Question:

Calculate the entropy change of surrounding if 2 moles of \( \text{H}_2 \) and 1 mole of \( \text{O}_2 \) gas combine to form 2 moles of liquid water by releasing 525 kJ heat to surrounding at constant pressure and at 300 K.

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Surroundings gain entropy when the system releases heat (exothermic).
Updated On: May 14, 2026
  • \( 1700 \text{ J K}^{-1} \)
  • \( 1750 \text{ J K}^{-1} \)
  • \( 1800 \text{ J K}^{-1} \)
  • \( 1650 \text{ J K}^{-1} \)
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The Correct Option is B

Solution and Explanation


Step 1: Concept

Entropy change of the surroundings (\(\Delta S_{surr}\)) is the heat absorbed by the surroundings divided by the temperature.

Step 2: Meaning

Since heat is released by the system, it is absorbed by the surroundings (\(Q_{surr} = -Q_{sys}\)).

Step 3: Analysis

\(\Delta S_{surr} = \frac{Q_{surr}}{T} = \frac{525 \text{ kJ}}{300 \text{ K}} = \frac{525000 \text{ J}}{300 \text{ K}}\).

Step 4: Conclusion

\(\Delta S_{surr} = 1750 \text{ J K}^{-1}\). Final Answer: (B)
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