equilibrium constant, Kc
I have no idea how to go about this question could you please help:
e.g [NH3]2/[N2]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
e.g [NH3]2/[N2]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
Original post by going2fail
I have no idea how to go about this question could you please help:
e.g [NH3]2/[N2]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
e.g [NH3]2/[N2]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.
Original post by theJoyfulGeek
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.
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):
Unparseable latex formula:
[br]$3H_2 + N_2 \rightarrow 2NH_3$\\[br]$K_c = \frac{[NH_3]^2}{[H_2]^3[N_2]}$\\[br]$[NH_3]^2 = K_c * [N_2]* [H_2]^3$\\[br]$[NH_3] = \sqrt{3.01 * 0.25 * 0.15^3}$\\[br]
So the answer is 0.0503953... or 0.05 to 1 sf.
for the first step wouldn't the nitrogen conc. be : 0.5mol/0.5dm3 = 1mol/dm3
and the hydrogen con is 0.3/0.dm3 = 0.6 mp/dm3 ?
and if so that would change the whole of part 2.
Original post by theJoyfulGeek
Yep - good point! You're right (I managed to multiply instead of divide - go me!). So you just have to replace 0.25 with 1.0 and 0.15 with 0.6.
And that would change the numbers in part two, but not the algebra thankfully!
Edit - sorry about that @going2fail! I probably should've checked...
And that would change the numbers in part two, but not the algebra thankfully!
Edit - sorry about that @going2fail! I probably should've checked...
Thank you so much this has been really helpful
!!!Original post by theJoyfulGeek
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
EDIT- THESE NUMBERS ARE INCORRECT. I PROBABLY SHOULD HAVE CHECKED. THEY ARE ACTUALLY 1.0 AND 0.6. SORRY!
Part 2 (I'll use LaTeX formatting):
So the answer is 0.0503953... or 0.05 to 1 sf.
Nitrogen concentration = 0.5 mol / 0.5dm3 = 0.25 mol/dm3
Hydrogen conc. = 0.3 mol / 0.5dm3 = 0.15 mol/dm3
EDIT- THESE NUMBERS ARE INCORRECT. I PROBABLY SHOULD HAVE CHECKED. THEY ARE ACTUALLY 1.0 AND 0.6. SORRY!
Part 2 (I'll use LaTeX formatting):
Unparseable latex formula:
[br]$3H_2 + N_2 \rightarrow 2NH_3$\\[br]$K_c = \frac{[NH_3]^2}{[H_2]^3[N_2]}$\\[br]$[NH_3]^2 = K_c * [N_2]* [H_2]^3$\\[br]$[NH_3] = \sqrt{3.01 * 0.25 * 0.15^3}$\\[br]
So the answer is 0.0503953... or 0.05 to 1 sf.
so if change the numbers the answer should by o.81 rounded to 2 s.f?
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