[butters]

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"

[opti384]

First think about which molecules should cancel out in order to

get 1/2  N 2(g) +   1/2 O 2(g) ->  NO(g)

[butters]

I know the concept, but I am simply stuck.
Here's what I have so far:

1/2N₂ + 1/2H₂  NH₃          H* = -45

NH₃ + 5/4O₂  NO + 6/4H₂O          H* = -226.55

H₂ + 1/2O₂    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)

[opti384]

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.

[butters]

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...

[opti384]

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  NO + 6/4 H2O

and

H2 + 1/2 O2  H2O

Just cancel out the rest.

[butters]

Okay, this is somewhat simpler, is this what you meant by whole numbers?

.5N₂ + 1.5H₂  NH₃

NH₃ + 1.25O₂  NO + 1.5 H₂O

H₂ + .5O₂  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?

[opti384]

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.

[butters]

So that would leave .5H₂ in the first reaction..
So it's:

.5N₂ + .5H₂  
1.25O₂   NO + .5H₂O
.5O₂

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!

[opti384]

.5N2 + 1.5H2  + 1.25O2     NO + 1.5 H2O

H2 + .5O2     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     H2O. The sign of H* will be changed of course.

[butters]

Hint: you might try to reverse H2 + .5O2    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..

[opti384]

1.     .5N2 + 1.5H2  + 1.25O2      NO + 1.5 H2O

2.       H2 + .5O2      H2O

If you reverse the second reaction it becomes

H2O  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.

[butters]

Yeah, I got it now.
I did:
.5N2 + .5N2  NO + 5H2O     H = -30.65
.5H2O  + .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...