Question

For a gaseous reaction involving the complete combustion of isobutane:

• A. $△H=△E$
• B. $△H>△E$
• C. $△H<△E$
• D. none of these

Answer 1

The correct option is C $△H<△E$

Solution:- (C) $\Delta H<\Delta E$

Combustion of isobutane-

${C_4{H}_{10}}_{\left(g\right)}+\frac{13}{2}{O_2}_{\left(g\right)}⟶4{C{O}_2}_{\left(g\right)}+5{H_{2}O}_{\left(l\right)}$

For the above reaction-

$\Delta n_g=n_P-n_R=4-(\frac{13}{2}+1)=\frac{-7}{2}=-ve$

As we know that,

$\Delta H=\Delta E+\Delta n_{g}RT$

$\because \Delta n_g<0$

$\therefore \Delta H<\Delta E$