Question

The difference between $\Delta H$ and $\Delta E$ on a molar basis for the combustion of n-octane at ${25}^o$C would be:

• A. $-13.6$ kJ
• B. $-1.14$ kJ
• C. $-11.15$ kJ
• D. $+11.15$ kJ

Answer 1

Answer: (C)

$-11.15$ kJ

Let us consider the problem:

Combustion of n-octane is

$C_8{H}_{18}\left(l\right)+\frac{25}{2}{O}_2\to 8C{O}_2\left(g\right)+9H_{2}O$

Hence, the relation between $\Delta H$ and $\Delta H$ is

$\Delta H=\Delta E+\left(RT\Delta n\right)$

Implies that

$\Delta H-\Delta E=\left(RT\Delta n\right)$

Therefore

$\Delta n=n_p-n_r=9+8-\frac{25}{2}=4.5$

Hence the answer is

$\Delta H-\Delta E=\left(8.314\times 298\times 4.5\right)$

$\therefore \Delta H=11149.07J$

$\therefore \Delta H=11.15KJ$