Calculations, Molar Heats of Combustion

[vrod21]

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Anyone have any tips or step by step instructions for calculating heats of combustion of alkanes/alkenes

[gigglinJess]

Split the the information you have up into;
- Initial Temperature
- Final Temperature
- Mass of water heated
- Specific heat of water (4.18 x 10^3 J kg^-1 K^-1)

Now use this information in your formula
ΔH = -mcΔT

where
m = mass of water
c = 4.18 x 10^3
ΔT = change in temperature (final temp - initial temp)

Hope this helps

[andreas_]

I call that 'ΔH = -mcΔT' equation Q=m.c.ΔT. I have no idea if it's any different but its just the one I have in my text book.

when you find Q you can then get ΔH with
Q/no of mol

that way I can find the mass required of fuel to heat the water. There have been a couple of questions that are structured so that you have to work in reverse I've found in past papers.

[Fortify]

Molar Heat of Combustion = Delta H / Moles of fuel burnt.

[boxhunter91]

Quick question guys. When finding Delta H do you times by 4.18x10^3 or just 4.18? Because I keep getting extra 1000J and i'd assume it is from their?

[Top Secret]

Originally Posted by boxhunter91

Quick question guys. When finding Delta H do you times by 4.18x10^3 or just 4.18? Because I keep getting extra 1000J and i'd assume it is from their?

Answer could be in kJ...

Check your working, and see if all the symbols (ml, kg, K etc) are standard and correct.

[jetblack2007]

You have to make sure that the units match up. kg, K etc. (even though there really isn't much of a difference between K and °C)

[untouchablecuz]

to avoid playing around the value for specific heat capacity (i.e. to keep it a constant 4.18), remember it this way:

if the value for m we use is in grams, then the value for ΔH will be in joules

if the value for m we use is in kilograms, then the value for ΔH will be in kilojoules

[untouchablecuz]

Originally Posted by jetblack2007

(even though there really isn't much of a difference between K and °C)

nah, there woudn't be because the equation uses the change in the temperature, which will be the same regardless of whether it was in °C or K

[boxhunter91]

In 2001 Paper the molar heat of combustion question i got an answer of:
-837,235,826.1 Joules/mol.
I did this by delta H/n
so: -250x4.18x10^3x40/(2.3/46.068)
The answer says -838kj/mol
Is my answer essentially the same thing?

[jetblack2007]

Originally Posted by boxhunter91

In 2001 Paper the molar heat of combustion question i got an answer of:
-837,235,826.1 Joules/mol.
I did this by delta H/n
so: -250x4.18x10^3x40/(2.3/46.068)
The answer says -838kj/mol
Is my answer essentially the same thing?

Yes, but writing it in kJ is prettier. You can convert afterwards. Don't forget to round to the least amount of significant figures in the data.

[boxhunter91]

Thanks alot jet, top secret and untouchable.
Really appreciate it.

[Aerath]

Originally Posted by boxhunter91

Thanks alot jet, top secret and untouchable.
Really appreciate it.

Yeah, significant figures is important - although, it's probably only one question in the entire exam.

[Aquawhite]

Originally Posted by Aerath

Yeah, significant figures is important - although, it's probably only one question in the entire exam.

It will only be one question that it'll happen. The exam may have up to 4-5 questions which want you to round to a certain amount of significant figures... You will only be penalised on one of these questions as only one will actually be testing that as a mark if it's wrong.

Same idea for maths.

P.S. I love doing Molar Heat calculations... they're simple and fun to do with a practical task.