Question

A 2.20g simple of quinone , C6H4O2 , IS burned a bomb calorimeter whose total heat capacity is 7.854KJ/°C . The temperature of the calorimeter increases from 23.44°C to 30.57°C . What is the heat combustion per gram of quinone ? Per mole of quinone ?

Answer 1

Step-by-step explanation:

Here, we want to calculate the heat of combustion per gram of quinone

By principle of thermodynamics,

The heat lost by burning the quinone = heat gained by the calorimeter

The temperature rise is:

`\Delta T\text{ = 30.57 - 23.44 = 7.13 }\degree C`

The heat of combustion per gram can be calculated as:

`Q\text{ = }(C*\Delta T)/(m)`

where C is the calorimeter heat capacity and m is the mass of the quinone

Thus,we have this as:

`Q\text{ = }(7.854*7.13)/(2.2)=-25.454KJ\text{ /g.}\degree C`

The heat is negative as it is exothermic since it is given off

To get the heat of combustion per mole, we have to multiply the heat of combustion per gram by the molar mass of quinone

The molar mass of quinone is 108 g/mol

So, we have the heat of combustion per mole calculated as:

`Q\text{ = }(7.854*7.13)/((2.2)/(108))\text{ = -2,749.043 KJ/mol.}\degree C`