Consider the following equilibrium: 2NOCl(g) 2NO(g)+ Cl₂(g) With K = 1.6 $\times$10^-⁵.1.00 mole of pure NOCl and 0.964 mole of pure Cl₂ are placed in a 1.00-L container.Calculate the equilibrium concentration of Cl₂(g).

Question

Quizplus · Accepted Answer

We can start by writing the equilibrium expression and plugging in the given values:

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

K = 1.6x10^-10 mol/L

[NOCl] = 1.00 mol / 1.00 L = 1.00 M

[Cl2] = 0.964 mol / 1.00 L = 0.964 M

[NO] = (2K[NOCl]^2 / [Cl2])^(1/2) = 0.022 M

Now we can use the fact that Cl2 has a coefficient of 1 in the balanced equation, which means that at equilibrium its concentration is the same as its initial concentration. Therefore:

[Cl2] (at equilibrium) = 0.964 M

So the equilibrium concentration of Cl2 is 0.964 M, which is closest to answer choice B of 0.966 M.

Consider the Following Equilibrium: 2NOCl(g) 2NO(g)+ Cl2(g) with K ×\times×

Consider the Following Equilibrium: 2NOCl(g) 2NO(g)+ Cl₂(g)
with K $\times$