Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: butters on October 30, 2010, 10:01:16 PM
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Guys, I can't figure this out. I need some help.
Please do this step by step, so I can learn how to do this.
Enthalpy changes for the following reactions can be determined experimentally.
N 2(g) + 3 H 2(g) -> 2 NH3(g) H* = -91.8 kJ
4 NH 3(g) + 5 O 2(g) -> 4 NO (g) + 6 H20( g) H* = -906.2 kJ
H 2(g) + 1/2 O 2(g) -> H20( g) H* = -241.8 kJ
Use these values to determine the enthalpy change for the formation of NO(g) from the elements (an enthalpy that cannot be measured directly because the reaction is reactant-favored) . 1/2 N 2(g) + 1/2 O 2(g) -> NO(g) H* = ?
When I do it, the values don't all cancel out, and my answer is also way off.
The correct answer is: "90.3 kj"
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First think about which molecules should cancel out in order to
get 1/2 N 2(g) + 1/2 O 2(g) -> NO(g)
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I know the concept, but I am simply stuck.
Here's what I have so far:
1/2N2 + 1/2H2 :rarrow: NH3 H* = -45
NH3 + 5/4O2 :rarrow: NO + 6/4H2O H* = -226.55
H2 + 1/2O2 :rarrow: H2O H* = -241.8 kJ
Only the NH3 cancels out in my equation. Where did I go wrong.
(only one example of this type of problem is given in my text book, so my understanding of this is minimal)
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Well, I'd rather use whole numbers because they are easy to add and subtract.
Anyway NH3 cancels out if you add the first and second reactions.
But there are errors in the first reaction. Look carefully at 1/2H2 and H* = -45.
Then all you have to do is cancel out H2 and H2O.
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Oh right, I forgot to put a 3 there.
So it's 1/2x3H2
And you are saying use whole numbers, so then I will multiply the equation whose enthalpy I want and make it a whole number?
Can I cancel out let's say H2O with H2 and leave the O intact?
You have to tell me how to do this, because the book doesn't really explain it all that well...
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It's fine using fractions but to me whole numbers look more simple. :)
Now you will have
1/2N2 + 3/2 H2 + 5/4 O2 :rarrow: NO + 6/4 H2O
and
H2 + 1/2 O2 :rarrow: H2O
Just cancel out the rest.
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Okay, this is somewhat simpler, is this what you meant by whole numbers?
.5N2 + 1.5H2 :rarrow: NH3
NH3 + 1.25O2 :rarrow: NO + 1.5 H2O
H2 + .5O2 :rarrow: H2O
The underlined terms are what I need, but only NH3 seems to cancel out.
I guess, I'm sort of unsure with the cancellation process and rules in general. For example, can I cancel out the 1.5H2 in the first equation with 1.5H2 from 1.5H2O in the second equation?
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Think of it as mathematical equations.
If you add the first and the second reaction the NH3 cancels out because there is NH3 on both sides.
For example, can I cancel out the 1.5H2 in the first equation with 1.5H2 from 1.5H2O in the second equation?
NO.
You can only cancel out the same molecules.
Hint: 1.5 H2 can cancled out from the H2 in the third reaction.
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So that would leave .5H2 in the first reaction..
So it's:
.5N2 + .5H2 :rarrow:
1.25O2 :rarrow: NO + .5H2O
.5O2 :rarrow:
But then, if I cancel out the O2 in the third reaction through the 2.25O2 in the second reaction, I won't have any O2 left.. This is very frustrating!
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.5N2 + 1.5H2 + 1.25O2 :rarrow: NO + 1.5 H2O
H2 + .5O2 :rarrow: H2O
After you cancel out NH3 you'll have above.
Now as you did before to cancel out NH3 you should match the quotients of H2 and H2O to cancel out.
Hint: you might try to reverse H2 + .5O2 :rarrow: H2O. The sign of H* will be changed of course.
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Hint: you might try to reverse H2 + .5O2 :rarrow: H2O. The sign of H* will be changed of course.
If I reverse it, wont' I lose the .5O2? And don't I need that .5O2 to complete the equation?
...Sigh.. I think you should just show me how to do this. It is the first time I have encountered a problems such as this, and I won't learn until I see an example of it..
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1. .5N2 + 1.5H2 + 1.25O2 :rarrow: NO + 1.5 H2O
2. H2 + .5O2 :rarrow: H2O
If you reverse the second reaction it becomes
H2O :rarrow: H2 + .5O2
and then we have H2O on the left and 1.5 H2O on the right. Just match the quotients and try adding the reaction.
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Yeah, I got it now.
I did:
.5N2 + .5N2 :rarrow: NO + 5H2O H = -30.65
.5H2O :rarrow: + .5H2 + .25O2 H = +120.9
and it gives:
.5N2 +.5H2 + .25O2 H = 90.25
I hadn't realized that you could only cancel out reactants with the products...