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(B) The entropy change of the surrounding $(\Delta S_{surr})$ is given by the formula: $\Delta S_{surr} = \frac{-q_{sys}}{T}$.
Given that the reactio

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$1750$

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(B) The entropy change of the surrounding $(\Delta S_{surr})$ is given by the formula: $\Delta S_{surr} = \frac{-q_{sys}}{T}$.
Given that the reaction releases heat,$q_{sys} = -525 \ kJ = -525000 \ J$.
Therefore,the heat absorbed by the surrounding is $q_{surr} = +525000 \ J$.
The temperature $T = 300 \ K$.
Substituting the values: $\Delta S_{surr} = \frac{525000 \ J}{300 \ K} = 1750 \ J \ K^{-1}$.

Calculate the entropy change of the surrounding if $2 \ moles$ of $H_2$ and $1 \ mole$ of $O_2$ gas combine to form $2 \ moles$ of liquid water by releasing $525 \ kJ$ of heat to the surrounding at constant pressure and at $300 \ K$. (in $J \ K^{-1}$)

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