Consider the following equilibrium: 2NOCl(g) ⇌ 2 NO(g) + Cl2(g) Kc = 1.6 x 10-5 1.00...

Question

Question

Consider the following equilibrium:

2NOCl(g) ⇌ 2 NO(g) + Cl₂(g) K_c = 1.6 x 10^-5

1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in a 1.00 L container. Calculate the equilibrium concentration of NO(g)

science chemistry

Answer 1

Answer #1

concentration of NOCl [NOCl] = no of moles/volume in L

= 1/1 = 1M

concentration of Cl2 [Cl2] = no of moles/volume in L

= 1/1 = 1M

2NOCl(g) ⇌ 2 NO(g) + Cl₂(g)

I 1 0 0

C -2x 2x x

E 1-2x 2x x

Kc = [NO]^2[Cl2]/[NOCl]^2

1.6*10^-5 = (2x)^2*x/(1-2x)^2

1.6*10^-5 *(1-2x)^2 = 4x^3

x = 0.01554

[NO] = 2x = 2*0.01554 = 0.03108M >>>>answer

Answer 2

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