Calculate the enthalpy change for the conversion of ethene $C_{2} H_{4}$ to ethane $C_{2} H_{6}$ (only magnitude in nearest integer in kj/mol).

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Solution

Given reaction
$R : C_{2} H_{4} + H_{2} \to C_{2} H_{6}; Δ H = ?$

From the given table, enthalpies of the following reactions are
$R_{1} : C + O_{2} \to C O_{2}; Δ H_{1} = - 394 k J / m o l$

$R_{2} : H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H_{2} = - 286 k J / m o l$

$R_{5} : C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H_{3} = - 1561 k J / m o l$

$R_{6} : C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O; Δ H_{3} = - 1393 k J / m o l$

And

$R = - R_{5} + R_{6} + R_{2}$

$\Rightarrow Δ H = - Δ H_{5} + Δ H_{6} + Δ H_{2} = 1561 - 1393 - 286 = - 118 k J / m o l$

Answer = 118

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