How Do You Calculate the Resonance Energy of Benzene?

[utkarshakash]

Homework Statement

Determine the resonance energy of gaseous benzene from the following data
ΔH^{o}_{diss}(H,g)=435.9kJmol^{-1}, ΔH^{o}_{sub} C_{graphite}=718.4kJmol^{-1}<br /> ΔH^{o}_{f} Benzene(g)=82.9kJmol^{-1}

ε_{C-H}=416.3kJmol^{-1},ε_{C-C}=331.4kJmol^{-1},ε_{C=C}=591.3kJmol^{-1}

Homework Equations

The Attempt at a Solution

I have the following reaction

6C+3H_{2} \rightarrow C_{6}H_{6}

To calculate ΔH
\left[ 6 \times ΔH^{o}_{sub} C_{graphite} + 6 \times ΔH^{o}_{diss}(H,g) \right] - \left[ 6 \times ε_{C-H} + 3 \times ε_{C-C} + 3 \times ε_{C=C} \right]

Now resonance energy can be calculated by
R.E. = ΔH_{observed}-ΔH_{calculated}

Substituting the values does not give me the correct answer.

 

[AGNuke]

To calculate ΔH
\left[ 6 \times ΔH^{o}_{sub} C_{graphite} + 3 \times ΔH^{o}_{diss}(H,g) \right] - \left[ 6 \times ε_{C-H} + 3 \times ε_{C-C} + 3 \times ε_{C=C} \right]

I believe you are dissociating 3 H-H bonds.

 

[utkarshakash]

To calculate ΔH
\left[ 6 \times ΔH^{o}_{sub} C_{graphite} + 3 \times ΔH^{o}_{diss}(H,g) \right] - \left[ 6 \times ε_{C-H} + 3 \times ε_{C-C} + 3 \times ε_{C=C} \right]

I believe you are dissociating 3 H-H bonds.

OK So I got my answer(not exactly but very close) but the answer given in my book is positive unlike mine.

 

[AGNuke]

Its because you did mistake there again.

R.E. = H(calculated) - H(observed)