∆ H > ∆ E ∆ H = ∆ E ∆ H < ∆ E ∆ H = 0 and ∆ E = ∞ Solution: Equation of combustion of isobutane will be as follows: C 4 H 10 ( g ) + 13 2 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 5 H 2 O ( g ) ∵ ∆ H = ∆ E + ∆ n R T where, ∆ n = no. of gaseous moles (products) − ∆ n = ( 4 + 5 ) − ( 1 + 13 2 ) number of moles of gaseous reactants. ∆ n = + 3 2 ( + v e ) Since, ∆ n is + ve ∆ H must be greater than ∆ E . ∆ E " class="fm-inline"> ∴ ∆ H > ∆ E" >
