00:01
In this problem, we've been given the value of k, you've been given a reaction, some initial conditions, and we've been asked to find the partial pressures of the reactants and products at equilibrium.
00:14
So let's say we have 4 .5 atmospheres of n204, so at equilibrium, some amount of it is going to use it to form n -o -2.
00:21
Since the equilibrium constant is equals to the partial pressure of n -2 square divided by partial pressure of n204, we can plug in these values at equilibrium into this equation and equate it to 0 .25, a value that we already have.
00:38
When we solve for x, we get 0 .498.
00:41
Using this value of x, we can calculate the partial pressure of n204, which is 4 .002, and n02, which is 0 .996.
00:51
Now, in the second case, we have been given that there is only 9 atmospheres of n02 initially.
00:59
And the reaction kind of proceeds in the opposite direction.
01:04
But as long as we keep our concentration straight and our value of kp straight, which is we are still using it as a forward reaction and not changing the value of k.
01:21
We can solve this equation that we get and get four atmospheres.
01:28
So actually there's two answers, four and five point zero...