Answer:
3,16x10⁻³M
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g). The kc is defined as:
kc = [NO]² / [N₂] [O₂] = 4,10x10⁻⁴ <em>(1)</em>
If you add in a 1,0L container 0,25 mol of N₂ and 0,10 mol of O₂, concentrations in equilibrium will be:
[N₂] = 0,25M - x
[O₂] = 0,10M - x
[NO] = 2x
Replacing in (1):
[2X]² / [0,25-x] [0,10-x] = 4,10x10⁻⁴
[2X]² / 0,025 - 0,35x + x²= 4,10x10⁻⁴
4X² = 4,10x10⁻⁴x² - 1,435x10⁻⁴x + 1,025x10⁻⁵
3,99959x² + 1,435x10⁻⁴x - 1,025x10⁻⁵ = 0
Solving for x:
x = -0,0016 (<em>Wrong answer, there is no negative concentrations</em>)
x = 0,00158 (<em>Right answer</em>)
As molar concentration of NO in equilibrium is 2x:
[NO] = 2x = 2×0,00158 = <em>3,16x10⁻³M</em>
I hope it helps!
