TiffanyL1G
Posts: 61
Joined: Fri Sep 28, 2018 12:22 am

In the 6th Edition textbook problem 11.39, the problem gives the reaction 2BrCl(g) + H2(g) $\rightleftharpoons$ Br2(g) + 2HCl(g) and hints for us to use Table 11.2 for help. From there I noticed we had to add two different equations to form the given reaction. How exactly do you add the two equations from Table 11.2 to make that reaction? Do we cross multiply the products and reactants of the two equilibrium constants? Or am I approaching this problem incorrectly?

Ray Guo 4C
Posts: 90
Joined: Fri Sep 28, 2018 12:15 am

You can multiply the equilibrium constants of the two reactions to get that of the third reaction.

Kate_Santoso_4F
Posts: 72
Joined: Fri Sep 28, 2018 12:29 am

Here is the rule based on the textbook:
If a chemical equation can be expressed as the sum of two or more chemical equations: The equilibrium constant for the overall reaction is the product of the equilibrium constants for the component reactions.
When you add the two reactions, Cl2 is present as a product in one equation and a reactant in another equation. Therefore, it cancels out to give you the overall reaction and you just multiple the two equilibrium constants for the two reactions together to find the K for the overall reaction.

105085381
Posts: 44
Joined: Fri Sep 28, 2018 12:15 am