Question

Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:

N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.

Answer 1

[N₂] = 1.1M

Step-by-step explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M