Question

If enthalpy of hydrogenation of $( C )_{6}( H )_{6}( l )$ into $( C )_{6}( H )_{12}( l )$ is $-205 \,k \,J$ and resonance energy of $( C )_{6}( H )_{6}( l )$ is $-152\, kJ\, mol ^{-1}$ then enthalpy of hydrogenation of

(l) to $( C )_{6}( H )_{12}( l )$ is? [Assume enthalpies of vapourisation of all liquids involved are equal].

• A. - 535.5 kJ/mol
• B. - 238 kJ/mol
• C. - 357 kJ/mol
• D. - 119 kJ/mol

Answer 1

Answer: (D)

$ C _{6}H _{6}(l)+3H _{2}( g ) \rightarrow C _{6} H _{12}(l)$
$\Delta H _{ hyd \left( C_{6} H _{6}\right) l}=-205 \,kJ$
Benzene on reactant side so add resonance energy of benzene
with sign. Heat evolved for hydrogenation of three double bonds
$=-205-152=-357 \,kJ\, mol { }^{-1}$
$\Delta H _{ hyd }^{\circ}=\frac{-357}{3}=-119 \,kJ /$ mole