Question

If enthalpies of formation for $C _{2} H _{4}(g), CO _{2( g )}$ and $H _{2} O _{(1)}$ at $25^{\circ} C$ and $1\, atm$ pressure are $52,-394$ and - $286\, kJ / mol$ respectively, then enthalpy of combustion of $C _{2} H _{4}( g )$ will be

• A. - 141.2 kJ/mol
• B. #ERROR!
• C. #ERROR!
• D. - 1412 kJ/mol

Answer 1

Answer: (D)

- 1412 kJ/mol

Answer 2

$2 C (s)+2 H _{2} \rightarrow C _{2} H _{4}(g) ; \Delta H _{1}=52\, KJ / mole$...(1)

$C (s)+ O _{2}(g) \rightarrow CO _{2}(g), \Delta H _{2}=-394\, KJ / mole$...(2)

$H _{2}(g)+1 / 2 O _{2}(g) \rightarrow H _{2} O (l), \Delta H _{3}=-286\, KJ / mole$...(3)

The combustion of $C _{2} H _{4}$ can be written as :

$C _{2} H _{4}( g ) \rightarrow 2 C ( s )+2 H _{2}, \Delta H _{1}=-52\, KJ / mole$
$2 C (s)+2 O _{2}(g) \rightarrow 2 CO _{2}(g), \Delta H _{2}=-2 \times 394$
$2 H _{2}(g)+ O _{2}(g) \rightarrow 2 H _{2} O (l), \Delta H _{3}=-2 \times 286$
$\Rightarrow \Delta H =-52-2 \times 394-2 \times 286=-1412\, KJ / mole$