Calculating Kp

Need help with question 6 as I have only 0.25 and 0.75 as the initial moles of the reactants due to the 1:3 ratio. However, the moles are just 1 and 3. Why is that because I thought the ratio add up to 100%.

Thanks

(edited 7 years ago)

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Reply 1

alow

What do you mean by "the moles are just 1 and 3"?

Reply 2

coconut64

Original post by alow

What do you mean by "the moles are just 1 and 3"?

the question wants you to assume that the initial moles the reactants are 1 mole and 3 moles. Thanks

Reply 3

alow

Original post by coconut64

the question wants you to assume that the initial moles the reactants are 1 mole and 3 moles. Thanks

Does it? It only gives a ratio.

Reply 4

coconut64

Original post by alow

Does it? It only gives a ratio.

That's what I thought but here is the explanation: 'Assuming no product gases (eg NH3) are there to start with (the question hintsat this by saying “left to equilibrate”) we can say that the initial mixture consists of 1 mol of N2(g) and 3 mol of H2(g) as this is the initial ratio'

Reply 5

alow

Original post by coconut64

Yeah that's because it's the easiest amounts to deal with. You can use any amount you want though, as long as the 1:3 ratio holds.

You could use 0.0073 mol Nitrogen and 0.0219 mol Hydrogen if you want. That would be stupid, but it would get you the same answer.

Reply 6

coconut64

Original post by alow

But if I am using this to work out Kp, it will affect the end result since I used 0.25 and 0.75, which are smaller than 1 and 3.

Reply 7

alow

Original post by coconut64

But if I am using this to work out Kp, it will affect the end result since I used 0.25 and 0.75, which are smaller than 1 and 3.

But your amount of produced ammonia will also be lower to account for that.

Reply 8

coconut64

Original post by alow

But your amount of produced ammonia will also be lower to account for that.

The question says that 15% of ammonia is produced, can I just say that it is 0.15 mol?

Reply 9

alow

Original post by coconut64

The question says that 15% of ammonia is produced, can I just say that it is 0.15 mol?

No, write down the equation for production of ammonia from hydrogen and nitrogen gases.

0.15mol cannot be 15% of the total.

Reply 10

Tuffyandtab

You can start with any amount of hydrogen and nitrogen as long as it's a 1:3 ratio. So 1mol of hydrogen and 3mol of nitrogen or 271.5mol of hydrogen and 814.5mol of nitrogen! It does not matter

Reply 11

coconut64

Original post by Tuffyandtab

You can start with any amount of hydrogen and nitrogen as long as it's a 1:3 ratio. So 1mol of hydrogen and 3mol of nitrogen or 271.5mol of hydrogen and 814.5mol of nitrogen! It does not matter

I used 0.25 and 0.75 as my 1:3 ratio. I though for ammonia you just find out 15% of 1 which is 0.15. Thanks

Reply 12

coconut64

Original post by alow

No, write down the equation for production of ammonia from hydrogen and nitrogen gases.

0.15mol cannot be 15% of the total.

So the equation would be N2+ 3H2 -> 2NH3
I'm not sure where to go from here...

Reply 13

alow

Original post by coconut64

So the equation would be N2+ 3H2 -> 2NH3
I'm not sure where to go from here...

You need to notice that when you react 4 moles of gas together, you get 2 back out.

This means your total amount will decrease as the reaction goes to equilibrium. So there won't be 0.15mol of ammonia as 0.15mol will be more than 15%.

Reply 14

coconut64

Original post by alow

Actually, I need a min

(edited 7 years ago)

Reply 15

Tuffyandtab

I agree i think but I haven't done this topic before so bare with (I'm going onto 2nd year a level chemistry).
n2 + 3h2 --> 2nh3
mole fraction of nh3 = 0.15
mole fraction of n2 = 0.25 - (0.15/4) = 0.2125
mole fraction of h2 = 0.75 - 3(0.15/4) = 0.6375
Pa of nh3 = 1500
Pa of n2 = 2125
Pa of h2 = 6375
Kp = (1500)^2/(2125) x (6375)^3 = 4.08679653e-9 I hope this helps. Happy to explain

Original post by coconut64

I used 0.25 and 0.75 as my 1:3 ratio. I though for ammonia you just find out 15% of 1 which is 0.15. Thanks

Reply 16

alow

Original post by coconut64

Actually, I need a min

Does it consider the change in amounts? Where do they get 6 moles of gas from?

Look at the second question here and its answer, it's basically the same thing but they don't give you a final answer as all of the pressures at equilibrium are unknown:

http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Chemical_Equilibria/Calculating_an_Equilibrium_Constant,_Kp,_with_Partial_Pressures

Reply 17

coconut64

Original post by alow

But that question only gives you the partial pressure whereas the question I'm doing now gives you hints about the inital moles and moles at equilibrium here is the explanation

(edited 7 years ago)

Reply 18

alow

Original post by coconut64

But that question only gives you the partial pressure whereas the question I'm doing now gives you hints about the inital moles and moles at equilibrium

You know there's a 1:3 ratio at the beginning and you know the pressure is 10000kPa. Use this information to draw a table like the one in the example.

Reply 19

coconut64

Original post by alow

You know there's a 1:3 ratio at the beginning and you know the pressure is 10000kPa. Use this information to draw a table like the one in the example.

I have actually done that already but I don't know how to work out the mole of ammonia so my equilibrium moles for the reactants are wrong. If I am using the values 0.25 and 0.75 for my 1:3 ratio, it wouldn't work as 3/2 of 0.9 is 1.35