Question

# A 1.93 g sample of caffeine (C8H10N4O2) burns in a constant-volume calorimeter that has a heat...

A 1.93 g sample of caffeine (C8H10N4O2) burns in a constant-volume calorimeter that has a heat capacity of 7.81 kJ/K. The temperature increases from 297.65 K to 303.96 K.

Part A: Determine the heat (qv) associated with this reaction.

Clue: Note that the heat capacity here is for an object (the calorimeter) rather than a material. So it does not have any mass or mole units associated with it. Be sure you have the sign correct for q!

Part B: Use the data above to find ΔE for the combustion of one mole of caffeine.

Clue: Is heat being absorbed or released in the combustion reaction?

PART A: Heat associated with the reaction = heat gained by calorimeter = heat capacity of calorimeter× increase in temperature

so; heat of reaction = (7.81 kj/ K)× (303.96-297.65)K = 49.28 kj

Part B:

Amount of caffeine : 1.93 grams

Molar mass of caffeine: 194.19 g/ mole

So; number of moles of caffeine in reaction = 1.93/194.19mole= 0.009939 mole

So; 0.009939 mole of caffeine reacts to produce 49.28 kj heat

So; heat from 1 mole of caffeine is (49.28/0.009939) kj = 4958 kj

The temperature of calorimeter increases. This is because heat is released from 1 mole of caffeine reaction.

So; delE = -4958 kj ( negative sign indicates that heat isxrejeased)

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