The enthalpy of combustion of benzoic acid `(C_(6)H_(5)COOH)` at `298 K` and `1 atm` pressure is `-2546.0 "kJ "mol"^(-1)`. What is `DeltaU` for the reaction?

Explanation

To find the internal energy change (ΔU) for the combustion of benzoic acid, we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔU):

\[ \Delta H = \Delta U + \Delta N RT \]

Where:
- ΔH = Enthalpy change of the reaction
- ΔU = Internal energy change of the reaction
- ΔN = Change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants)
- R = Universal gas constant (8.314 J/mol·K)
- T = Temperature in Kelvin

### Step-by-Step Solution:

1. **Identify the given values**:
- Enthalpy of combustion of benzoic acid (ΔH) = -2546.0 kJ/mol
- Temperature (T) = 298 K
- Universal gas constant (R) = 8.314 J/mol·K = 0.008314 kJ/mol·K (to match units)

2. **Write the balanced combustion reaction for benzoic acid**:
The combustion of benzoic acid (C₆H₅COOH) in excess oxygen produces carbon dioxide (CO₂) and water (H₂O). The balanced reaction is:
\[ C_6H_5COOH(s) + 7O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) \]

3. **Calculate ΔN**:
- Gaseous products: 6 (from 6 CO₂) + 0 (from H₂O, which is liquid) = 6 moles of gas
- Gaseous reactants: 7 (from 7 O₂) = 7 moles of gas
- Change in gaseous moles (ΔN) = moles of gaseous products - moles of gaseous reactants
\[ \Delta N = 6 - 7 = -1 \]

4. **Substitute values into the equation**:
Rearranging the equation gives us:
\[ \Delta U = \Delta H - \Delta N RT \]
Substituting the known values:
\[ \Delta U = -2546.0 \text{ kJ/mol} - (-1)(0.008314 \text{ kJ/mol·K})(298 \text{ K}) \]

5. **Calculate the term involving RT**:
\[ RT = 0.008314 \times 298 = 2.478572 \text{ kJ/mol} \]

6. **Final calculation of ΔU**:
\[ \Delta U = -2546.0 \text{ kJ/mol} + 2.478572 \text{ kJ/mol} \]
\[ \Delta U \approx -2543.521428 \text{ kJ/mol} \]
Rounding to three significant figures, we get:
\[ \Delta U \approx -2544.0 \text{ kJ/mol} \]

### Final Answer:
\[ \Delta U \approx -2544.0 \text{ kJ/mol} \]