Question

Consider the following reaction A(g) + 3B(g) \( \longrightarrow \) 2C(g); \( \Delta H^\ominus = -24 \) kJ. At \(25 \, ^\circ\text{C}\) if \( \Delta G^\ominus \) of the reaction is -9 kJ, the standard entropy change (in JK\(^{-1}\)) of the same reaction at same temperature is

• A. -5.33
• B. -50.33
• C. -500.33
• D. -0.533

Hint:

Gibbs-Helmholtz equation: \( \Delta G = \Delta H - T\Delta S \). For standard conditions: \( \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \). Rearrange to find \( \Delta S^\ominus = \frac{\Delta H^\ominus - \Delta G^\ominus}{T} \). Ensure units are consistent: - \( \Delta H^\ominus \) and \( \Delta G^\ominus \) are usually in kJ/mol or J/mol. - \( T \) must be in Kelvin (K). \( T(K) = T(^\circ C) + 273.15 \) (or 273). - \( \Delta S^\ominus \) will then be in J/mol��K or kJ/mol��K. Question asks for JK\(^{-1}\), implying per mole if not specified (standard changes usually are).

Answer 1

The Correct Option is B

The relationship between Gibbs free energy change (\( \Delta G^\ominus \)), enthalpy change (\( \Delta H^\ominus \)), and entropy change (\( \Delta S^\ominus \)) at a constant temperature T is given by the Gibbs-Helmholtz equation: \[ \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \] Given: \( \Delta H^\ominus = -24 \) kJ \( = -24000 \) J.
\( \Delta G^\ominus = -9 \) kJ \( = -9000 \) J.
Temperature \( T = 25 \, ^\circ\text{C} = 25 + 273.
15 = 298.
15 \) K.
(Using 298 K for simplicity is common if precision matches options).
Let's use T = 298 K.
We need to find \( \Delta S^\ominus \) in JK\(^{-1}\).
Rearrange the equation to solve for \( \Delta S^\ominus \): \[ T\Delta S^\ominus = \Delta H^\ominus - \Delta G^\ominus \] \[ \Delta S^\ominus = \frac{\Delta H^\ominus - \Delta G^\ominus}{T} \] Substitute the values: \[ \Delta S^\ominus = \frac{(-24000 \, \text{J}) - (-9000 \, \text{J})}{298 \, \text{K}} \] \[ \Delta S^\ominus = \frac{-24000 + 9000}{298} \, \text{J K}^{-1} \] \[ \Delta S^\ominus = \frac{-15000}{298} \, \text{J K}^{-1} \] Calculate \( \frac{-15000}{298} \): \( 15000 \div 298 \approx 15000 \div 300 = 50 \).
More precisely: \( 15000 / 298 \approx 50.
33557.
.
.
\) So, \( \Delta S^\ominus \approx -50.
33557.
.
.
\, \text{J K}^{-1} \).
Rounding to two decimal places, \( \Delta S^\ominus \approx -50.
34 \, \text{J K}^{-1} \).
Option (2) is -50.
33.
This is very close.
The slight difference might be from using T=298.
15 K or rounding.
If T=298.
15 K: \( \Delta S^\ominus = \frac{-15000}{298.
15} \approx -50.
309.
.
.
\, \text{J K}^{-1} \).
This is also very close to -50.
33.
The options suggest the calculation intended \(15000/298 \approx 50.
33\).