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5H.3 7th edition

Posted: Wed Jan 09, 2019 2:32 pm
by Katie Frei 1L
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

Posted: Wed Jan 09, 2019 2:57 pm
by Anusha 1H
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!

Re: 5H.3 7th edition

Posted: Wed Jan 09, 2019 3:08 pm
by Katie Frei 1L
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!