MCQ
For the reaction, $X_2O_{4(l)} \to 2XO_{2(g)}$
$\Delta U = 2.1\ kcal, \, \Delta S = 20\, cal\, K^{-1}$ at $300\, K$ Hence, $\Delta G$ is ............ $\mathrm{kcal}$
- A$2.7$
- ✓$- 2.7$
- C$9.3$
- D$- 9.3$
✓
Answer
Correct option: B.
$- 2.7$
b
$\Delta H=\Delta U+\Delta n_{g} R T$
$\Delta H=\Delta U+\Delta n_{g} R T$
Given, $\Delta U=2.1 \,\mathrm{kcal}, \Delta n_{g}=2$
$R=2 \times 10^{-3} \,\mathrm{kcal}, T=300\, \mathrm{K}$
$\therefore \Delta H=2.1+2 \times 2 \times 10^{-3} \times 300=3.3\, \mathrm{kcal}$
Again, $\Delta G=\Delta H-T \Delta S$
Given, $\Delta S=20 \times 10^{-3} \,\mathrm{kcal} \,\mathrm{K}^{-1}$
On putting the values of $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ in the equation.
we get
$\Delta G=3.3-300 \times 20 \times 10^{-3}$
$=3.3-6 \times 10^{3} \times 10^{-3}$
$=-2.7\; \mathrm{kcal}$
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