Chemistry Question 2?

Question

A 0.1964 g sample of quinine (c6H4O2) is burned in a bomb calorimeter that has a heat capacity of 1.56 kj/K. The termperature of the colorimeter increases by 3.2k Calculate the heat of combustion of quinine per gram and per mole.

Please help me, i'm clueless!

Answer 1

The temperature rise seen in the calorimiter is the energy released by combustion of the sample. They state that it takes 1.56kj to raise the temperature by one K. Since your sample raised it by 3.2K your sample released 3.2 x 1.56kj or ~ 5.0 kj. Your sample weighed 0.1964g, so a whole gram of the quinine would release 5.0kj x 1/0.1964 or 25.4kj/g.

To get the heat released per mole multiply the heat released per gram by the molar weight for quinine in grams:

Quinine is: (6 x 12.01) + (4 x 1.01) + (2 x 16.00) =

108.1g/mole

25.4kj/g x 108.1g/mole = 2750kj/mole

Answer 2

The heat output is found from the equation:

qcal=CdeltaT

Where C is the heat capacity of the calorimeter.

The heat output then is:

qcal=(1.56 kJ/K)(3.2K)
qcal=4.992 kJ

Because, this is a calorimeter, no heat is allowed to escape, and according to the First Law.

qrxn=-qcal

The heat given off by the combustion is, thus:

qrxn=-4.992 kJ

The heat of combustion per gram is:

qrxn/m = -4.992 kJ/0.1964 g = 25.42 kJ/g

The heat of combustion per mole is:

qrxn/n = -4.992 kJ/ (0.1964 g/108.092g/mol))

qrxn/n = -4.992 kJ/ 1.817e-3 mole

qrxn/n = -2747 kJ/mol