HBr(g) ⇌ H2(g) + Br2(g) Kc = ?
Hey there Lynsae. Here is the solution to your question. This is a classic problem where you must rearrange a chemical equation in order to obtain a new equilibrium constant value for your new chemical equation. This was a simple one. Notice that HBr was initially a product, and then became a reactant for the second equation. This means that the reaction had to be reversed in order for the products to now become reactants (same thing for the reactants that now became products).
Once you reverse the initial equation, you'll get the exact same reaction! You still must show that you reversed the initial equation when it comes to Kc. So once you reverse the equation, you must find the reciprocal of your original Kc. This is done only when the reaction is reversed. This is why my Kc1 or my initial Kc was placed under a 1. That will give me my new Kc or Kc2.
Hope this helped! Please let us know if you have any other questions :)
Thank you so much! I knew it would be fairly simple but I just couldn't figure it out. I'm really understanding most of the equilibrium chapter which is nice :)