Answer to Question #146698 in Physical Chemistry for Atta ur Rehman

Question

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.

Answer 1

H_2(g)+I_2(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

K_c= [HI]²/{[H₂] [I₂]}

   = (0.316)²/ {0.042×0.042}

   = 56.61