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Q.
For the reaction, $X_{2} O _{4}(l) \longrightarrow 2 XO _{2}(g)$, $\Delta U=2.1\, kcal , \Delta S=20\, cal\, K ^{-1}$ at $300\, K$. Hence, $\Delta G$ is
Thermodynamics
Solution:
As we know that,
$\Delta H=\Delta U+\Delta n_{g} R T$
where, $\Delta U=$ change in internal energy
$\Delta n_{g}=$ number of moles of gaseous products
- number of moles of gaseous reactants $=2-0=2$
$R=$ gas constant $=2\, cal$
But, $\Delta U=2.1\, kcal$
$=2.1 \times 10^{3} cal\,\,\left[\because 1\, kcal =10^{3} cal \right]$
$\therefore \Delta H=\left(2.1 \times 10^{3}\right)+(2 \times 2 \times 300)=3300\, cal$
Now, $\Delta G=\Delta H-T \Delta S$
$\Rightarrow \Delta G=(3300)-(300 \times 20)$
$\Rightarrow \Delta G=-2700\, cal$
$\therefore \Delta G=-2.7\, kcal$