Question

If the standard enthalpy change (\(\Delta H^o\)) for a certain reaction at 298 K and constant pressure is -1860 kJ mol\(^{-1}\), the standard entropy change (\(\Delta S^o\)) of the same reaction is -550 J K\(^{-1}\) mol\(^{-1}\), which of the following statements is correct?

• A. \( \Delta S^o + \Delta H^o = -7692 \, \text{J mol}^{-1} \, \text{K}^{-1} \), the reaction is spontaneous
• B. \( \Delta S^o + \Delta H^o = -7692 \, \text{J mol}^{-1} \, \text{K}^{-1} \), the reaction is non-spontaneous
• C. \( \Delta S^o + \Delta H^o = +5692 \, \text{J mol}^{-1} \, \text{K}^{-1} \), the reaction is spontaneous
• D. \( \Delta S^o + \Delta H^o = +5692 \, \text{J mol}^{-1} \, \text{K}^{-1} \), the reaction is non-spontaneous

Hint:

For spontaneity, check if the Gibbs free energy (\(\Delta G^o\)) is negative. Use \(\Delta G^o = \Delta H^o - T \Delta S^o\) to calculate.

Answer 1

The Correct Option is C

We are given: \[ \Delta H^o = -1860 \, \text{kJ/mol} = -1860 \times 10^3 \, \text{J/mol} \] and \[ \Delta S^o = -550 \, \text{J/mol/K} \] To find the free energy change (\(\Delta G^o\)) at 298 K, we use the following relation: \[ \Delta G^o = \Delta H^o - T \Delta S^o \] Substituting the given values: \[ \Delta G^o = (-1860 \times 10^3) - 298 \times (-550) \] \[ \Delta G^o = -1860 \times 10^3 + 163900 \] \[ \Delta G^o = -1694900 \, \text{J/mol} \] The reaction is spontaneous if \(\Delta G^o\) is negative, which it is, hence the reaction is spontaneous. Thus, the correct statement is option (3).

Answer 2

Step 1: Understand the given data.
- Standard enthalpy change, \(\Delta H^o = -1860 \, \text{kJ mol}^{-1} = -1,860,000 \, \text{J mol}^{-1}\)
- Standard entropy change, \(\Delta S^o = -550 \, \text{J K}^{-1} \text{mol}^{-1}\)
- Temperature, \(T = 298 \, \text{K}\)

Step 2: Calculate the Gibbs free energy change \(\Delta G^o\).
\[ \Delta G^o = \Delta H^o - T \Delta S^o \] Substitute values:
\[ \Delta G^o = (-1,860,000) - 298 \times (-550) = -1,860,000 + 163,900 = -1,696,100 \, \text{J mol}^{-1} \]

Step 3: Calculate \(\Delta S^o + \frac{\Delta H^o}{T}\).
\[ \Delta S^o + \frac{\Delta H^o}{T} = -550 + \frac{-1,860,000}{298} = -550 - 6,242.95 = -6,792.95 \, \text{J mol}^{-1} \text{K}^{-1} \] (Note: This does not match the given answer, so the problem statement might have a typographical or interpretation error.)

Step 4: Conclusion.
- Given \(\Delta H^o < 0\) and \(\Delta S^o < 0\), the spontaneity depends on temperature.
- Since \(\Delta G^o\) is very negative, the reaction is spontaneous at 298 K.
- The sum \(\Delta S^o + \Delta H^o\) as provided in the question seems inconsistent with the data.

Final note: The reaction is spontaneous at 298 K due to negative \(\Delta G^o\).