To determine the enthalpy of combustion for benzoic acid, follow these steps:
1. Identify the energy released and the number of moles reacted:
- Energy released = 49,936.64 joules
- Moles of benzoic acid reacted = 0.015475 moles
2. Convert the energy released from joules to kilojoules (since enthalpy is typically expressed in kilojoules per mole):
- Energy in kilojoules = [tex]\( \frac{49,936.64 \text{ joules}}{1000} = 49.93664 \text{ kilojoules} \)[/tex]
3. Calculate the enthalpy change (ΔH) for the combustion reaction:
[tex]\[
\Delta H_{\text{rxn}} = \frac{\text{Energy released in kJ}}{\text{Moles reacted}} = \frac{49.93664 \text{ kJ}}{0.015475 \text{ mol}}
\][/tex]
4. Perform the division to find the enthalpy of combustion:
[tex]\[
\Delta H_{\text{rxn}} = \frac{49.93664 \text{ kJ}}{0.015475 \text{ mol}} \approx 3226.923 \text{ }\frac{\text{kJ}}{\text{mol}}
\][/tex]
5. Consider the sign of the enthalpy change:
Since energy is released in the reaction, the enthalpy change for combustion is negative.
6. Express the enthalpy of combustion to three significant figures:
[tex]\[
\Delta H_{\text{rxn}} = -3226.923 \text{ }\frac{\text{kJ}}{\text{mol}} \approx -3226.923 \text{ }\frac{\text{kJ}}{\text{mol}}
\][/tex]
Rounded to three significant figures, the answer is:
[tex]\[
\boxed{-3226.923 \text{ }\frac{\text{kJ}}{\text{mol}}}
\][/tex]
So, the enthalpy of combustion for benzoic acid is [tex]\( -3226.923 \text{ }\frac{\text{kJ}}{\text{mol}} \)[/tex] with the correct sign and magnitude.
