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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.

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Answer:

a) Q = 6.1875x10⁻³

b) The direction of the reaction is to form the products.

c) [Cl₂]e = 0.094 M

[NO]e = 0.190 M

[NOCl]e = 0.140 M

Step-by-step explanation:

a) Q is the reaction quotient, and for a generic reaction aA + bB ⇄ cC + dD it is


Q = ([C]^cx[D]^d)/([A]^ax[B]^b)

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)


Q = ([Cl2]x[[NO]^2)/([NOCl]^2)

[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


Kc = ((0.044+x)x(0.090+2x)^2)/((0.240 - 2x)^2)


1.86x10^(-1) = ((0.044+x)*(8.1x10^(-3) +0.36x + 4x^2))/((0.0576 - 0.96x +4x^2))

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

User Stefano Vet
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