Question
Consider the following reaction:
$$\begin{matrix} & 2\text{ }A(\text{ }g) + B(\text{ }g) \rightarrow 2D(\text{ }g) \\ & \end{matrix}$$
$${\Delta U^{\ominus} = - 10\text{ }kJ{\text{ }mol}^{- 1}\text{~}\text{and}\text{~}\Delta S^{\ominus} = - 44{JK}^{- 1}\text{~}\text{at}\text{~}298\text{ }K. }{ Identify\ the\ correct\ option\ with\ \Delta G^{\ominus}}$$ for the reaction and spontaneity of the reaction at 298 K .
(Given: $R = 8.31\text{ }J{\text{ }mol}^{- 1}{\text{ }K}^{- 1}$ )
Options
Solution
Sol. [Key Concept: Gibbs energy ΔG = ΔH − TΔS decides spontaneity]
$${2\text{ }A(\text{ }g) + B(g) \rightarrow 2D(g) } \begin{matrix} \Delta U^{\circ} & \ = - 10\text{ }kJ/mol \\ \Delta{\text{ }S}^{\circ} & \ = - 44\text{ }J/K \\ \Delta H^{\circ} & \ = \Delta U^{\circ} + \Delta n_{g}RT \\ & \ \\ & \ \\ & \ \\ & \ \\ & \ \end{matrix}$$
$$\Delta H^{\circ} = - 10 - \frac{1 \times 298 \times (8.31)}{1000} = - 10 - 2.48 = - 12.48\text{ }kJ/mol$$
$$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta{\text{ }S}^{\circ}$$
$$= - 12.48 - \frac{298 \times ( - 44)}{1000}$$
$$= - 12.48 + 13.112$$
$${= + 0.632\text{ }kJ/mol }{ Since\ \Delta G^{\circ}}$$ comes out to be positive, so given process is non-spontaneous.
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