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  • Calculate the resonance energy of gaseous benzene (in kJ/mol) from the following data. <img alt="\\BE(C - H) = 416.3\ \textup{kJmol}^{-1} \\BE(C - C) = 331.4\ \textup{kJmol}^{-1} \\BE(C =

Calculate the resonance energy of gaseous benzene (in kJ/mol) from the following data.

\\BE(C - H) = 416.3\ \textup{kJmol}^{-1} \\BE(C - C) = 331.4\ \textup{kJmol}^{-1} \\BE(C = C) = 591.1\ \textup{kJmol}^{-1} \\ \Delta_{\textup{sub}} H(C, \textup{graphite}) =718.4\ \textup{kJmol}^{-1} \\ \Delta_{\textup{BE}} H(H_2, \textup{g}) =435.9\ \textup{kJmol}^{-1}\\ \Delta_{f} H^o( \textup{benzene, g} ) =82.9\ \textup{kJmol}^{-1}

Option: 1

269.9


Option: 2

-269.9


Option: 3

352.8


Option: 4

-352.8


Answers (1)

best_answer

To compute resonance energy, we compare the calculated value of \Delta_fH^\ominus(\textup{benzene, g}) with the given one. To calculate \Delta_fH^\ominus(\textup{benzene, g}), we add the following reactions.

\mathrm{(a)\ 6C(graphite) \longrightarrow 6C(g), \Delta H= 6\times 718.4}

\mathrm{(b)\ 3H_2(g) \longrightarrow 6H(g), \Delta H= 3\times 435.9}

\mathrm{(c)\ 6C(g)+6H(g) \longrightarrow C_6 H_6(g), \Delta H= -3\times BE_{(C-C)}-3\times BE_{(C=C)}-6\times BE_{(C-H)}}

Adding (a+b+c)

The resultant equation will be 

\mathrm{6C(graphite) + 3 H_2(g) \longrightarrow C_6H_6(g)}  

\mathrm{\Delta_fH^o = -(3BE_{(C-C)} + 3BE_{(C=C)} + 6BE_{(C-H)}) + (6\times 718.4 + 3\times 435.9)}

putting the value of the respective bond energy values

\mathrm{\Delta_fH^o = -(3\times331.4 + 3\times 591.1 + 6\times 416.3) + (6\times 718.4 + 3\times 435.9)}

\mathrm{\Delta_fH^o = 352.8\ kJ\ mole^-}

We have given, 

\mathrm{\Delta_fH^o(\textup{benzene, g}) = 82.9\ kJ\ mol^-}

The difference  gives us the resonance energy, which is equal to

\mathrm{Resonance\ Energy=\Delta_fH^o(benzene, g)_{theoretical} - \Delta_fH^o\({benzene, g})_{actual}}

\therefore \mathrm{Resonance\ Energy=352.8-82.9 = 269.9\ kJ\ mol^-}

Hence, option number (1) is correct

Posted by

manish

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