Propan-1-ol (C3H7OH\text{C}_3\text{H}_7\text{OH}C3H7OH) is a liquid at room temperature and pressure. The equation for the complete combustion of propan-1-ol is:
C3H7OH(l)+4.5O2(g)→3CO2(g)+4H2O(l) \text{C}_3\text{H}_7\text{OH}(\text{l}) + 4.5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l}) C3H7OH(l)+4.5O2(g)→3CO2(g)+4H2O(l)Propan-1-ol is burned in a spirit burner in an experiment to determine its ΔH\Delta HΔH of combustion. The heat from the burner is used to heat a beaker containing water.
State what the symbol ΔH\Delta HΔH represents.
The table shows the results of the experiment:
| Variable | Value |
|---|---|
| mass of water heated | 220 g220\text{ g}220 g |
| mass of propan-1-ol burned | 1.35 g1.35\text{ g}1.35 g |
| initial temperature of water | 19.8 ∘C19.8\ ^\circ\text{C}19.8 ∘C |
| final temperature of water | 51.3 ∘C51.3\ ^\circ\text{C}51.3 ∘C |
Use the following formula to calculate the heat produced (in J) when 1.35 g1.35\text{ g}1.35 g of propan-1-ol is burned in this experiment:
heat produced (J)=mass of water (g)×4.2×temperature rise of water (∘C) \text{heat produced (J)} = \text{mass of water (g)} \times 4.2 \times \text{temperature rise of water (}^\circ\text{C)} heat produced (J)=mass of water (g)×4.2×temperature rise of water (∘C)A student uses the value from part (b) to calculate ΔH\Delta HΔH for the combustion of propan-1-ol as −1294 kJ/mol-1294\text{ kJ/mol}−1294 kJ/mol. The data book value is −2021 kJ/mol-2021\text{ kJ/mol}−2021 kJ/mol.
What is the significance of the negative sign for ΔH\Delta HΔH?
The student notices that at the end of the experiment, the bottom of the beaker is covered in black soot (carbon). Suggest how this soot is formed.
Explain how the formation of the soot may account for the difference between the experimental value of ΔH\Delta HΔH and the value in the data book.
Suggest one other reason why the two ΔH\Delta HΔH values are different.
