11.39 Combination of Reactions

[jgreynoso 2J]

For this problem, I understand how they got K for the reaction, but am unsure about the reaction addition done. I am curious and do not know fully why the addition of the two separate reactions, as seen on page 269 of the solutions manual, resulted in the reaction of the problem. What happened to the Cl2(g) from the products of the first equation and the Cl2(g) from the reactants of the second equation?

Thank you

[Ilana Golub 1A]

I am attaching two videos, but here is a short summary.

We know that our final equilibrium equation must be 2BrCl(g) + H2(g) <---> Br2(g) + 2HCl(g)

From Table 11.2, we can pull out the two equations
2BrCl(g) <---> Br2(g) + Cl2(g), K1=377
and
H2(g) + Cl2(g) <---> 2HCl(g), K2=4.0x10^31

If we "add" the two equations together, we get essentially
2BrCl(g) + H2(g) + Cl2(g) <---> Br2(g) + 2HCl(g) + Cl2(g)

The chlorine gases cancel out because one is on the reactant and one is on the product side, yielding
2BrCl(g) + H2(g) <---> Br2(g) + 2HCl(g)

Now, because we "added" the two equilibrium equations, we must multiply the two respective K constants.
Hence, K = K1 x K2 = 1.5x10^34

Why do we multiply the two K constants?
Take the reaction 2BrCl(g) <---> Br2(g) + Cl2(g), K1=377 for example.
If asked to calculate the K constant for Br2(g) + Cl2(g) <---> 2BrCl(g), one would simply find (K1)^-1 = (377)^-1 = 1/377
Effectively, we multiply the entire first equation by a negative one to yield its reverse: Br2(g) + Cl2(g) <---> 2BrCl(g)

So, if multiplying an equilibrium equation by a given number requires that K be set to a given power...
adding two equilibrium equations requires that the two K's be multiplied.


Part 1:

IMG_6089.MOV [ 7.41 MiB | Viewed 705 times ]

[Ilana Golub 1A]

Part 2:

IMG_6091.MOV [ 13.44 MiB | Viewed 703 times ]

Hope this helps:)