Question

Calculate the amount of heat evolved during the complete combustion of $100ml$ of liquid benzene from the following data:

• A. 18 gm o graphite on complete combustion evolve 590 KJ heat
• B. 15889 KJ heat is required to dissociate all the molecules of 1 litre water into $H_2$ and $O_2$
• C. The heat of formation of liquid benzene is 50 Kj/mol
• D. Density of $C_6{H}_6\left(l\right)=0.87gm/ml$

Answer 1

The correct option is A 18 gm o graphite on complete combustion evolve 590 KJ heat

Heat evolved in combustion of benzene $\left(C_6{H}_6\right)$ is given by the following equation:

$C_6{H}_6+\frac{15}{2}{O}_2⟶3H_{2}O+6C{O}_2$

$100ml$ of liquid benzene= density $\times 100ml$

$=\left(0.87gm/ml\right)\times 100ml=87gm$

$C_{\left(s\right)}+O_2⟶{C{O}_2}_{\left(g\right)}\to \left(1\right),\Delta H_1=590kJ\times \frac{2}{3}=-393.3kJ/mol$

${2H_{2}O}_{\left(l\right)}⟶{2H_2}_{\left(g\right)}+{O_2}_{\left(g\right)}\to \left(2\right),$ $\Delta H_2=\frac{15889KJ/mol}{\frac{1000}{2\times 18}}=572kJ/mol$

$6C_{\left(s\right)}+{3H_2}_{\left(g\right)}⟶{C_6{H}_6}_{\left(l\right)}\to \left(3\right),\Delta H_3=50kJ/mol$

Now applying $\left(1\right)\times 6-\left(1\right)\times \frac{3}{2}-3$ we will get our desired equation.

$\Delta H=-393.3\times 6-572\times \frac{3}{2}-50$

$=-3267.8kJ/mol$

Molecular weight of $C_6{H}_6=78gm/mol$

$\therefore 78gm$ of $C_6{H}_6$ evolves $3267.8kJ$ amount of heat.

$\therefore 87gm$ of $C_6{H}_6$ evolves= $\frac{3267.8}{78}\times 87=3644.8kJ$