In Part A, you were given the equilibrium pressures, which could be plugged dire

Question

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 713?C :

H2(g)+I2(g)?2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.25M. The equilibrium concentration of I2 is 0.0300M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Express your answer numerically.

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 460 C. At equilibrium, the concentrations were measured and the following results obtained: Partial Pressure (atm) 0.840 1.30 Gas CO Cl2 COCl20.210

Explanation / Answer

The following reaction was performed in a sealed vessel at 713?C :

H2(g)+I2(g)?2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.25M. The equilibrium concentration of I2 is 0.0300M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Get Kc, given

K = [HI]^2 /([H2][I2))

initially

[H2] = 3.9

[I2] = 2.25

[HI] = 0

in equilbirium

[H2] = 3.9 - x

[I2] = 2.25 - x = 0.03

[HI] = 0 + 2x

get x

x = 2.25-0.03 = 2.22

[H2] = 3.9 - 2.22 = 1.68

[I2] = 2.25 - x = 0.03

[HI] = 0 + 2*2.22 = 4.44

get

Kc = (4.44^2) / (0.03*1.68)

Kc = 391.14

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