## Homework Help 11.39

Andrew Uesugi 3I
Posts: 38
Joined: Wed Sep 21, 2016 2:56 pm

### Homework Help 11.39

Hello.

So I'm looking at question 11.39, and they showed me the calculations. But I don't exactly understand what is going on in this question. An explanation is appreciated.

From,

Andrew

Vivian Wang 3J
Posts: 29
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 1 time

### Re: Homework Help 11.39

Table 11.2 contains the K values for the following equations at 300 K:

H2 (g) + Cl2 (g) <==> 2HCl(g), K1 = 4.0 x 1031 = $\frac{(P_{H_{2}})(P_{Cl_{2}})}{(P_{HCl})^{2}}$

2BCl(g) <==> Br2 (g) + Cl2 (g), K2 = 377 = $\frac{(P_{BCl})^{2}}{(P_{Br_{2}})(P_{Cl_{2}})}$

To get the equation the problem has (2BrCl(g) + H2 (g) <==> Br2 (g) + 2HCl(g)), add the two equations above, and the Cl2 (g) will cancel as it is on either side of the equation.
As for the equilibrium constant equations, you will want to multiply them to get the equilibrium constant for the final equation, which will be:
Kfinal = $\frac{(P_{H_{2}})(P_{BCl})^{2}}{(P_{HCl})^{2}(P_{Br_{2}})}$

Kfinal = $\frac{(P_{H_{2}})(P_{Cl_{2}})}{(P_{HCl})^{2}}$ x $\frac{(P_{BCl})^{2}}{(P_{Br_{2}})(P_{Cl_{2}})}$ = $\frac{(P_{H_{2}})(P_{BCl})^{2}}{(P_{HCl})^{2}(P_{Br_{2}})}$ = K1 * K2

To get the equilibrium constant of the final equation, simply multiply K1 and K2.

Kfinal = (4.0 x 1031)(377) = 1.5 x 1034