Question

For the reaction, $X_2{O}_{4\left(l\right)}\to 2X{O}_{2\left(g\right)}$

$\Delta U=2.1kcal,\Delta S=20cal{K}^{-1}at300K$Hence, $\Delta G$ is:

• A. $2.7kcal$
• B. $-2.7kcal$
• C. $9.3kcal$
• D. $-9.3kcal$

Answer 1

The correct option is B $-2.7kcal$

$\Delta H=\Delta U+\Delta n_{g}RT$
Given,
$\Delta U=2.1kcal,\Delta n_g=2,$
$R=2\times {10}^{-3}kcal,T=300K$
$\therefore \Delta H=2.1+2\times 2\times {10}^{-3}\times 300=3.3kcal$
Now,
$\Delta G=\Delta H-T\Delta S$
Given,
$\Delta S=20\times {10}^{-3}kcal{K}^{-1}$

On putting the values of $\Delta H$ and $\Delta S$ in the equation,

we get

$\Delta G=3.3-300\times 20\times {10}^{-3}$
$\Delta G=3.3-6\times {10}^3\times {10}^{-3}$
$\Delta G=-2.7kcal$