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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: cloudy1290 on June 22, 2019, 10:28:25 AM

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 33x 9x 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?

Equilibrium equation should contain some powers.

Equilibrium equation should contain some powers.
power?

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

Write down a general equilibrium equation. It should be a cubic one.
K = [NH3]^2 / [H2]^3*[N2]^3 ?
TBH, i don't understand

example

example
So it becomes K = 4x^2 / 27(x9)(x1)^3 then?

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 N_{2} is left and 2 moles of NH_{3} 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

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