Calculate the entropy change in the surroundi | vedclass

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(A) The formation of $1 \ mol$ of $H_2O_{(l)}$ releases $286 \ kJ \ mol^{-1}$ of heat,meaning $q_{sys} = -286 \ kJ \ mol^{-1}$.Since the process is ex

Vedclass · Accepted Answer

$959.73 \ J \ K^{-1} \ mol^{-1}$

Vedclass · Answer

(A) The formation of $1 \ mol$ of $H_2O_{(l)}$ releases $286 \ kJ \ mol^{-1}$ of heat,meaning $q_{sys} = -286 \ kJ \ mol^{-1}$.Since the process is exothermic,the surroundings absorb this heat: $q_{surr} = +286 \ kJ \ mol^{-1} = +286000 \ J \ mol^{-1}$.Standard conditions imply a temperature of $T = 298 \ K$.The entropy change for the surroundings is given by $\Delta S_{surr} = \frac{q_{surr}}{T}$.$\Delta S_{surr} = \frac{286000 \ J \ mol^{-1}}{298 \ K} = 959.73 \ J \ K^{-1} \ mol^{-1}$.

Calculate the entropy change in the surroundings when $1.00 \ mol$ of $H_2O_{(l)}$ is formed under standard conditions. Given: $\Delta_f H^{\theta} = -286 \ kJ \ mol^{-1}$.

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