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Heat of Reaction

The Δ Ht∘ f...

Question

The $Δ H_{t}^{\circ}$ for $C O_{2 (g)}, C O_{(g)}$ and $H_{2} O_{(g)}$ are $- 393.5, - 110.5$ and $- 241.8 k J m o l^{- 1}$ respectively the standard enthalpy change (in $K J$ ) for the reaction $C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}$ is:

A

524.1

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B

41.2

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C

262.5

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D

- 41.2

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Solution

The correct option is B $41.2$
Given that,
$\begin{matrix} C O_{2}, Δ H = - 393.5 k J C O, Δ H = - 110.5 k J H_{2} O, Δ H = - 241.8 k J C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)} \end{matrix}$
$Δ H^{0}$ for the reaction $C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}$
$Δ H^{0} = Δ H^{0} f$ (product) $- Δ H^{0} f$ (reaction)
$C O_{2}, Δ H = - 393.5 k J$
$C O, Δ H^{0} = Δ H^{0} f$
$H_{2} O, Δ H = - 241.8 k J$
$C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}$
$Δ H^{0} = [Δ H^{0} f (C O) + Δ H^{0} f (H_{2} O)] - [Δ H^{0} f (C O_{2}) + Δ H^{0} f (H_{2})]$
$Δ H = [(- 110.5) + (- 241.8)] - [(- 393.5) + 0]$
$Δ H = [- 352.3] - [- 393.5]$
$Δ H = 41.2 k J$
$= 41.2$
Then,
OPtion $B$ is correct answer.

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