Problem 57 - 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 \(105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The temperature at which the reaction will be in equilibrium is (a) \(450 \mathrm{~K}\) (b) \(300 \mathrm{~K}\) (c) \(285.7 \mathrm{~K}\) (d) \(273 \mathrm{~K}\)

Step-by-Step Solution

Answer

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

Step 1: Understand Given Data

We are provided with the change in enthalpy, \( \Delta H = 30 \, \text{kJ mol}^{-1} \) and the change in entropy, \( \Delta S = 105 \, \text{J K}^{-1} \text{mol}^{-1} \). We need to convert \( \Delta H \) into \( \text{J} \text{mol}^{-1} \), so \( \Delta H = 30,000 \, \text{J mol}^{-1} \).

Step 2: Use Gibbs Free Energy Equation

The equation \( \Delta G = \Delta H - T \Delta S \) is used to calculate the Gibbs Free Energy. For equilibrium, \( \Delta G = 0 \), which implies \( \Delta H = T \Delta S \).

Step 3: Solve for Temperature

Rearrange the equation to solve for temperature: \( T = \frac{\Delta H}{\Delta S} \). Substitute in the given values: \( T = \frac{30,000}{105} \approx 285.7 \, \text{K} \).

Step 4: Match Temperature with Options

The calculated temperature at equilibrium, \( 285.7 \, \text{K} \), corresponds with option (c).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a crucial concept in chemical thermodynamics. It represents the heat content change in a system during a constant pressure process. In the exercise, the enthalpy change for the reaction from bromine and chlorine to form bromine chloride is given as \( 30 \, \text{kJ mol}^{-1} \).
This value indicates how much energy is absorbed or released. In this case, since the value is positive, it signifies an endothermic reaction where the system absorbs heat. Understanding whether a reaction is endothermic or exothermic helps predict how energy flow occurs during the reaction.
The unit of \( \Delta H \) can vary based on the scale of measurement but is typically reported in kilojoules per mole \( \text{kJ}\,\text{mol}^{-1} \) or joules per mole \( \text{J} \, \text{mol}^{-1} \). Always remember to keep units consistent when performing thermodynamic calculations.

Entropy Change

Entropy change \( \Delta S \) measures the degree of disorder or randomness in a system. It is a fundamental principle of thermodynamics that systems tend to move towards higher entropy. In this exercise, \( \Delta S \) for the reaction is given as \( 105 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \).
When the entropy change is positive, as it is here, the products of the reaction are at a higher state of disorder than the reactants. This often occurs when gases are involved, as they usually have higher entropy due to increased molecular movement and freedom.
Keep in mind that the units of entropy are typically presented in joules per Kelvin per mole \( \text{J}\, \text{K}^{-1} \, \text{mol}^{-1}\). Like enthalpy, it is essential to maintain consistent units throughout your calculations.

Gibbs Free Energy

Gibbs Free Energy, represented by \( \Delta G \), is a powerful concept that determines whether a reaction is spontaneous. It's derived using the equation \( \Delta G = \Delta H - T \Delta S \), where \( T \) is the absolute temperature in Kelvin.
If \( \Delta G \) is negative, the reaction is spontaneous, proceeding without any external energy input. A positive \( \Delta G \) denotes non-spontaneity. For reactions at equilibrium, \( \Delta G \) equals zero, implying no net change in the system's free energy.
In the given exercise, to find the equilibrium temperature, \( \Delta G = 0 \) leads to the equation \( \Delta H = T \Delta S \). Solving for \( T \) reveals the temperature where the reaction reaches equilibrium. Calculating correctly here is vital for understanding reaction dynamics.

Equilibrium Temperature

Equilibrium temperature is the temperature at which a chemical reaction reaches a state of balance, where the forward and reverse reactions occur at the same rate. This is when \( \Delta G \) becomes zero, indicating no net change in Gibbs Free Energy.
For the exercise, the equilibrium temperature is calculated using the formula \( T = \frac{\Delta H}{\Delta S} \). By substituting \( \Delta H = 30,000 \, \text{J mol}^{-1} \) and \( \Delta S = 105 \, \text{J}\, \text{K}^{-1} \text{mol}^{-1} \), the equilibrium temperature is found to be \( 285.7 \, \text{K} \).
This specific temperature is crucial because it marks the point where the system doesn't favor either the forward or backward reaction, thus achieving equilibrium. Understanding equilibrium temperature provides insights into how conditions affect chemical reactions and system behavior.