The Student Room Group

Working out Kc Equilibruim

14936405121431407299880.jpg
Can someone please explain this question to me please ive worked out the correct moles for NO which is 0.2 but i have got O2 wrong i got 0.175 when it should be 0.4 and i dont get why
Reply 1
Original post by Robertpotter1999
14936405121431407299880.jpg
Can someone please explain this question to me please ive worked out the correct moles for NO which is 0.2 but i have got O2 wrong i got 0.175 when it should be 0.4 and i dont get why


At equilibrium you are reacting 75% of the 'starting moles' NO which = 0.20 - correct. However 0.2 is moles at equilibrium. You need to use the moles of NO reacting to get to equilibrium - 0.60.

It's a 2:1 mole ratio also so actually you have to take 0.60 moles of NO (the moles reacting off) then divide that value by 2 (2:1 mole ratio). Which gives you 0.30 moles of O2 reacting to get to equilibrium.

So to find the actual moles at equilibrium we do:

0.70 - 0.3 = 0.40 (because you have 0.3 moles reacting off from the 0.70 starting moles of O2.)

I think it helps to lay it out like this:

Equation X + Y --------Z

Starting Moles: (under each species)
Change**:
Equilibrium Moles: '' ''

Then fill in each of the mole values as you go along, do your working out beneath that 'table'.


Change** = S +- X where X refers to the moles reacting. S = Starting moles

X can vary due to mole ratios so you can have -x/2 (like above) or -3x if the ratio is 1:3.


Note: With Products is usually +nX (plus) because you usually don't have products at the start or starting moles so e.g +3x for products

X in the example above is 0.60 and the mole ratio is 1:1 NO:NO2 so it's 0.60 moles of NO2 at equilibrium

i.e Change products = +1(0.60)

Confirm - Look at the equilibrium moles: 0.2 + 0.4 = 0.60 moles of NO2
Hope that helps
(edited 6 years ago)
Original post by _NMcC_
At equilibrium you are reacting 75% of the 'starting moles' NO which = 0.20 - correct. However 0.2 is moles at equilibrium. You need to use the moles of NO reacting to get to equilibrium - 0.60.

It's a 2:1 mole ratio also so actually you have to take 0.60 moles of NO (the moles reacting off) then divide that value by 2 (2:1 mole ratio). Which gives you 0.30 moles of O2 reacting to get to equilibrium.

So to find the actual moles at equilibrium we do:

0.70 - 0.3 = 0.40 (because you have 0.3 moles reacting off from the 0.70 starting moles of O2.)

I think it helps to lay it out like this:

Equation X + Y --------Z

Starting Moles: (under each species)
Change**:
Equilibrium Moles: '' ''

Then fill in each of the mole values as you go along, do your working out beneath that 'table'.


Change** = S +- X where X refers to the moles reacting. S = Starting moles

X can vary due to mole ratios so you can have -x/2 (like above) or -3x if the ratio is 1:3.


Note: With Products is usually +nX (plus) because you usually don't have products at the start or starting moles so e.g +3x for products

X in the example above is 0.60 and the mole ratio is 1:1 NO:NO2 so it's 0.60 moles of NO2 at equilibrium

i.e Change products = +1(0.60)

Confirm - Look at the equilibrium moles: 0.2 + 0.4 = 0.60 moles of NO2
Hope that helps


Thank you so much! I litro figured that out 5 mins after i posted this!
Reply 3
Original post by Robertpotter1999
Thank you so much! I litro figured that out 5 mins after i posted this!


No probs, It's not very obvious tbh, you do have to think logically about what is actually happening in the system and then represent that in your calculations.
Original post by _NMcC_
No probs, It's not very obvious tbh, you do have to think logically about what is actually happening in the system and then represent that in your calculations.


Yes! That exactly what i do not do i always miss out that step and get tge wrong answers i know how hard chemistry is for me so if i find something i normally find hard easy i need to go back over it

Quick Reply

Latest