I'm studying AS Chemistry via distance learning and I'm struggling with one of the questions that came up in our weekly quiz but doesn't seem to be in any of the textbooks I have.
When 1 g of propane, C3H8 (Calor gas), is burned completely, 50 kJ are transferred. What value does this give for the standard enthalpy of combustion of propane?
Any ideas? I know what the answer is, I just have no idea how to get to the answer. There just doesn't seem to be enough information in the question but I may just be being incredibly stupid.
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- Thread Starter
- 22-02-2016 01:21
- 22-02-2016 02:08
Well, the standard enthalpy of combustion is defined as the enthalpy change that occurs when one mole of a substance is burned completely in excess oxygen with all substances in their standard states under standard conditions (100kPa, 298K).
One mole of propane is 44g (Mr of C3H8 = 3(12) + 8(1) = 44).
Therefore, 1g of propane is 1/44th of a mole of propane. So if there is 50kJ of energy released when 1/44th of one mole of propane is burned, if you burn one mole of propane you will release 44 times more energy. Therefore the enthalpy change, ΔH, of the reaction = 44 x 50kJ = -2200kJ mol-1 (the sign is negative since energy is released in combustion, ie it is exothermic).
Hope that helps.