Question

18. Propane (CH) burns in oxygen to form carbon dioxide and water. A2.00 sample of propane was burned bomb calorimeter whose total heat capacity is 15.77 kJ/K. The temperature of the calorimeter apparatus Inc 6.34 K Calculate the heat of combustion per mole of propane. (a) -221 x 10' /mol (b)-5.00 K TO' 'mol (c)-1.00 x 10' /mol (d) -2.21 x 10' mol (e)-4.41 x 10' mol

Answer 1

the d option is correct . Explanation is shown below

since we have to find the heat of combustion of propane.

The combustion of propane produces CO2 and H2O.

C3H8 + 5O2 -------------> 3CO2 + 4H2O.

Since, the combustion takes place in bomb calorimeter we have ,

The heat capacity of bomb calorimeter = 15.77 KJ/K

Rise in temperature = 6.34 K

Molar mass of compunds = 44.1 g/mol / 2 gm

Therefore,

The heat of combustion = heat of capacity of calorimeter * Rise in temperature * molar mass of compund.

= 15.77 KJ/K *6.34 K * 44.1 g/ mol / 2 gm

=- 2204.59 KJ/mol

=- 2.20459*10^3 KJ/mol

= -2.21 * 10^3 KJ/mol.

therefore the d option is correct.