At 350 K, Kc = 0.142 for the reaction
2BrCl(g) *) Br2(g) + Cl2(g)
An equilibrium mixture at this temperature
contains equal concentrations of bromine and
chlorine, 0.067 mol/L. What is the equilib-
rium concentration of BrCl?
Answer in units of M.
Anonymous, check your 2-11-11,12:00am post.
2BrCl ==> Br2 + Cl2
K = 0.142 = (Br2)(Cl2)/(BrCl)^2
Substitute and solve for the only unknown you have; i.e., (BrCl)
To find the equilibrium concentration of BrCl, we can use the given equilibrium constant, Kc, and the concentrations of bromine and chlorine in the equilibrium mixture at 350 K.
The balanced equation for the reaction is:
2BrCl(g) ↔ Br2(g) + Cl2(g)
Let's denote the equilibrium concentration of BrCl as [BrCl]. Since there are equal concentrations of bromine and chlorine, the concentration of bromine, [Br2], and the concentration of chlorine, [Cl2], are both equal to 0.067 mol/L.
Using the equilibrium constant expression for the reaction:
Kc = [Br2][Cl2] / [BrCl]^2
Substituting the given values into the expression:
0.142 = (0.067)(0.067) / [BrCl]^2
To find [BrCl]^2, we can rearrange the equation:
[BrCl]^2 = (0.067)(0.067) / 0.142
Calculating the right side:
[BrCl]^2 = 0.0002208
Taking the square root of both sides, we get:
[BrCl] = √0.0002208
Calculating the square root:
[BrCl] ≈ 0.0149 M
Therefore, the equilibrium concentration of BrCl is approximately 0.0149 M.