Hess's Law of Hydrazine

Question

Hydrazine, a compound of rocket fuel, undergoes combustion to yield N2 and H2O:

N2H4(l) + O2(g) → N2(g) + 2 H2O(l)

(b) Use the following information to set up a Hess’s law cycle, and then calculate ? H° for the combustion reaction. You will need to use fractional coefficients for some equations.

2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(l) ? H° = - 1011.2 kJ
N2O(g) + 3 H2(g) → N2H4(l) + H2O(l) ? H° = - 317.2 kJ
4 NH3(g) + O2(g) → 2 N2H4(l) + 2 H2O(l) ? H° = - 286.0 kJ
H2O(l) ? H°f = - 285.8 kJ/mol

(c) How much heat is released on combustion of 100.0 g of hydrazine?

Answer 1

I think of Hess's Law like a puzzle game! You have to do whatever you need to get the result that you want.

This is the reaction that you are trying to find:

N₂H₄(l) + O₂(g) --> N₂(g) + 2 H₂O(l)

Using these reactions that we can manipulate:

2 NH₃(g) + 3 N₂O(g) --> 4 N₂(g) + 3 H₂O(l) H = -1011.2 kJ

N₂O(g) + 3 H₂(g) --> N₂H₄(l) + H₂O(l) H = -317.2 kJ

4 NH₃(g) + O₂(g) --> 2 N₂H₄(l) + 2 H₂O(l) H = -286.0 kJ

H₂(g) + 1/2 O₂(g) --> H₂O(l) H_f = -285.8 kJ/mol

Let's take this one reaction at a time.

2 NH₃(g) + 3 N₂O(g) --> 4 N₂(g) + 3 H₂O(l) H = -1011.2 kJ

This needs to be divided by 4 to have just N₂ in the products like the reaction we are looking for.

1/2 NH₃(g) + 3/4 N₂O(g) --> N₂(g) + 3/4 H₂O(l) H = -252.8 kJ

We need to flip and multiply the second reaction by 3/4:

N₂O(g) + 3 H₂(g) --> N₂H₄(l) + H2O(l) H = -317.2 kJ

3/4 N₂H₄(l) + 3/4 H₂O(l) --> 3/4 N²O(g) + 9/4 H₂(g) H = +237.9 kJ

For Equation 3, we need to flip and divide by 8:

4 NH₃(g) + O₂(g) --> 2 N₂H₄(l) + 2 H₂O(l) H = -286.0 kJ

1/4 N₂H₄(l) + 1/4 H₂O(l) --> 1/2 NH₃(g) + 1/8 O₂(g) H = +35.75 kJ

The formation of water equation is:

H₂(g) + 1/2 O₂(g) --> H₂O(l) H_f = -285.8 kJ/mol

It needs to be multiplied by 9/4:

9/4 H₂(g) + 9/8 O₂(g) --> 9/4 H₂O(l) H = -643.05 kJ

Let's look at the reactions together now:

1/2 NH₃(g) + 3/4 N₂O(g) --> N₂(g) + 3/4 H₂O(l)

3/4 N₂H₄(l) + 3/4 H₂O(l) --> 3/4 N₂O(g) + 9/4 H₂(g)

1/4 N₂H₄(l) + 1/4 H₂O(l) --> 1/2 NH₃(g) + 1/8 O₂(g)

9/4 H₂(g) + 9/8 O₂(g) --> 9/4 H₂O(l)

NH₃, N₂O, and H₂ appear on both sides and cancel. Now let's look at the trickier ones:

4/4 N₂H₄ + 1/4 H₂O + 9/8 O₂ --> N₂ + 1/8 O₂ + 9/4 H₂O

After cancelling 1/8 O2 and 1/4 H2O from both sides:

N₂H₄ + O₂ --> N₂ + 2 H₂O

That was a lot, but we got the correct equation. Now we can add our delta H values:

-252.8 + 237.9 + 35.75 - 643.05 = -622.2 kJ

There are a lot of places to make a silly algebra mistake in this one, so be sure to take your time and write it all out.

I hope this helps make chemistry easier!