Question

∆E of combustion of isobutylene is – X kJ mol⁻¹. The value of
∆H is :
(a) = ∆E (b) > ∆E
(c) = 0 (d) < ∆E

Answer 1

answer : option (d) < ∆E

given, ∆E of combustion of isobutylene is -X kJ/mol.

first see chemical reaction of isobutylene...

C4H8(g) + 6O2(g)⇒4CO2(g) + 4H2O

now ∆n_g = number of moles of gaseous products - number of moles of gaseous reactants

= (4 -6 - 1) = -3

so, ∆n_gRT = -3RT

from first law of thermodynamics,

∆H = ∆E + ∆n_gRT

⇒∆H = ∆E - 3RT

here it is clear that ∆H < ∆E

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Answer 2

∆E of combustion of isobutylene is ∆H  < ∆E kJ mol⁻¹

Step by step explanation:

"The letter "H" stands for a  enthalpy of system. Enthalpy refers to the sum of a system's internal energy plus the product from the pressure and volume of the system."

                             $\bold{\Delta H^{o}=\Delta HE^{o}+\Delta n_{g}RT}$......................(1)

The combustion of isobutylene is as follows.

(in attachment)

Number of moles $\Delta n_{g}$ = 4 - 7 = -3

Substitute the $\Delta n_{g}$ in equation (1)

If we get

$\bold{\Delta H^{o} < \Delta E^{o}}$

Therefore, ∆E of combustion of isobutylene is ∆H  < ∆E kJ mol⁻¹

Hence, correct option is "d".