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) \(-142.16 \mathrm{~kJ} /\) mole (b) \(-71.08 \mathrm{~kJ} /\) mole (c) \(-35.54 \mathrm{~kJ} / \mathrm{mole}\) (d) none of these

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We are given equilibrium constants at two different temperatures for a reaction, and we need to find the enthalpy change of the reaction. This requires using the van't Hoff equation, which relates the change in equilibrium constant with temperature to the enthalpy change of the reaction.

Step 2: Apply the Van't Hoff Equation

The van't Hoff equation is \( \ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \), where \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \) respectively, and \( R \) is the universal gas constant \( 8.314 \, \text{J/mol K} \). In this case, \( K_1 = 18.5 \), \( K_2 = 9.25 \), \( T_1 = 925 + 273 = 1198 \, \text{K} \), and \( T_2 = 1000 + 273 = 1273 \, \text{K} \).

Step 3: Calculate the Natural Log of the Equilibrium Constant Ratio

Calculate \( \ln \left(\frac{K_2}{K_1}\right) = \ln \left(\frac{9.25}{18.5}\right) = \ln(0.5) \approx -0.693 \).

Step 4: Calculate the Temperature Inverse Difference

Compute \( \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{1273} - \frac{1}{1198} \approx 7.736 \times 10^{-5} \, \text{K}^{-1} \).

Step 5: Solve for Enthalpy Change (\(\Delta H\))

Using the Van't Hoff equation, solve for \( \Delta H \): \( -0.693 = -\frac{\Delta H}{8.314} \times 7.736 \times 10^{-5} \). Rearrange to find \( \Delta H = \frac{0.693 \times 8.314}{7.736 \times 10^{-5}} \approx -71,080 \, \text{J/mol} = -71.08 \, \text{kJ/mol} \).

Step 6: Choose the Correct Answer

The calculated change in enthalpy is \(-71.08 \, \text{kJ/mol}\). This matches option (b).

Key Concepts

equilibrium constant

In the realm of chemical reactions, the equilibrium constant, denoted as \( K \), plays a critical role. It tells us about the proportions of reactants and products at equilibrium. An equilibrium state is when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products no longer change.

• A large \( K \) implies a high concentration of products at equilibrium.
• A small \( K \) means the reactants are more favored at equilibrium.

Understanding \( K \) helps predict the direction in which a reaction mixture must shift to reach equilibrium. The values given at different temperatures show the nature of the reaction changes with temperature, as seen with \( K = 18.5 \) at 925 K and \( 9.25 \) at 1000 K, indicating temperature sensitivity.

van't Hoff equation

The van't Hoff equation connects the change in the equilibrium constant with temperature to the reaction's enthalpy change. It is a specific tool that chemists use to predict how \( K \) will change as temperature shifts. The equation is:\[\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]Here’s what to consider:

• \( \Delta H \) is the change in enthalpy or heat of the reaction.
• \( R \) is the universal gas constant.
• The temperatures \( T_1 \) and \( T_2 \) are in Kelvin.

Using the equation helps determine whether a reaction absorbs or releases heat, leading to insights on the reaction mechanism.

reaction enthalpy

Reaction enthalpy, \( \Delta H \), reflects the heat change during a chemical reaction at constant pressure. It indicates whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).

• A negative \( \Delta H \) value, like \( -71.08 \, \text{kJ/mol} \), means the reaction is exothermic.
• It is calculated using the van't Hoff equation when equilibrium constant values and temperature are provided.

Knowing \( \Delta H \) is crucial for understanding the energy profile of a reaction, enabling predictions of reaction spontaneity under varying conditions.

chemical equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions balance each other. At equilibrium, the concentration of reactants and products remain constant because the processes occur simultaneously at the same rate. This state can change if pressure, temperature, or concentrations are altered, illustrating Le Chatelier's principle.

• It is dynamic, meaning the reactions don't stop but occur at the same rate in both directions.
• Shifts in equilibrium, due to temperature changes, influence the equilibrium constant value \( K \).

Understanding equilibrium is essential for controlling reactions in industrial and laboratory settings.

temperature dependence

Temperature plays a pivotal role in chemical reactions, affecting both reaction rates and equilibrium. The dependence of equilibrium constant \( K \) on temperature can be analyzed using the van't Hoff equation, highlighting the intricate relationship:

• An increase in temperature can shift an equilibrium position, either favoring reactants or products depending on the reaction enthalpy.
• Endothermic reactions generally have a higher \( K \) value at increased temperatures.
• Exothermic reactions see a decrease in \( K \) with a rise in temperature.

Thus, temperature control is a vital aspect in chemical processes, from synthesis to product stability.