Chemists studied the formation of phosgene by sealing 1.04 atm of carbon monoxid

Question

Chemists studied the formation of phosgene by sealing 1.04 atm of carbon monoxide and 1.27 atm of chlorine in a reactor at a certain temperature. The pressure dropped smoothly to 1.39 atm as the system reached equilbrium. Calculate Kp (in atm-1) for co(g) + cl2(g) find the final pressure of each gas. Consider 1.20 mol of carbon monoxide and 4.00 mol of chlorine sealed in a 3.00 L container at 476 degree C. The equilibrium constant, Kc, is 2.50 (in M-1) for CO(g) + Cl2(g) COCI2(g) Calculate the equilibrium molar concentration of CO. Answer: Hint, you need to find the equilibrium concentrations of all the gases. Note, that only one of the solutions to the quadratic equation makes sense (the other gives at least one negative concentration).

Explanation / Answer

Question 1)

Pressure of carbon monoxide – 1.04 atm

Pressure of chlorine = 1.27 atm

Final pressure of the system at equilibrium = 1.39 atm

Lets use Reaction to set up ICE chart for the partial pressure.

            CO(g) +      Cl2(g) -- > COCl2 (g)

I           1.04 atm            1.27 atm         0

C          -x                     -x                     +x

E          (1.04 atm –x)   (1.27atm -x)     x

Calculation of x :

Total pressure of the system = 1.39 atm = P(CO) + P(Cl2) + P(COCl2)

(1.04 atm –x) + (1.27atm -x) + x= 1.39 atm

Lets simplify the equation

1.04 atm – x + 1.27 atm – x + x = 1.39 atm

2.31 atm – 1.39 atm = x

x = 0.92 atm

Calculation of partial pressure of each gas :

P(CO) = 1.04 atm –x = 1.04 atm – 0.92 atm = 0.12 atm

P(Cl2) = 1.27 atm - x= 1.27 atm – 0.92 atm = 0.35 atm

P(COCl2) = x = 0.92 atm

Calculation of kp

Kp = P(COCl2) / (P(CO) x P(Cl2) )

= 0.92 atm / ( 0.12 atm x 0.35 atm )

= 21.90

Hence kp of this reaction is 21.90

Question 2)

Given :

Mol of CO = 1.20 , mol Cl2 = 4.00 , Volume of container = 3.00 L

T = 476 deg C = 476 deg C + 273.15 = 749.15 K

Kc = 2.50

Solution :

Lets calculate concentration of reactant :

Concentration of CO (g) = n / V in L = 1.20 mol/ 3.0 L = 0.4 M

Concentration of Cl2 = 4.00 mol / 3.0 L = 1.33 M

Lets set up ICE chart

CO (g)            +          Cl2 (g) --- > COCl2 (g)

I           0.4 M                         1.33                 0

C          -x                                 -x                     +x

E            (0.4-x)                         (1.33-x)         x

Kc expression for this reaction :

Kc = [COCl2 ]/ [CO] [Cl2]

2.50 = x / ( 0.4 –x ) ( 1.33 – x)

Lets simplify this equation.

2.50 ( 0.4 –x ) ( 1.33 – x) = x

2.5 x2 – 5.32 x + 1.33 = 0

Now we got quadratic equation.

Lets solve for x

After solving above quadratic equation we got

x = 0.288 and x = 1.84

Since x = 1.84 is more than concentration of reactants that is not possible so we reject that value and use x = 0.288

Lets plug this value to get concentration of each species at equilibrium

[ CO ]= 0.4 – x = 0.4 – 0.288 M = 0.112 M

[Cl2]= 1.33 – 0.288 = 1.042 M

[COCl2] = 0.288 M

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