A student uses a simple calorimetry apparatus to measure the temperature rise of water when propan-1-ol, CH3CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}CH3CH2CH2OH, is burned.
There are several reasons why the rise in temperature is less than the theoretical value.
One reason is the incomplete combustion of propan-1-ol to form only carbon monoxide and water. Write the balanced chemical equation for this incomplete combustion.
State another reason why the temperature rise of the water is less than expected.
The student records the following results:
Calculate the heat energy change (QQQ), in joules, using the expression:
Q=m×4.18×ΔT Q = m \times 4.18 \times \Delta T Q=m×4.18×ΔTwhere mmm is the mass of water in grams and ΔT\Delta TΔT represents the increase in temperature.
The relative molecular mass of propan-1-ol is 60.060.060.0. Use this information and your value for QQQ to calculate the molar enthalpy change, ΔH\Delta HΔH, for the combustion of propan-1-ol. Give your answer in kJ/mol\text{kJ/mol}kJ/mol and include the appropriate sign.
The student is given this table of average bond energies:
| Bond | Average bond energy (kJ/mol\text{kJ/mol}kJ/mol) |
|---|---|
| C−H\text{C}-\text{H}C−H | 413 |
| C−C\text{C}-\text{C}C−C | 347 |
| C−O\text{C}-\text{O}C−O | 358 |
| O−H\text{O}-\text{H}O−H | 464 |
| O=O\text{O}=\text{O}O=O | 498 |
| C=O\text{C}=\text{O}C=O | 805 |
The equation for the complete combustion of propan-1-ol is:
CH3CH2CH2OH+4.5O2→3CO2+4H2O \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + 4.5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} CH3CH2CH2OH+4.5O2→3CO2+4H2OUse this equation and the bond energies in the table to calculate a theoretical value for the molar enthalpy change, ΔH\Delta HΔH, for the combustion of propan-1-ol in kJ/mol\text{kJ/mol}kJ/mol.
