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

Question

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

Step-by-Step Solution

Answer

ΔG° is -16.64 KJ, which matches option (d).

Step 1: Understand the Relationship between ΔG° and K_eq

The change in Gibbs free energy (ΔG°) for a reaction at standard conditions is related to the equilibrium constant (K_eq) at a given temperature by the equation:\[\Delta G^{\circ} = -RT \ln{K_{eq}}\]where \(R\) is the universal gas constant \(8.314 \text{ J/mol K}\) and \(T\) is the temperature in Kelvin.

Step 2: Convert Units for ΔG°

Since we need to express ΔG° in kilojoules (KJ), we will multiply the result from the ΔG° equation by 0.001 to convert from joules to kilojoules as our gas constant is in joules.

Step 3: Insert Known Values into the Equation

Substitute the known values into the equation. Given:- \(R = 8.314 \text{ J/mol K}\)- \(T = 298 \text{ K}\)- \(K_{eq} = 826 \)Thus, the equation becomes:\[\Delta G^{\circ} = -(8.314 \text{ J/mol K}) \times 298 \text{ K} \times \ln{826}\]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm of 826:\[\ln{826} \approx 6.716\]

Step 5: Calculate ΔG° Using the Equation

Replace \(\ln{826}\) in the equation and calculate:\[\Delta G^{\circ} = -(8.314) \times 298 \times 6.716\]Calculate step-by-step:\[\Delta G^{\circ} = -8.314 \times 298 = -2477.772\]Then,\[-2477.772 \times 6.716 = -16640.7968 \text{ J/mol}\]

Step 6: Convert J/mol to KJ/mol

Since we need \(\Delta G^{\circ}\) in KJ/mol, convert the calculated value from joules to kilojoules:\[-16640.7968 \text{ J/mol} \times 0.001 = -16.64 \text{ KJ/mol}\]

Step 7: Verify the Correct Answer

The value calculated is \(-16.64 \text{ KJ/mol}\). Look at the multiple-choice options and find that the correct answer is option (d), which corresponds to \(-16.64 \text{ KJ/mol}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K_{eq} \), is a key concept in chemical thermodynamics. It provides a snapshot of the balance between the products and reactants in a reversible chemical reaction at a given temperature. Essentially, it tells us which direction the reaction favors under standard conditions. For instance:

• If \( K_{eq} > 1 \), the reaction tends to favor the formation of products. More products are formed at equilibrium.
• If \( K_{eq} < 1 \), the reaction leans towards reactants, indicating that less of the product is formed.
• If \( K_{eq} = 1 \), the concentrations of products and reactants are roughly equal, suggesting a balanced reaction.

Understanding \( K_{eq} \) is crucial because it links directly to Gibbs Free Energy \( (\Delta G^\circ) \) through the equation \( \Delta G^\circ = -RT \ln K_{eq} \) . A positive \( \Delta G^\circ \) suggests a non-spontaneous reaction under standard conditions, meanwhile a negative \( \Delta G^\circ \) indicates a spontaneous process.

Natural Logarithm

The natural logarithm, represented by \( \ln \), is a mathematical function that is frequently used in calculus and applications of chemistry and physics. Unlike other logarithms, which might base off 10 or another integer, the natural logarithm uses Euler's number (approximately 2.718) as its base. This special logarithm is powerful because of its relationship with growth processes, decay, and many natural phenomena.
The connection to the Gibbs Free Energy equation arises because the distribution of particles across energy levels often follows an exponential pattern. Consequently, when determining the change in Gibbs Free Energy for a reaction, you use \( \ln K_{eq} \). This value helps capture the relation between reaction equilibrium and the spontaneity of a reaction.
For instance, in the exercise, calculating \( \ln{826} \) resulted in approximately 6.716, which played a critical role in determining \( \Delta G^\circ \). Understanding how to efficiently compute the natural logarithm enables you to solve thermodynamic equations involved in chemical equilibria.

Universal Gas Constant

The Universal Gas Constant, symbolized as \( R \), is a fundamental constant in chemistry and physics. It shows up in a variety of equations, making it one of the most versatile constants. Its value is 8.314 J/mol K, representing how it can connect energy (manifests in joules)with molar quantities (moles) and temperature (Kelvin).
In the context of Gibbs Free Energy calculations, \( R \) is used in the equation \( \Delta G^\circ = -RT \ln K_{eq} \). This illustrates how energy change in a reaction (represented as \( \Delta G^\circ \)) is related to temperature and the equilibrium state of the molecules involved.

• Temperature Dependency: Increasing temperature can affect \( \Delta G^\circ \), thereby influencing the spontaneity of a reaction.
• Energy Measurements: Utilizing the value of \( R \) as a bridge between different units ensures consistency in calculations.

The universal applicability of \( R \) makes it an indispensable part of understanding how reactions behave under varying conditions, linking temperature, pressure, and quantities of gas into a cohesive framework.