Question

For the gaseous reaction involving complete combustion of isobutane (assuming all products and reactants are in gaseous state), the relation between $\Delta H$ and $\Delta E$ will be:

• A. $\Delta H>\Delta E$
• B. $\Delta H=\Delta E$
• C. $\Delta H<\Delta E$
• D. $\Delta H=0$ and $\Delta E=\infty$

Answer 1

The correct option is A $\Delta H>\Delta E$

Equation of combustion of isobutane will be as follows:

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

$\because \Delta H=\Delta E+\Delta nRT$

where,

$\Delta n=$ number of gaseous moles (products)

$\Delta n=(4+5)-(1+\frac{13}{2})$ number of moles of gaseous reactants.

$\Delta n=+\frac{3}{2}(+ve)$

Since, $\Delta n$ is $+ve$

$\Delta H$ must be greater than $\Delta E$.

$\therefore \Delta H>\Delta E$