

Step 3: Rearranging the Equation

We rearrange the equation \( \Delta G = 0 \) to solve for \( T \):\[ \Delta H = T \Delta S \] \[ T = \frac{\Delta H}{\Delta S} \]

Step 4: Unit Consistency

Ensure that \( \Delta H \) and \( \Delta S \) are in the same units. Convert \( \Delta H \) from kJ/mol to J/mol by multiplying by 1000: \[ \Delta H = 30 \times 1000 = 30000 \text{ J/mol} \]

Step 5: Calculate the Temperature

Substitute the values into the rearranged equation to find \( T \):\[ T = \frac{30000 \text{ J/mol}}{105 \text{ J/mol K}} \approx 285.7 \text{ K} \]

Step 6: Check Answer Against Options

The calculated equilibrium temperature is 285.7 K. This corresponds to option (c) in the given choices.

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of the total heat content of a system undergoing a chemical reaction at constant pressure. It helps us understand whether a reaction absorbs or releases heat. When a reaction takes place, the enthalpy change can either be positive or negative.

• Positive \( \Delta H \) indicates that the reaction is endothermic (absorbs heat).
• Negative \( \Delta H \) denotes an exothermic reaction (releases heat).

In the context of our exercise, the reaction \( \mathrm{Br}_2(1) + \mathrm{Cl}_2(g) \rightarrow 2 \mathrm{BrCl}(g) \) has an enthalpy change of \( 30 \text{ kJ/mol} \). This shows that the reaction is endothermic, meaning heat is absorbed from the surroundings during the transformation from reactants to products. Understanding enthalpy change is vital for predicting how temperature changes affect reaction directions and rates.

Entropy Change

Entropy change, represented by \( \Delta S \), quantifies the disorder or randomness in a system as it undergoes a transformation. When a chemical reaction occurs, if the system's disorder increases, \( \Delta S \) is positive, indicating the products have greater randomness than the reactants. Conversely, a negative \( \Delta S \) signals a decrease in disorder.

• \( + \Delta S \): Increase in disorder or randomness.
• \( - \Delta S \): Decrease in disorder or randomness.

In the given reaction, \( \Delta S = 105 \text{ J/mol K} \) indicates that the reaction leads to an increase in disorder. This increase in entropy suggests that the gases are more spread out or have higher freedom to move. Entropy plays a crucial role in determining the temperature-dependent spontaneity of reactions and contributes significantly to the Gibbs free energy equation.

Equilibrium Temperature

The equilibrium temperature of a reaction is the point where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products. At equilibrium, the Gibbs free energy change, \( \Delta G \), is zero. The relation between enthalpy change, entropy change, and the equilibrium temperature is given by the formula: \[ T = \frac{\Delta H}{\Delta S} \]Here, \( T \) is the temperature at which \( \Delta G = 0 \). For the provided reaction, substituting \( \Delta H = 30000 \text{ J/mol} \) and \( \Delta S = 105 \text{ J/mol K} \) into the formula gives an equilibrium temperature of approximately 285.7 K. This calculation shows the temperature at which the reaction is perfectly balanced and emphasizes the importance of both \( \Delta H \) and \( \Delta S \) in determining reaction behavior with changing temperatures.