Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) is \(18.5\) at 925 and \(9.25\) at 1000 respectively. What is the enthalpy of the reaction? (a) \(-142.16 \mathrm{~kJ} / \mathrm{mole}\) (b) \(-71.08 \mathrm{~kJ} / \mathrm{mole}\) (c) \(-35.54 \mathrm{~kJ} /\) mole (d) none of these

Step-by-Step Solution

Answer

The enthalpy of the reaction is \(-71.08 \text{ kJ/mol}\) (option b).

Step 1: Understand the Problem

We need to find the enthalpy change (ΔH) of a reaction given equilibrium constants \( K \) at two different temperatures. The reaction is \( \mathrm{H}_2(\mathrm{g}) + \mathrm{S}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2\mathrm{S}(\mathrm{g}) \). The equilibrium constants are provided: \( K = 18.5 \) at 925 K and \( K = 9.25 \) at 1000 K. We will use the van 't Hoff equation to solve for ΔH.

Step 2: Apply the van 't Hoff Equation

The van 't Hoff equation relates the change in the equilibrium constant with temperature to the enthalpy change: \[ \ln \frac{K_2}{K_1} = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \( K_1 = 18.5 \), \( K_2 = 9.25 \), \( T_1 = 925 \) K, \( T_2 = 1000 \) K, and \( R = 8.314 \) J/mol·K.

Step 3: Calculate the Natural Log of K Ratio

Calculate \( \ln (K_2/K_1) \): \[ \ln \left( \frac{9.25}{18.5} \right) = \ln(0.5) \approx -0.6931 \]

Step 4: Calculate the Inverse Temperature Difference

Find \( \frac{1}{T_2} - \frac{1}{T_1} \): \[ \frac{1}{1000} - \frac{1}{925} = 0.001 - 0.001081 = -0.000081 \text{ K}^{-1} \]

Step 5: Solve for ΔH

Rearrange the van 't Hoff equation to solve for \( \Delta H \): \[ \Delta H = -\frac{R \cdot \ln (K_2/K_1)}{\frac{1}{T_2} - \frac{1}{T_1}} \] Plug in the numbers:\[ \Delta H = -\frac{8.314 \times (-0.6931)}{-0.000081} \approx -71082.9 \text{ J/mol} \approx -71.08 \text{ kJ/mol} \]

Step 6: Identify the Correct Answer

The calculated enthalpy change \( \Delta H \) is \(-71.08 \text{ kJ/mol} \). Therefore, the correct answer is option (b) \(-71.08 \text{ kJ/mol}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial concept in chemistry that describes the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. This is a state where the forward and backward reaction rates are equal, resulting in no net change in concentration of reactants and products. The constant is unique for each reaction at a specific temperature.

• For gases, the equilibrium constant is often expressed in terms of partial pressures, known as \( K_p \).
• For solutions, it is expressed in terms of molar concentrations, denoted as \( K_c \).

The value of \( K \) indicates the extent of the reaction:

• A large \( K \) (much greater than 1) means the reaction favors the formation of products at equilibrium.
• A small \( K \) (much less than 1) indicates the reaction favors the reactants.

In this exercise, the change in equilibrium constant over a range of temperatures is leveraged to calculate the enthalpy change for the reaction.

Van 't Hoff Equation

The van 't Hoff equation is a key mathematical expression used to relate the change in the equilibrium constant \( K \) with the change in temperature. It serves to understand how a change in temperature affects the position of equilibrium for a reaction. The equation is given by: \[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right), \] where \( \Delta H \) is the enthalpy change, \( R \) is the ideal gas constant, and \( T_1 \) and \( T_2 \) are the initial and final temperatures respectively in kelvins.

• This equation illustrates that the equilibrium constant changes exponentially with temperature.
• If \( \Delta H \) is positive (endothermic reaction), an increase in temperature increases \( K \).
• If \( \Delta H \) is negative (exothermic reaction), an increase in temperature decreases \( K \).

In this exercise, the van 't Hoff equation was applied using the known equilibrium constants and temperatures to solve for \( \Delta H \), which was found to be \(-71.08 \text{ kJ/mol}\). This corresponds to the exothermic nature of the reaction.

Temperature Dependence of Equilibrium

Temperature plays a vital role in the position of chemical equilibrium. According to Le Chatelier’s principle, if the temperature of an exothermic reaction is increased, the system will shift to absorb the added heat, favoring the endothermic direction (reverse reaction). Conversely, for endothermic reactions, an increase in temperature favors the forward reaction.

• This temperature dependence is also quantitatively described by the van 't Hoff equation.
• For the given reaction, as temperature increased from 925 K to 1000 K, the equilibrium constant decreased from 18.5 to 9.25.

This decrease in \( K \) with rising temperature further emphasizes the exothermic nature of the reaction. As the reaction releases heat, increasing the temperature shifts the equilibrium towards the reactants, hence leading to a lower value of \( K \). Understanding how temperature affects equilibrium constants is essential in predicting the concentrations of reactants and products at equilibrium under different conditions.