A 0.92-g sample of caffeine C8H10N4O2 burns in a constant-volume calorimeter that has a heat capacity of 7.85 kJ/K. The temperature increases from 298.15 K to 303.19 K. What is the molar heat of combustion of caffeine (in kJ).
2007-10-17 19:49:51 · 2 answers · asked by fpsfynest 1 in Science & Mathematics ➔ Chemistry
Heat released = mcT, where m=mass, c=heat capacity, T=temperature change
So Heat released
= 0.92 x 7.85 x (303.19-298.15)
= 36.399 kJ (unit in kJ since your heat capacity given is in kJ/K)
Relative molar mass of caffeine
= 8(12.0) + 10(1.0) + 4(14.0) + 2(16.0)
= 194
Amount of caffeine used
= 0.92/194
= 0.00474 mol
Molar heat of combustion
= - Heat released /amount used
= - 36.399 / 0.00474
= - 7679.1 kJ/mol (remember the negative sign in front since heat is released during combustion)
2007-10-17 20:28:53 · answer #1 · answered by musical_bell 3 · 0⤊ 0⤋
1. multiply the heat capacity by the temperature rise.
2. work out the Mr of caffeine.
3. work out the moles of caffeine.
4. divide answer 1 by answer 3.
2007-10-18 03:01:14 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋