Consider 2 NOCl(g) \Longleftrightarrow⟺ 2 NO(g) + Cl2 (g) At 25oC under conditions other than equilibrium, there are 1.20 moles of NOCl , 0.450 moles of NO, and 0.220 moles of Cl2 in a 5.00 L flask. Kc = 1.86 x 10-1. SHOW WORK FOR CREDIT. A. Calculate Q. B. Predict the direction of the reaction. C. Calculate equilibrium concentrations of all species present. The equilibrium concentration of Cl2 is 0.0490 M.

Which is the same equation for Kc, but in Kc expressions, the concentrations are in the equilibrium. Q is calculated at any time. So, for the reaction given

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

[tex]Q = \frac{[Cl2]x[[NO]^2}{[NOCl]^2}[/tex]

[Cl₂] = 0.220/5.00 = 0.044 M

[NO] = 0.450/5.00 = 0.090 M

[NOCl] = 1.20/5.00 = 0.240 M

Q = (0.044)x(0.090)²/(0.240)²

Q = 6.1875x10⁻³

b) Q < Kc, which means that there are fewer products to what are needed to the equilibrium. So the direction of the reaction is to form the products.

c)

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

0.240 0.090 0.044 Initial

-2x +2x +x Reacts (stoichiometry is 2:2:1)

0.240-2x 0.090+2x 0.044+x Equilibrium

[tex]Kc = \frac{(0.044+x)x(0.090+2x)^2}{(0.240 - 2x)^2}[/tex]

[tex]1.86x10^{-1} = \frac{(0.044+x)*(8.1x10^{-3} +0.36x + 4x^2)}{(0.0576 - 0.96x +4x^2)}[/tex]

3.564x10⁻⁴+0.01584x+0.176x²+8.1x10⁻³x+0.36x²+4x³ = 0.010714-0.17856x+0.744x²

4x³ - 0.208x² + 0.2025x - 0.01036 = 0

Solving this third grade equation in a computer program:

x = 0.05 M

So:

[Cl₂]e = 0.044 + 0.05 = 0.094 M

[NO]e = 0.090 + 2x0.05 = 0.190 M

[NOCl]e = 0.240 - 2x0.05 = 0.140 M