# Calculate the value of K_p for the reaction "H"_2(g) + "Cl"_2(g) rightleftharpoons 2"HCl"(g), given the following reactions and their K_p?

## May 7, 2018

Warning! Long Answer. K_text(p) = 4.4 × 10^"-4"

#### Explanation:

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 $\text{2HCl(g)}$ 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 $\text{H"_2} \left(g\right)$ 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 $\text{CH"_2"Cl"_2"(g)}$ 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"