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Ethanol can undergo decomposition to form two sets of products. If the molar ratio of $C_{2} H_{4} to {CH}_{3} CHO is 8 : 1$ in a set of product gases, then the energy involved in the decomposition of 1 mole of ethanol is :
$C_{2} H_{5} OH (g) C_{2} H_{4} (g) + H_{2} O (g); Δ H^{\circ} = 45.54 kJ$ (1)
$C_{2} H_{5} OH (g) {CH}_{3} CHO (g) + H_{2} (g); Δ H^{\circ} = 68.91 kJ$ (2)

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$48.137 k J$

$57.22 k J$

$65.98 kJ$

$48.46 k J$

answer is B.

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Detailed Solution

We have,

$C_{2} H_{5} OH (g) C_{2} H_{4} (g) + H_{2} O (g); Δ H^{\circ} = 45.54 kJ$ (1)

$C_{2} H_{5} OH (g) {CH}_{3} CHO (g) + H_{2} (g); Δ H^{\circ} = 68.91 kJ$ (2)

Energy used in the breakdown of ethanol equals the sum of the energies used in the first and second reactions.

i.e.

${ΔH}_{rxn} = {ΔH}_{1} + {ΔH}_{2}$

according to question

${ΔH}_{rxn} = \frac{8 \times {ΔH}_{1} + 1 \times {ΔH}_{2}}{9}$

${ΔH}_{rxn} = \frac{8 \times 45.5 + 1 \times 68.91}{9}$

$= 48.13 kJ$

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