Question

The standard enthalpies of combustion of $C_6{H}_6\left(l\right),$ $C_{\left(graphite\right)}$ and $H_2\left(g\right)$ are respectively $-3270$ $kJ$ ${mol}^{-1},$ $-394$ $kJ$ ${mol}^{-1}$ and $-286$ $kJ$ ${mol}^{-1}$. What is the standard enthalpy of formation of $C_6{H}_6\left(l\right)$ in $kJ$ ${mol}^{-1}$?

• A. $-48$
• B. $+48$
• C. $-480$
• D. $+480$
• E. $-72$

Answer 1

Answer: (B)

$+48$

Given,
(i) $C_6{H}_6+\frac{15}{2}{O}_2⟶6C{O}_2+3H_{2}O;$ $\Delta H=-3270$ $kJ$ ${mol}^{-1}$

(ii) $C\left(graphite\right)+O_2⟶C{O}_2;$ $\Delta H=-394$ $kJ$ ${mol}^{-1}$

(iii) $H_2+\frac{1}{2}{O}_2⟶H_{2}O;$ $\Delta H=-286$ $kJ$ ${mol}^{-1}$

Multiplication of equation (ii) by $6$ and equation (iii) by $3$, gives

(iv) $6C\left(s\right)+6O_2⟶6C{O}_2;$ $\Delta H=-394\times 6$ $kJ$ ${mol}^{-1}$ $=-2364$ $kJ$ ${mol}^{-1}$

(v) $3H_2+\frac{3}{2}{O}_2⟶3H_{2}O;$ $\Delta H=-286\times 3$ $kJ$ ${mol}^{-1}$ $=-858$ $kJ$ ${mol}^{-1}$

On inverting equation (i), we get

(vi) $6C{O}_2+3H_{2}O⟶C_6{H}_6+\frac{15}{2}{O}_2;$ $\Delta H=+3270$ $kJ$ ${mol}^{-1}$

Addition of equation (iv), (v) and (vi) gives

$6C\left(s\right)+3H_2⟶C_6{H}_6;$ $\Delta H=+3270+(-2364-858)=+48$ $kJ$ ${mol}^{-1}$

Thus, the standard enthalpy of formation of $C_6{H}_6,$ $\left({\Delta }_f{H}_{C_6{H}_6}\right)$ is $+48$ $kJ$ ${mol}^{-1}$.