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#1
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5 years ago
#2
do you have the answer? is it -312?
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#3
(Original post by Nightinwind)
do you have the answer? is it -312?
No I don't have the answer, I am really trying to work it out?

Do I use PV = nRT? (the gas law)
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#4
(Original post by Nightinwind)
do you have the answer? is it -312?
What was your method for this answer? I don't know if it's correct, but you might be right!
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5 years ago
#5
I looked up my notes for Hess's law as I don't quite understand combustion etc. to do with enthalpy, but here is the answer which I got. (attached)

Incase you're wondering where the bit comes from in which the arrows are pointing to, well thats what happens when it combusts and the reason as to why I didn't state where abouts the Oxygen comes from is because it burns in an excess anyway, so it isn't relevant in the equation as it doesn't affect the reaction with or without it being included.
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#6
(Original post by ThomasOakes)
I looked up my notes for Hess's law as I don't quite understand combustion etc. to do with enthalpy, but here is the answer which I got. (attached)

Incase you're wondering where the bit comes from in which the arrows are pointing to, well thats what happens when it combusts and the reason as to why I didn't state where abouts the Oxygen comes from is because it burns in an excess anyway, so it isn't relevant in the equation as it doesn't affect the reaction with or without it being included.
-86 is correct with regards to Hess's Law, and was the answer to part A.

But this question is for part B, and the 6g must have some relation to the working out
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5 years ago
#7
(Original post by Sam00)
-86 is correct with regards to Hess's Law, and was the answer to part A.

But this question is for part B, and the 6g must have some relation to the working out
I see!

What specification is this question for and also, what does the question say word for word from the start, so from part (a) to (b)
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#8
(Original post by ThomasOakes)
I see!

What specification is this question for and also, what does the question say word for word from the start, so from part (a) to (b)
It's just a generic revision question paper I was given at college.

Here is the full question which I am still working on: https://www.flickr.com/photos/140547...posted-public/
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#9
Actually I think the equation I need to use is q = mc(deltaT)

Still not sure though
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5 years ago
#10
(Original post by Sam00)
Actually I think the equation I need to use is q = mc(deltaT)

Still not sure though
Okay so I've found it out. Basically, we are told that 1 mole of C2H6 produces -1560Kj/mol
So, from this we know that 1 mole of C2H6 is 30g as (12x2)+6=30
From this we now know that 30g of C2H6 gives out -1560Kj/mol.
So to get 6g we just divide by 5, so 6g of C2H6 gives out -312Kj/mol.
1
5 years ago
#11
6*(-1560)/30 = -315 kJ
1
#12
(Original post by ThomasOakes)
Okay so I've found it out. Basically, we are told that 1 mole of C2H6 produces -1560Kj/mol
So, from this we know that 1 mole of C2H6 is 30g as (12x2)+6=30
From this we now know that 30g of C2H6 gives out -1560Kj/mol.
So to get 6g we just divide by 5, so 6g of C2H6 gives out -312Kj/mol.
Thanks
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5 years ago
#13
(Original post by ThomasOakes)
Okay so I've found it out. Basically, we are told that 1 mole of C2H6 produces -1560Kj/mol
So, from this we know that 1 mole of C2H6 is 30g as (12x2)+6=30
From this we now know that 30g of C2H6 gives out -1560Kj/mol.
So to get 6g we just divide by 5, so 6g of C2H6 gives out -312Kj/mol.

You mean just kJ?
0
5 years ago
#14
(Original post by B_9710)
You mean just kJ?
I'm pretty sure it's kJ/mol as you still have just one mole, instead of 30g you have 6g so it should be the same units
0
5 years ago
#15
(Original post by B_9710)
You mean just kJ?
I'm pretty sure it's kJ/mol as you still have just one mole, instead of 30g you have 6g so it should be the same units
0
5 years ago
#16
You did -1560 kJ/mol multiplied by 0.2 mol

kJ/mol x mol leaves just kJ.

-1560 kJ/mol means that one mole of ethane releases 1560kJ of energy when combusted.
0.2 moles releases 312kJ of energy, kJ being a unit of energy. Hence the solution is 312kJ.
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5 years ago
#17
(Original post by ThomasOakes)
I'm pretty sure it's kJ/mol as you still have just one mole, instead of 30g you have 6g so it should be the same units
The question asks for the energy given out. The units for energy are J (or kJ) not kJmol-1. Also the answer can't be negative as it is asking for how much energy was given out - which is the magnitude of the energy change so your answer must be positive not negative.
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