The reaction 2 NO(g) ⇌ N2(g) + O2(g) has a Kc value of 2400 at 2000 K. If 0.850 M each of N2 and O2 are initially present in a 3.00-L vessel, calculate the equilibrium concentrations of NO, N2, and O2 ​

[N2] = [O2] = 0.841M

And [NO] = 0.00172M

Explanation:

The equilibrium constant of this reaction, Kc, is:

Kc = 2400 = [N2] [O2] / [NO]²

Where [] are the equilibrium concentration of each specie.

The initial concentration of [N2] = [O2] = 0.850M. The equilibrium will shift to the left in order to produce NO. The equilibrium concentrations are:

[N2] = [O2] = 0.850M - X

And [NO] = 2X

Replacing:

2400 = [0.850-X]²  / [2X]²

9600X² = 0.7225 - 1.7 X + X²

0 = 0.7225 - 1.7 X - 9599X²

Solving for X:

X = -0.0088M. False solution, there is no negative concentrations.

X = 0.00859M. Right solution.

Replacing:

[N2] = [O2] = 0.850M - 0.00859M

And [NO] = 2*0.00859M

[N2] = [O2] = 0.841M

And [NO] = 0.00172M

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