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JEE MainChemistryModeratePhysical Chemistry

E0E^{0} of combustion of isobutylene is -X kJ mol^-1. The
value of ΔH∘\Delta H^{\circ} is

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(CH3)2C=CH2(g)+6O2(g)→4CO2(g)+4H2O(l)\left( {CH}_{3} \right)_{2}C = {CH}_{2}(g) + 6O_{2}(g) \rightarrow 4{CO}_{2}(g) + 4H_{2}O(l)
ng=4−6=−2ng = 4 - 6 = - 2 (i.e, negative) 
We know that H=ΔE+ΔngRTH = \Delta E + \Delta ngRT
=ΔE−(Δng)RT(∵Δng=−ve)= \Delta E - (\Delta ng)RT(\because\Delta ng = - ve)
∴ΔH<ΔE\therefore\Delta H < \Delta E

Source: Grass-G / CH-11 Chemistry Paper-2 1Nov New Jee main.docx

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