Question

For the reaction ;
$A.(l) \to 2B(g)$
$\Delta U = 2.1\, kcal, \,\Delta S = 20 \,cal\, K^{-1} \,at \,300 \,K.$
Hence $\Delta$G in kcal is__________.

Answer 1

$A\left(\ell\right) \to 2B\left(g\right)$
$\Delta U=2.1 kcal, \Delta s=20 cal ^{K-1} at 300 k$
$\Delta H=\Delta U+\Delta n_{g} RT
\Delta G=\Delta H-T\Delta S$
$\Delta G=\Delta U+\Delta n_{g}RT-T\Delta S$
$=2.1+\frac{2\times2\times300}{1000}-\frac{300\times20}{1000}$
$\left(R=2 cal\, k^{-1}mol^{-1}\right)$
$=2.1+1.2-6=-2.70 \, Kcal/mol$