Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

$- 890 k J m o l^{- 1}$

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$- 809 k J m o l^{- 1}$

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$809 k J m o l^{- 1}$

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$- 971 k J m o l^{- 1}$

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Solution

The correct option is B
$- 809 k J m o l^{- 1}$

$C H_{4 (g)} + 2 O_{2 (g)} \to 2 C O_{2 (g)} + 2 H_{2} O_{ℓ} . Δ H = - 890 k J$ ...(i)
$2 H_{2} O_{(ℓ)} \to 2 H_{2} O_{(g),} Δ H = 2 \times 40.5 = 81 k J$ ...(ii)
$(i) + (i i), C H_{4 (g)} + 2 O_{2 (g)} \to 2 C O_{2 (g)} + 2 H_{2} O_{(g)^{'}} Δ H = - 890 + 81 = - 809 k J$

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Similar questions

Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

Q. Calculate the enthalpy change for the following reaction:

{C H}_{4} (g) + 2 O_{2} (g) ⟶ {C O}_{2} (g) + 2 H_{2} O (l)

Given, enthalpies of formation of

{C H}_{4}, {C O}_{2}

and

H_{2} O

are

74.8 k J {m o l}^{- 1}, - 393.5 k J {m o l}^{- 1}, - 286 k J {m o l}^{- 1}

respectively.

Calculate the enthalpy change for the following reaction
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
Given enthalpies of formation of $C H_{4}, C O_{2} & H_{2} O$ are $- 74.8 K J m o l^{- 1}, - 393.5 K J m o l^{- 1}$ and $- 286.2 K J m o l^{- 1}$ respectively.