Question

Ok, so have this problem I'm trying to solve....

What is the missing ΔH?

2H2O (l) → 2H2 (g) + O2 (g), ΔH = 572 kJ
2N2  (g) + 3H2 (g) → 2NH3 (g), ΔH = –138 kJ
2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l), ΔH = ?
2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH = 99.0 kJ

Now I know I need to reverse some of the equations as well as balance the equations. Here's what I have so far:

2H2 (g) + O2 (g) → 2H2O (l) , ΔH = 572 kJ
2N2  (g) + 3H2 (g) → 2NH3 (g), ΔH = –138 kJ
2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH = 99.0 kJ

Can anyone walk me through this problem the rest of the way?

Answer 1

1)2x ( 2H2 (g) + O2 (g) → 2H2O (l)), ΔH1 = -572 x 2 kJ = -1144 kJ
2) 2N2 (g) + 3H2 (g) → 2NH3 (g), ΔH2 = –138 kJ
3) 2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH3 = 99.0 kJ
------------------------------------------------
4) 2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l), ΔH = ΔH1+ ΔH2 + ΔH3

Answer 2

1) 2H2O (l) → 2H2 (g) + O2 (g), ΔH1 = 572 kJ
2) 2N2 (g) + 3H2 (g) → 2NH3 (g), ΔH2 = –138 kJ
3) 2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH3 = 99.0 kJ
------------------------------------------------
4) 2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l), ΔH4 = ?

1) reverse the reaction and multiply by 2, the ΔH1 change the sign and multiply by 2
the other 2 equation remain the same.
you can cancel the 2N2 from the reactants of eq.2) wiht the product of eq 3)
cancel the O2 from eq 1) with the O2 of the products of Eq 3)

Answer 3

1)2x ( 2H2 (g) + O2 (g) → 2H2O (l)), ΔH5 = -572 x 2 kJ = -1144 kJ
2) 2N2 (g) + 3H2 (g) → 2NH3 (g), ΔH2 = –138 kJ
3) 2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH3 = 99.0 kJ
------------------------------------------------
4) 2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l), ΔH4 = ΔH5+ ΔH2 + ΔH3
ΔH4 = 2 x (-ΔH1)+ ΔH2 + ΔH3

Answer 4

Ok, so the sign for equation one changed because the equation was reversed, correct? Now I believe the purpose for multiplying 2 is to balance the equation? Just asking to make sure I understand.
As for the cancellation they should now be:
3NH3 (g) → 2NH3 (g)
2NO2 (g) → 2O2 (g)

Would that be correct?

Answer 5

the first part is correct.

this part NO
As for the cancellation they should now be:
3NH3 (g) → 2NH3 (g)
2NO2 (g) → 2O2 (g)

Answer 6

Wait a second. I missed the fact that you multiplied it. So actually it should be:

4H2 (g) → 4H2O
3H2 (g) → 2NH3
2NO2 (g)

Because O2 would then be 2O2 which would allow the bottom equation to cancel out with it.

Answer 7

Ok so let me try this again. When you multiply the first equation by 2 will get
4H2 (g) + 2O2 (g) → 4H2O (I)

After that you can cancel the 2O2 in first equation with the 2O2 in the third equation. You also cancel the 2N2 in the second equation with the 2N2 in the third equation. Once those are canceled out it leaves you with

4H2 (g) → 4H2O
3H2 (g) → 2NH3
2NO2 (g)

Which when added together makes

2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l)

Which is the same as our equation with the missing enthaply. Therefore, multiplying -572 x 2 which gives me -1144 kJ I will be able to add the three together to find the missing enthaply. Is that correct?

Sorry if I'm slow to catching on.

Answer 8

That is perfect!!,

the only thing is the simplification we are doing in the equations it is assuming that we are going to add the equations. Otherwise it is not totally correct the partial equations that you write down as

4H2 (g) → 4H2O
3H2 (g) → 2NH3
2NO2 (g)

To be academically correct you add everything and then simplify in the final equation if you have the same compound in the reactant as well in the products

1)2x ( 2H2 (g) + O2 (g) → 2H2O (l)), ΔH5 = -572 x 2 kJ = -1144 kJ
2) 2N2 (g) + 3H2 (g) → 2NH3 (g), ΔH2 = –138 kJ
3) 2NO2 (g) → 2N2 (g) + 2O2 (g), ΔH3 = 99.0 kJ
------------------------------------------------
4) 2NO2 (g) + 7H2 (g) → 2NH3 (g) + 4H2O (l), ΔH4 = ΔH5+ ΔH2 + ΔH3
ΔH4 = 2 x (-ΔH1)+ ΔH2 + ΔH3


4H2 (g) + 2O2 (g)+2N2 (g) + 3H2 (g) + 2NO2 (g) → 4H2O (l) + 2NH3 (g) +2N2 (g) + 2O2 (g)

then you cancel

Answer 9

Ahh ok. I'll keep that in mind. Thank you so much! You've been a great help.