Question

Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for this reaction is at the temperature of the flask. Calculate the equilibrium molarity of . Round your answer to two decimal places.

Answer 1

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:

N₂(g) +O₂(g) ⇄ 2 NO(g)

The equilibrium constant K for this reaction is 7.70 at the temperature of the flask.  Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

    N₂(g) +O₂(g) ⇄ 2 NO(g)

I    1.2        0              2.8

C  +x         +x            -2x

E  1.2+x      x           2.8 - 2x

The equilibrium constant (K) is:

$K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}$

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M