At 298 K, the heats of formation of CCl $_{4}$ (g), H $_{2}$ O (g), CO $_{2}$ (g) and HCl (g) are -25, -57, -94 and -22
K.cal mol $^{- 1}$ respectively. The $Δ H$ of the reaction:
$C C l_{4} (g) + 2 H_{2} O \to C O_{2} (g) + 4 H C l (g)$

-43 K.cal

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+72 K.cal

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14 K.cal

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-125 K.cal

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Solution

The correct option is A -43 K.cal
$Δ H_{r e a c t i o n} = Δ H_{f} (C O_{2}) + 4 Δ H_{f} (H C l)$ $- Δ H_{f} (C C l_{4}) - 2 \times Δ H_{f} (H_{2} O)$
$= - 94 + 4 (- 22) - (- 25)$ $- 2 \times (- 57)$
$= - 94 - 88 + 25 + 114$
$=$ -43 Kcal mol $^{- 1}$

Option A is correct.

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