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Chemistry · JEE | Thermodynamics (Chemical Thermodynamics)

हिन्दी

Calculate 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 \( 298 \text{ K} \).

  1. A
    -5530 J K⁻¹ mol⁻¹
  2. B
    +5530 J K⁻¹ mol⁻¹
  3. C
    +5.53 J K⁻¹ mol⁻¹
  4. D
    -5.53 J K⁻¹ mol⁻¹
View Answer & Solution

Correct answer · Option B

+5530 J K⁻¹ mol⁻¹

Explanation

Using \( \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} = \frac{1648000 \text{ J mol}^{-1}}{298 \text{ K}} = +5530 \text{ J K}^{-1} \text{ mol}^{-1} \).
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