Question

At a certain temperature, the equilibrium constant Kc for this reaction is 53.3

H2 + I2 ---> 2HI
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at this temperature, 0.500 mol of H2 AND 0.500 mol of I2 were placed in 1.00L container to react. What concentration of HI is present at equilibrium?

[HI]= ?M

Answer 1

Initial concentration of H₂ = initial concentration of I₂ = number of moles / volume of the solution in L

                                                                            = 0.500 mol / 1.00 L

                                                                            = 0.500 M

                             H2    +    I2        2 HI

initial conc            0.500     0.500          0

change                    -c          -c           +2c

Equb conc           0.500-c   0.500-c        2c

Equilibrium constant , K = [HI]² / ([H₂][I₂])

                            53.3 = (2c)2 / [(0.500-c)(0.500-c)]

   2c / (0.500-c) = 7.3

                               2c = 3.65 - 7.3 c

                                c = 0.392 M

Therefore the equilibrium concentration of HI = 2c = 2x0.392 = 0.784 M