Question

Using the three formation reactions given, calculate the enthalpy of this reaction:

N2H4(l) + 2H2O2(g) → N2(g) + 4H2O(l)
N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ΔH = -622.3 kJ
H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.8 kJ
H2O2(l) → H2(g) + O2(g) ΔH = 187.8 kJ

-818.3 kJ

-1095.9 kJ

818.3 kJ

1095.8 kJ

Answer 1

@whpalmer4

Answer 2

you've got to play around with the three reactions below. sometimes you may have to reverse them and when you do this you change the sign of the enthalpy, this is something that's hard to explain without actually doing the question and takes practice so i'll show you how to do it.

Answer 3

So we need to get to this reaction by using the three reactions below.

\[N2H4_{l} + 2H2O2_{g} → N2_{g} + 4H2O_{l}\]

Answer 4

This really took me a while to write but I want to show you how to do this step by step so you get whats' involved in this. this is what I mean when I say you need to really play around with the formulas first. hopefully I got this lol.

We take a look at all of our reactions and compare them to the one written above.

This looks okay because if you notice all the products and reactants are where we need them:

1. N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ΔH = -622.3 kJ


See this here below? we look at the top reaction and we find that O2 and H2 aren't in our final equation so we need to get rid of them.

but you see something here? H2 is in the reactants and products in these two reactions. whenever we have two of the same compounds in two reactions that are in opposite places like H2 is in the products in one equation and H2 is in the reactants in the other, we can eliminate them.

then we combine 2. and 3.

2, H2(g)+1/2O2(g) --> H2O(l) (ΔH = -285.8 kJ)
3. H2O2(l) → H2(g) + O2(g) + ΔH = 187.8 kJ

add those two enthalpies for reaction. and we get -98 kj for the two reactions.

and we get to get this equation below 4.

Next step is to combine equations 4. and 5. to get 1.

4. H2O2(l) --> H2O + 0.5O2(g) -98kj
5. N2H4(l) + O2(g) → N2(g) + 2 H2O(l) -622.3kj

let's compare them this is what we need below

N2H4l+2H2O2g→N2g+4H2Ol

now all we need to do is multiply the top equation by 2 right and we get this.

let's compare what we have now we see that we can eliminate o2 on both sides because they are on opposite sides. and because we have 2 moles of h2O on each side we can add them together.

4. 2H2O2 ---> 2H2O + O2(G) = -196 kj
5. N2H4(L) + O2(g) --> N2 + 2H2O -622.3kj

This gives us

N2H4(l) + 2H2O2 --> 4H2O + N2

now we find that this is the same as what we started with so we got the answer right

N2H4l+2H2O2g→N2g+4H2Ol

now all we need to do is add the enthalpies -196 + -622.3 and we get our total enthalpy for this reaction

Answer 5

@Sydx as you can see problems like this take practice just take a look at what I did make sure you understand what's going on first.