00:01
The equilibrium constant can change when a reaction is also changed.
00:05
If, say, we have a reaction that is described as a forming b, if the reaction is reversed, such that we would have b forming a.
00:19
So by the way, the equilibrium concept for the original equation is k, then for the reverse reaction, the equilibrium constant will be 1 over k.
00:29
Now if the reaction is multiplied by n such that we would have n a forming nb, the equilibrium constant will be raised to the same factor n.
00:44
So in this problem, we're given with two equations that are at equilibrium, namely the reaction between nitrogen and oxygen gas to form 2no with a kc, which is the equilibrium concept in terms of concentration as 4 .3 times 10 to the negative.
01:01
25 the second equation being the decomposition of no into or rather the reaction of and o with oxygen gas to form 2 and o2 with kc that is equal to 6 .4 times 10 to the power of 9 we want to know the equilibrium constant for our desired reaction described as you have 4no forming n2 and 2 and 2.
01:49
So we will be doing something for the given 2 reactions.
01:55
Let's name them first...