[Phoenix]

In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
      N2(g) + 3H2(g) == 2NH3(g)

How is the answer 6.6 mol-2 dm6?

I tried change in equilibrium divided by the reaction vessel to find each concentration of each gas, then substitute it into the Kc expression. I had no success.

[AWK]

Enter the data into the formula correctly.

[Phoenix]

Could you give me a nudge in the right direction?? I'm abit stuck

[AWK]

Show your work.

[Phoenix]

Thank you for your reply.

N2 +         H2  2NH3
9 mol        27mol             18mol

reaction vessel 10dm3-concentration from mols-
9/10           27/10               18/10

 therefore Concs are:  0.9             2.7                  1.8

put into Kc expression     

1.8 x 1.8/ 0.9 x ( 2.7x 2.7 x 2.7)
=
70.86

[AWK]

 

It was found that two thirds of the nitrogen and hydrogen were converted into ammonia

[Phoenix]

Thank you for that...

However, if I use
N2 +         H2  2NH3
6 mol        18mol             18mol

reaction vessel 10dm3-concentration from mols-
6/10           18/10               18/10

 therefore Concs are:  0.6             1.8                  1.8

put into Kc expression     

1.8 x 1.8/ 0.6 x ( 1.87x 1.7 x 1.7)
=
31.49

Have i missed out a step?

[AWK]

Some amounts od H₂ and N₂ were left - from that reacted ammonia was formed (usually ICE tables are prepared)

[Phoenix]

Thank you for your response.
I have tried looking up ICE tables- I understand about products losing and gaining when changing equilibrium- but the answers are still not working out for me.
My ICE table must be incorrect.

N2 +         H2  2NH3
6 mol        18mol             18mol

reaction vessel 10dm3-concentration from mols-
6/10           18/10               18/10

 Initial: Concs are:  0.6             1.8                  1.8
Change:          0.6-1.8           1.8-(1.8x3)         1.8
Equilibrium          -0.12            -0.36                1.8

put into Kc expression     

gives an incorrect number...

[AWK]

Equilibrium          -0.12            -0.36                1.8

All concentrations are still wrong.
eg: 0.9 - (2/3)·0.9=

[Phoenix]

Thank you for your reply.
 I have tried using the new examples, however I am not getting this.
Could I ask you for the method.
All your help is appreciated.

[AWK]

It was found that two thirds of the nitrogen and hydrogen were converted into ammonia.

Just read carefully
Nitrogen and hydrogen form ammonia - how may moles or what is ammonia concentration?

[Phoenix]

For the reacting amounts of ammonia there are 2mols.
Initial /mol= o

Is it saying there is 2/3 of 2 mols?

[AWK]

Ammonia is as much as formed from a portion of nitrogen and hydrogen. You have all the data needed to calculate the amount of ammonia after reaching equilibrium. Since you give hydrogen with nitrogen, there is, of course, no ammonia at the beginning.

[Phoenix]

Okay,
I have tried...
0.9- 2/3.0.0.9= 0.23
2.7- 3x 2/3=2.0

2/3 of ammonia = 24mols??
I have crunched this through the ICE and Kc and still incorrect.