Question

The equilibrium concentrations of the species in the reaction A + B \(\rightleftharpoons\) C + D are 2, 3, 10 and 6 mol L\(^{-1}\) respectively at 300 K. \(\Delta\)G\(^{0}\) for the reaction is
(R = 2 cal/mol K)

• A. \(-13.73\) cal
• B. \(1372.60\) cal
• C. \(-137.26\) cal
• D. \(-1381.80\) cal
• E.

Hint:

Remember that \(\ln(10) \approx 2.303\).
In multiple-choice questions, if \(K_{c}>1\), \(\Delta G^{0}\) must be negative.
Check the units of R (cal vs Joules) to match the options.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to calculate the standard Gibbs free energy change (\(\Delta G^{0}\)) for the given chemical reaction at 300 K.
We are provided with the equilibrium concentrations of reactants and products, the temperature, and the gas constant \(R\).

Step 2: Key Formula or Approach:
The standard Gibbs free energy change is related to the equilibrium constant (\(K_{c}\)) by the formula:
\[ \Delta G^{0} = -RT \ln K_{c} \] Where:
\(R\) is the gas constant (\(2 \text{ cal/mol K}\))
\(T\) is the absolute temperature (\(300 \text{ K}\))
\(K_{c}\) is the equilibrium constant based on molar concentrations.

Step 3: Detailed Explanation:
First, we find the equilibrium constant \(K_{c}\) for the reaction:
\[ \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \] \[ K_{c} = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} \] Substituting the given equilibrium concentrations:
\[ K_{c} = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10 \] Now, calculate \(\Delta G^{0}\):
\[ \Delta G^{0} = -R \times T \times \ln(K_{c}) \] Using the relation \(\ln(x) \approx 2.303 \log_{10}(x)\):
\[ \Delta G^{0} = -2 \times 300 \times 2.303 \times \log_{10}(10) \] Since \(\log_{10}(10) = 1\):
\[ \Delta G^{0} = -600 \times 2.303 \] \[ \Delta G^{0} = -1381.8 \text{ cal} \]

Step 4: Final Answer:
The standard Gibbs free energy change for the reaction is \(-1381.80\) cal.