For the reaction, `C_(7)H_(8)(I) + 9O_(2)(g) rarr 7 CO_(2)(g) + 4H_(2)O(I)`, the calculated heat of reaction is 232 kJ `mol^(-1)` and observed heat of reaction is 50.4 kJ `mol^(-1)`, then the resonance energy is

Options

A. `-182.2 kJ mol^(-1)`
B. `+182.2 kJ mol^(-1)`
C. 172 kJ `mol^(-1)`
D. `-172 kJ mol^(-1)`

Answer

To find the resonance energy for the reaction \( C_7H_8 (l) + 9 O_2 (g) \rightarrow 7 CO_2 (g) + 4 H_2O (l) \), we will use the formula for resonance energy, which is given by:

\[ \text{Resonance Energy} = \text{Observed Heat of Reaction} - \text{Calculated Heat of Reaction} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Calculated heat of reaction = 232 kJ/mol
- Observed heat of reaction = 50.4 kJ/mol

2. **Substitute the values into the resonance energy formula**:
\[ \text{Resonance Energy} = 50.4 \, \text{kJ/mol} - 232 \, \text{kJ/mol} \]

3. **Perform the subtraction**:
\[ \text{Resonance Energy} = 50.4 - 232 = -181.6 \, \text{kJ/mol} \]

4. **Interpret the result**:
The negative value of the resonance energy indicates that the compound is stabilized due to resonance.

### Final Answer:
The resonance energy is approximately \(-181.6 \, \text{kJ/mol}\).

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