MCQ
At $25^{\circ} C$ and 1 atm pressure, the enthalpies of combustion are as given below:

Substance $H _{2}$ $C ($ graphite $)$ $C _{2} H _{6}( g )$
$\frac{\Delta_{ C } H ^{\Theta}}{ kJmol ^{-1}}$ $-286.0$ $-394.0$ $-1560.0$

The enthalpy of formation of ethane is ........ .

  • A
    $+54.0\, kJ \,mol ^{-1}$
  • B
    $-68.0 \,kJ \,mol ^{-1}$
  • $-86.0\, kJ \,mol ^{-1}$
  • D
    $+97.0\, kJ \,mol ^{-1}$

Answer

Correct option: C.
$-86.0\, kJ \,mol ^{-1}$
c
$C _{2} H _{6}( g )+\frac{7}{2} O _{2}( g ) \rightarrow 2 CO _{2}( g )+3 H _{2} O (\ell)$

$\Delta_{ C } H \left( C _{2} H _{6}\right)=2 \Delta_{ f } H CO _{2}( g )+3 \Delta_{ f } H \left( H _{2} O , \ell\right)$

$-\Delta_{ f } H ^{\prime}\left( C _{2} H _{6}, g \right)$

$-1560=2(-394)+3(-286)-\Delta_{ f } H \left( C _{2} H _{6}, g \right)$

$\Delta_{ f } H \left( C _{2} H _{6}, g \right)=-86 \,kJ / mole$

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