Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in m
ID: 913336 • Letter: P
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?
Explanation / Answer
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