Question

What is the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of ethylene, hydrogen and ethane are $-1410.0,-286.2$ and $-1560.6 \,k \,J \,mol^{-1}$ respectively at $298 \,K$ ?

• A. 135.6
• B. -135.6
• C. 125.6
• D. $-125.6

Answer 1

Answer: (B)

Given, (i) $C _{2} H _{4}(g)+3 O _{2}(g) \rightarrow 2 CO _{2}(g)+2 H _{2} O (l)$
$\Delta H=-1410\, kJ \,mol ^{-1}$ (ii)
$H _{2}(g)+\frac{1}{2} O _{2}(g) \rightarrow H _{2} O (l) ;$
$\Delta H=-286.2 \,k J\, m o l^{-1}$ (iii)
$C _{2} H _{6}(g)+3 \frac{1}{2} O _{2}(g) \rightarrow 2 CO _{2}(g)+3 H _{2} O (l)$
$\Delta H=-1560.6 \,kJ\, mol ^{-1}$
Required reaction is
$C _{2} H _{4}+ H _{2}(g) \rightarrow C _{6} H _{6}(g)$,
$\Delta H=?$
Equation (i) + Equation (ii) - Equation (iii) gives
$C _{2} H _{4}+ H _{2}(g) \rightarrow C _{2} H _{6}(g)$,
$\Delta H=-1410.0+(-286.2)-(-1560.6)$
$=-135.6 \,kJ\, mol ^{-1}$