Find Mole % of N2H4 in Mixture: Homework Statement

[Saitama]

Homework Statement

A mixture of NH₃ and N₂H₄ is placed in a sealed container at 300K. The total pressure is 0.5atm. The container is heated to 1200K, when both substances decompose completely according to the equations,
2NH₃--->N₂+3H₂
N₂H₄--->N₂+2H₂
After decomposition is complete, the total pressure at 1200K is found to be 4.5atm. Find the mole percent of N₂H₄ in the original mixture.

Homework Equations

The Attempt at a Solution

I am not getting any idea on how to start.
Please give me some ideas on beginning with it.

Thanks! :)

 

[Borek]

Assume the initial partial pressures of both were p0 and p1.

Can you express both initial and final pressures as functions of p0 and p1?

 

[Saitama]

Assume the initial partial pressures of both were p0 and p1.

Can you express both initial and final pressures as functions of p0 and p1?

I can express initial pressure like this:-
p0= 0.5 x X_NH3
p1= 0.5 x X_N2H4

X_NH3--> mole fraction of NH3
X_N2H4---->mole fraction of N2H4.

But i don't have the moles. :(

 

[Borek]

Why not p0 + p1 = 0.5 atm?

You don't need moles - you can express molar fraction using just total and partial pressures.

 

[Saitama]

Why not p0 + p1 = 0.5 atm?

You don't need moles - you can express molar fraction using just total and partial pressures.

Yes, we can do p0+p1=0.5 atm.

But why then we need molar fraction here?

 

[Borek]

Molar fraction, mole percent - same thing, just expressed in slightly different way. That's what you have to calculate, aren't you?

 

[Saitama]

Molar fraction, mole percent - same thing, just expressed in slightly different way. That's what you have to calculate, aren't you?

Yeah I have to calculate that.

I am having problems finding the final pressure. The gases has changed to N2 and H2. What should i do now?

 

[Borek]

You can find ratio between initial and final number of moles from the stoichiometry. A volume doesn't change, number of moles is directly related to the pressure.

 

[Saitama]

Sorry Borek. I did not replied to this post for a long time. :)

I worked on your hints and they proved to be very useful. I was only stuck at the stoichiometric calculation which was the easiest part. :P

Here's how i solved it:-

I assumed moles of NH3 as 'a' and that of N2H4 as 'b'.

Then using stoichiometry i found the total moles of N2 as $\frac{a+2b}{2}$ and of H2 as $\frac{3a+4b}{2}$.

Before decomposition, using ideal gas equation:-
(0.5V)/(300R)=(a+b)

After decomposition, using ideal gas equation:-
(4.5V)/(1200R)=(2a+3b)

After solving, i get a=3b.

Mole fraction of N2H4= (b)/(a+b)=b/4b=0.25

Since we are asked to find mole percent, therefore 0.25*100=25%.

Thanks for the help Borek.