[La-Lu]

Hey there I was wondering if someone could help here!

I'm sure the answer to this problem is very simple but I did do A-levels so it's all new to me  I just started uni and I'm kind of struggling...

So:

Calculate the standard enthalpy (in kJ mol-1) of the oxidation of ammonia from the following information:
∆Hf0 values (kJ mol-1):
NH3(g) -46.11;
H2O(g) -241.82:
NO(g) +90.25
Reaction is
4NH3(g) + 5O2(g) -- > 4NO(g)+ 6H2O(g)


a) -1267.0   b) -1090.0   c) -1274.0   d) -905.5   e) -1996.0

How on earth am I supposed to do this?  I know it's got something to do with Hess's law but without the enthalpy for the whole reaction I really don't know where to start...

Please someone that feels like reviewing this little thing five minutes for my salvation!
Many thanks! 

[magician4]

my problem is NOT how to solve this, but is how to give you a hint without violating forum rules by solving this for you instead ...

regards

Ingo

[sjb]

Hey there I was wondering if someone could help here!

I'm sure the answer to this problem is very simple but I did do A-levels so it's all new to me  I just started uni and I'm kind of struggling...

So:

Calculate the standard enthalpy (in kJ mol^-1) of the oxidation of ammonia from the following information:
∆Hf0 values (kJ mol-1):
NH₃(g) -46.11;
H₂O(g) -241.82:
NO(g) +90.25
Reaction is
4NH₂(g) + 5O₂(g)  4NO(g)+ 6H₂O(g)


a) -1267.0   b) -1090.0   c) -1274.0   d) -905.5   e) -1996.0

How on earth am I supposed to do this?  I know it's got something to do with Hess's law but without the enthalpy for the whole reaction I really don't know where to start...

Please someone that feels like reviewing this little thing five minutes for my salvation!
Many thanks! 

So, what would the ΔH be for the reaction NH₃(g)  1/2N₂ (g) + 3/2H₂ (g)

[magician4]

hmmm...let's try this..



let's say you wanted to barter

you have apples worth  3$/piece , water "for free" per liter ... and you wanted to trade this for bananas ( 2 $ a piece) and oranges (1 $ a piece)

you agree to do it like this:

you trade 3 apples and 10 L of water of yours for 4 bananas plus 2 apples

calculate your net loss/gain in $

[La-Lu]

hmmm...let's try this..



let's say you wanted to barter

you have apples worth  3$/piece , water "for free" per liter ... and you wanted to trade this for bananas ( 2 $ a piece) and oranges (1 $ a piece)

you agree to do it like this:

you trade 3 apples and 10 L of water of yours for 4 bananas plus 2 apples

calculate your net loss/gain in $

you sell $9 worth of apples and get 14$ worth fruit?

Btw thanks for the replies 

[La-Lu]

hmmm...let's try this..



let's say you wanted to barter

you have apples worth  3$/piece , water "for free" per liter ... and you wanted to trade this for bananas ( 2 $ a piece) and oranges (1 $ a piece)

you agree to do it like this:

you trade 3 apples and 10 L of water of yours for 4 bananas plus 2 apples

calculate your net loss/gain in $

hmmm...let's try this..



let's say you wanted to barter

you have apples worth  3$/piece , water "for free" per liter ... and you wanted to trade this for bananas ( 2 $ a piece) and oranges (1 $ a piece)

you agree to do it like this:

you trade 3 apples and 10 L of water of yours for 4 bananas plus 2 apples

calculate your net loss/gain in $

you sell $9 worth of apples and get 14$ worth fruit?

Btw thanks for the replies 

OMG I JUST DID THIS

so basically I was doing random calculations and I got

(6x(-241.82))-(4x90.25)-(4x(-46.11)) = 905.48

but could someone explain why I have to subtract the ∆Hf0 values of NH3 and NO from H2O? why would I do that?

[magician4]

you sell $9 worth of apples and get 14$ worth fruit?

your problems seem to go deeper than just understanding chemistry..

lol but how am I supposed to calculate (...)

the "prices" for each item are given to you, and the "trade agreement" too
(think what the analogy of "water for free " might be!)

... and if you still can't see the analogy in all of this , I can't help it....

btw.:

if the reaction has NH2 and H2O...

how comes  you did correct you original error in your first posting all by yourself, rightly replacing NH₂ with NH₃
... but still insist that there was a problem with NH₂
what NH₂ ? there is no NH₂ here at all to worry about!

but could someone explain why I have to subtract the ∆Hf0 values of NH3 and NO from H2O? why would I do that?

calculate the "value" of what you gain in total and what you have to have to "pay" in total, and do a subtraction, for heavens sake !

regards

Ingo