multiplied reaction

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Ami_Pant_4G
Posts: 106
Joined: Sat Aug 24, 2019 12:17 am

multiplied reaction

Postby Ami_Pant_4G » Thu Jan 09, 2020 7:27 pm

If a reaction is multiplied, for example 2N2(g) + 6H2(g) <-> 4NH3(g) from N2(g) + 3H2(g) <-> 2NH3(g), why would the K value be raised to the second power? If the coefficients of the balanced equation can be reduced by some factor why would you need to raise K to some power? Thanks in advance.

805312064
Posts: 50
Joined: Thu Jul 11, 2019 12:16 am

Re: multiplied reaction

Postby 805312064 » Thu Jan 09, 2020 7:32 pm

If you find K for the original problem, you get something along the lines of K= [NH3}^2 / [N2][H2]^3
The K value for the other problem is K = [NH3}^4 / [N2]^2[H2]^6 which can be simplified to {[NH3}^2 / [N2][H2]^3}^2
Therefore, the K value of the other problem is K^2 of the original problem. I hope this helps!

Ryan Yee 1J
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

Re: multiplied reaction

Postby Ryan Yee 1J » Thu Jan 09, 2020 7:42 pm

Since we take the products over reactants raised to their stoichiometric powers, a multiple of the entire equation would cause for an increase in the same number power for each reactant and product. So if we were to multiply that equation's coefficients by 3 then the resulting K would be K^3

DanielTalebzadehShoushtari2A
Posts: 53
Joined: Fri Aug 02, 2019 12:16 am

Re: multiplied reaction

Postby DanielTalebzadehShoushtari2A » Thu Jan 09, 2020 7:46 pm

The above explanations make sense mathematically (thank you for posting them) but I would like to ask:
If each K value is valid for its respective reaction (multiplied or unmultiplied), how can there be 2 different valid K values for what is essentially the same reaction?


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