The enthalpy change for the reaction $C_{3} H_{8} (g) + H_{2} (g) \to C_{2} H_{6} (g) + C H_{4} (g)$ at $25^{o}$ is -55.7 kJ / mol . Calculate the enthalpy of combustion of $C_{2} H_{6} (g)$ . The enthalpy of combustion of propane is $- 2220 k J m o l^{- 1}$ and enthalpy of $H_{2}$ and $C H_{4}$ are -285.8 KJ/mol and -890 KJ/mol respectively.

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Solution

${C_{3} H_{8}}_{(g)} + {H_{2}}_{(g)} \to {C_{2} H_{6}}_{(g)} + {C H_{4}}_{(g)}$
$Δ H = Δ H_{P} - Δ H_{R}$
$= (Δ H_{C_{2} H_{6}} + Δ H_{C H_{4}}) - (Δ H_{C_{3} H_{8}} + Δ H_{H_{2}})$
$\Rightarrow - 55.7 = Δ H_{C_{2} H_{6}} + (- 890) - [(- 2220) + (- 285.8)]$
$= Δ H_{C_{2} H_{6}} + 1615.8$
$\Rightarrow Δ H_{C_{2} H_{6}} = - 55.7 - 1615.8 = - 1671.5 k J$
Enthalpy of combustion of $C_{2} H_{6}$ is $- 1671.5 k J$ .

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Q. The enthalpy change for the reaction

C_{3} H_{8} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g) + {C H}_{4} (g)

at 25C is -55.7 KJ/mol. Calculate the enthalpy of combustion of

C_{2} H_{6} (g)

. The enthalpy of combustion of

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{C H}_{4}

are -285.8 and -890.0 KJ/mol respectively. Enthalpy of combustion of propane is -2220

K J {m o l}^{- 1}

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 :
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