Question

Consider the reaction:

N₂ + 3H₂ ↔ 2NH₃

            Suppose 13.85 g N₂ and 1.88 g H₂ are combined in a 2.50 L flask. At equilibrium, we find that the flask still contains 7.00 g N₂.

            a.         Calculate the equilibrium concentrations of N₂, H₂, and NH₃.

            b.         Calculate K_C.

Answer 1

Calculate mass to mol

N2 = 13.85 g / 28 g/mol => 0.4946 mol

H2 = 1.88 g / 2 g/mol => 0.94 mol

N2 = 7 g/ 28 g/mol = > 0.25 mol

Calculate the molarity

Molarity = moles of solute / Ltr of solution

N2 = 0.4946 mol / 2.5 Ltr = 0.1978 M

H2 = 0.94 mol / 2.5 Ltr = 0.376 M

N2 = 0.25 mol / 2.5 Ltr = 0.1 M

N2. + 3H2. --------> 2NH3

Initial. 0.1978. 0.376. 0

Change. -x. -3x. +2x

Equilibrium 0.1978-x. 0.376-3x. +2x

0.1978-x = 0.1

X = 0.1978 - 0.1

X = 0.0978 M

A) at equilibrium N2 = 0.1 M

at equilibrium H2 = 0.376-(3*0.0978)

= 0.0826 M

at equilibrium NH3 = 2*0.0978

= 0.1976 M

B) Kc = [NH3]^2 / [N2] [H2]^3

Kc = 0.1976^2 / ( 0.1 * 0.0826^3)

Kc = 692.84