Problem 32 - A Complete Resource Book in Chemistry for JEE Main

Question

find \(\Delta \mathrm{G}^{\circ}\) for the reaction given below? \(\frac{1}{2} \mathrm{~A}+\frac{3}{2} \mathrm{~B} \rightleftharpoons \mathrm{C}_{\mathrm{g}}\) \(\mathrm{K}_{e q}=826 \mathrm{~atm}^{-1}\) at \(298 \mathrm{~K}\) (a) \(-8.32 \mathrm{KJ}\) (b) \(8.32 \mathrm{KJ}\) (c) \(16.64 \mathrm{KJ}\) (d) \(-16.64 \mathrm{KJ}\)

Step-by-Step Solution

Answer

The value of \( \Delta G^{\circ} \) is \(-16.64 \text{ kJ/mol}\), which corresponds to option (d).

Step 1: Understand the Gibbs Free Energy Formula

The Gibbs free energy change for a reaction \( \Delta G^{\circ} \) at a constant temperature can be calculated from the equilibrium constant \( K_{eq} \) using the formula: \[ \Delta G^{\circ} = -RT \ln K_{eq} \]where \( R \) is the universal gas constant \( 8.314 \frac{J}{mol \cdot K} \) and \( T \) is the temperature in Kelvin.

Step 2: Convert Constants to Appropriate Units

Before calculating, ensure all constants are in the correct units. The gas constant \( R = 8.314 \frac{J}{mol \cdot K} = 0.008314 \frac{kJ}{mol \cdot K} \). Temperature is already given as \( 298 \) K.

Step 3: Substitute Values into the Formula

Substitute the known values into the Gibbs free energy equation: \[ \Delta G^{\circ} = -(0.008314 \frac{kJ}{mol \cdot K})(298 K) \ln(826) \].

Step 4: Calculate the Natural Logarithm

Calculate \( \ln(826) \) using a calculator. - \( \ln(826) \approx 6.717 \).

Step 5: Complete the Calculation

Now complete the multiplication: \[ \Delta G^{\circ} = -(0.008314 \times 298 \times 6.717) \approx -16.638 \text{kJ/mol} \].

Step 6: Round and Choose the Closest Answer

Round \( \Delta G^{\circ} \approx -16.64 \text{kJ/mol} \). This matches option (d) according to the significant figures of the given choices.

Key Concepts

Gibbs Free Energy

The Gibbs Free Energy (\( \Delta G^{\circ} \)) is a significant concept in thermodynamics that helps predict whether a chemical reaction will occur spontaneously under constant temperature and pressure. Simply put, it tells us about the energy available in a system to do work when temperature and pressure are held constant. The sign of \( \Delta G^{\circ} \) is critical:

• If \( \Delta G^{\circ} \) is negative, the reaction proceeds spontaneously in the forward direction.
• If \( \Delta G^{\circ} \) is positive, the reverse reaction is favored, indicating that the forward reaction is non-spontaneous under the given conditions.
• If \( \Delta G^{\circ} \) is zero, the system is at equilibrium, meaning both reactions occur at the same rate.

In this exercise, we find \( \Delta G^{\circ} \) using the formula \( \Delta G^{\circ} = -RT \ln K_{eq} \). Here, \( R \) stands for the universal gas constant, and \( T \) is the temperature. Understanding this concept is critical in thermodynamics, as it links with other important properties like enthalpy \((\Delta H)\) and entropy \((\Delta S)\). A deeper study of Gibbs Free Energy can reveal even more insights into reaction spontaneity and chemical equilibrium.

Equilibrium Constant

The Equilibrium Constant \( K_{eq} \) is crucial in understanding chemical equilibria. It is a measure derived from the concentrations of reactants and products in a reaction at equilibrium. A large \( K_{eq} \) value, like 826 in this exercise, suggests that the products are favored at equilibrium, thus more product molecules are formed compared to the reactants.

The relationship between \( \Delta G^{\circ} \) and \( K_{eq} \) is given by the equation \( \Delta G^{\circ} = -RT \ln K_{eq} \). This formula helps us determine the favorability of the reaction direction. When \( K_{eq} > 1 \), \( \Delta G^{\circ} \) is negative, meaning the reaction is spontaneous under standard conditions. Conversely, when \( K_{eq} < 1 \), \( \Delta G^{\circ} \) is positive, showing a non-spontaneous process. Hence, \( K_{eq} \) serves as a direct indicator of the direction in which the equilibrium lies. Understanding the equilibrium constant helps chemists predict the yields of reactions, which is vital for industrial chemical processes.

Chemical Equilibrium

Chemical Equilibrium refers to the state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time. While the term 'equilibrium' suggests stillness, it is truly a dynamic balance.

At equilibrium, neither reactants nor products are being used up or generated, leading to what we observe as a stable condition. However, molecules continue to interact and react at molecular levels, maintaining a delicate balance. For the reaction \( \frac{1}{2}\mathrm{~A}+\frac{3}{2} \mathrm{~B} \rightleftharpoons \mathrm{C}_{\mathrm{g}} \), the state of equilibrium is a result of equal formation and decomposition of products and reactants, highlighted by the equilibrium constant \( K_{eq} = 826 \).

Chemical equilibrium ensures no net change in concentrations. Understanding it is crucial for fields such as chemical engineering and materials science, where maintaining specific reaction conditions is a key component of successful operations.

Energy Calculations

Energy Calculations in thermodynamics involve determining the amount of energy associated with chemical processes. Using the Gibbs Free Energy equation, \( \Delta G^{\circ} = -RT \ln K_{eq} \), we can predict the spontaneity of reactions. This formula connects temperature \( T \), the equilibrium constant \( K_{eq} \), and the gas constant \( R \).

Calculating \( \Delta G^{\circ} \) involves substituting these constants and calculating the natural logarithm of \( K_{eq} \). The significance of these calculations is profound—they provide insights into which reactions will occur under standard conditions and how much energy is available from the reaction.

• Use \( R = 8.314 \frac{J}{mol \cdot K} \) or converted to \( 0.008314 \frac{kJ}{mol \cdot K} \).
• Ensure temperature \( T \) is in Kelvin.
• Substitute into the formula to find \( \Delta G^{\circ} \).

Such calculations are fundamental in chemistry, helping design processes that maximize energy efficiency and minimize waste. By understanding Energy Calculations, scientists and engineers can better control and predict reactions for various applications.