Calculate the enthalpy change for the isomerization reaction as given :
${CH}_{2} = CH - {CH}_{2} - \underset{(A)}{CH} = CH - CH = {CH}_{2}$ $\underset{Δ}{\overset{{NaNH}_{2}}{⟶}} {CH}_{2} = CH - \underset{(B)}{CH} = CH - CH = CH - {CH}_{3}$
Use the following data :
${ΔH}_{f}^{\circ} (C_{2} H_{4}) = - 2275 .5 kJ / mol$
${ΔH}_{f}^{\circ} (C_{2} H_{6}) = - 2839 .2 kJ / mol$
Resonance energy of A = 50 kJ/mol
Resonance energy of B = 70 kJ/mol

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-20,000 J/mol

-1622.6 kJ/mol

-1642.6 kJ/mol

-1692.6 kJ/mol

answer is D.

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Detailed Solution

Given data is,

${ΔH}_{f}^{\circ} (C_{2} H_{4}) = - 2275 .5 kJ / mol$

${ΔH}_{f}^{\circ} (C_{2} H_{6}) = - 2839 .2 kJ / mol$

Resonance energy of A = 50 kJ/mol

Resonance energy of B = 70 kJ/mol
The enthalpy change for the isomerization reaction -20,000 J/mol for the given reaction.

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