For the reaction $2A_{(g)} + B_{(g)} \to 2D_{(g)}$,given $\Delta U^{\theta} = -10.5 \ kJ$ and $\Delta S^{\theta} = -44.1 \ J \ K^{-1}$. Calculate $\Delta G^{\theta}$ for the reaction at $298 \ K$ and predict whether the reaction may occur spontaneously.
(N/A) For the reaction $2A_{(g)} + B_{(g)} \to 2D_{(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = 2 - (2 + 1) = -1 \ mol$.
First,calculate $\Delta H^{\theta}$ using the relation $\Delta H^{\theta} = \Delta U^{\theta} + \Delta n_{g} R T$:
$\Delta H^{\theta} = -10.5 \ kJ + (-1 \ mol)(8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1})(298 \ K)$
$\Delta H^{\theta} = -10.5 \ kJ - 2.477 \ kJ = -12.977 \ kJ$.
Now,calculate $\Delta G^{\theta}$ using $\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta}$:
$\Delta G^{\theta} = -12.977 \ kJ - (298 \ K)(-44.1 \times 10^{-3} \ kJ \ K^{-1})$
$\Delta G^{\theta} = -12.977 \ kJ + 13.142 \ kJ = +0.165 \ kJ$.
Since $\Delta G^{\theta} > 0$,the reaction is non-spontaneous at $298 \ K$.
First,calculate $\Delta H^{\theta}$ using the relation $\Delta H^{\theta} = \Delta U^{\theta} + \Delta n_{g} R T$:
$\Delta H^{\theta} = -10.5 \ kJ + (-1 \ mol)(8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1})(298 \ K)$
$\Delta H^{\theta} = -10.5 \ kJ - 2.477 \ kJ = -12.977 \ kJ$.
Now,calculate $\Delta G^{\theta}$ using $\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta}$:
$\Delta G^{\theta} = -12.977 \ kJ - (298 \ K)(-44.1 \times 10^{-3} \ kJ \ K^{-1})$
$\Delta G^{\theta} = -12.977 \ kJ + 13.142 \ kJ = +0.165 \ kJ$.
Since $\Delta G^{\theta} > 0$,the reaction is non-spontaneous at $298 \ K$.
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