How Do You Calculate the Resonance Energy of Benzene?

[utkarshakash]

Homework Statement

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

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

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)

 

[DeeNos]

I think there may be some missing information or some assumptions made in the given data. It would be helpful to know the source of the data and any other relevant information. Additionally, it is important to note that resonance energy is not a directly measurable quantity and can vary depending on the method used to calculate it. Therefore, the exact value of resonance energy may differ depending on the approach taken.