Question

A 2.250 −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 22.04 ∘C to 29.33 ∘C.

A)What is the heat of combustion per gram of quinone?

B)What is the heat of combustion per mole of quinone?

Answer 1

heat released by reaction = Ccalorimeter * delta T

= 7.845 KJ/oC * (29.33 - 22.04) oC

= 57.190 KJ

Since this is heat released, it must be negative

So,

Q = - 57.190 KJ

A)

Heat of combustion = Q / Mass

= - 57.190 KJ / 2.250 g

= -25.42 KJ/g

Answer: -25.42 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)= 2.250 g

number of mol of C6H4O2,

n = mass of C6H4O2/molar mass of C6H4O2

=(2.25 g)/(108.092 g/mol)

= 2.082*10^-2 mol

Heat of combustion = Q / Mass

= - 57.190 KJ / (2.082*10^-2 mol)

= -2747 KJ/mol

Answer: -2747 KJ/mol