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

Question

The enthalpy and entropy change for the reaction, \(\mathrm{Br}_{2}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrCl}(\mathrm{g})\) 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 approximately 285.7 K, which is option (c).

Step 1: Understand the Problem

We are given changes in enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) for a chemical reaction. We need to find the temperature at which the reaction is at equilibrium. This involves using the Gibbs free energy equation.

Step 2: Gibbs Free Energy Equation at Equilibrium

The Gibbs free energy equation is \(\Delta G = \Delta H - T\Delta S\). At equilibrium, \(\Delta G = 0\). Therefore, the equation becomes \(0 = \Delta H - T\Delta S\) or \(T = \frac{\Delta H}{\Delta S}\).

Step 3: Plug in the Values

Use the given values: \(\Delta H = 30\, \mathrm{kJ}\, \mathrm{mol}^{-1} = 30000\, \mathrm{J}\, \mathrm{mol}^{-1}\) and \(\Delta S = 105\, \mathrm{J}\, \mathrm{mol}^{-1}\, \mathrm{K}^{-1}\). Substitute these into the temperature equation: \(T = \frac{30000}{105}\).

Step 4: Calculate the Temperature

Divide the enthalpy by the entropy to find the temperature: \(T = \frac{30000}{105} \approx 285.7\, \mathrm{K}\).

Step 5: Select the Correct Option

The calculated equilibrium temperature is approximately \(285.7\), which corresponds to option (c) in the given choices.

Key Concepts

Understanding Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measurement of the heat released or absorbed during a chemical reaction. This value indicates whether a reaction is endothermic or exothermic.

• An endothermic reaction absorbs heat, and \( \Delta H \) is positive.
• An exothermic reaction releases heat, and \( \Delta H \) is negative.

For the reaction \( \text{Br}_2(\text{l}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{BrCl}(\text{g}) \), the enthalpy change of 30 kJ/mol implies the reaction absorbs heat from the surroundings. Knowing \( \Delta H \) helps predict how a reaction behaves with temperature changes.

Exploring Entropy Change

Entropy change, \( \Delta S \), measures the degree of disorder or randomness in the system. During a chemical reaction, the system's entropy can increase or decrease.

• If \( \Delta S \) is positive, the system becomes more disordered.
• If \( \Delta S \) is negative, the system becomes more ordered.

For the given reaction, \( \Delta S = 105 \text{ J mol}^{-1} \text{ K}^{-1} \), indicating an increase in disorder. The generation of more gas molecules from liquid and gas reactants generally leads to greater entropy, reflecting increased randomness.

Calculating Equilibrium Temperature

The equilibrium temperature of a reaction is the temperature at which the forward and reverse reactions occur at equal rates, resulting in no net change in the concentration of reactants and products. To find this, we use the Gibbs Free Energy equation \( \Delta G = \Delta H - T\Delta S \). At equilibrium, \( \Delta G = 0 \), giving the equation:
\[ T = \frac{\Delta H}{\Delta S} \]
By substituting \( \Delta H = 30000 \text{ J mol}^{-1} \) and \( \Delta S = 105 \text{ J mol}^{-1} \text{ K}^{-1} \), we calculate:
\[ T = \frac{30000}{105} \approx 285.7 \text{ K} \]
This temperature is the point where chemical equilibrium is achieved for the reaction.

The Dynamics of Chemical Reactions

Chemical reactions involve breaking and forming bonds between atoms, transforming reactants into products. During this process, energy conversion plays a critical role, governed by enthalpy and entropy changes.

• Reactions can be spontaneous, non-spontaneous, or in equilibrium.
• Spontaneity depends on the total Gibbs Free Energy change.

Understanding the role of \( \Delta H \) and \( \Delta S \) helps forecast the conditions under which a reaction is favorable. A reaction will naturally move toward the state with lower free energy, explaining why we need to reach a particular temperature to achieve equilibrium.