Question

If enthalpies of formation for $C_2{H}_4\left(g\right),C{O}_2\left(g\right)$ and $H_{2}O\left(I\right)$ at ${25}^0$C and $1$ atm pressure are $52;-394$ and $-286kJ/mol$ respectively, then enthalpy of combustion of $C_2{H}_4\left(g\right)$ will be?

Answer 1

Solution:-

Combustion of $C_2{H}_4$-

${C_2{H}_4}_{\left(g\right)}+3{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

For the above reaction-

$\Delta H={\sum \Delta H_f}_P-{\sum \Delta H_f}_R$

$\Delta H=(2\times (-394)+2\times (-286))-(52+0)$

$\Delta H=-1360+52=-1308kJ/mol$

Hence the enthapy of combustion of ${C_2{H}_4}_{\left(g\right)}$ is $-1308kJ/mol$.