Question

A 1.148 gram sample of benzoic acid is burned in an excess of oxygen gas in a bomb calorimeter. The temperature of the calorimeter rises from 24.96 to 30.25 degrees Celsius. The heat of combustion of benzoic acid is -26.42 kJ/g. In a second experiment, a 0.895 gram powdered coal sample is burned in the same calorimeter assembly. The temperature rises from 24.98 to 29.73 degrees Celsius. How many kilograms of this coal would have burned to liberate 1.00E^9 kJ of heat.

Answer 1

Answer on Question #52881 – Chemistry – General chemistry

Question:

A 1.148 gram sample of benzoic acid is burned in an excess of oxygen gas in a bomb calorimeter. The temperature of the calorimeter rises from 24.96 to 30.25 degrees Celsius. The heat of combustion of benzoic acid is -26.42 kJ/g. In a second experiment, a 0.895 gram powdered coal sample is burned in the same calorimeter assembly. The temperature rises from 24.98 to 29.73 degrees Celsius. How many kilograms of this coal would have burned to liberate 1.00E^9 kJ of heat.

Solution:

Heat released after combustion of benzoic acid is: $q_{1} = m_{1} \times \Delta H_{1} = 1.148 \, \text{g} \times (-26.42 \, \text{kJ/mol}) = 30.33 \, \text{kJ}$, where $\Delta H_{1}$ – the enthalpy of combustion of benzoic acid.

If $\Delta T_{1}$ is the first change of the temperature, the calorimeter constant is:

$$K = q_{1} / \Delta T_{1} = 30330 \, \text{J} / 5.29 \, \text{K} = 5733 \, \text{J/K}$$

Burned coal gives: $\Delta T_{2} = 29.73 \, \text{K} - 24.98 \, \text{K} = 4.75 \, \text{K}$, thus $q_{2} = \Delta T_{2} \times K$ and $\Delta H_{2} = \Delta T_{2} \times K / m_{2}$, where $\Delta H_{2}$ – the enthalpy of combustion of the coal.

$$\Delta H_{2} = 4.75 \, \text{K} \times 5733 \, \text{J/K} / 0.895 \, \text{g} = 30426.54 \, \text{J/g}$$

The mass of the coal needed to liberate $10^{9}$ kJ is: $m = q / \Delta H_{2} = 10^{12}$ J / 30426.54 J/g = 32870 kg

Answer: 32870 kg

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