What is the change in entropy of the surround | vedclass

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(A) The change in entropy of the surroundings is given by the formula: $\Delta S_{	ext{surr}} = -\frac{\Delta H_{	ext{sys}}}{T}$.Given: $\Delta H_{\

Vedclass · Accepted Answer

$959.7 	ext{ J K}^{-1} 	ext{ mol}^{-1}$

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(A) The change in entropy of the surroundings is given by the formula: $\Delta S_{	ext{surr}} = -\frac{\Delta H_{	ext{sys}}}{T}$.Given: $\Delta H_{	ext{sys}} = -286 	ext{ kJ mol}^{-1} = -286000 	ext{ J mol}^{-1}$ and $T = 298 	ext{ K}$.Substituting the values: $\Delta S_{	ext{surr}} = -\frac{-286000 	ext{ J mol}^{-1}}{298 	ext{ K}}$.$\Delta S_{	ext{surr}} = \frac{286000}{298} 	ext{ J K}^{-1} 	ext{ mol}^{-1} \approx 959.73 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.Thus, the correct option is $A$.

What is the change in entropy of the surroundings for the reaction, $H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l)$ at $298 \text{ K}$, if the standard enthalpy of formation of water is $-286 \text{ kJ mol}^{-1}$?

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