A bomb calorimeter can be used for accurate determination of the heat change during combustion of a fuel - AQA - A-Level Chemistry - Question 5 - 2020 - Paper 3

For octane, we again use the relationship between heat and temperature change:

1. Using the previously calculated value for C_cal, we can find the heat change:

   $q = C_{cal} \times \Delta T$

   Where:

   • C_cal = 6.52 kJ K^{-1} (not using the correct value from question 05.1)
   • ΔT = 12.2 °C

   Plugging in the values:

   $q = 6.52 \text{ kJ K}^{-1} \times 12.2 \text{ K} = 79.7 \text{ kJ}$

2. Now calculate for the moles of octane:

   $\text{Moles of octane} = \frac{2.00 \text{ g}}{114.0 \text{ g mol}^{-1}} = 0.0175 \text{ mol}$

   The heat change per mole will be:

   $\text{Heat change per mole} = \frac{79.7 \text{ kJ}}{0.0175 \text{ mol}} \approx 4543.4 \text{ kJ mol}^{-1}$

The heat change calculated from the bomb calorimeter experiment is not an enthalpy change because the bomb calorimeter operates at constant volume. Enthalpy change requires constant pressure, while the bomb calorimeter measures the heat released at constant volume, hence it reflects internal energy change rather than enthalpy.

The percentage uncertainty can be calculated using the formula:

$\text{Percentage Uncertainty} = \frac{\text{Uncertainty}}{\text{Measured Value}} \times 100$

1. Given the uncertainty is ±0.1 °C and the measured temperature change is 12.2 °C:

   $\text{Percentage Uncertainty} = \frac{0.1}{12.2} \times 100 \approx 0.82\%$

To decrease the percentage uncertainty while using the same thermometer, one could perform the experiment at a higher temperature change. A larger temperature change would reduce the relative impact of the thermometer's uncertainty, thereby minimizing the percentage uncertainty in the measurement.