Questions: A 0.1964 g sample of quinone (C6H4O2) is burned in a bomb calorimeter that has a heat capacity of 1.56 kJ / °C. The temperature of the calorimeter increases b 3.2°C. Calculate the heat of combustion (kJ / mole) of quinone. Ans: -2745 kJ / mol or to 2 sigfigs -2700 kJ / mol

Solution Steps

The heat absorbed by the calorimeter can be calculated using the formula: \[ q = C \Delta T \] where \( C \) is the heat capacity of the calorimeter and \( \Delta T \) is the change in temperature.

Given: \[ C = 1.56 \, \text{kJ/}^{\circ}\text{C} \] \[ \Delta T = 3.2 \, ^{\circ}\text{C} \]

Substitute the values: \[ q = 1.56 \times 3.2 = 4.992 \, \text{kJ} \]

The molar mass of quinone (\(\text{C}_6\text{H}_4\text{O}_2\)) is calculated as follows: \[ \text{Molar mass of C}_6\text{H}_4\text{O}_2 = 6 \times 12.01 + 4 \times 1.008 + 2 \times 16.00 = 108.14 \, \text{g/mol} \]

Given the mass of quinone burned: \[ \text{Mass} = 0.1964 \, \text{g} \]

Calculate the moles of quinone: \[ \text{Moles of quinone} = \frac{0.1964 \, \text{g}}{108.14 \, \text{g/mol}} = 0.001816 \, \text{mol} \]

The heat of combustion per mole of quinone is calculated by dividing the total heat absorbed by the moles of quinone burned: \[ \text{Heat of combustion} = \frac{q}{\text{moles of quinone}} = \frac{4.992 \, \text{kJ}}{0.001816 \, \text{mol}} = 2748 \, \text{kJ/mol} \]

Since the heat of combustion is typically expressed as a negative value (indicating an exothermic reaction): \[ \text{Heat of combustion} = -2748 \, \text{kJ/mol} \]

Final Answer