Question

A student places 7mol of ammonia in 14L flask. At equilibrium 0.2M ammonium was found in the flask.what is the equilibrium constant for this reaction? Use mole ratio to fill all the blank boxes initial, change and equilibrium and use the equilibrium concentration to solve for equilibrium constant.

Answer 1

[AI-generated answer — Bot GPT 3.5]

The balanced chemical equation for the reaction is:

NH3 (g) ⇌ NH4+ (aq) + OH- (aq)

Initially, the student placed 7 mol of ammonia in a 14 L flask, so the initial concentration of ammonia is:
[ NH3 ] = 7 mol / 14 L = 0.5 M

At equilibrium, the concentration of ammonium ions (NH4+) is 0.2 M.

Let x be the change in concentration of ammonia at equilibrium.

The equilibrium concentrations are given by:
[ NH3 ] = 0.5 - x
[ NH4+ ] = x
[ OH- ] = x

The equilibrium constant, Kc, is given by:
Kc = [ NH4+ ] [ OH- ] / [ NH3 ]

Substitute the equilibrium concentrations into the equation:
Kc = (0.2) (0.2) / 0.5

Kc = 0.04 / 0.5
Kc = 0.08

Therefore, the equilibrium constant for this reaction is 0.08.