Equation of combustion of isobutane will be as follows:
$C _{4} H _{10}( g )+\frac{13}{2} O _{2}( g ) \rightarrow 4 CO _{2}( g )+5 H _{2} O ( g )$
$\because \Delta H =\Delta E +\Delta nRT$
where,
$\Delta n =$ number of gaseous moles (products)
$\Delta n =(4+5)-\left(1+\frac{13}{2}\right)$ number of moles of gaseous reactants.
$\Delta n =+\frac{3}{2}(+ ve )$
Since, $\Delta n$ is + ve
$\Delta H$ must be greater than $\Delta E$.
$\therefore \Delta H > \Delta E$