Question

If 0.25 mol of br2 and 0.55 mol of cl2 are introduced into a 3.0-l container at 400 k, what will be the equilibrium concentration of br2

Answer 1

Answer is:  the equilibrium concentration of Br₂ is 0,02 mol/L.
Chemical reaction: Br₂ + Cl₂ → 2BrCl.
Kc = 7,0.
c₀(Br₂) = 0,25 mol ÷ 3 L.
c₀(Br₂) = 0,083 mol/L.
c₀(Cl₂) = 0,55 mol ÷ 3 L.
c₀(Cl₂) = 0,183 mol/L.
Kc = c(BrCl)² ÷ c(Br₂) · c(Cl₂).
7 = (2x)² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
7 = 4x² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
Solve quadratic equation: x = 0,063 mol/L.
c(Br₂) = 0,083 mol/L - 0,063 mol/L = 0,02 mol/L.

Answer 2

Answer:

0.0194 M

Explanation:

1. Find molarity

0.25 Br2 moles / 3.00 L = 0.0833 M

0.55 Cl2 moles / 3.00 L = 0.18333 M

2. Does the reaction go to the left or the right? Since K = 7 > 1, the reaction goes to the products. That means you are adding x to the products.

3. Create the Qc equilibrium expression

$\frac{[BrCl]^2)}{[Br2][Cl2]}$

4. Create an ICE Chart

ICE CHART

1Br2       +       1Cl2       ⇄       2 BrCl

0.0833          0.1833                    0

- 1x                   -1x                         +2x

0.0833-x        0.1833-x                 2x

Plug in your new values of concentration into your equilibrium expression, and solve for x. Thus, your quadratic equation is

7 = $\frac{[2x]^2}{(0.0833-x)(0.1833-x)}$)

Using a graphing calculator,

x = 0.0638, x = 0.558

Use whichever x will not make your concentration negative!

Br2 = 0.0833 - x

Qc of Br2 = 0.0833 - 0.06384 = 0.01946 M