Question

Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g)+Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 459 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.760 Cl2 1.12 COCl2 0.250 What is the equilibrium constant, Kp, of this reaction? Kp = 0.294

Deriving concentrations from data In Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K. In Part B however, you will be given initial concentrations and only one equilibrium concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula for K. Part B The following reaction was performed in a sealed vessel at 779 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.20M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

Solutions :-

Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g)+Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 459 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.760 Cl2 1.12 COCl2 0.250 What is the equilibrium constant, Kp, of this reaction? Kp = 0.294

Solution :-

Using the equilibrium pressures we can calculate the kp as follows

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

Kp equation for the reaction is as follows.

Kp = [COCl2]/[CO][Cl2]

Kp = [0.250]/[0.760][1.12]

Kp = 0.294

Deriving concentrations from data In Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K. In Part B however, you will be given initial concentrations and only one equilibrium concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula for K. Part B The following reaction was performed in a sealed vessel at 779 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.20M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Solution :-

Lets make the ICE table for the react ion

             H2(g)                +         I2(g) < -------- > 2HI(g)

   3.90 M                               2.20 M                      0

-x                                          -x                              +2x

3.90 –x                              2.20-x = 0.0800           2x

Using the equilibrium concentration of the I2 we can find the value of x as follows

2.20-x = 0.0800

X= 2.20 – 0.0800 = 2.12 M

Now lets find the equilibrium concetrations of the H2 and HI

H2 = 3.90 – x = 3.90 M – 2.12 M = 1.78 M

HI = 2x = 2*2.12 M = 4.24 M

Now lets write the Kc equation

Kc = [HI]^2 /[H2][I2]

     = [ 4.24]^2/[1.78][0.0800]

     = 126

So the equilibrium constant Kc for the reaction is 126