Calculate the entropy change $(\Delta S)$ per mole for the following reactions:\ (i) Combustion of

Question

Calculate the entropy change $(\Delta S)$ per mole for the following reactions:\ (i) Combustion of hydrogen in a fuel cell at 298 K 
$ \begin{aligned} & \mathrm{H}_{2(g)}+1 / 2 \mathrm{O}_{2(g)} ightarrow \mathrm{H}_{2} \mathrm{O}_{(g)} \ & \Delta H=-241.6 \mathrm{~kJ}, \Delta G=-228.4 \mathrm{~kJ} \end{aligned} $
(ii) Vaporisation of methanol at its normal boiling point, $\mathrm{CH}_{3} \mathrm{OH}_{(l)} ightarrow \mathrm{CH}_{3} \mathrm{OH}_{(g)} \Delta H_{	ext {vap }}=23.9 \mathrm{~kJ}$, boiling point $=338 \mathrm{~K}$\ (iii) Fusion of ice at its normal melting point, Ice → Water\ $\Delta H_{	ext {fusion }}=6.025 \mathrm{~kJ}$, melting point $=0^{\circ} \mathrm{C}$\

Professor PC Thomas Classes · Answer

(a) : (i) According to the reaction,
$
\Delta G=\Delta H-T \Delta S 	ext { or } \Delta S=\frac{\Delta H-\Delta G}{T}
$
Given that $\Delta G=-228.4 \mathrm{~kJ}, \Delta H=-241.6 \mathrm{~kJ}$ and $T=298 \mathrm{~K}$
Substituting the values, we get
$
\begin{aligned}
& \Delta S=\frac{(-241.6 \mathrm{~kJ})-(-228.4 \mathrm{~kJ})}{298 \mathrm{~K}}=\frac{-132}{298} \
& =-0.443 \mathrm{~kJ} \mathrm{~K}^{-1}=-443 \mathrm{~J} \mathrm{~K}^{-1}
\end{aligned}
$
(ii) Vaporisation represents a state of equilibrium between the liquid and the vapour state of the substance. At this state, $\Delta G=0$, so,
$
\Delta H=T \Delta S 	ext { or } \Delta S=\frac{\Delta H}{T}
$
Here, $\Delta H=23.9 \mathrm{~kJ} ; T=338 \mathrm{~K}$
$
	herefore \quad \Delta S=\frac{23.9 \mathrm{~kJ}}{338 \mathrm{~K}}=0.070 \mathrm{~kJ} \mathrm{~K}^{-1}=70 \mathrm{JK}^{-1}
$
(iii) Fusion represents a state of equilibrium between ice and water. At this state, $\Delta G=0$ so,
$
\Delta H=T \Delta S 	ext { or } \Delta S=\frac{\Delta H}{T}
$
Here, $\Delta H=6.025 \mathrm{~kJ}$ and $T=273 \mathrm{~K}$
$
\Delta S=\frac{6.025 \mathrm{~kJ}}{273 \mathrm{~K}}=0.0220 \mathrm{~kJ} \mathrm{~K}^{-1}=22 \mathrm{~J} \mathrm{~K}^{-1}
$