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Molar and Specific Heat Capacity

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Question

Calculate the enthalpy of hydrogenation for the reaction , $C_{2} H_{4} + H_{2} ⟶ C_{2} H_{6}$ . Given that the enthalpy of combustion of Ethane is $C_{2} H_{4}; C_{λ} H_{6} . & H_{2} a r e - 140.1 K g, - 1550 K g & - 286.0 K g | m o l$

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Solution

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O Δ H_{1} = - 140.1 k g / m o l$ .
$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O Δ H_{2} = - 15550 k g / m o l$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O Δ H_{3} = - 286 k g / m o l$
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O Δ H_{1} = - 1401$
$2 C O_{2} + 3 H_{2} O \to C_{2} H_{6} + \frac{7}{2} O_{2} Δ H_{2} = 1550$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O Δ H_{3} = - 286$
$C_{2} H_{9} + H \to C_{2} + H_{6}$
$Δ H = - 140.1 + 1350 - 286 = - 1123.9$

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