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Heat of Formation

The Δf H0 for...

Question

The $Δ_{f} H^{0}$ for $C O_{2} (g), C O (g) a n d H_{2} O (g)$ are -393.5, -110.5 and -241.8 $k J m o l^{- 1}$ respectively. The standard enthalpy change (in $k J m o l^{- 1}$ ) for the reaction
$C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (g)$ is :

524.1

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+ 41.2

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-262.5

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- 41.2

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Solution

The correct option is B + 41.2
Given, $Δ_{f} H^{0} (C O_{2}) = - 393.5 k J m o l^{- 1}$
$Δ_{f} H^{0} (C O) = - 110.5 k J m o l^{- 1}$
$Δ_{f} H^{0} (H_{2}) = - 241.8 k J m o l^{- 1}$
$C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} (g)$
$Δ_{r} H$ is related to bond enthalpies of reactants and products in the gas phase reaction as
$Δ_{r} H = \sum Δ_{f} H^{0} (p r o d u c t s) - \sum Δ_{f} H^{0} (r e a c t a n t s) = Δ_{f} H^{0} (C O) + Δ_{f} H^{0} (H_{2}) - Δ_{f} H^{0} (C O_{2}) - Δ_{f} H^{0} (H_{2}) = - 110.5 - 241.8 - (- 393.5) = + 41.20 k J m o l^{- 1}$
hence the standard enthalpy change for the given reaction is $+ 41.20 k J m o l^{- 1}$

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Similar questions

T h e △ H_{f}^{0}

C O_{2} (g), C O (g) a n d H_{2} O a r e - 393.5, - 110.5 a n d - 241.8 k J m o l^{- 1}

C O_{2} + H_{2} ⟶ C O + H_{2} O

T h e △ H_{f}^{0}

C O_{2} (g), C O (g) a n d H_{2} O a r e - 393.5, - 110.5 a n d - 241.8 k J m o l^{- 1}

C O_{2} + H_{2} ⟶ C O + H_{2} O

Δ H_{f}^{o}

C O_{2}

C O

H_{2} O (l)

^{- 1}

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Δ_{f} H^{o}

C O_{2} (g), C O (g)

H_{2} O (l)

k J m o l^{- 1}

k J m o l^{- 1}

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Δ_{f} H^{\circ}

C O_{2} (g), C O (g) a n d H_{2} O (g)

- 393.5, - 110.5

- 241.8 k J m o l^{- 1}

C O_{2} (g) + H_{2} (g) ⟶ C O (g) + H_{2} O (g)

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