[Fish200398]

2.00 mol N2 and 5.00 mol H2 were mixed together, the pressure at equilibrium is 1.01 x 10^6 Pa.  Mole fraction of NH3 is 0.25
N2 + 3H2 = 2NH3 +92 kJ
A) how much heat evolved?
B) what is the partial pressure of N2 equilibrium?

I dont know how to find the heat evolved.
Is it mol of NH3 *92Kj ?
Mol of NH3 at equlibrium will be
2x mol?
So to find 2x is
0.25 = 2x/(7-4x) ?
But my answer is wrong.
How to find the heat evolved and the partial pressure?

[AWK]

+92 kJ

Are you sure concerning sign and units?

Add also ammonia in denominator

[Fish200398]

Yes. It said +92kJ

[AWK]

-92 kJ/mol

[Fish200398]

It is said
2 mole of N2 and 5 mole of H2 were mixed in the present of proper catalyst and maintained at certain temperature. The following reaction occured in the gas mixture and then arrived at the equilibrium where the total pressure was 1.01 x 10^6 Pa . The mole fraction of NH3 was found to be 2.05 x 10^-1

N2 + 3H2 = 2NH3 +92kJ
A) how much heat was evolved through this reaction
B) what was the partial pressure of N2 at equilibrium

[AWK]

The problem probably comes from the internet, not from the book. Will you get heat in kJ if you multiply 2x moles by 92 kJ. The heat released has the sign "-".

[Borek]

I dont know how to find the heat evolved.
Is it mol of NH3 *92Kj ?

I understand AWK hints are rather cryptic and can be difficult to understand, but he is pointing you at the right direction. Heat of the reaction has units of kJ/mol. That means you need to know how many moles of ammonia were produced to calculate the heat/ Look at the reaction equation - when there are 2 moles of ammonia produced 92 kJ of energy evolves. How much energy evolves when one mole of ammonia is produced? When four moles of ammonia are produced?

Once you get that right problem becomes that of finding the equilibrium. You did the right thing here:

0.25 = 2x/(7-4x)

but you in the latter post you wrote 0.205, not 0.25, so that can be a source of problem.

Once you know what x is finding the molar fraction of nitrogen should be trivial, use exactly the same approach you used to derive the above formula.

[Fish200398]

0.25=2x/(7-4x)
x=0.583
2x=1.166
So the heat evolved is 92×1.166 = 107.3 kJ?
But its wrong..

[AWK]

You have:
2-x moles of nitrogen
5-3x moles of hydrogen
and 2x moles of ammonia

mole fraction of NH₃ is:
2x/(2-x + 5-3x + 2x) = 2x/(7-2x)=0.25

92 kJ is the heat released in the reaction for 2 moles of ammonia (since standard enthalpy of formation for ammonia is -46 kJ/mol)

[Fish200398]

Thank you.