Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The value of \(\log _{10} K\) for a reaction \(A \rightleftharpoons B\) is \(\left(\right.\) Given : \(\Delta_{r} H_{298 \mathrm{~K}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{r} S_{298 \mathrm{~K}}^{\circ}\) \(=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\); \(2.303 \times 8.314 \times 298=5705)\) (a) 5 (b) 10 (c) 95 (d) 100

Step-by-Step Solution

Answer

The value of \( \log_{10} K \) is 10 (option b).

Step 1: Understanding the Relation between Gibbs Free Energy and Equilibrium Constant

The relationship between the Gibbs free energy change \( \Delta G^{\circ} \) and the equilibrium constant \( K \) at a given temperature \( T \) is given by the equation:\[ \Delta G^{\circ} = -RT \ln K \].For calculations in base 10, we use \( \ln K = \log K \times 2.303 \). Therefore, if we want \( \log K \), we will use:\[ \Delta G^{\circ} = -2.303 \cdot RT \cdot \log K \].We'll use this formula to find \( \log_{10} K \) using the provided \( \Delta G^{\circ} \).

Step 2: Calculating the Change in Gibbs Free Energy

To find \( \Delta G^{\circ} \), we use the equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \].Substituting the given values, \( \Delta H^{\circ} = -54.07 \text{ kJ/mol} = -54070 \text{ J/mol} \), \( \Delta S^{\circ} = 10 \text{ J/K/mol} \), and \( T = 298 \text{ K} \), we have:\[ \Delta G^{\circ} = -54070 - 298 \times 10 \].Calculating this gives:\[ \Delta G^{\circ} = -54070 - 2980 = -57050 \text{ J/mol} \].

Step 3: Solving for Logarithm of Equilibrium Constant

Using \( \Delta G^{\circ} = -2.303 \cdot RT \cdot \log K \) from Step 1, we rearrange to solve for \( \log K \):\[ \log K = \frac{-\Delta G^{\circ}}{2.303 \times R \times T} \].Substituting \( \Delta G^{\circ} = -57050 \text{ J/mol} \), \( R = 8.314 \text{ J/K/mol} \), and \( T = 298 \text{ K} \):\[ \log K = \frac{57050}{2.303 \times 8.314 \times 298} \].According to the problem, \( 2.303 \times 8.314 \times 298 = 5705 \), so:\[ \log K = \frac{57050}{5705} \approx 10 \].

Step 4: Choosing the Correct Answer

The calculated value of \( \log_{10} K \) is approximately 10, matching the given option (b) in the problem.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a value that helps to understand the extent to which a chemical reaction occurs. It reflects the ratio of product to reactant concentrations at equilibrium. A high \( K \) value indicates that at equilibrium, a large amount of reactants have been converted into products. Conversely, a low \( K \) means that few reactants convert to products.
Understanding \( K \) involves knowing how it is derived. The relationship between Gibbs Free Energy, \( \Delta G^{\circ} \), and \( K \) is crucial. It is given by the equation \( \Delta G^{\circ} = -RT \ln K \).
Here's what each term means:

• \( \Delta G^{\circ} \) is the standard change in Gibbs free energy.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

When the reaction reaches equilibrium, \( \Delta G^{\circ} \) becomes zero, meaning there is no net change in free energy, as the energy required to form products from reactants is balanced by the energy to reform reactants from products. Thus, knowing \( K \) helps predict the direction in which the reaction naturally proceeds.

Entropy Change

Entropy, symbolized as \( S \), is a measure of the degree of disorder or randomness in a system. In chemistry, it helps describe how energy is dispersed within molecules in a reaction. A positive entropy change, \( \Delta S^{\circ} \), means the products of the reaction are more disordered than the reactants.
This concept is important when calculating Gibbs Free Energy through the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \).
When applying this in our exercise:

• We have \( \Delta S^{\circ} = 10 \text{ J/K/mol} \), indicating an increase in entropy, or disorder, with product formation.
• The term \( T\Delta S^{\circ} \) is the temperature times the entropy change, which reflects how temperature influences entropy's contribution to the system's free energy.

In general, higher entropy suggests the system favors a more spontaneous reaction, as systems naturally tend towards greater disorder.

Enthalpy Change

Enthalpy, represented as \( H \), is a measure of the total energy of a thermodynamic system. It reflects the system's internal energy plus the product of its pressure and volume. In reactions, changes in enthalpy, \( \Delta H^{\circ} \), give insight into whether a reaction absorbs or releases heat.
In our exercise, we see the given \( \Delta H^{\circ} = -54.07 \text{ kJ/mol} \). The negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.
The relationship between enthalpy and Gibbs Free Energy is represented by the formula \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \). Here's what happens:

• If \( \Delta H^{\circ} \) is negative, the system is losing heat, generally making \( \Delta G^{\circ} \) negative, favoring product formation.
• It plays a key role in determining spontaneity along with entropy change—a more negative \( \Delta H^{\circ} \) usually makes reactions more likely to occur spontaneously.

Thus, understanding \( \Delta H^{\circ} \) helps predict whether a reaction involves endothermic (heat-absorbing) or exothermic (heat-releasing) transformations.