## 7th Edition 5H.3

Ryan Danis 1J
Posts: 66
Joined: Fri Sep 28, 2018 12:18 am

### 7th Edition 5H.3

This question says ‘Use the information in table 5G.2 to determine the value of K at 300 K for the reaction 2BrCl(g)+H2(g) <-> Br2(g)+2HCl(g).

I am confused about how to combine reactions in table 5G2 to determine the value of K. Thanks! I

Samantha Ito 2E
Posts: 64
Joined: Fri Sep 28, 2018 12:29 am

### Re: 7th Edition 5H.3

Using table 5G.2, you will notice that K=377 for 2BrCl(g) <-> Br2(g) + Cl2(g) at 300K and K=4.0*10^31 for H2(g) + Cl2(g) <-> HCl(g) at 300K. You are then able to multiple these 2 K values together in order to get the K value of the combined reaction.

Andre_Galenchik_2L
Posts: 69
Joined: Fri Sep 28, 2018 12:23 am

### Re: 7th Edition 5H.3

As you can see by the reactions Samantha pointed out, when they are added together the Cl2 cancel out, making the reaction asked about in question 5H.3. When you add reactions, the K of the new reaction is the product of the K's of the reactions that were added together.

Aili Ye 4L
Posts: 58
Joined: Fri Sep 28, 2018 12:16 am

### Re: 7th Edition 5H.3

Since that equation is the addition of two previous equations whose K eqs are given, the K eq value of the total equation is the product of the two previous Keqs.