Q.In the reaction,

[!FORMULA] $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq),$

if the concentration of $OH^-$ ions is decreased by $\frac{1}{4}$ times, then the equilibrium concentration of $Fe^{3+}$ will

Using Kc = [NH3]^2 / ([N2][H2]^3) with the given equilibrium concentrations, Kc works out to 1.0 x 10^3 (mol^-2 L^2).

The reaction for formation of ammonia is:

N2(g) + 3H2(g) <=> 2NH3(g)

The equilibrium constant expression (law of mass action) is:

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

Given:

[NH3] = 1.8 x 10^-2 M

[N2] = 1.2 x 10^-2 M

[H2] = 3 x 10^-2 M

Step 1 — numerator:

[NH3]^2 = (1.8 x 10^-2)^2 = 3.24 x 10^-4

Step 2 — denominator:

[H2]^3 = (3 x 10^-2)^3 = 27 x 10^-6 = 2.7 x 10^-5

[N2] x [H2]^3 = (1.2 x 10^-2) x (2.7 x 10^-5) = 3.24 x 10^-7

…

In an equilibrium constant expression, the exponents are the stoichiometric coefficients of that species, with products in the numerator and reactants in the denominator; reading Kc = [NH3]^4[O2]^5 / ([NO]^4[H2O]^6) this way and checking atom balance gives 4NO + 6H2O ⇌ 4NH3 + 5O2.

For a general equilibrium a A + b B ⇌ c C + d D, the equilibrium constant is Kc = [C]^c[D]^d / ([A]^a[B]^b) — products (right side of the equation) appear in the numerator, reactants (left side) in the denominator, each raised to its stoichiometric coefficient.

Here, Kc = [NH3]^4 [O2]^5 / ([NO]^4 [H2O]^6), so:

• Products: 4 NH3 and 5 O2
• Reactants: 4 NO and 6 H2O

Proposed equation: 4NO + 6H2O ⇌ 4NH3 + 5O2

Check atom balance: …