NEET 2023 Chemistry Gibbs Free Energy Numerical Question

To solve the problem, we need to calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the reaction given the equilibrium concentrations of the species involved.

The reaction is:

$A + B \rightleftharpoons C + D$

with the equilibrium concentrations provided as follows:

• $[A] = 2 \, \text{mol L}^{-1}$
• $[B] = 3 \, \text{mol L}^{-1}$
• $[C] = 10 \, \text{mol L}^{-1}$
• $[D] = 6 \, \text{mol L}^{-1}$

Step 1: Write the expression for the equilibrium constant ($K_{eq}$)

The equilibrium constant $K_{eq}$ for the reaction can be expressed in terms of the equilibrium concentrations:

$K_{eq} = \frac{[C][D]}{[A][B]}$

Substituting the given concentrations into this equation:

$$K_{eq} = \frac{(10)(6)}{(2)(3)} = \frac{60}{6} = 10$$

Step 2: Calculate $\Delta G^\circ$

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation:

$\Delta G^\circ = -RT \ln K_{eq}$

where:

• $R$ is the universal gas constant (given as $2 \, \text{cal/mol K}$),
• $T$ is the absolute temperature (given as $300 \, \text{K}$),
• $K_{eq}$ is the equilibrium constant we just calculated.

Substituting the values:

1. Calculate $\ln K_{eq}$:

   • Since $K_{eq} = 10$: $\ln(10) \approx 2.302$

2. Substitute into the equation for $\Delta G^\circ$:

$$\Delta G^\circ = - (2 \, \text{cal/mol K}) \times (300 \, \text{K}) \times \ln(10)$$ $$\Delta G^\circ = - (2) \times (300) \times (2.302) = - 1381.80 \, \text{cal}$$

Thus, the calculated value for $\Delta G^\circ$ is $-1381.80 \, \text{cal}$.

Conclusion

The correct answer is B) -1381.80 cal.

Clarification of Other Options:

• A) -137.26 cal: This value is incorrect because it does not correspond to our calculated value. It might result from a calculation error or misunderstanding of the logarithmic relationship.
• C) -13.73 cal: This too is incorrect and likely results from miscalculating the logarithm or the Gibbs free energy equation.
• D) 1372.60 cal: This option is also incorrect. It is a positive value which contradicts the expected negative Gibbs free energy change for a spontaneous reaction at equilibrium.

Thus, the final answer to the problem is B) -1381.80 cal.