Question

A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) If the heat capacity of the calorimeter and contents was 10.5 kJ / oC and the temperature of the calorimeter rose from 25.00 to 53.13, (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.

Answer 1

The calorimeter absorbed -295.365 kJ, and the ∆H for the combustion of 1 mole of benzene is -3271.07 kJ/mol. The calculation considered the temperature change, the heat capacity of the bomb calorimeter, and the number of moles of benzene burned.

Step-by-step explanation:

To determine the ∆H of the combustion of benzene in a bomb calorimeter, we use the measured temperature change and the heat capacity of the calorimeter. Since the temperature increased from 25.00 °C to 53.13 °C, we have a ∆T of 53.13 °C - 25.00 °C = 28.13 °C.

Using the formula q = Ccalorimeter × ∆T, where q is the heat absorbed by the calorimeter, and given that the heat capacity Ccalorimeter is 10.5 kJ/°C, we can calculate the total heat absorbed by the calorimeter during the reaction as follows:

q = 10.5 kJ/°C × 28.13 °C = 295.365 kJ.

Since the calorimeter absorbs the heat released by the reaction, the total heat released by the combustion of benzene sample is q = -295.365 kJ. To find the ∆H per mole of benzene, we need to divide this by the number of moles of benzene burned.

The molar mass of benzene (C6H6) is 78.11 g/mol. For a 7.05 g sample, the number of moles is:

n = mass / molar mass = 7.05 g / 78.11 g/mol = 0.0903 mol

Since the reaction involves the combustion of 2 moles of benzene, the ∆H for the reaction is for 2 moles. We thus have:

∆H = q / (n/2) = -295.365 kJ / (0.0903 mol / 2) = -6542.14 kJ/mol (for 2 moles of benzene)

Therefore, the ∆H for the combustion of 1 mole of benzene is -3271.07 kJ/mol.

Answer 2

(1) ΔH of the reaction:

• Total: 295.4 kJ
• Molar: 3,270 kJ/mol

(2) Calorimeter + content: the hardware + water

(3) Type of process: exothermic.

Step-by-step explanation:

(1) what is the ΔH of the reaction?

The enthalpy of the reaction will be equal to the heat absorbed by the calorimeter.

The heat absorbed by the calorimeter will be the product of the heat capacity of the calorimeter by the increase of temperature:

• ΔHrxn = Heat capacity × ΔT

• ΔHrxn = 10.5 kJ/ºC × (53.13ºC - 25.00ºC) = 295.365kJ ≈ 295.4 kJ

Taht is the total ΔH of the reactions but generally they are reported on a molar basis.

That means that you must divide by the number of moles of the sample.

The molar mass of benzen is 78.11g/mol

Then:

• number of moles = 7.05 g / (78.11 g/mol) = 0.090257 mol

• ΔHrxn = 3,272KJ/mol

Rounding to 3 significant figures:

• ΔHrxn = 3,270 kJ/mol

(2) Describe the calorimeter + contents:

The calorimeter is a bomb calorimeter, which consists of an insulated container, made with resistant materials, with water within it. The materials must be resistant because inside will be performed vigorous reactions, which produce large amount of energy; normally combustion reactions.

When the reaction occurs the water and the calorimeter will absorb the heat of the reaction and both will increase their temperature.

That is why the bomb calorimeters are calibrated to determine the heat capacity of the calorimeter and the content.

The calorimeter must be equipped whith a small device to ignite the reaction, which is part of the hardware.

(3) the type of process.

The process is exothermic because the reaction, like any combusion reaction, is releasing energy, which is what makes that the temperature of the calorimeter and its content increase.