Question

In: Chemistry

Be sure to answer all parts. At 1280°C the equilibrium constant Kc for the reaction Br2(g)...

Be sure to answer all parts.

At 1280°C the equilibrium constant K_c for the reaction

Br₂(g) ⇌ 2Br(g)

is 1.1 ×10⁻³. If the initial concentrations are [Br₂] = 0.0320 M and [Br] = 0.0290 M, calculate the concentrations of these two species at equilibrium.

[Br₂]_eq =

[Br]_eq =

Expert Solution

Initially,

Qc = [Br]^2 / [Br2]

Qc = (0.0290)^2 / (0.0320)

Qc = 2.63*10^-2

Since Qc is greater than Kc, the reaction will move to left

Br2 (g) <—> 2 Br (g)

0.0320 0.0290 (initial)

0.0320+x 0.0290-2x (equilibrium)

at equilibrium,

Kc = [Br]^2 / [Br2]

1.1*10^-3 = (0.0290 - 2x)^2 / (0.0320+x)

3.52*10^-5 + 1.1*10^-3 x = 8.41*10^-4 + 4*x^2 - 0.116*x

4*x^2 - 0.1171*x + 8.058*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 4.0

b = -0.1171

c = 8.058*10^-4

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 8.196*10^-4

roots are :

x = 1.822*10^-2 and x = 1.106*10^-2

x can't be 1.822*10^-2 as this will make the concentration negative.so,

x = 1.106*10^-2

At equilibrium:

[Br2]eq = 0.0320 + x = 0.0320 + 1.106*10^-2 = 2.09*10^-2 M

[Br]eq = 0.0290 - 2x = 0.0290 - 2*(1.106*10^-2) = 6.88*10^-3 M

queen_honey_blossom answered 2 years ago

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