Working out Kc Equilibruim

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#1

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
0
3 years ago
#2
(Original post by Robertpotter1999)

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
0
#3
(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!
0
3 years ago
#4
(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.
0
#5
(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
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