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Using bond energy data, calculate heat of formation of isoprene.Given, C-H, H-H, C-C, C=C and C(s)$\to$C(g) respectively as 98.8 kcal, 104 kcal, 83kcal, 147kcal, 171kcal.(a) -21 kcal(b) 21 kcal(c) 40 kcal(d) 50 kcal

Last updated date: 18th Jun 2024
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Answer
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Hint: As we know, a chemical reaction involves breaking and forming of chemical bonds. During the formation of the bond, energy is released and during the breaking of a bond, energy is absorbed.
So, the enthalpy changes involving gaseous reactants and gaseous products having covalent bonds can be calculated with the help of bond enthalpies of reactants and products using the following formula.
Heat of Reaction=$\sum{\Delta {{\text{H}}^{\text{o}}}}(\text{reactant bonds)-}\sum{\Delta {{\text{H}}^{\text{o}}}}(\text{products bonds)}$

Complete step-by-step answer:
As it is mentioned in the hint, the bonds between atoms may break, reform or both to either absorb or release energy during chemical reaction. This results in a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy and the change in enthalpy that results from a chemical reaction is the enthalpy of reaction. It is generally written as$\Delta {{\text{H}}_{\text{rxn}}}$.
Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the energy of the reactant bonds:
$\Delta {{\text{H}}_{\text{rxn}}}$=potential energy of product bonds-potential energy of reactant bonds.
As it is given in the question:
$\text{C-H=98}\text{.8 kcal}$
$\text{H-H=104 kcal}$
$\text{C-C=83 kcal}$
$\text{C=C =147 kcal}$
$\text{C(s)}\to \text{C(g)=171 kcal}$
Reaction:

$\text{5x }\!\![\!\!\text{ C(s)}\to \text{C(g) }\!\!]\!\!\text{ =5x171 kcal=855 kcal}$
$4{{\text{H}}_{2}}$$\Rightarrow 4\text{x}104=416\text{kcal}$
$\sum{\text{BE(}5\text{C}+4{{\text{H}}_{2}})}\Rightarrow 855+416$
$\sum{\text{BE=C-H}\Rightarrow \text{8x98}\text{.8=790}\text{.4}}$
$\Rightarrow \text{C-C=2x83=166 kcal}$
$\Rightarrow \text{C=C= 2x147=294 kcal}$
$\Delta {{\text{H}}_{\text{f}}}=\Delta \text{HB}{{\text{E}}_{\text{r}}}-\Delta \text{HB}{{\text{E}}_{\text{p}}}$= (855+416)-(790.4+166+294)
=1271-1250.4
=20.6 kcal
The closest option from the answer obtained is option (b) 21 kcal.

Note: Please note that the bond energy that is given in the question is for single bonds. Always multiply the stoichiometric coefficient to have the right bond energy for the specific compounds. This can often be a very lengthy calculation if there are a lot of bonds, so you should practice to just look and calculate for the bonds that are changing as in the example.