7.5 - Magnitude of the Equilibrium Constant

Overview

Every equilibrium constant is a ratio, with products on top and reactants on the bottom. That makes the size of K a quick read on what the equilibrium mixture looks like. A huge K means the reaction runs essentially to completion, a tiny K means it barely gets started, and a K near 1 means you'll find plenty of both. On the AP exam, reading K this way tells you what to expect before you calculate anything, and which shortcut to use when you do.

What the size of K tells you

For the general reaction

a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}

, the equilibrium expression from Topic 7.3 is

K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

The numerator holds the products and the denominator holds the reactants, each measured at equilibrium. (For

K_p

, use partial pressures instead. The reasoning is the same.)

K much greater than 1: the numerator is much bigger than the denominator, so the equilibrium mixture is mostly products. The reaction is product-favored. When K is very large, the reaction goes essentially to completion: so little reactant is left that, for most purposes, you can treat it as zero.

K much less than 1: the denominator wins, so the equilibrium mixture is mostly reactants. The reaction is reactant-favored. When K is very small, the reaction barely proceeds, and only a trace of product forms.

K close to 1: neither side dominates, and the equilibrium mixture has significant amounts of both reactants and products.

There's no official cutoff for "very large" or "very small." A common rule of thumb: K above about

10^3

means essentially complete, and K below about

10^{-3}

means the reaction barely proceeds.

Seeing it in a one-to-one reaction

The cleanest case is a reaction with one reactant particle and one product particle,

\mathrm{A} \rightleftharpoons \mathrm{B}

, where

K = [\mathrm{B}]/[\mathrm{A}]

. Here K is literally the ratio of product to reactant at equilibrium. If you start with pure A, the fraction that ends up as B is

K/(1 + K)

K	[B] : [A] at equilibrium	Percent of A converted to B
$10^{-3}$	1 : 1,000	about 0.1%
$10^{-1}$	1 : 10	about 9%
1	1 : 1	50%
10	10 : 1	about 91%
$10^{3}$	1,000 : 1	about 99.9%

Notice how lopsided it gets. At

K = 10^3

, only about 1 molecule in 1,000 is still A. At

K = 10^{10}

, it's 1 in 10 billion. The graphs show three such reactions, each starting with 1.00 M A and no B.

7.5 - Magnitude of the Equilibrium Constant

All three reactions reach equilibrium, and then their concentrations stop changing. What differs is where they stop: almost all the way to B, halfway, or barely past the start.

Example 1: Reading K values

Three different reactions, each of the form

\mathrm{A(g) \rightleftharpoons B(g)}

, reach equilibrium at the same temperature. Reaction 1 has

K_c = 3.5 \times 10^{5}

, reaction 2 has

K_c = 0.80

, and reaction 3 has

K_c = 2.0 \times 10^{-6}

. Each flask starts with pure A.

(a) Which reaction goes essentially to completion, and which barely proceeds? Reaction 1 goes essentially to completion, because its K is far above

10^3

. Reaction 3 barely proceeds, because its K is far below

10^{-3}

. Reaction 2 ends with significant amounts of both A and B.

(b) What is the ratio [B]/[A] at equilibrium in each? It's just K. Reaction 1 has 350,000 molecules of B for every molecule of A. Reaction 3 has 2 molecules of B for every million molecules of A.

(c) What percent of A is converted in reaction 2?

\frac{K}{1 + K} \times 100\% = \frac{0.80}{1.80} \times 100\% = 44\%

Real reactions span a huge range

Measured K values run from far below

10^{-30}

to far above

10^{30}

. A few examples at 25 °C:

Reaction	K at 25 °C	What you find at equilibrium
$\mathrm{2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)}$	$K_p \approx 10^{80}$	Essentially all water
$\mathrm{H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)}$	$K_p \approx 2 \times 10^{33}$	Essentially all HCl
$\mathrm{H_3O^{+}(aq) + OH^{-}(aq) \rightleftharpoons 2H_2O(l)}$	$K = 1.0 \times 10^{14}$	Neutralization goes to completion
$\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$	$K_p \approx 0.15$	Both gases present
$\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons H_3O^{+}(aq) + CH_3COO^{-}(aq)}$	$K_a = 1.8 \times 10^{-5}$	Mostly un-ionized acetic acid
$\mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)}$	$K_{sp} = 1.8 \times 10^{-10}$	Almost none of the solid dissolves
$\mathrm{2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)}$	$K_w = 1.0 \times 10^{-14}$	Almost every water molecule intact
$\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}$	$K_p \approx 5 \times 10^{-31}$	Essentially no NO

You'll meet

K_{sp}

in Topic 7.11 and

K_a

and

K_w

in Unit 8. They're all equilibrium constants, so everything on this page applies to them. The last row explains why the air around you, about 78% nitrogen and 21% oxygen, doesn't turn into nitrogen monoxide: K is so small that essentially none forms at room temperature.

Very large K: treat it like stoichiometry

When K is very large, assume the reaction goes to completion, just like a limiting-reactant problem from Topic 4.5. Then, if the question asks for it, use K to find the tiny amount of the limiting reactant that's still there. This is easy to miss: students often set up a long equilibrium calculation and get the product concentration wrong, when the large K is telling them the reaction is basically finished.

Find the limiting reactant and let it react completely. This gives you the product concentrations and the leftover excess reactant.

Let a small amount $x$ of product go back to reactants. Because K is huge, $x$ is tiny, so it doesn't change the large concentrations.

Solve the K expression for $x$ , which is the concentration of the limiting reactant at equilibrium.

Example 2: A reaction that goes essentially to completion (AP-style)

Carbon monoxide and chlorine combine to form phosgene,

\mathrm{COCl_2}

\mathrm{CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)}

At a certain temperature,

K_c = 4.6 \times 10^{9}

. A 2.00 L flask is filled with 0.400 mol of CO and 0.300 mol of

\mathrm{Cl_2}

and held at that temperature until it reaches equilibrium.

(a) Starting concentrations.

[\mathrm{CO}]_0 = \frac{0.400\ \mathrm{mol}}{2.00\ \mathrm{L}} = 0.200\ \mathrm{M} \qquad [\mathrm{Cl_2}]_0 = \frac{0.300\ \mathrm{mol}}{2.00\ \mathrm{L}} = 0.150\ \mathrm{M}

(b) Explain why you can treat the reaction as complete.

K_c

is very large (

4.6 \times 10^{9} \gg 1

), so at equilibrium the ratio of product to reactants is enormous. Nearly all of the limiting reactant must end up as product. The reactants combine 1 : 1, so

\mathrm{Cl_2}

(0.150 M) runs out before CO (0.200 M):

\mathrm{Cl_2}

is the limiting reactant.

(c) Let it react completely, then back off by x.

	$[\mathrm{CO}]$ (M)	$[\mathrm{Cl_2}]$ (M)	$[\mathrm{COCl_2}]$ (M)
Initial	0.200	0.150	0
After reacting completely	0.050	0	0.150
Change (back off)	$+x$	$+x$	$-x$
Equilibrium	$0.050 + x \approx 0.050$	$x$	$0.150 - x \approx 0.150$

(d) Solve for the chlorine that's left.

K_c = \frac{[\mathrm{COCl_2}]}{[\mathrm{CO}][\mathrm{Cl_2}]} = \frac{0.150}{(0.050)(x)} = 4.6 \times 10^{9}

x = \frac{0.150}{(0.050)(4.6 \times 10^{9})} = 6.5 \times 10^{-10}\ \mathrm{M}

At equilibrium:

[\mathrm{COCl_2}] = 0.150

[\mathrm{CO}] = 0.050

M, and

[\mathrm{Cl_2}] = 6.5 \times 10^{-10}

M. The chlorine isn't literally zero, but it's less than a billionth of a mole per liter, so the reaction is complete for any practical purpose.

(e) Check.

x

is far too small to change 0.050 or 0.150, so dropping it in those terms was fine. Substituting back:

0.150 / [(0.050)(6.5 \times 10^{-10})] = 4.6 \times 10^9

, which matches

K_c

Very small K: almost nothing happens

When K is very small, the opposite shortcut works. So little reactant turns into product that the reactant concentrations barely change. Topic 7.7 turns this into the small-x approximation, along with a check that tells you whether it's allowed.

Example 3: How much does water ionize?

Water molecules transfer protons to each other:

\mathrm{2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)} \qquad K_w = 1.0 \times 10^{-14} \text{ at } 25\ ^\circ\mathrm{C}

Water is a pure liquid, so it's left out of the expression (Topic 7.3). In pure water, the two ions form in a 1 : 1 ratio, so call each concentration

x

K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = x^2 = 1.0 \times 10^{-14} \quad \Rightarrow \quad x = 1.0 \times 10^{-7}\ \mathrm{M}

Compare that with the concentration of water itself. A liter of water has a mass of about

1.00 \times 10^3

g, and the molar mass of water is

\mathcal{M} = 18.02

g/mol:

[\mathrm{H_2O}] = \frac{1.00 \times 10^{3}\ \mathrm{g} \div 18.02\ \mathrm{g/mol}}{1.00\ \mathrm{L}} = 55.5\ \mathrm{M}

\text{fraction ionized} = \frac{1.0 \times 10^{-7}\ \mathrm{M}}{55.5\ \mathrm{M}} = 1.8 \times 10^{-9}

Only about 2 of every billion water molecules have handed off a proton at any moment. That's what a K of

10^{-14}

looks like: the reaction really does happen (pure water conducts electricity very slightly), but it barely proceeds.

K tells you how far, not how fast

K describes the equilibrium state, the place a reaction ends up. It says nothing about how long the reaction takes to get there. Speed is kinetics (Unit 5), and it depends on the activation energy, not on K.

A catalyst speeds up the forward and reverse reactions equally. It helps a system reach equilibrium sooner, but it doesn't change K or the equilibrium composition.

Example 4: Evaluating a claim (AP-style)

A student seals hydrogen and oxygen gas in a steel container at 25 °C. A week later, no water has formed. The student claims: "This reaction must have a small K, because almost no product formed." Evaluate the claim.

Answer: The claim is incorrect. For

\mathrm{2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)}

K_p \approx 10^{80}

, so at equilibrium the mixture would be almost entirely water. The container simply isn't at equilibrium: with almost no product,

Q \approx 0

, far below K (Topic 7.7). The reaction is just extremely slow at 25 °C because its activation energy is high, so almost no collisions succeed. The observation is evidence about the rate, not about K. A spark or a platinum catalyst gets the reaction going (with a spark, explosively), and the system then heads to its equilibrium, which lies almost entirely on the product side.

K depends on the equation and the temperature

K also changes with temperature (Topic 7.10) and with the way the equation is written (Topic 7.6). Reversing an equation turns K into

1/K

, so the breakdown of phosgene, the reverse of Example 2, has

K_c = 1/(4.6 \times 10^{9}) = 2.2 \times 10^{-10}

. A very large K one way is a very small K the other way.

Common mistakes

Thinking a large K means a fast reaction. K tells you how far a reaction goes, not how fast.

Thinking a small K means no product at all. Some product always forms; there's just very little of it.

Setting up a full equilibrium calculation when K is huge. Let the reaction go to completion first, then use K to find the trace of leftover reactant.

Assuming $K = 1$ means equal concentrations of everything. That's only true for a one-to-one reaction like $\mathrm{A} \rightleftharpoons \mathrm{B}$ .

Comparing K values of reactions with different forms as if they were percentages. Comparisons are safest between reactions of the same type, like the $K_a$ values of two acids.

Forgetting that a K value belongs to one temperature and one way of writing the equation.

Back Next