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# Kc example watch

1. 4HCl + O2 = 2Cl2 + 2H2O

0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
At eqm only 0.04 mol of HCl was present. Find Kc

I said it was 164025 mol-1 dm3

But the book said it was 1.6 mol-1 dm3

I am fairly sure they are wrong
2. (Original post by jsmith6131)
4HCl + O2 = 2Cl2 + 2H2O

0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
At eqm only 0.04 mol of HCl was present. Find Kc

I said it was 164025 mol-1 dm3

But the book said it was 1.6 mol-1 dm3

I am fairly sure they are wrong
Strange.. I got the answer to be 41006.25 mol^-1dm^3
3. are you sure you used the correct mol ratio?
4. (Original post by jsmith6131)
are you sure you used the correct mol ratio?
Yeah.. what values did you use?
5. (Original post by jsmith6131)
4HCl + O2 = 2Cl2 + 2H2O

0.4 mol of HCl and 0.1 mol of O2 were placed in a 4 dm3 vessel
At eqm only 0.04 mol of HCl was present. Find Kc

I said it was 164025 mol-1 dm3

But the book said it was 1.6 mol-1 dm3

I am fairly sure they are wrong
initially

4HCl + O2 = 2Cl2 + 2H2O
0.4 ..... 0.1 ..... 0.0 .....0.0

If we assume it's water vapour and it takes part...

at equilibrium

4HCl + O2 = 2Cl2 + 2H2O
0.04 ..... 0.01 ..... 0.18 .....0.18

Concentrations in 4 litre vessel

4HCl + O2 = 2Cl2 + 2H2O
0.01 ..... 0.0025 ..... 0.045 .....0.045

equilibrium law

kc = [Cl2]2[H2O]2/[HCl]4[O2]

kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

kc = 164025

I agree with you
6. ok thanks
7. (Original post by NutterFrutter)
Yeah.. what values did you use?

(Original post by charco)
initially

4HCl + O2 = 2Cl2 + 2H2O
0.4 ..... 0.1 ..... 0.0 .....0.0

If we assume it's water vapour and it takes part...

at equilibrium

4HCl + O2 = 2Cl2 + 2H2O
0.04 ..... 0.01 ..... 0.18 .....0.18

Concentrations in 4 litre vessel

4HCl + O2 = 2Cl2 + 2H2O
0.01 ..... 0.0025 ..... 0.045 .....0.045

equilibrium law

kc = [Cl2]2[H2O]2/[HCl]4[O2]

kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

kc = 164025

I agree with you

I agree but I got values closer to nutter when I used 0.225 instead of 0.18 because I tried to make the number of moles equate which seemed wrong.
8. (Original post by charco)
initially

4HCl + O2 = 2Cl2 + 2H2O
0.4 ..... 0.1 ..... 0.0 .....0.0

If we assume it's water vapour and it takes part...

at equilibrium

4HCl + O2 = 2Cl2 + 2H2O
0.04 ..... 0.01 ..... 0.18 .....0.18

Concentrations in 4 litre vessel

4HCl + O2 = 2Cl2 + 2H2O
0.01 ..... 0.0025 ..... 0.045 .....0.045

equilibrium law

kc = [Cl2]2[H2O]2/[HCl]4[O2]

kc = (0.002025 * 0.002025)/1x10-8 * 0.0025

kc = 164025

I agree with you
I see my mistake.

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