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Enthalpy of Combustion

The enthalpie...

Question

The enthalpies of combustion of $C_{2} H_{4} (g)$ , $C_{2} H_{6} (g)$ and $H_{2}$ are -1409.5 kJ, -1558.3 kJ and - 285.6 kJ, respectively . Calculate the enthalpy of hydrogenation of enthylene

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Solution

${C_{2} H_{4}}_{(g)} + 3 {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} + 2 {H_{2} O}_{(l)}; Δ H = - 1409.5 k J . . . . . (1)$
${C_{2} H_{6}}_{(g)} + \frac{7}{2} {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} + 3 {H_{2} O}_{(l)}; Δ H = - 1558.3 k J$
$2 {C O_{2}}_{(g)} + 3 {H_{2} O}_{(l)} ⟶ {C_{2} H_{6}}_{(g)} + \frac{7}{2} {O_{2}}_{(g)}; Δ H = 1558.3 k J . . . . . (2)$
${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.6 k J . . . . . (3)$
Adding ${e q}^{n} (1), (2) & (3)$ , we get
${C_{2} H_{4}}_{(g)} + {H_{2}}_{(g)} ⟶ {C_{2} H_{6}}_{(g)}; Δ H = - 136.8 k J$
Hence the enthalpy of hydrogenation of ethylene is $- 136.8 k J$ .

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