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?
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
A 2.300 −g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat...
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