Question

The enthalpy of combustion at $ 25{}^\circ C $ of $ {{H}_{2}}, $ cyclohexane $ ({{C}_{6}}{{H}_{12}}) $ and cyclohexene $ ({{C}_{6}}{{H}_{10}}) $ $ ({{C}_{6}}{{H}_{10}}) $ are - 241, - 3920 and - 3800 kJ/mol respectively. The heat of hydrogenation of cyclohexene is:

• A. -121 kJ/mol
• B. +121 kJ/mol
• C. -242 kJ/mol
• D. +242 kJ/mol

Answer 1

Answer: (A)

Use Hess's law to solve the problem. It states that total heat changes during a reaction are independent of path.
$ {{H}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}{{H}_{2}}O $ $ \Delta H=-241\,kJ $ .. (i)
$ {{C}_{6}}{{H}_{10}}+\frac{17}{2}{{O}_{2}}\xrightarrow{{}}6C{{O}_{2}}+5{{H}_{2}}O $
$ \Delta H=-3800\text{ }kJ $ ...(ii)
$ {{C}_{6}}{{H}_{12}}+9{{O}_{2}}\xrightarrow{{}}6C{{O}_{2}}+6{{H}_{2}}O $
$ \Delta H=-3920\text{ }kJ $ ...(iii)
$ \underset{Cyclohexene}{\mathop{{{C}_{6}}{{H}_{10}}}}\,+{{H}_{2}}\xrightarrow{{}}{{C}_{6}}{{H}_{12}} $
$ \Delta H=? $
Adding Eqs. (i) and (ii) and subtracting Eq (iii) we get,
$ {{C}_{6}}{{H}_{10}}+{{H}_{2}}\xrightarrow{{}}{{C}_{6}}{{H}_{12}} $
$ H=-241-3800-(-3920) $
$ =-4041+3920~ $
$ =-121kJ $