1. Tardigrade
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  3. Chemistry
  4. If enthalpy of hydrogenation of mathrmC6 mathrmH6( mathrml) into mathrmC6 mathrmH12 (l) is -205 mathrm~kJ and resonance energy of mathrmC6 mathrmH6( mathrml) is -152 mathrm~kJ mathrm~mol-1 then enthalpy of <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/c-6fq8xdze7aks.png /> (l) to ( C )6( H )12( l ) is? [Assume enthalpiies of vapourisation of all liquids involved are equal].

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Q. If enthalpy of hydrogenation of $\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$ into $\mathrm{C}_6 \mathrm{H}_{12}$ (l) is $-205 \mathrm{~kJ}$ and resonance energy of $\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$ is $-152 \mathrm{~kJ} \mathrm{~mol}^{-1}$ then enthalpy of
Question
(l) to $( C )_6( H )_{12}( l )$ is?
[Assume enthalpiies of vapourisation of all liquids involved are equal].

NTA AbhyasNTA Abhyas 2020Thermodynamics

Solution:

$\begin{array}{c}\mathrm{C}_6 \mathrm{H}_6(l)+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_6 \mathrm{H}_{12}(l) \\ \Delta \mathrm{H}_{\mathrm{hyd}\left(\mathrm{C}_6 \mathrm{H}_6\right) l}^{\circ}=-205 \mathrm{~kJ}\end{array}$
Benzene on reactant side so add resonance energy of benzene with sign. Heat evolved for hydrogenation of three double bonds
$=-205-152=-357 kJ mol ^{-1}$
$\Delta H _{ \text{hyd }}^{\circ}=\frac{-357}{3}=-119 kJ /$ mole