Problem 177 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard Gibbs energy change at \(300 \mathrm{~K}\) for the reaction \(2 \mathrm{~A}=\mathrm{B}+\mathrm{C}\) is \(2494.2 \mathrm{~J}\). At a given time, the composition of the reaction mixture is \(\mathrm{A}=\frac{1}{2}[\mathrm{~B}]=2\) and \([\mathrm{C}]=\frac{1}{2}\) The reaction proceeds in the : \([\mathrm{R}=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol}, \mathrm{e}=2.718\) (a) Forward direction because \(\mathrm{Q}>\mathrm{K}_{\mathrm{C}}\) (b) Reverse direction because \(\mathrm{Q}>\mathrm{K}_{\mathrm{c}}\) (c) Forward direction because \(Q

Step-by-Step Solution

Answer

The reaction proceeds in the reverse direction because \(Q > K_C\).

Step 1: Write the Reaction Quotient Expression

The reaction given is \(2 \mathrm{~A} = \mathrm{B} + \mathrm{C}\). The reaction quotient \(Q\) is expressed as:\[ Q = \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2} \]

Step 2: Plug in Concentrations

Using the provided concentrations, \([\mathrm{A}] = 2\), \([\mathrm{B}] = 4\), and \([\mathrm{C}] = \frac{1}{2}\), calculate \(Q\):\[ Q = \frac{4 \times \frac{1}{2}}{2^2} = \frac{2}{4} = \frac{1}{2} \]

Step 3: Calculate Equilibrium Constant \(K_C\)

Use the standard Gibbs energy change \(\Delta G^0 = 2494.2 \text{ J}\) and the equation:\[ \Delta G^0 = -RT \ln K_C \]Re-arrange to solve for \(K_C\):\[ \ln K_C = -\frac{\Delta G^0}{RT} \]\[ K_C = e^{ -\frac{2494.2}{8.314 \times 300} } \]

Step 4: Calculate \(K_C\) Numerically

Substitute the values \(R = 8.314 \text{ J/K/mol}\), \(T = 300 \text{ K}\) to find \(K_C\):\[ K_C = e^{ -\frac{2494.2}{2494.2}} \approx e^{-1} \approx 0.368 \]

Step 5: Compare \(Q\) and \(K_C\)

Compare the calculated \(Q\) and \(K_C\):- \(Q = \frac{1}{2} = 0.5\)- \(K_C \approx 0.368\)Since \(Q > K_C\), the reaction will proceed in the reverse direction.

Key Concepts

Reaction Quotient

The Reaction Quotient, often denoted as \( Q \), is a way to measure the relative amounts of products and reactants in a chemical reaction mixture at any point in time. It's helpful to predict how the reaction will proceed to reach equilibrium. For the reaction \( 2 \mathrm{A} = \mathrm{B} + \mathrm{C} \), the expression for \( Q \) is given by:\[Q = \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2}\]Plugging in the given concentrations \([\mathrm{A}] = 2\), \([\mathrm{B}] = 4\), and \([\mathrm{C}] = \frac{1}{2}\), we calculate \( Q \) as:- \( Q = \frac{4 \times \frac{1}{2}}{2^2} = \frac{2}{4} = \frac{1}{2} \)This value of \( Q \) is then compared with the equilibrium constant to determine the reaction direction.

Equilibrium Constant

The Equilibrium Constant, represented by \( K_C \), is a fundamental value that indicates the ratio of the concentrations of products to reactants when a reaction is at equilibrium. Unlike \( Q \), which can vary throughout the reaction, \( K_C \) is constant at a given temperature for a particular reaction. For this reaction, \( K_C \) is calculated using the Gibbs free energy change formula:\[ \Delta G^0 = -RT \ln K_C\]Here, \( \Delta G^0 \) is the standard Gibbs free energy change, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

• \( \Delta G^0 = 2494.2 \text{ J} \)
• \( R = 8.314 \text{ J/K/mol} \)
• \( T = 300 \text{ K} \)

Re-arranging gives:\[ \ln K_C = -\frac{2494.2}{8.314 \times 300}\]Calculating this yields a \( K_C \) of approximately 0.368. This value reflects the balance point of the reaction at this temperature.

Gibbs Free Energy

Gibbs Free Energy, noted as \( \Delta G \), provides crucial insight into a reaction's spontaneity and equilibrium position. The standard Gibbs free energy change \( \Delta G^0 \) helps calculate \( K_C \) under standard conditions using:\[ \Delta G^0 = -RT \ln K_C\]For any reaction mixture, \( \Delta G \) relates to \( \Delta G^0 \) and the reaction quotient \( Q \):\[ \Delta G = \Delta G^0 + RT \ln Q\]

• If \( \Delta G < 0 \), the reaction proceeds forward spontaneously.
• If \( \Delta G > 0 \), the reaction tends to move in reverse.
• When \( \Delta G = 0 \), the system is at equilibrium.

Consequently, Gibbs Free Energy not only defines reaction equilibrium but also evaluates whether the reaction will proceed forward or backwards, given the current conditions.

Reaction Direction

To determine the direction in which a reaction proceeds, compare the Reaction Quotient \( Q \) with the Equilibrium Constant \( K_C \). This comparison reveals whether the reaction mixture is "left" or "right" of equilibrium.

• If \( Q < K_C \): The reaction moves forward (to the right) to produce more products.
• If \( Q > K_C \): The reaction proceeds in reverse (to the left) to form more reactants.
• If \( Q = K_C \): The reaction is at equilibrium, and no net movement is observed.

For the exercise given, \( Q \) was calculated to be 0.5, and \( K_C \) was approximately 0.368. Since \( Q > K_C \), excess products are present initially, driving the reaction backwards to form more reactants. Therefore, the reaction proceeds in the reverse direction, aligning with the comparison strategy of \( Q \) and \( K_C \).