The standard enthalpies of formation at 300 K for ${CCl}_{4} (l), H_{2} O (g), {CO}_{2} (g)$ and $HCl (g)$ are $- 107, - 242, - 394$ and $- 93 kJ {mol}^{- 1},$ respectively. The $Δ U^{\circ}$ at 300 K for the reaction ${CCl}_{4} (l) + 2 H_{2} O (g) \to {CO}_{2} (g) + 4 HCl (g)$ is

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$- 175 kJ {mol}^{- 1}$

$- 170 kJ {mol}^{- 1}$

$- 182.5 kJ {mol}^{- 1}$

$182.5 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

For the given equation

$\begin{array}{l} Δ H = Δ_{f} H ({CO}_{2}, g) + 4 Δ_{f} H (HCl, g) - Δ_{f} H ({CCl}_{4}, l) - 2 Δ_{f} H (H_{2} O, g) \\ = [- 394 + 4 (- 93) - (- 107) - 2 (- 242)] kJ {mol}^{- 1} \\ = - 175 kJ {mol}^{- 1} \\ Δ U = Δ H - (Δ v_{g}) R T = - 175 kJ {mol}^{- 1} - (3) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1}) (300 K) \\ = - 175 kJ {mol}^{- 1} - 7.48 kJ {mol}^{- 1} = - 182.5 kJ {mol}^{- 1} \end{array}$