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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Phoenix on May 04, 2020, 08:12:02 AM

Title: Kc and gases
Post by: Phoenix on May 04, 2020, 08:12:02 AM
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.
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 08:31:09 AM
Enter the data into the formula correctly.
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 08:32:46 AM
Could you give me a nudge in the right direction?? I'm abit stuck
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 08:33:22 AM
Show your work.
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 09:37:49 AM
Thank you for your reply.

N2 +         H2  ::equil:: 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





Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 09:49:50 AM
 
Quote
It was found that two thirds of the nitrogen and hydrogen were converted into ammonia
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 09:53:42 AM
Thank you for that...

However, if I use
N2 +         H2  ::equil:: 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?
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 09:57:52 AM
Some amounts od H2 and N2 were left - from that reacted ammonia was formed (usually ICE tables are prepared)
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 11:18:18 AM
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  ::equil:: 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...
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 11:46:06 AM
Quote
Equilibrium          -0.12            -0.36                1.8
All concentrations are still wrong.
eg: 0.9 - (2/3)·0.9=
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 11:58:07 AM
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.
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 12:11:31 PM
Quote
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?
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 12:19:24 PM
For the reacting amounts of ammonia there are 2mols.
Initial /mol= o

Is it saying there is 2/3 of 2 mols?
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 12:25:40 PM
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.
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 12:44:16 PM
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.
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 12:50:13 PM
(2/3)·0.9 = 1.8/3 = ?
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 01:07:01 PM
Much appreciated for you staying with this...
SO the change is 0.6.

this then gives in kc expression
 0.6x0.6/ 0.3x(1.2x 1.2x1.2)
which results incorrectly.... :(
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 01:22:15 PM
Read stoichiometry of reaction: N2 gives xNH3, and calculate correctly hydrogen!
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 01:35:55 PM
still no success I'm afraid...
kc = 1.2x1.2/ 0.3x (0.6x0.6x0.6)
Title: Re: Kc and gases
Post by: AWK on May 04, 2020, 02:31:55 PM
2.7 - (2/3)·2.7 = xH2
It's high time to learn how to do simple calculations.
Title: Re: Kc and gases
Post by: Phoenix on May 04, 2020, 02:42:50 PM
I have been down that route before and used 1.8.

Thank you for your patience.
Enjoy the rest of your day.