In: Chemistry
Be sure to answer all parts.
At 1280°C the equilibrium
constant Kc 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 =
Initially,
Qc = [Br]^2 / [Br2]
Qc = (0.0290)^2 / (0.0320)
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