Calculate the heat of formation (of one mole) benzene $(C_6H_6 (l))$ from acetylen (ethyne $C_2H_2(g))$ using the following data of combustion enthalpies: $\Delta H_c (C_6H_6)$ $=$ $-3273 \;\frac{kJ}{mol}$ $\Delta H_c (C_2H_2)$ $=$ $-1299 \;\frac{kJ}{mol}$
(a) $C_2H_2(g)$ $+$ $\frac{5}{2}O_2(g)$ $\longrightarrow$ $2CO_2(g)$ $+$ $H_2O(l)$ $\Delta H_a$ $ = $ $\color{red}-\color{black}1299\;kJ$ (b) $C_6H_6(l)$ $+$ $\frac{15}{2}O_2(g)$ $\longrightarrow$ $6CO_2(g)$ $+$ $3H_2O(l)$ $\Delta H_b$ $=$ $\color{red}-\color{black}1509\;kJ$
3(a) $3C_2H_2(g)$ $+$ $\frac{15}{2}O_2(g)$ $\longrightarrow$ $6CO_2(g)+3H_2O(l)$ $3\Delta H_a$ -(b) $6CO_2(g)$ $+$ $3H_2O(l)$ $\longrightarrow$ $C_6H_6(l)$ $+$ $\frac{15}{2}O_2(g)$ $-\Delta H_b $
3(a)-(b) $3C_2H_2(g)$ $+$ $\frac{15}{2}O_2(g)$ $+$ $6CO_2(g)$ $+$ $3H_2O(l)$ $\longrightarrow$ $6CO_2(g)$ $+$ $3H_2O(l)$ $+$ $C_6H_6(l)$ $+$ $\frac{15}{2}O_2(g)$ $3\Delta H_a$ $-$ $\Delta H_b $ (c)=3(a)-(b) $3C_2H_2(g)$ $\longrightarrow$ $C_6H_6(l)$ $\Delta H_c$ $=$ $3\Delta H_a-\Delta H_b$ $=$ $-2388\;kJ$
The reaction is exothermic!