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. 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] = M [CO] = M [Br_2] = M

Explanation / Answer

The equilibrium reaction is

COBr2 (g) <=====> CO (g) + Br2 (g); K = 2.21 at 383 K.

(a) K = [CO][Br2]/[COBr2]

Number of mole(s) of reactant = 1.

Number of mole(s) of product(s) = 2.

Change in number of moles in going from the reactant side to the product side = 2 – 1 = 1

The volume of the reaction flask is halved from 16.8 L to 8.4 L. Decrease in volume is accompanied by an increase in the pressure. Pressure and the number of moles are directly related as per the ideal gas law: P*V = n*R*T where P = pressure of the gas, V = volume of the gas, T = absolute temperature of the gas and n= number of moles of the gas.

Since the given reaction is accompanied by an increase in the number of moles, reducing volume (which translates to increasing pressure) will shift the equilibrium in a direction where the effect of increased pressure is nullified. The reactant side contains less moles of the gas and hence reducing the volume will favour the reverse reaction.

Ans: The reaction will proceed to the left.

(b) Calculate the moles of the gaseous reactant and products from the given concentrations and volume in part (a).

Moles of COBr2 = (volume of the reaction vessel)*[COBr2] = (16.8 L)*(1.93*10-2 mol/L) = 0.32424 mole.

Moles of CO = (16.8 L)*(0.207 mol/L) = 3.4776 mole.

Moles of Br2 = (16.8 L)*(0.207 mol/L) = 3.4776 mole.

The volume is reduced, hence the molar concentrations will be reduced (number of moles remains the same).

Calculate the new molar concentrations:

[COBr2]new = (number of moles)/(new volume) = (0.32424 mole)/(8.40 L) = 0.0386 M

[CO]new = (3.4776 mole)/(8.40 L) = 0.414 M

[Br2]new = (3.4776 mole)/(8.40 L) = 0.414 M

These are the new initial concentrations; set up the ICE chart (use the reverse reaction since the reverse reaction is favoured) to find out the new equilibrium concentrations:

CO (g) + Br2 (g) <=====> COBr2 (g)

initial                                       0.414      0.414                    0.0386

change                                      -x           -x                           +x

equilibrium                           (0.414 – x)(0.414 – x)        (0.0386 + x)

K’ = 1/K = [COBr2]/[CO][Br2] = (0.0386 + x)/(0.414 – x)2

===> 1/2.21 = (0.0386 + x)/(0.414)2 (assume small dissociation)

===> 0.4525 = (0.0386 + x)/0.171369

===> 0.0386 + x = 0.077557

===> x = 0.03895 0.0389

The equilibrium concentrations are [COBr2] = (0.0386 + 0.0389) = 0.0775 M

[CO] = [Br2] = (0.414 – 0.0389) = 0.3751 M (ans).

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