Question

The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentrations are [Br2] = 0.0345 M and [Br] = 0.0416 M, calculate the concentrations of these species at equilibrium.

Answer 1

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

             Br2(g)    ⇌  2Br(g)

Initially  0.0345M  0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the  backward direction

 The I.C.E table is as follows:

                       Br2(g)    ⇌  2Br(g)

I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M