Question

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

Substance	$H_2$	C(graphite)	$C_2H_6(g)$
$\frac{\Delta_cH^{\ominus}}{Kj\,mol^{-1}}$	$-286.0$	$-394.0$	T-1560.0

The enthalpy of formation of ethane is

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

Answer 1

Answer: (C)

$C _{2} H _{6}( g )+\frac{7}{2} O _{2}( g ) \rightarrow 2 CO _{2}( g )+3 H _{2} O (l )$
$\Delta_{ C } H \left( C _{2} H _{6}\right)=2 \Delta_{ f } H CO _{2}( g )+3 \Delta_{ f } H \left( H _{2} O , \ell\right)$
$-\Delta_{ f } H \left( C _{2} H _{6}, g \right)$
$-1560=2(-394)+3(-286)-\Delta_{ f } H \left( C _{2} H _{6}, g \right)$
$\Delta_{ f } H \left( C _{2} H _{6}, g \right)=-86 \,kJ / mole$