Question

A 2.300 −g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854kJ/∘C. The temperature of the calorimeter increases from 23.84 ∘C to 31.29 ∘C

Part A

What is the heat of combustion per gram of quinone?

Part B

What is the heat of combustion per mole of quinone?

Answer 1

A)

Heat change,

Q = C*delta T

= 7.854 KJ/oC * (31.29 - 23.84) oC

= 7.854 KJ/oC * 7.45 oC

= 58.512 KJ

mass = 2.300 g

So,

heat of combustion = Q/mass

= 58.512 KJ / 2.300 g

= 25.44 KJ/g

Answer: 25.44 KJ/g

B)

Molar mass of C6H4O2,

MM = 6*MM(C) + 4*MM(H) + 2*MM(O)

= 6*12.01 + 4*1.008 + 2*16.0

= 108.092 g/mol

mass(C6H4O2)= 2.300 g

use:

number of mol of C6H4O2,

n = mass of C6H4O2/molar mass of C6H4O2

=(2.3 g)/(1.081*10^2 g/mol)

= 2.128*10^-2 mol

So,

heat of combustion = Q/mol

= 58.512 KJ / (2.128*10^-2 mol)

= 2.750*10^3 KJ/mol

Answer: 2.750*10^3 KJ/mol