Question:
For the reaction;
$\mathrm{A}(\mathrm{l}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})$
$\Delta U=2.1 \mathrm{kcal}, \Delta S=20 \mathrm{cal} K^{-1}$ at $300 \mathrm{~K}$
Hence $\Delta \mathrm{G}$ in $\mathrm{kcal}$ is ___________________ .
Solution:
$(-2.70)$
$\Delta \mathrm{U}=2.1 \mathrm{kcal}=2.1 \times 10^{3} \mathrm{cal}$
$\Delta \mathrm{n}_{\mathrm{g}}=2$
$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta n_{\mathrm{g}} \mathrm{RT}$
$=2.1 \times 10^{3}+2 \times 2 \times 300$
$=2100+1200$
$=3300 \mathrm{cal}$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
$=3300-300 \times 20$
$=3300-6000$
$=-2700 \mathrm{cals}$
$=-2.7 \mathrm{kcal}$