# equilibrium constant, Kc

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Thread starter 1 day ago
#1
e.g [NH3]2/[N2][H2]3 where [ ] is the concentration in mol dm-3
1. Calculate the concentrations if the flask has a volume of 0.5dm3 and there are 0.50 moles of nitrogen and 0.30 moles of hydrogen
2. If Kc has a value of 3.01, calculate (NH3)

If you could do it step by step that would be really helpful thank you
1
1 day ago
#2
(Original post by going2fail)
e.g [NH3]2/[N2][H2]3 where [ ] is the concentration in mol dm-3
1. Calculate the concentrations if the flask has a volume of 0.5dm3 and there are 0.50 moles of nitrogen and 0.30 moles of hydrogen
2. If Kc has a value of 3.01, calculate (NH3)

If you could do it step by step that would be really helpful thank you
Step 1 - you have amounts and volumes, that is a GCSE calculation.
Step 2 - you have the equation for Kc and in step 1 you worked out two of the concentrations, a little algebra later and you'll have the third concentration.
0
1 day ago
#3
(Original post by going2fail)
e.g [NH3]2/[N2][H2]3 where [ ] is the concentration in mol dm-3
1. Calculate the concentrations if the flask has a volume of 0.5dm3 and there are 0.50 moles of nitrogen and 0.30 moles of hydrogen
2. If Kc has a value of 3.01, calculate (NH3)

If you could do it step by step that would be really helpful thank you
My step by step solution (sorry!):
Nitrogen concentration = 0.5 mol / 0.5dm3 = 0.25 mol/dm3
Hydrogen conc. = 0.3 mol / 0.5dm3 = 0.15 mol/dm3

Part 2 (I'll use LaTeX formatting):

So the answer is 0.0503953... or 0.05 to 1 sf.
Last edited by theJoyfulGeek; 1 day ago
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