Question

If enthalpies of formation for $C_2{H}_4\left(g\right),{CO}_2\left(g\right)$ and $H_{2}O\left(l\right)$ at ${25}^{o}C$ and $1$ atm pressure be $52,-394$ and $-286kJ$ ${mol}^{-1}$ respectively, enthalpy of combustion of $C_2{H}_4\left(g\right)$ will be:

• A. $+141.2kJ$ ${mol}^{-1}$
• B. $+1412kJ$ ${mol}^{-1}$
• C. $-141.2kJ$ ${mol}^{-1}$
• D. $-1412kJ$ ${mol}^{-1}$

Answer 1

The correct option is D $-1412kJ$ ${mol}^{-1}$

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

${\Delta }_{f}H\left(C_2{H}_4\left(g\right)\right)=52KJ/mol$

${\Delta }_{f}H\left(O_2\left(g\right)\right)=0KJ/mol$ (standard state)

${\Delta }_{f}H\left({CO}_2\left(g\right)\right)=-394KJ/mol$

${\Delta }_{f}H\left(H_{2}O\left(l\right)\right)=-286KJ/mol$

Enthalpy of combustion of $C_2{H}_4\left(g\right)=\sum {\Delta }_{f}H\left(products\right)-\sum {\Delta }_{f}H\left(reactants\right)$

$=[2\times (-394)+2\times (-286)]-52$

$=-1412KJ/mol$