5H.3 7th edition

Moderators: Chem_Mod, Chem_Admin

Katie Frei 1L
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

5H.3 7th edition

Postby Katie Frei 1L » Wed Jan 09, 2019 2:32 pm

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!

Anusha 1H
Posts: 65
Joined: Fri Sep 28, 2018 12:15 am

Re: 5H.3 7th edition

Postby Anusha 1H » Wed Jan 09, 2019 2:57 pm

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!

Katie Frei 1L
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

Re: 5H.3 7th edition

Postby Katie Frei 1L » Wed Jan 09, 2019 3:08 pm

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!


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 4 guests