Question

A 1.00 mol sample of hydrogen iodide vapor is placed in a 1.00 L vessel at 460oC. It decomposes to form 0.11 moles each of hydrogen gas and iodine vapor. What is Kc for this reaction at this temperature? Report your answer to 2 s.f. and use scientific notation

Answer 1

we know that

concentration = moles / volume (L)

in this case

volume = 1 L

so

concentration = moles

initially

[HI] = 1

now

consider the reaction

2HI --> H2 + I2

using ICE table

at equilibrium

[HI] = 1 -2x

[H2] = x

[I2] = x

given

at equilibrium conc of H2 is 0.11

so

x = 0.11

so

at equilibrium

[H2] = [I2] = x = 0.11

[HI] = 1 -2x = 1 -0.22 = 0.78

now

2HI ---> H2 + I2

the equilibrium constant is given by

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

so

Kc= [0.11] [0.11] / [0.78]^2

Kc = 0.01988

so

the value of Kc for this reaction is 0.02