Answer: The equilibrium concentration of NO is 0.0034 M
Explanation:
We are given:
Initial moles of nitrogen gas = 0.20 moles
Initial moles of oxygen gas = 0.15 moles
Volume of container = 1.0 L
We know that:
[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, [tex]\text{Molarity of }N_2=\frac{0.20}{1}=0.20M[/tex]
[tex]\text{Molarity of }O_2=\frac{0.15}{1}=0.15M[/tex]
The given chemical reaction follows:
[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g)[/tex]
Initial: 0.20 0.15
At eqllm: 0.20-x 0.15-x 2x
The expression of [tex]K_c[/tex] follows:
[tex]K_c=\frac{[NO]^2}{[N_2]\times [O_2]}[/tex]
We are given:
[tex]K_c=4.10\times 10^{-4}[/tex]
Putting values in above equation, we get:
[tex]4.10\times 10^{-4}=\frac{(2x)^2}{(0.20-x)\times (0.15-x)}\\\\x=-0.0018,0.0017[/tex]
Neglecting the negative value of 'x' because concentration cannot be negative.
So, equilibrium concentration of NO = 2 x = (2 × 0.0017) = 0.0034 M
Hence, the equilibrium concentration of NO is 0.0034 M
