Question

For a certain reaction at 300 K, \( K = 10 \), then \( \Delta G^\circ \) for the same reaction is \[- \, \_ \times 10^{-1} \, \text{kJ mol}^{-1}.\](Given \( R = 8.314 \, \text{JK}^{-1} \text{mol}^{-1} \))

Answer 1

Correct Answer: 57

Given: 

We are asked to calculate \( \Delta G^\circ \) using the formula:

\[ \Delta G^\circ = -RT \ln K \]

Step 1: Substituting the known values into the equation:

Given that \( R = 8.314 \, \text{J/molK} \), \( T = 300 \, \text{K} \), and \( K = 10 \), we substitute these values into the equation:

\[ \Delta G^\circ = -8.314 \times 300 \ln(10) \]

Step 2: Simplifying the equation:

Calculating the logarithm of 10:

\[ \Delta G^\circ = -8.314 \times 300 \times 2.3026 \]

This simplifies to:

\[ \Delta G^\circ = -5744.14 \, \text{J/mol} \]

Step 3: Converting to kJ/mol:

To express the result in kJ/mol, divide by 1000:

\[ \Delta G^\circ = -5.744 \, \text{kJ/mol} \quad \Rightarrow \quad \Delta G^\circ = 57.44 \times 10^{-1} \, \text{kJ/mol} \]

Final Answer:

The correct answer is \( 57 \, \text{kJ/mol} \).

Answer 2

Step-by-step Calculation:
The standard Gibbs free energy change \( \Delta G^\circ \) for a reaction is related to the equilibrium constant \( K \) by the equation:
\[\Delta G^\circ = -RT \ln K\]
where:
\( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \) (Universal gas constant)
\( T = 300 \, \text{K} \)
\( K = 10 \)
Substituting the values into the equation:
\[\Delta G^\circ = -8.314 \times 300 \times \ln(10)\]
We know that \( \ln(10) \approx 2.303 \).
Therefore:
\[\Delta G^\circ = -8.314 \times 300 \times 2.303\]
\[\Delta G^\circ = -8.314 \times 690.9 \approx -5730 \, \text{J mol}^{-1}\]
Converting to \( \text{kJ mol}^{-1} \):
\[\Delta G^\circ = -5.73 \, \text{kJ mol}^{-1}\]
Expressing in the required format:
\[\Delta G^\circ = -57 \times 10^{-1} \, \text{kJ mol}^{-1}\]
Conclusion: The value of \( \Delta G^\circ \) for the reaction is \( -57 \times 10^{-1} \, \text{kJ mol}^{-1} \).