Calculate Delta S_{total} for the following r | vedclass

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(A) The entropy change of the surroundings is given by $\Delta S_{surr} = \frac{-\Delta H}{T}$.
Given $\Delta H = 28.1 \ kJ \ mol^{-1} = 28100 \ J \

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$15.1 \ J \ K^{-1} \ mol^{-1}$

Vedclass · Answer

(A) The entropy change of the surroundings is given by $\Delta S_{surr} = \frac{-\Delta H}{T}$.
Given $\Delta H = 28.1 \ kJ \ mol^{-1} = 28100 \ J \ mol^{-1}$ and $T = 300 \ K$.
$\Delta S_{surr} = \frac{-28100 \ J \ mol^{-1}}{300 \ K} = -93.67 \ J \ K^{-1} \ mol^{-1}$.
The total entropy change is $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
$\Delta S_{total} = 108.7 \ J \ K^{-1} \ mol^{-1} + (-93.67 \ J \ K^{-1} \ mol^{-1}) = 15.03 \ J \ K^{-1} \ mol^{-1}$.
Rounding to one decimal place,we get $15.1 \ J \ K^{-1} \ mol^{-1}$.

Calculate $\Delta S_{total}$ for the following reaction at $300 \ K$.
$NH_4NO_{3(s)} \longrightarrow NH_4^+{(aq)} + NO_3^-{(aq)}$ $(\Delta H = 28.1 \ kJ \ mol^{-1}, \Delta S_{sys} = 108.7 \ J \ K^{-1} \ mol^{-1})$

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Calculate $\Delta S_{total}$ for the following reaction at $300 \ K$. $NH_4NO_{3(s)} \longrightarrow NH_4^+{(aq)} + NO_3^-{(aq)}$ $(\Delta H = 28.1 \ kJ \ mol^{-1}, \Delta S_{sys} = 108.7 \ J \ K^{-1} \ mol^{-1})$