Question

Find the value of the equilibrium constant Kc at 460°C for the reaction-
½ H2(g) + ½ I2(g)
HI(g)
given the following data: A 4.50-mol sample of HI is placed in a 1.00-L vessel at 460°C, and the reaction system is allowed to come to equilibrium. The HI partially decomposes, forming 0.343 mol H2 at equilibrium.
A) 0.0123 B) 0.0081 C) 0.0309 D) 11.1 E) 5.69

Answer 1

Initial concentration of HI = number of moles / volume in L

                                    = 4.50 mol / 1.00 L

                                    = 4.50 M

                           HI(g)        ½ H₂(g) + ½ I₂(g)

initial moles    4.50    0    0

change -a +a/2    +a/2

Equb moles    4.50-a a/2    a/2

Given Equilibrium moles of H₂ is = a/ 2 = 0.343 mol

                                                           a = 0.686 mol

So Equilibrium moles of HI = 4.50 - a = 4.50 - 0.686 = 3.814 moles

   Equilibrium concentration of HI = number of moles / volume in L

                                                = 3.814 mol / 1.0 L

                                                = 3.814 M

Equilibrium concentration of H₂ & I₂ = 0.343 mol / 1.0L

                                                    = 0.343 M

Equilibrium constant , K = ([H₂(g)]^½ [I₂(g)]^½ ) / [HI]

                                    = (0.343^½ x 0.343 ^½) / 3.814

                                    = 0.0899