Question

How many moles of H2 and HI will be present at equilbrium if 0.2980 mol HI are placed into a 1-L flask and allowed to react at a temperature where for the following reaction takes place: 2HI(g) → H2(g) + 12(g) K = 10.8 mol H2 mol HI

Answer 1

Sol :-

Given moles of HI = 0.2980 mol

ICE table of HI is :

.......................................2HI (g) <----------------------> H2 (g)..........................+..........................I2 (g)

Initial (I)..........................0.2980 mol..........................0.0 mol.....................................................0.0 mol

Change (C).......................-2α....................................+α...............................................................+α

Equilibrium (E)..............(0.2980-2α) mol....................α mol.........................................................α mol

Here, α = Degree of dissociation of HI

Equilibrium constant (K) is equal to the ratio of molar concentration of products to the molar concentration of reactants raise to power their stoichiometric coefficient at equilibrium stage of the reaction.

Expression of K is :

K = [H2].[I2] / [HI]2

10.8 = α2/(0.2980-2α)

10.8.(0.2980-2α) = α2

α2 + 21.6 α - 3.2184 = 0

By solving this quadratic equation :

α = 0.1480

So,

Equilibrium moles of H2 = α = 0.1480 mol

and

Equilibrium moles of HI = (0.2980-2α) mol = 0.2980 - 2 x 0.148 = 0.0020 mol