Question

At 25°C and 1 atm pressure,the enthalpies of combustion are as given below:$Substance$	$H_2$	$C(graphite)$	$C_2H_6(g)$
$\frac{D_cH^{\theta}}{kJmol^{-1}}$	$-286.0$	$-394.0$	-1560.0

The enthalpy of formation of ethane is

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

Answer 1

Answer: (C)

–86.0 kJ $mol^{–1}$

Answer 2

$2C( graphite) + 3H_2(g) → C_2H_6(g)$

$Δ H_r = +1560 +2(-394)+3(-286)$

= $86.0 \;kj \;mol ^{-1}$

Enthalpy of formation of $C_2H_6(g) = -86.0 \;kj \;mol^{-1}$