We have here a disguised **Hess' Law problem** using equilibrium constants.

We have three equations:

#bb(1.) color(white)(m)"COCl"_2"(g)" + "H"_2"O(l)" ⇌ "CH"_2"Cl"_2"(g)" + "O"_2"(g)"; color(white)(mmmmmmm)K_text(p₁) = 0.015#

#bb(2.) color(white)(m)"4HCl(g)" + "O"_2"(g)" ⇌ 2"H"_2"O(l)" + "2Cl"_2"(g)";color(white)(mmmmmmmmll)K_text(p₂) = 6800#

#bb(3.) color(white)(m)"2CH"_2"Cl"_2"(g)" + "2H"_2"(g)" + "3O"_2"(g)" ⇌ 2"COCl"_2"(g)" + "4H"_2"O(l)"; K_text(p₃) = 5.8#

From these, we must construct the target equation.

#"H"_2"(g)" + "O"_2"(g)" ⇌ "2HCl(g)"; K_text(p) = ?#

The target equation has #"2HCl(g)"# oh the right, so we reverse Equation **2.** and halve it.

When we reverse an equation, we **take the reciprocal** of its equilibrium constant.

When we halve an equation, we take **the square root** of its equilibrium constant.

#bb(4.) color(white)(m)"H"_2"O(l)" + "Cl"_2"(g)" ⇌ "2HCl(g)" + "½O"_2"(g)"; K_5 = 1/sqrtK_text(p₂)#

The target equation has #"H"_2"(g)# on the left.

We halve Equation **3.**

#bb(5.) color(white)(m)"CH"_2"Cl"_2"(g)" + "H"_2"(g)" + "³/₂O"_2"(g)" ⇌ "COCl"_2"(g)" + "2H"_2"O(l)" ;K_5 = sqrtK_text(p₃)#

Equation **5.** has #"CH"_2"Cl"_2"(g)"# on the left, and that is not in the target equation.

We re-write Equation **1.**

#bb(6.) color(white)(m)"COCl"_2"(g)" + "H"_2"O(l)" ⇌ "CH"_2"Cl"_2"(g)" + "O"_2"(g)"; K_6 = K_text(p₁)#

Now, we add Equations **4.**, **5.**, and **6.**, cancelling terms that appear on opposite sides of the equation.

When you add equations, you **multiply** their equilibrium constants.

#bb(4.) color(white)(m)color(red)(cancel(color(black)("H"_2"O(l)"))) + "Cl"_2"(g)" ⇌ "2HCl(g)" + color(red)(cancel(color(black)("½O"_2"(g)"))); color(white)(mmmmmmmml)K_4 = 1/sqrtK_text(p₂)#

#bb(5.) color(white)(m)color(red)(cancel(color(black)("CH"_2"Cl"_2"(g)"))) + "H"_2"(g)" + color(red)(cancel(color(black)("³/₂O"_2"(g)"))) ⇌ color(red)(cancel(color(black)("COCl"_2"(g)"))) + color(red)(cancel(color(black)("2H"_2"O(l)"))) ;K_5"="sqrtK_text(p₃)#

#bb(6.) ul(color(white)(m)color(red)(cancel(color(black)("COCl"_2"(g)"))) + color(red)(cancel(color(black)("H"_2"O(l)"))) ⇌ color(red)(cancel(color(black)("CH"_2"Cl"_2"(g)"))) + color(red)(cancel(color(black)("O"_2"(g)"))); color(white)(mmmmmml)K_6 = K_text(p₁))#

#color(white)(mml)"H"_2"(g)" + "Cl"_2"(g)" ⇌ "2HCl(g)"; color(white)(mmmmmmmmmmmmmm)K_text(p) = K_4K_5K_6#

#K_text(p) = K_text(p₁)sqrt(K_text(p₃)/K_text(p₂)) = 0.015sqrt(5.8/6800) = 0.015sqrt(8.53 × 10^"-4") = 0.015 × 0.0292 = 4.4 × 10^"-4"#