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Find $Δ G^{\circ}$ and $Δ H^{\circ}$ for the reaction
$CO (g) + \frac{1}{2} O_{2} (g) ⟶ {CO}_{2} (g)$ at $300 K$
When the standard entropy is $- 0.094 kJ {mol}^{- 1} K^{- 1}$ . The standard Gibbs free energies of formation for ${CO}_{2}$ and $CO$ are $- 394.4$ and $- 137.2 kJ {mol}^{- 1}$ respectively.

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$\begin{aligned} Δ G^{\circ} = 257.2 kJ / mol, Δ H^{\circ} = 285.4 kJ / mol \end{aligned}$

$Δ G^{\circ} = 514.4 kJ / mol, Δ H^{\circ} = - 570.8 kJ / mol$

$\begin{aligned} Δ G^{\circ} = + 514.4 kJ / mol, Δ H^{\circ} = 570.8 kJ / mol \end{aligned}$

$Δ G^{\circ} = - 257.2 kJ / mol, Δ H^{\circ} = - 285.4 kJ / mol$

answer is D.

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Detailed Solution

$Δ G_{R x^{n}}^{\circ} = - 394.4 + 137.2 Δ G_{R x^{n}}^{\circ} = - 257.2 kJ / mol$

$Δ H^{\circ} = - 257.2 - 300 \times 0.094 Δ H^{\circ} = - 285.4 kJ / mol$

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