For the reaction;

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:

$(-2.70)$

$\Delta \mathrm{U}=2.1 \mathrm{kcal}=2.1 \times 10^{3} \mathrm{cal}$

$\Delta \mathrm{n}_{\mathrm{g}}=2$

$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta n_{\mathrm{g}} \mathrm{RT}$

$=2.1 \times 10^{3}+2 \times 2 \times 300$

$=2100+1200$

$=3300 \mathrm{cal}$

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

$=3300-300 \times 20$

$=3300-6000$

$=-2700 \mathrm{cals}$

$=-2.7 \mathrm{kcal}$

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