Question

What is the enthalpy change for the isomerization reaction:
$C{H_2} = CH - C{H_2} - CH = CH - CH = C{H_2}(A){\text{ }}\xrightarrow[\Delta ]{{NaN{H_2}}}{\text{ C}}{{\text{H}}_2} = CH - CH = CH - CH = CH - C{H_2}(B)$
Magnitude of resonance energies of A and B are $50$ and $70{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ respectively. The enthalpies of formation of A and B are $ - 2275.2$ and $ - 2839.2{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ respectively.
$(i){\text{ - 584 kJ}}$
$(ii){\text{ - 564 kJ}}$
$(iii){\text{ - 544 kJ}}$
$(iv){\text{ - 20 kJ}}$

Answer 1

Hint: 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:
${\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,
${\Delta _f}H{\text{ = }}\sum\limits_{}^{} {{\Delta _f}H(products)} {\text{ - }}\sum\limits_{}^{} {{\Delta _f}H(reac{\text{tant}})} $ ________________$(1)$$(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,
$\Delta {H_f}(C{H_2} = CH - C{H_2} - CH = CH - CH = C{H_2})(A){\text{ = - 2275}}{\text{.2 kJ mo}}{{\text{l}}^{ - 1}}{\text{ }}$
Also we can write for product B,
$\Delta {H_f}(C{H_2} = CH - CH = CH - CH = CH - C{H_2})(B){\text{ = - 2839}}{\text{.2 kJ mo}}{{\text{l}}^{ - 1}}{\text{ }}$
Using the above two values and substituting the equation $(1)$ we get the following result,
${\Delta _f}H{\text{ = }}\sum\limits_{}^{} {{\Delta _f}H(products)} {\text{ - }}\sum\limits_{}^{} {{\Delta _f}H(reac{\text{tant}})} $
${\Delta _f}H{\text{ = }}\left[ { - 2839.2{\text{ - ( - 2275}}{\text{.2)}}} \right]$
${\Delta _f}H{\text{ = - 564 kJ mo}}{{\text{l}}^{ - 1}}$
Hence we get the change in enthalpy for the above reaction. Therefore the correct option is $(ii)$.

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.