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, 0.069 mol/L. What is the equilib-
rium concentration of BrCl?
Answer in units of M
..........2BrCl ==> Br2 + Cl2
This done the same way as the earlier problem with SO2 --> SO3. These are equilibrium concns; therefore, substitute into Kc and calculate the one unknown. Post your work if you have trouble.
To find the equilibrium concentration of BrCl, we can use the equilibrium constant expression (Kc) and the given concentrations of bromine and chlorine.
The equilibrium constant expression for the given reaction is:
Kc = [Br2] x [Cl2] / [BrCl]^2
Given that Kc = 0.142, [Br2] = [Cl2] = 0.069 mol/L, and we need to find [BrCl].
Let's substitute the given values into the equation:
0.142 = (0.069) x (0.069) / [BrCl]^2
Simplifying the equation:
0.142 = 0.004761 / [BrCl]^2
Now, cross-multiply:
[BrCl]^2 = 0.004761 / 0.142
Divide both sides of the equation by the quotient:
[BrCl]^2 = 0.0335
To find [BrCl], take the square root of both sides of the equation:
[BrCl] = √(0.0335)
Calculating the square root:
[BrCl] ≈ 0.183 M
Therefore, the equilibrium concentration of BrCl is approximately 0.183 M.