Question:
At 300 K and constant pressure, the following data is obtained for the reaction:
\[
4M(s) + 3O_2(g) \rightarrow 2M_2O_3(s)
\]
Given:
\[
\Delta H^\circ = -1548 \, \text{kJ mol}^{-1}, \quad \Delta S_{\text{sys}} = -550 \, \text{J K}^{-1}\text{mol}^{-1}
\]
What is the value of $\Delta S_{\text{surr}}$ in J K$^{-1}$ mol$^{-1}$?
At 300 K and constant pressure, the following data is obtained for the reaction:
\[
4M(s) + 3O_2(g) \rightarrow 2M_2O_3(s)
\]
Given:
\[
\Delta H^\circ = -1548 \, \text{kJ mol}^{-1}, \quad \Delta S_{\text{sys}} = -550 \, \text{J K}^{-1}\text{mol}^{-1}
\]
What is the value of $\Delta S_{\text{surr}}$ in J K$^{-1}$ mol$^{-1}$?
Show Hint
For any process: \(\Delta S_{\text{surr}} = -\Delta H/T\) at constant pressure and temperature.
Updated On: Jul 18, 2026
- $+$4644
- $+$5160
- $-$4644
- $-$5160
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understand the concept of surroundings entropy change.
The entropy change of the surroundings is directly related to the enthalpy change of the system at constant pressure. This comes from the thermodynamic relation that heat released or absorbed by the system is exchanged with the surroundings. Hence, surroundings entropy depends on enthalpy change and temperature.
Step 2: Write the fundamental relation.
At constant pressure, the entropy change of surroundings is given by: \[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} \] The negative sign indicates that heat lost by the system is gained by the surroundings and vice versa.
Step 3: Convert enthalpy into consistent units.
Given: \[ \Delta H^\circ = -1548 \, \text{kJ mol}^{-1} \] Convert into joules: \[ -1548 \times 10^3 = -1548000 \, \text{J mol}^{-1} \]
Step 4: Substitute values into formula.
Temperature: \[ T = 300 \, K \] So, \[ \Delta S_{\text{surr}} = -\frac{-1548000}{300} \]
Step 5: Perform numerical calculation.
\[ \Delta S_{\text{surr}} = \frac{1548000}{300} = 5160 \, \text{J K}^{-1}\text{mol}^{-1} \] The positive sign indicates that the surroundings gain entropy because the reaction is exothermic.
Step 6: Final conclusion.
Thus, the entropy change of surroundings is: \[ \boxed{+5160 \, \text{J K}^{-1}\text{mol}^{-1}} \]
The entropy change of the surroundings is directly related to the enthalpy change of the system at constant pressure. This comes from the thermodynamic relation that heat released or absorbed by the system is exchanged with the surroundings. Hence, surroundings entropy depends on enthalpy change and temperature.
Step 2: Write the fundamental relation.
At constant pressure, the entropy change of surroundings is given by: \[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} \] The negative sign indicates that heat lost by the system is gained by the surroundings and vice versa.
Step 3: Convert enthalpy into consistent units.
Given: \[ \Delta H^\circ = -1548 \, \text{kJ mol}^{-1} \] Convert into joules: \[ -1548 \times 10^3 = -1548000 \, \text{J mol}^{-1} \]
Step 4: Substitute values into formula.
Temperature: \[ T = 300 \, K \] So, \[ \Delta S_{\text{surr}} = -\frac{-1548000}{300} \]
Step 5: Perform numerical calculation.
\[ \Delta S_{\text{surr}} = \frac{1548000}{300} = 5160 \, \text{J K}^{-1}\text{mol}^{-1} \] The positive sign indicates that the surroundings gain entropy because the reaction is exothermic.
Step 6: Final conclusion.
Thus, the entropy change of surroundings is: \[ \boxed{+5160 \, \text{J K}^{-1}\text{mol}^{-1}} \]
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