Question

For the given reaction:
\[ \mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)} \] The following data table is provided at \(600 \, \mathrm{K}\):
& \(\Delta_f H \, (\mathrm{kJ/mol})\) & \(S_m \, (\mathrm{J\,K^{-1}\,mol^{-1}})\)
\(\mathrm{X_2}\) & 80 & 140
\(\mathrm{Y_2}\) & 8 & 250
\(\mathrm{XY}\) & 42 & 200
\endtabular Calculate \(\Delta_r G\) at \(600 \, \mathrm{K}\).

• A. \(-10 \, \mathrm{kJ/mol}\)
• B. \(-100 \, \mathrm{kJ/mol}\)
• C. \(-2 \, \mathrm{kJ/mol}\)
• D. \(+2 \, \mathrm{kJ/mol}\)

Hint:

For thermodynamics questions, always use \(\Delta G = \Delta H - T\Delta S\). Be careful with units: if \(\Delta H\) is in \(\mathrm{kJ/mol}\) and entropy is in \(\mathrm{J\,K^{-1}\,mol^{-1}}\), convert \(T\Delta S\) into \(\mathrm{kJ/mol}\) before subtraction.

Answer 1

The Correct Option is A

Step 1: Write the formula for Gibbs free energy change.
For a reaction, Gibbs free energy change is calculated by the relation:
\[ \Delta_r G = \Delta_r H - T\Delta_r S \] So, we first need to calculate the reaction enthalpy change \(\Delta_r H\) and the reaction entropy change \(\Delta_r S\), and then substitute \(T = 600 \, \mathrm{K}\).

Step 2: Calculate the enthalpy change of reaction \(\Delta_r H\).
Given reaction is:
\[ \mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)} \] Using the relation:
\[ \Delta_r H = \sum \nu \Delta_f H(\text{products}) - \sum \nu \Delta_f H(\text{reactants}) \] Substituting the given values:
\[ \Delta_r H = \left[2 \times \Delta_f H(\mathrm{XY})\right] - \left[\Delta_f H(\mathrm{X_2}) + \Delta_f H(\mathrm{Y_2})\right] \] \[ = \left[2 \times 42\right] - \left[80 + 8\right] \] \[ = 84 - 88 \] \[ = -4 \, \mathrm{kJ/mol} \] Thus, the reaction enthalpy change is:
\[ \Delta_r H = -4 \, \mathrm{kJ/mol} \]
Step 3: Calculate the entropy change of reaction \(\Delta_r S\).
Using the relation:
\[ \Delta_r S = \sum \nu S_m(\text{products}) - \sum \nu S_m(\text{reactants}) \] Substituting the given values:
\[ \Delta_r S = \left[2 \times S_m(\mathrm{XY})\right] - \left[S_m(\mathrm{X_2}) + S_m(\mathrm{Y_2})\right] \] \[ = \left[2 \times 200\right] - \left[140 + 250\right] \] \[ = 400 - 390 \] \[ = 10 \, \mathrm{J\,K^{-1}\,mol^{-1}} \] Thus,
\[ \Delta_r S = 10 \, \mathrm{J\,K^{-1}\,mol^{-1}} \]
Step 4: Convert entropy term into proper units.
Since \(\Delta_r H\) is in \(\mathrm{kJ/mol}\), the term \(T\Delta_r S\) must also be converted into \(\mathrm{kJ/mol}\).
First calculate:
\[ T\Delta_r S = 600 \times 10 = 6000 \, \mathrm{J/mol} \] Now convert joules to kilojoules:
\[ 6000 \, \mathrm{J/mol} = 6 \, \mathrm{kJ/mol} \] So,
\[ T\Delta_r S = 6 \, \mathrm{kJ/mol} \]
Step 5: Calculate \(\Delta_r G\).
Now use:
\[ \Delta_r G = \Delta_r H - T\Delta_r S \] \[ \Delta_r G = -4 - 6 \] \[ \Delta_r G = -10 \, \mathrm{kJ/mol} \]
Step 6: Compare with the options.
• (A) \(-10 \, \mathrm{kJ/mol}\): Correct. This matches the calculated value. • (B) \(-100 \, \mathrm{kJ/mol}\): Incorrect. This is much larger in magnitude than the actual value. • (C) \(-2 \, \mathrm{kJ/mol}\): Incorrect. This ignores the proper contribution of the entropy term. • (D) \(+2 \, \mathrm{kJ/mol}\): Incorrect. The sign is wrong because the overall Gibbs free energy change is negative.
Step 7: Conclusion.
Therefore, the Gibbs free energy change for the reaction at \(600 \, \mathrm{K}\) is:
\[ \Delta_r G = -10 \, \mathrm{kJ/mol} \] Final Answer: \(-10 \, \mathrm{kJ/mol}\)