5H.3 7th edition
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5H.3 7th edition
Does anyone know why for problem 5H.3 of the 7th edition of the textbook, the answer is the result of multiplying the two equilibrium constants of two chemical equations from the table 5G.2? I am confused why in order to find K of the chemical equation given, I would multiply the K's of the other two chemical equations. Thank you!
Re: 5H.3 7th edition
it helps if you write out the K values for all of the equations
2BrCl >>> Br2 + Cl2
H2 +Cl2 >>> 2HCl
-----------------------
2BrCl + H2 >>> Br2 + 2HCl
K1 = [Br2][Cl2]/[BrCl]^2
K2 = [HCl]^2/[H2][Cl2]
K3 = [Br2][HCl]^2/[H2][BrCl]^2 = K1 x K2
Hope this helps!
2BrCl >>> Br2 + Cl2
H2 +Cl2 >>> 2HCl
-----------------------
2BrCl + H2 >>> Br2 + 2HCl
K1 = [Br2][Cl2]/[BrCl]^2
K2 = [HCl]^2/[H2][Cl2]
K3 = [Br2][HCl]^2/[H2][BrCl]^2 = K1 x K2
Hope this helps!
-
- Posts: 64
- Joined: Fri Sep 28, 2018 12:27 am
Re: 5H.3 7th edition
Anusha 1C wrote:it helps if you write out the K values for all of the equations
2BrCl >>> Br2 + Cl2
H2 +Cl2 >>> 2HCl
-----------------------
2BrCl + H2 >>> Br2 + 2HCl
K1 = [Br2][Cl2]/[BrCl]^2
K2 = [HCl]^2/[H2][Cl2]
K3 = [Br2][HCl]^2/[H2][BrCl]^2 = K1 x K2
Hope this helps!
Thank you so much, your explanation really helps!
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