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oct chemistry help !!

An industrial chemist carries out some research into the NO/O2/NO2 equilibrium used in Stage 2 of the manufacture of nitric acid.
• The chemist mixes together 0.80mol NO(g) and 0.70 mol of O2(g) in a container with a volume of 2.0dm3.
• The chemist heats the mixture and allows it to stand at constant temperature to reach equilibrium.
The container is kept under pressure so that the total volume is maintained at 2.0 dm3.
• At equilibrium, 75% of the NO has reacted.

The question is to work out the KC value.. I'm really stuck :frown:

I got the moles at equilibrium of NO2 as 0.60… meaning a 0.20 difference. But when you take 0.20 from 0.70 you get 0.50, but the mark scheme is stating the moles of 02 at equilibrium are 0.40. I'm probably doing this wrong, or being really stupid but any help would really be appreciated

Reply 1

lorobolorolo

Original post by ellie2996

Hey, you nearly got it right.

2NO + O₂ → 2NO₂

I first found the moles of NO that reacted as 0.6. So that means 0.6 moles of NO₂ are present at equilibrium.

If 0.6 of of NO reacted, it must mean that 0.2 moles of NO are left at equilibria (0.8-0.6=0.2)

Now all you need to find is the equilibrium moles of oxygen. You know that 0.6moles of NO reacted, and from the equation you can see that only half the moles of oxygen would have reacted. So 0.3 moles of oxygen reacted. Which leaves 0.4 moles of oxygen (0.7-0.3=0.4)

Now you just put it into the Kc equation.
So Kc = [NO₂]² divided by [NO]² [O₂]

make sure when you put the values into the equation, you divide by the volume to find the concentrations.

Hope I helped :smile: