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Topic: Thermodynamics  (Read 4734 times)

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Offline Noor

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Thermodynamics
« on: October 26, 2007, 01:04:23 AM »
Ok I've done the question and want to check if I am on track or not.

Q. A 0.006141 mol sample of an organic compound was burned in oxygen in a bomb calorimeter. THe temperature of the calorimeter increased from 23.1 degress to 27.1 degrees. If the heat capacity of the calorimeter is 4.90 kJ(degrees)-1, then what is the constant volume heat of combustion of this compound, in kJ/mol[Remember to include a "+" or "-" sign in your answer, as appropriate)

Have your answer with three significant figures.

A. Here's what I've done,

since delta T = q comb x moles / heat capacity

(27.1 - 23.1) = q comb x 0.006141/ 4.9

q comb = 3191.828 kJ/mol

but my answer is not with three signi. figures?? and where am i supposed to include the + or - sign??

I am confused. HELP PLEASEEEE!!!!

Offline Noor

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Re: Thermodynamics
« Reply #1 on: October 26, 2007, 11:53:01 AM »
*Ignore me, I am impatient*!!!

Is what I am doing right??

Offline LQ43

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Re: Thermodynamics
« Reply #2 on: October 26, 2007, 12:59:00 PM »
no its not, you're confusing the heat that the calorimeter absorbs with the heat of combustion of the compound (they are equal but opposite in sign)

calculate q for the heat of the absorption of the calorimeter

q = heat capacity x delta T

this is equal but opposite in sign to the heat of combustion of your organic compound

-(q from above) = heat of combustion for .006141mol of compound

calculate the heat PER mole of compound

Offline Noor

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Re: Thermodynamics
« Reply #3 on: October 27, 2007, 07:46:04 PM »
ok so does that mean I DIVIDE 19.6 kJ (4 degrees x 4.90 kJ(degrees)(-1) ) by the number of moles??

That will give me the same answer: 3191.828 kJ/mol

but if I MULTIPLY, it gives me 0.1203636

so which one is it??

Offline LQ43

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Re: Thermodynamics
« Reply #4 on: October 27, 2007, 08:52:56 PM »
yes you divide, and you get the same answer,

but the way you had calculated it before was confusing the heat absorbed by the calorimeter with the heat of comb of the compound and so lost sight of the + or - of the heat of comb.

do you know what the sign is now? (by the way any reaction that releases heat has this sign and is called exothermic)

sorry if I may have confused you before but it seemed that your focus was the math and not the chemistry ;)

Offline Noor

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Re: Thermodynamics
« Reply #5 on: October 27, 2007, 09:14:47 PM »
Hi LQ43

I am actually trying to understand it, but on the system (where we're supposed to submit our answers), it tells me that it's wrong!!! What do they mean by 3 sig. figures? do they mean decimal places??because as you see, my answer is 4 sig figures!

And the sign is negative, yes?

Offline LQ43

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Re: Thermodynamics
« Reply #6 on: October 27, 2007, 09:20:07 PM »
yes its neg  :)

change 3191.828 kJ/mol to scientific notation

3.191828 x 10^3 kJ/mol

round this to 3 sig figs..

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