Question

Difference between $\Delta H$ and $\Delta E$ for the combustion of liquid benzene at ${27}^{\circ }C$ is:

• A. $7.48kJ$
• B. $3.47kJ$
• C. $14.86kJ$
• D. $5.73kJ$

Answer 1

The correct option is B $3.47kJ$

Solution:- (B) $3.47kJ$

Combustion of benzene-

${C_6{H}_6}_{\left(l\right)}+\frac{15}{2}{O_2}_{\left(g\right)}⟶6{C{O}_2}_{\left(g\right)}+3{H_{2}O}_{\left(l\right)}$

$\Delta n_g=n_P-n_R=6-\frac{15}{2}=\frac{-3}{2}$

Given:- $T=27℃=(27+273)=300K$

As we know that,

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

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

$\Rightarrow \Delta H-\Delta E=(-\frac{3}{2})\times 8.314\times {10}^{-3}\times 300=-3.73kJ\approx -3.47kJ$

Hence the difference between $\Delta H$ and $\Delta E$ for the combustion of liquid benzene is $3.47kJ$.