Calculate the resonance energy of gaseous benzene from the given data
`DeltaH_(C-H)= 416, DeltaH_(C-C)=331 , DeltaH_(C=C)= 591 , DeltaH_("sub")(C, "graphite")=718`
`DeltaH_(diss)(H_(2),g)=436` and `DeltaH_(f)("benzene",g)=83` (all unit in `kJ "mole"^(-1)`)

Options

A. 354
B. 271
C. 437
D. 83

Explanation

To calculate the resonance energy of gaseous benzene, we will follow these steps:

### Step 1: Write the formation reaction of benzene
The formation of gaseous benzene from graphite and hydrogen can be represented as:
\[ 6 \text{C (graphite)} + 3 \text{H}_2 (g) \rightarrow \text{C}_6\text{H}_6 (g) \]

### Step 2: Calculate the enthalpy change for the formation reaction (\( \Delta H_1 \))
Using the bond enthalpies provided:
- Each benzene molecule has 6 C-H bonds and 3 C-C bonds (considering resonance structures).
- The bond enthalpy for C-H is 416 kJ/mol, C-C is 331 kJ/mol, and C=C is 591 kJ/mol.

The enthalpy change for breaking the bonds in benzene can be calculated as:
\[ \Delta H_1 = -\left(6 \times \Delta H_{C-H} + 3 \times \Delta H_{C-C}\right) \]
Substituting the values:
\[ \Delta H_1 = -\left(6 \times 416 + 3 \times 331\right) \]
\[ \Delta H_1 = -\left(2496 + 993\right) = -3489 \text{ kJ/mol} \]

### Step 3: Calculate the enthalpy change for sublimation of graphite (\( \Delta H_2 \))
The sublimation of graphite to gaseous carbon can be represented as:
\[ 6 \text{C (graphite)} \rightarrow 6 \text{C (g)} \]
The enthalpy change for this reaction is:
\[ \Delta H_2 = 6 \times \Delta H_{\text{sub}}(C, \text{graphite}) = 6 \times 718 = 4308 \text{ kJ/mol} \]

### Step 4: Calculate the enthalpy change for dissociation of hydrogen (\( \Delta H_3 \))
The dissociation of hydrogen gas to form atomic hydrogen can be represented as:
\[ 3 \text{H}_2 (g) \rightarrow 6 \text{H (g)} \]
The enthalpy change for this reaction is:
\[ \Delta H_3 = 3 \times \Delta H_{\text{diss}}(H_2) = 3 \times 436 = 1308 \text{ kJ/mol} \]

### Step 5: Combine the enthalpy changes to find the overall reaction enthalpy
Now, we can combine the enthalpy changes:
\[ \Delta H_{\text{formation}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H_{\text{formation}} = -3489 + 4308 + 1308 = 1127 \text{ kJ/mol} \]

### Step 6: Calculate the resonance energy
The resonance energy can be calculated using the formula:
\[ \text{Resonance Energy} = \Delta H_f^{\text{calculated}} - \Delta H_f^{\text{theoretical}} \]
Where:
- \( \Delta H_f^{\text{calculated}} = 1127 \text{ kJ/mol} \)
- \( \Delta H_f^{\text{theoretical}} = 83 \text{ kJ/mol} \)

Substituting the values:
\[ \text{Resonance Energy} = 1127 - 83 = 1044 \text{ kJ/mol} \]

### Final Answer
The resonance energy of gaseous benzene is **1044 kJ/mol**.