[cloudy1290]

Given moles of H2 = 3 mol, moles of N2 = 9 mol for equation, 3H2 + N2 --> 2NH3, and the container containing both of them has 3L of volume

after Equilibrium NH3 gives 50% mole fraction(Means that after reaction is complete, the number of moles of NH3 formed from reaction occupies 50% of the total number of moles of all gases in the container) , the total volume of H2 and N2 in container before reaction is 3L.

1. Find mole of H2, N2 and NH3 after equilibrium
2. Calculate the heat. Given that delta H = -91 kJ
3. Calculate K
my solution :

  3H2 + N2 ----> 2NH3

I   3      9

C -3x    -x           +2x

E 3-3x  9-x           2x

then 2x/(2x+12−3x−x) = 1/2 then x = 2

but it's impossible because moles of H2 after equilibrium will be -3

so what's my mistake?

[AWK]

Equilibrium equation should contain some powers.

[cloudy1290]

Equilibrium equation should contain some powers.

power?

[AWK]

Write down a general equilibrium equation. It should be a cubic one.

[cloudy1290]

Write down a general equilibrium equation. It should be a cubic one.

K = [NH3]^2 / [H2]^3*[N2]^3  ?

TBH, i don't understand

[AWK]

 example

[cloudy1290]

example

So it becomes K = 4x^2 / 27(x-9)(x-1)^3 then?

[AWK]

Now nothing can be calculated since your numerical data are wrong.

Given moles of H2 = 3 mol, moles of N2 = 9 mol for equation, 3H2 + N2 --> 2NH3

Assuming complete reaction for hydrogen: 8 moles of N₂ is left and 2 moles of NH₃ are formed.
It means that the maximum number of moles of ammonia equal to 20 % moles of all gases after the reaction.

the number of moles of NH3 formed from reaction occupies 50% of the total number of moles of all gases in the container

[cloudy1290]

I see, but how to calculate question 2 ; Calculate the heat. Given that delta H = -91 kJ. Could you tell me the method in case data is right