Answer:
The concentration in equilibrium of NO is 0,550M.
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO
The equilibrium constant is defined as:
k = [NO]² / [N₂][O₂] <em>(1)</em>
Replacing for the concentrations in equilibrium:
k = (0,400M)² / (0,200M)(0,200M)
<em>k = 4,000</em>
If you add more NO until 0,700M, the equilibrium concentrations will be:
[NO] = 0,700M-2x
[N₂] = 0,200M+x
[O₂] = 0,200M+x
Replacing in (1)
4,000 = (0,700M-2x)² / (0,200M+x)²
4,000 = 4x²- 2,8x + 0,49 / x² + 0,4x + 0,04
4x² + 1,6x + 0,16 = 4x²- 2,8x + 0,49
4,4x = 0,33
x = 0,075M
That means that concentration in equilibrium of NO is:
[NO] = 0,700M - 2×0,075M = <em>0,550M</em>
I hope it helps!
