Chemistry · JEE | Thermodynamics (Chemical Thermodynamics)
हिन्दीCalculate the standard entropy change of the surroundings (\( \Delta S_{\text{surr}} \) ) when a reaction releases \( 1648 \text{ kJ mol}^{-1} \) of heat to the surroundings at \( 298 \text{ K} \).
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Correct answer · Option B
+5530 J K⁻¹ mol⁻¹
Explanation
Using \( \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} = \frac{1648000 \text{ J mol}^{-1}}{298 \text{ K}} = +5530 \text{ J K}^{-1} \text{ mol}^{-1} \).