For the reaction

Question:

For the reaction

$2 \mathrm{~A}_{(g)}+\mathrm{B}_{(g)} \rightarrow 2 \mathrm{D}_{(g)}$

$\Delta U^{\theta}=-10.5 \mathrm{~kJ}$ and $\Delta S^{\ominus}=-44.1 \mathrm{JK}^{-1}$.

Calculate $\Delta G^{\theta}$ for the reaction, and predict whether the reaction may occur spontaneously..

 

Solution:

For the given reaction,

$2 \mathrm{~A}_{(g)}+\mathrm{B}_{(g)} \rightarrow 2 \mathrm{D}_{(g)}$

$\Delta n_{a}=2-(3)$

= –1 mole

Substituting the value of $\Delta U^{\theta}$ in the expression of $\Delta H$ :

$\Delta H^{\theta}=\Delta U^{\theta}+\Delta n_{g} R T$

$=(-10.5 \mathrm{~kJ})-(-1)\left(8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K})$

$=-10.5 \mathrm{~kJ}-2.48 \mathrm{~kJ}$

$\Delta H^{\theta}=-12.98 \mathrm{~kJ}$

Substituting the values of $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ in the expression of $\Delta G^{\theta}$ :

$\Delta G^{\theta}=\Delta H^{\theta}-T \Delta S^{\theta}$

$=-12.98 \mathrm{~kJ}-(298 \mathrm{~K})\left(-44.1 \mathrm{JK}^{-1}\right)$

$=-12.98 k J+13.14 k J$

$\Delta G^{\theta}=+0.16 \mathrm{~kJ}$

Since $\Delta G^{\theta}$ for the reaction is positive, the reaction will not occur spontaneously.

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