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Question: What is the enthalpy change for the isomerization reaction: \(C{H_2} = CH - C{H_2} - CH = CH - CH ...

What is the enthalpy change for the isomerization reaction:
CH2=CH−CH2−CH=CH−CH=CH2(A) →ΔNaNH2 CH2=CH−CH=CH−CH=CH−CH2(B)C{H_2} = CH - C{H_2} - CH = CH - CH = C{H_2}(A){\text{ }}\xrightarrow[\Delta ]{{NaN{H_2}}}{\text{ C}}{{\text{H}}_2} = CH - CH = CH - CH = CH - C{H_2}(B)
Magnitude of resonance energies of A and B are 5050 and 70 kJ mol−170{\text{ kJ mo}}{{\text{l}}^{ - 1}} respectively. The enthalpies of formation of A and B are −2275.2 - 2275.2 and −2839.2 kJ mol−1 - 2839.2{\text{ kJ mo}}{{\text{l}}^{ - 1}} respectively.
(i) - 584 kJ(i){\text{ - 584 kJ}}
(ii) - 564 kJ(ii){\text{ - 564 kJ}}
(iii) - 544 kJ(iii){\text{ - 544 kJ}}
(iv) - 20 kJ(iv){\text{ - 20 kJ}}

Explanation

Solution

Enthalpy of formation is the amount of energy released or consumed while forming one mole of the substance. Here we are given with the enthalpies of formation of the molecule, A and B. For finding the enthalpy change for the above reaction we will subtract the enthalpies of formation of A and B.
Formula Used:
ΔfH = ∑ΔfH(products) - ∑ΔfH(reactant){\Delta _f}H{\text{ = }}\sum\limits_{}^{} {{\Delta _f}H(products)} {\text{ - }}\sum\limits_{}^{} {{\Delta _f}H(reac{\text{tant}})}

Complete answer:
Then enthalpy for formation of any substance may be defined as the amount of energy released or consumed when one mole of the substance is formed. When the enthalpy of formation is negative, it means a certain amount of energy is released during its formation. We can also say that the reaction is exothermic when the enthalpy of formative comes out to be negative. Here, we are given with the enthalpy of formation of A and B. For finding the change in enthalpy of formation we know that,
ΔfH = ∑ΔfH(products) - ∑ΔfH(reactant){\Delta _f}H{\text{ = }}\sum\limits_{}^{} {{\Delta _f}H(products)} {\text{ - }}\sum\limits_{}^{} {{\Delta _f}H(reac{\text{tant}})} ________________(1)(1) (i)(i)
Therefore the change in enthalpy of the reaction is the difference of enthalpy of formation of products and reactants. According to question, the reactant is compound A, while the product formed is compound B. Hence we can write that,
ΔHf(CH2=CH−CH2−CH=CH−CH=CH2)(A) = - 2275.2 kJ mol−1 \Delta {H_f}(C{H_2} = CH - C{H_2} - CH = CH - CH = C{H_2})(A){\text{ = - 2275}}{\text{.2 kJ mo}}{{\text{l}}^{ - 1}}{\text{ }}
Also we can write for product B,
ΔHf(CH2=CH−CH=CH−CH=CH−CH2)(B) = - 2839.2 kJ mol−1 \Delta {H_f}(C{H_2} = CH - CH = CH - CH = CH - C{H_2})(B){\text{ = - 2839}}{\text{.2 kJ mo}}{{\text{l}}^{ - 1}}{\text{ }}
Using the above two values and substituting the equation (1)(1) we get the following result,
ΔfH = ∑ΔfH(products) - ∑ΔfH(reactant){\Delta _f}H{\text{ = }}\sum\limits_{}^{} {{\Delta _f}H(products)} {\text{ - }}\sum\limits_{}^{} {{\Delta _f}H(reac{\text{tant}})}
ΔfH = [−2839.2 - ( - 2275.2)]{\Delta _f}H{\text{ = }}\left[ { - 2839.2{\text{ - ( - 2275}}{\text{.2)}}} \right]
ΔfH = - 564 kJ mol−1{\Delta _f}H{\text{ = - 564 kJ mo}}{{\text{l}}^{ - 1}}
Hence we get the change in enthalpy for the above reaction. Therefore the correct option is (ii)(ii).

Note:
The negative value of change enthalpy indicates that the reaction is exothermic in nature. For finding any change in enthalpy it is always recommended to take difference of enthalpy of products and reactants. Resonance energy has no role play in finding change in enthalpy of the reaction.