Question #146698

1.000 mole of H2 gas and 1.000 mole of I2 vapor are introduced into a 5.00-liter sealed flask. The mixture is heated to a certain temperature and the following reaction occurs until equilibrium is established.
H2(g) + I2(g) ⇄ 2HI(g) 2-1.58=0.42/2=0.21
At equilibrium, the mixture is found to contain 1.580 mole of HI. (a) What are the concentrations of H2, I2 and HI at equilibrium? (b) Calculate the equilibrium constant Kc.

Expert's answer

H2(g)+I2(g)=2Hl(g)

I (moles)        1        1        0

E(moles)       1-x    1-x       2x

                                1.580

Moles of H2 that reacted= 1.580/2

                                      =0.79 moles

Moles of I2 that reacted =0.79 moles

Moles of H2 at equilibrium= 1-0.79

                                          =0.21 moles

Moles of I2 at equilibrium= 0.21 moles

Concentration of H2 at equilibrium= 0.21/5

                                                      =0.042moldm-3

Concentration of I2 at equilibrium= 0.042moldm-3

Concentration of HI at equilibrium=1.580/5

                                                       =0.316 moldm-3

Kc= [HI]2/{[H2] [I2]}

   = (0.316)2/ {0.042×0.042}

   = 56.61



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