For the reaction ; $A_{(l)} \longrightarrow 2 B_{(g)}$\ $\Delta U=2.1 \mathrm{kcal}, \Delta S=20 \m

Question

For the reaction ; $A_{(l)} \longrightarrow 2 B_{(g)}$\ $\Delta U=2.1 \mathrm{kcal}, \Delta S=20 \mathrm{cal} \mathrm{K}^{-1}$ at 300 K .\ Hence, $\Delta G$ in kcal is $\qquad$ .\

Professor PC Thomas Classes · Answer

(-2.7) : We know that,\ $\Delta H=\Delta U+\Delta n_{g} R T$\ $=2.1 	imes 10^{3}+2(2)(300)$\ $=2100+1200=3300 \mathrm{cal}$\ $\Delta G=\Delta H-T \Delta S$ $$ \begin{aligned} & =3300-(300)(20)=3300-6000 \ & =-2700 \mathrm{cal}=-2.7 \mathrm{kcal} \end{aligned} $$