Calculating Equilibrium Concentrations

Question

Consider the equilibrium system involving the decomposition of nitrogen monoxide.

2NO(g) N2(g) + O2(g)	K =	[N2] [O2]	= 8.10×10-2 at 283 K
[NO]2

A flask originally contains 0.206 M nitrogen monoxide. Calculate the equilibrium concentrations of the three gases.

[NO]	=	M
[N2]	=	M
[O2]	=	M

Answer 1

[NO] = 0.180M, [O₂] = 0.0513M, [N₂] = 0.0513M

In summary, I put the concentrations in terms of x with x being the M lost by NO in the reaction. Then I plugged the values into the formula for K, used algebra to solve for x, and then plugged the x value into the concentrations that were in terms of x.

2NO (g) ⇔ N₂ (g) + O₂ (g)

K= ([B]^b[C]^c)/([A]^a) for reaction aA⇔bB+cC

In this case 0.081 = ([N₂][O₂])/([NO]²)

[NO]=0.206 - x

[O₂] = +2x

[N₂] = +2x

Plugging into the formula, we get 0.081 = 4x²/(0.042436 - 0.412x + x²)

Removing the denominator, we get: 0.081x² - 0.033372x + 0.003437316 = 4x²

Making one side zero: -3.919x² - 0.033372x + 0.003437316 = 0

Solving for x with the quadratic formula, we get x=0.0256625 as the only positive answer.

Recalling the concentrations in terms of x, we have:

[NO]=0.206 - x

[O₂] = +2x

[N₂] = +2x

Solving for these with respect to 3 significant figures from given 0.206M, we get:

[NO] = 0.180M

[O₂] = 0.0513M

[N₂] = 0.0513M