Question

A certain material's temperature increases by 1.0 C for every 1560J that it gains. A 0.1964g sample of quinone (molar mass= 108.1 g/mol) was burnt, and the surrounding material's temperature increased from 20.3C to 23.5C. Find the molar heat of reaction for the combustion of quinone.

Answer 1

1) The surrounding's temperature increased by 3.2ºC, So, the heat absorbed by the surroundings is:

q = 3.2C (1560 J/C) = 4992 J

So, the combustion of that small sample released -4992 J of heat.

To determine the molar heat of combustion, convert the mass of the sample into moles:
0.1964 g / 108.1 g/mol = 1.800X10^-3 moles

Now, divide the heat released by the sample by the moles of that sample:
4992 J / 1.800X10^-3 mol = 2.773 X 10^6 J/mol = 2.733 X 10^3 kJ/mol
(That seems awfully large to me, but I don't know what the actual value should be.)

or

1. Ans:
Q = 1560 J for every DT = 1oC
Mass of quinone, Mq = 0.1964g
Molar mass = 108.1g/mol
Nos of moles = 0.1964 / 108.1
= 1.817 mmol
= 0.001817 moles
T1 = 20.3oC
T2 = 23.5oC
Heat of combustion, Hcomb = Q required to raise the T of the material from T1 to T2
DT = [ 23.5 - 20.3 ] = 3.2oC
Q per DT = 1oC, is given as 1560 J/oC
Thus, for DT = 3.2oC,
Q = 1560 x 3.2 = 4992 J
Hcomb = Q = 4992 J

Thus, for 0.1964g of Quinone, Q = 4992 J
0.1964g = 0.001817 moles
Molar Hcomb = Q / nos of moles
Molar Hcomb = 4992 / 0.001817
Molar Hcomb = 2747385.801 J/mol

Molar Hcomb = 2747.386 kJ/mol