## Homework 11.39 Edition 6

Christine Chen 1H
Posts: 64
Joined: Fri Sep 28, 2018 12:18 am

### Homework 11.39 Edition 6

"Use the information in Table 11.2 to determine the value of K at 500K for the reaction 2NH3(g) + 3Br2(g) --><-- Br2(g) + 2HCl(g)."

In the solutions guide, it shows that we multiply K1 by K2. Why is that so?

Ibrahim Malik 1H
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am
Been upvoted: 1 time

### Re: Homework 11.39 Edition 6

You typed out the question for 11.40 instead of 11.39, is it possible you are looking at the wrong problem?

Brian Kwak 1D
Posts: 88
Joined: Fri Sep 28, 2018 12:17 am

### Re: Homework 11.39 Edition 6

If I am not mistaken it is because when you multiple K1 and K2 (the equilibrium constant equation for both of the given chemical reactions) you the equilibrium constant for the given equation after you do the algebra and cancel out the Cl2. It makes more sense when you write out the Equilibrium constant equations for both and multiple them.

Karyn Nguyen 1K
Posts: 72
Joined: Fri Apr 06, 2018 11:04 am

### Re: Homework 11.39 Edition 6

Brian Kwak 1D
Posts: 88
Joined: Fri Sep 28, 2018 12:17 am

### Re: Homework 11.39 Edition 6

We multiple it because we want to find the K(constant) for the given equation: 2BrCl(g) + H2(g) =(this is suppose to be the forward and reverse reaction) Br2(g) + 2 HCl(g). If you derive the Equilibrium constant for this equation it gives you: ([HCl]^2 [Br]^2)\([Brcl]^2 [H2]) If you add the equations for K1( H2(g) + Cl2(g) = 2HCl(g) ) and K2( 2BrCl(g) = Br2(g) + Cl2(g) ) and then derive the equilibrium constant you do not get the Equilibrium constant for the given equation and are therefor finding K for a different reaction. But if you multiple both reactions and derive the Equilibrium constant equation then you get the Equation for the Reaction you want. From there you can just Multiple the K values for K1 and K2 to find K3(the value for the question at hand).