The heat of hydrogenation of cyclohexene is $28.6 \, kcal$ and that of cyclohexadiene is slightly more than double,$(55.4 \, kcal)$. What will be the heat of hydrogenation of benzene,which contains three double bonds?
- AThree times that of cyclohexene $(85.8 \, kcal)$
- BSame as cyclohexene
- CSame as cyclohexadiene
- D$49.8 \, kcal$
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Similar Questions
Given the following data:
Reaction Energy Change (in $kJ$) $Li_{(s)} \to Li_{(g)}$ $161$ $Li_{(g)} \to Li^{+}_{(g)}$ $520$ $\frac{1}{2} F_{2(g)} \to F_{(g)}$ $77$ $F_{(g)} + e^- \to F^{-}_{(g)}$ (Electron gain enthalpy) $Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$ $-1047$ $Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$ $-617$
Based on the data provided,the value of electron gain enthalpy of fluorine would be $kJ\ mol^{-1}$.
| Reaction | Energy Change (in $kJ$) |
| $Li_{(s)} \to Li_{(g)}$ | $161$ |
| $Li_{(g)} \to Li^{+}_{(g)}$ | $520$ |
| $\frac{1}{2} F_{2(g)} \to F_{(g)}$ | $77$ |
| $F_{(g)} + e^- \to F^{-}_{(g)}$ | (Electron gain enthalpy) |
| $Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$ | $-1047$ |
| $Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$ | $-617$ |
Based on the data provided,the value of electron gain enthalpy of fluorine would be $kJ\ mol^{-1}$.
DifficultJEE MAIN 2013
View SolutionGiven the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.
Medium
View SolutionThe enthalpy of reaction for the reaction: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ is $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$. What will be the standard enthalpy of formation of $H_{2}O_{(l)}$?
Medium
View SolutionIf the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:
Medium
View SolutionCalculate the enthalpy of solution of potassium chloride $(KCl)$ if its lattice enthalpy $\Delta_{L} H = 700 \ kJ \ mol^{-1}$ and hydration enthalpy $\Delta_{hyd} H = -680 \ kJ \ mol^{-1}$.
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