Question

For the gaseous reaction involving the complete combustion of isobutane :

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

Answer 1

Answer: (B)

$\Delta H > \Delta E$

Answer 2

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 $\Delta E$. $\therefore \Delta H > \Delta E$