Predict and calculate the effect of volume change on an equilibrium system. Cons

Question

Predict and calculate the effect of volume change on an equilibrium system. Consider the equilibrium between COBr_2, CO and Br_2. COBr_2(g) CO(g) + Br_2(g) K = 2.21 at 383 K The reaction is allowed to reach equilibrium in a 16.8-L flask. At equilibrium, [COBr_2] - 1.93 times 10^-2 M, [CO] = 0.207 M and [Br_2] = 0.207 M. The equilibrium mixture is transferred to a 8.40-L flask. In which direction will the reaction proceed to reach equilibrium? Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 8.40-L flask. [COBr_2] = [CO] = [Br_2] =

Explanation / Answer

a)
Here volume is halfed
so, concentration of each will be doubled
[CoBr2] = 1.93*10^-2 * 2 = 3.86*10^-2 M = 0.0386 M
[CO] = 0.207 * 2 = 0.414 M
[Br2] = 0.207 * 2 = 0.414 M

Qc = [CO] [Br2]/[CoBr2]
= 0.414*0.414 / (3.86*10^-2)
= 4.44

since Qc > KC, the equilibrium will move to left

Answer: to the left

b)
CoBr2 (g) <---------> Co (g) + Br2 (g)
0.0386                           0.414       0.414   (initial)
0.0386 + x                0.414-x      0.414-x (at equilibrium)

Kc = (0.414-x)^2 / (0.0386+x)
2.21 = (0.1714 + x^2 - 0.828*x) /(0.0386+x)
0.0853 + 2.21*x =   0.1714 + x^2 - 0.828*x
x^2 - 3.038*x + 0.0861 = 0

solving above quadratic equation,
x = 3.01 M and x = 0.029 M

x can't be 3.01 M as it will make some concentration negative
so, x = 0.029 M

[CO] = 0.414-x = 0.414 - 0.029 = 0.385 M
[Br2] = 0.414-x = 0.414 - 0.029 = 0.385 M
[COBr2] = 0.0386 + x = 0.0386 + 0.029 = 0.068 M

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