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 reaction is in equilibrium at approximately 285.7 K.

Step 1: Understanding Gibbs Free Energy

To find the temperature at which a reaction is in equilibrium, we use the Gibbs free energy equation, \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \( \Delta G = 0 \). So the equation becomes:\[ 0 = \Delta H - T \Delta S \].This can be rearranged to find the temperature \( T \) as: \[ T = \frac{\Delta H}{\Delta S} \].

Step 2: Substituting Known Values into the Equation

Given that \( \Delta H = 30 \mathrm{~kJ} \mathrm{~mol}^{-1} \) and \( \Delta S = 105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \), we must ensure both are in the same units. Convert \( \Delta H \) to \( \mathrm{~J} \mathrm{~mol}^{-1} \):\[ \Delta H = 30 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \].Now substitute into the equation:\[ T = \frac{30 \times 10^3}{105} \].

Step 3: Calculating the Temperature

Perform the division to find the temperature at equilibrium:\[ T = \frac{30000}{105} \approx 285.71 \mathrm{~K} \].

Step 4: Selecting the Correct Answer

Now, compare the calculated temperature 285.71 K with the provided options: 450 K, 300 K, 285.7 K, and 273 K. The closest option is (c) 285.7 K.

Key Concepts

Enthalpy Change

In chemistry, understanding the concept of enthalpy change is key to analyzing reactions. Enthalpy change, denoted as \( \Delta H \), represents the heat content change of a system during a reaction. When a chemical reaction takes place, it may either absorb or release energy. This energy change will affect how the reaction proceeds.

Generally, an exothermic reaction releases heat, resulting in a negative \( \Delta H \), while an endothermic reaction absorbs heat, leading to a positive \( \Delta H \). In our example with bromine and chlorine reacting to form bromine chloride, \( \Delta H \) is 30 kJ/mol, indicating that the reaction absorbs heat. This means it's an endothermic reaction.

Knowing the enthalpy change allows chemists to determine how favorable the reaction is. It is crucial in calculations involving Gibbs Free Energy, which determines if a reaction can occur spontaneously.

Entropy Change

Entropy change, represented by \( \Delta S \), is another important aspect of chemical reactions. Entropy measures the disorder or randomness of a system. The concept is pivotal in understanding how reactions result in increased or decreased randomness.

In the given reaction, \( \Delta S \) is 105 J K\(^{-1}\) mol\(^{-1}\). A positive value of entropy indicates that the system has increased disorder after the reaction. Here, the entropy change suggests greater randomness as the number of gas molecules increases from 1 to 2, accounting for the formation of two moles of gaseous bromine chloride.

Entropy is integral to the Gibbs Free Energy equation and acts as a driving force for reactions, especially at higher temperatures where the \( T \Delta S \) term becomes significant, potentially promoting spontaneity in reactions despite enthalpy conditions.

Equilibrium Temperature

The equilibrium temperature is the temperature at which a reaction is at equilibrium, meaning that the forward and reverse reactions occur at the same rate. This is an important concept, as it indicates the precise temperature where a reaction neither absorbs nor releases energy.To find this temperature, we use the formula derived from setting the Gibbs Free Energy equation to zero at equilibrium:\[ T = \frac{\Delta H}{\Delta S} \]This equation essentially balances the enthalpy change and the entropy change. In our exercise, to find the equilibrium temperature, we need to substitute \( \Delta H = 30,000 \) J/mol and \( \Delta S = 105 \) J K\(^{-1}\) mol\(^{-1}\) into the equation. Doing the math, we find the equilibrium temperature to be approximately 285.71 K.

This concept helps us understand at what temperature a reaction is perfectly balanced and represents a key analytical tool in designing chemical processes that are efficient and controlled. Knowing the equilibrium temperature is essential for predicting whether reactions will proceed and to what extent under different conditions.