The heats of combustion of Carbon, Hydrogen and Ethylalcohol are – 395 KJ/mol,  - 285 KJ/mol  and  -1367 KJ/mol  respectively. The heat of formation of ethyl alcohol is ________ kJ

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}+{\text{3O}}_{\text{2}}\to {\text{2CO}}_{\text{2}}+{\text{3H}}_{\text{2}}\text{O}+\text{1367\hspace{0.17em}\hspace{0.17em}KJ}$

$\text{C}+{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+\text{393\hspace{0.17em}\hspace{0.17em}KJ}$

${\text{H}}_2+\frac{1}{2}{\text{O}}_{\text{2}}\to {\text{H}}_{\text{2}}\text{O}+\text{285\hspace{0.17em}\hspace{0.17em}KJ}$

$\therefore$ Heat of combustion of ethylalcohol = Total heat formation of products – Total heat of formation of reactants

$-1367=[2\left(\Delta H_{f_{C{O}_2}}\right)+3\left(\Delta H_{f_{H_{2}O}}\right)]-[\left(\Delta H_{f_{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\right)+\left(\Delta H_{f_{O_2}}\right)]$

$-1367=[2(-373)+(-285)]-[\Delta H_{f_{C_2{H}_{5}OH}}+Zero]$

$-1367=[-786-855]-\left[\Delta H_{f_{C_2{H}_{5}OH}}\right]$

$-1367=[-1641]-\left[\Delta H_{f_{C_2{H}_{5}OH}}\right]$

$\therefore \Delta H_{f_{C_2{H}_{5}OH}}=-1641+1367=-274KJ$