VI. Phosgene , COCl 2 , is a toxic gas that decomposes according to the reaction

Question

VI. Phosgene, COCl2, is a toxic gas that decomposes according to the reaction:

COCl2 (g) CO(g) + Cl2 (g)
At a certain temperature, an equilibrium mixture was found to contain the following molar concentrations: 0.078M

CO; 0.035M Cl2; and 0.50M COCl2. The value of Kc for this reaction at this temperature is 5.46×103 . Hint: the specific temperature is not required for this problem.

Determine: (a) how many mol/liter of CO must be removed to triple the equilibrium concentration of chlorine and (b) the final equilibrium concentration of CO. (15 pts.).

[RP]25o =

mol/L CO removed

= [CO]eq

Explanation / Answer

The balanced chemical equation for the equilibrium reaction is given as

COCl2 (g) <======> CO (g) + Cl2 (g)

Kc = [CO][Cl2]/[COCl2] = (0.078)*(0.035)/(0.50) = 5.46*10-3

Now, a stress is applied on the system at equilibrium such that the concentration of Cl2 is tripled, i.e, [Cl2]eq,new = 3*0.035 M = 0.105 M. Now, as per Le Chatilier principle, it is evident that when [Cl2] is increased, the reverse reaction, i.e, formation of COCl2 is enhanced. Hence, we shall reverse the reaction and determine the equilibrium constant for the reverse reaction as

Kc’ = 1/Kc = 1/(5.46*10-3) = 183.150

(a) Set up the ICE chart as below.

CO (g) + Cl2 (g) <=======> COCl2 (g)

initial                                    0.078      0.035                           0.50

change                                     -x                                              +x

equilibrium                         (0.078 – x) 0.105                       (0.50 + x)

Kc’ = [COCl2]/[CO][Cl2]

=====> 183.150 = (0.50 + x)/(0.078 – x)(0.105)

=====> 183.150*(0.078 – x)*(0.105) = 0.50 + x

=====> 19.23075*(0.078 – x) = 0.50 + x

=====> 1.4999985 – 19.23075x = 0.50 + x

=====> 0.9999985 = 19.23075x + x

=====> 0.9999985 = 20.23075x

=====> x = 0.9999985/20.23075 = 0.04943 0.049

The number of mol/L of CO removed from the system at equilibrium is 0.049 M (ans).

(b) The new equilibrium concentration of CO is (0.078 – 0.049) M = 0.029 M (ans).

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