Question

Suppose a 500. mL flask is filled with 2.0 mol of I_2 and 0.40 mol of HI. The following reaction becomes possible: H_2(g) + IK_2(g) 2HI(g) The equilibrium constant K for this reaction is 9.22 at the temperature of the flask. Calculate the equilibrium molarity if I_2.

Answer 1

Ans. Balanced reaction:       H_2(g) + I_2(g) ------------> 2 HI_(g)

            Equilibrium constant, Kc = [HI]² / ([H₂] [I₂])                  - equation 1

Given, volume of the flask = 0.5 mL

            Concentration of I₂ = moles of I₂ / Volume (in Liters) of flask

                                                = 2.0 mol/ 0.5 L

                                                = 4.0 mol/ L

                                                = 4.0 M

            Concentration of HI = moles of HI / Volume (in Liters) of flask

                                                = 0.4 mol/ 0.5 L

                                                = 0.8 mol/ L

                                                = 0.8 M

To establish equilibrium, there must be presence of H₂(g) in the flask. It only appears when the reaction takes pace in backward direction, i.e. 2HI dissociates to form H2 and I2. So, Create the ICE table for this as shown in picture.

2 HI (g) H2(g) 0.8 (-2x) +12(g) Initial (M) Change (M) Equilibrium (M) 0 +X +X 4 +X 4 + X 0.8-2x

Now,

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

            Or, 9.22 = [ (x) (4+x)] / (0.8 – 2x)² = (x² + 4x) / (4x² – 3.2x + 0.64)

Or, 9.22 x (4x² – 3.2x + 0.64) = (x² + 4x)

Or, 36.88x² -29.50x + 5.90 = x² + 4x

Or, 35.88x² -33.50x + 5.90 = 0

Solving the quadratic equation, we get two roots-

            X1 = 0.235     , X2 = 0.698

Note that the value of X can’t be greater than half of 0.8M because it will then make HI equilibrium concentration negative; that is, if X. 0.4, then (0.8-2x) will become negative. Concentrations can’t be negative. So, reject X2.

Thus, x = 0.235

Now,

Equilibrium concertation of I2 = 4+x = 4 + 0.235 = 4.235

Thus, molarity of I2 at equilibrium = 4.2 M