7.6 - Properties of the Equilibrium Constant

Overview

An equilibrium constant belongs to one specific balanced equation. Flip the equation, scale it, or add it to another one, and K changes in a predictable way. These rules let you find K for a reaction nobody measured, by building it out of reactions whose K values you know. They also let you write one overall equilibrium expression for a process that happens in several steps. Everything here works the same way for

K_c

K_p

, and the reaction quotient Q.

K goes with the equation, not the mixture

Picture a sealed flask of

\mathrm{N_2O_4}

and

\mathrm{NO_2}

at equilibrium, at the same temperature as in Topics 7.1 to 7.4. Each molecule in the diagram represents 0.0500 M, so

[\mathrm{N_2O_4}] = 5 \times 0.0500 = 0.250

M and

[\mathrm{NO_2}] = 6 \times 0.0500 = 0.300

7.6 - Properties of the Equilibrium Constant

The mixture is a physical fact. But you can describe it with more than one balanced equation, and each equation has its own K expression:

Equation	K expression	Value
$\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$	$\dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$	$\dfrac{(0.300)^2}{0.250} = 0.360$
$\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$	$\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$	$\dfrac{0.250}{(0.300)^2} = 2.78$
$\mathrm{\tfrac{1}{2}N_2O_4(g) \rightleftharpoons NO_2(g)}$	$\dfrac{[\mathrm{NO_2}]}{[\mathrm{N_2O_4}]^{1/2}}$	$\dfrac{0.300}{(0.250)^{1/2}} = 0.600$
$\mathrm{2N_2O_4(g) \rightleftharpoons 4NO_2(g)}$	$\dfrac{[\mathrm{NO_2}]^4}{[\mathrm{N_2O_4}]^2}$	$\dfrac{(0.300)^4}{(0.250)^2} = 0.130$

Same flask, four different K values. Nothing about the mixture changed, only the bookkeeping. That's why a K value means nothing unless you know which equation it goes with. The four values are also related:

2.78 = 1/0.360

0.600 = \sqrt{0.360}

, and

0.130 = (0.360)^2

. Those relationships are the rules below.

Rule 1: Reversing a reaction inverts K

When you reverse a reaction, products and reactants trade places, so the K expression flips upside down:

K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}

For

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

K_{\text{forward}} = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \qquad K_{\text{reverse}} = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} = \frac{1}{K_{\text{forward}}}

So a reaction with a very large K has a reverse reaction with a very small K. That fits Topic 7.5: if the products are strongly favored one way, they're still strongly favored when you write the equation backward, where they're now called reactants.

Rule 2: Multiplying the coefficients by c raises K to the power c

Multiply every coefficient in an equation by a factor

c

, and every exponent in the K expression gets multiplied by

c

too. That raises the whole expression to the power

c

K_{\text{new}} = \left(K_{\text{original}}\right)^c

Doubling

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

gives

\mathrm{2N_2O_4(g) \rightleftharpoons 4NO_2(g)}

K_{\text{new}} = \frac{[\mathrm{NO_2}]^4}{[\mathrm{N_2O_4}]^2} = \left(\frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}\right)^2 = K^2

The factor can be a fraction. Multiplying by

\tfrac{1}{2}

means taking the square root of K, and multiplying by

\tfrac{1}{3}

means taking the cube root. You never multiply K itself by

c

: doubling the equation squares K, it doesn't double it.

Example 1: Flipping and scaling one reaction

At 430 °C,

K_c = 54.3

for

\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}

Find

K_c

at the same temperature for each equation below.

(a)

\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2(g)}

. This is the reverse, so invert:

K_c = \frac{1}{54.3} = 0.0184

(b)

\mathrm{\tfrac{1}{2}H_2(g) + \tfrac{1}{2}I_2(g) \rightleftharpoons HI(g)}

. Every coefficient is multiplied by

\tfrac{1}{2}

, so take the square root:

K_c = (54.3)^{1/2} = 7.37

(c)

\mathrm{HI(g) \rightleftharpoons \tfrac{1}{2}H_2(g) + \tfrac{1}{2}I_2(g)}

. This is reversed and halved, so do both:

K_c = (54.3)^{-1/2} = \frac{1}{7.37} = 0.136

Each answer keeps 3 significant figures, the same as 54.3.

Rule 3: Adding reactions multiplies their K values

When you add two reactions to get an overall reaction, the overall K is the product of the individual K values:

K_{\text{overall}} = K_1 \times K_2

This is how you represent a multistep process with one overall equilibrium expression. Nitrogen monoxide reacts with oxygen in two steps:

\text{Step 1:} \quad \mathrm{2NO(g) \rightleftharpoons N_2O_2(g)} \qquad K_1 = \frac{[\mathrm{N_2O_2}]}{[\mathrm{NO}]^2}

\text{Step 2:} \quad \mathrm{N_2O_2(g) + O_2(g) \rightleftharpoons 2NO_2(g)} \qquad K_2 = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_2}][\mathrm{O_2}]}

Adding the steps cancels

\mathrm{N_2O_2}

, an intermediate that appears on both sides:

\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}

K_1 K_2 = \frac{[\mathrm{N_2O_2}]}{[\mathrm{NO}]^2} \times \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_2}][\mathrm{O_2}]} = \frac{[\mathrm{NO_2}]^2}{[\mathrm{NO}]^2[\mathrm{O_2}]} = K_{\text{overall}}

The

[\mathrm{N_2O_2}]

term cancels in the K expressions for the same reason

\mathrm{N_2O_2}

cancels in the equations. The setup should look familiar from Hess's law (Topic 6.9), with one key difference: when equations add, their

\Delta H

values add, but their K values multiply.

Change to the equation	Effect on K	Effect on $\Delta H$ (for comparison)
Reverse it	$K$ becomes $1/K$	$\Delta H$ changes sign
Multiply all coefficients by $c$	$K$ becomes $K^c$	$\Delta H$ is multiplied by $c$
Add two equations	$K_{\text{overall}} = K_1 \times K_2$	$\Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2$

Example 2: Building a target reaction (AP-style)

At a certain high temperature, these two reactions have the

K_p

values shown:

(1) \quad \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \qquad K_{p1} = 4.0 \times 10^{-9}

(2) \quad \mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)} \qquad K_{p2} = 2.5 \times 10^{-2}

Find

K_p

at the same temperature for the breakdown of nitrogen dioxide into its elements:

\mathrm{NO_2(g) \rightleftharpoons \tfrac{1}{2}N_2(g) + O_2(g)}

Step 1: Plan. In the target,

\mathrm{NO_2}

is a reactant and

\mathrm{N_2}

is a product, the opposite of the given reactions. NO doesn't appear in the target, so it has to cancel. The moves: add (1) and (2), reverse the sum, then multiply by

\tfrac{1}{2}

Step 2: Add (1) and (2). The 2 NO cancels:

\mathrm{N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g)} \qquad K = K_{p1} \times K_{p2} = (4.0 \times 10^{-9})(2.5 \times 10^{-2}) = 1.0 \times 10^{-10}

Step 3: Reverse it.

\mathrm{2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)} \qquad K = \frac{1}{1.0 \times 10^{-10}} = 1.0 \times 10^{10}

Step 4: Multiply by $\tfrac{1}{2}$ .

\mathrm{NO_2(g) \rightleftharpoons \tfrac{1}{2}N_2(g) + O_2(g)} \qquad K_p = \left(1.0 \times 10^{10}\right)^{1/2} = 1.0 \times 10^{5}

Answer:

K_p = 1.0 \times 10^{5}

. The order of the moves doesn't matter; any order gives

(K_{p1} K_{p2})^{-1/2}

. Since K is large, the breakdown of

\mathrm{NO_2}

into

\mathrm{N_2}

and

\mathrm{O_2}

is product-favored at this temperature (Topic 7.5), even if it's slow.

Two checks: both given values are

K_p

, so the answer is a

K_p

(never mix

K_c

and

K_p

in one combination), and both have 2 significant figures, so the answer does too.

Q follows the same rules

The reaction quotient Q has exactly the same form as K, just with the concentrations (or pressures) at any moment instead of at equilibrium. So every rule on this page applies to Q too: reversing the equation turns Q into

1/Q

, multiplying by

c

turns Q into

Q^c

, and adding equations multiplies their Q values.

Example 3: Same prediction, either direction

For

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

K_c = 0.360

at the temperature of the flask above. Another flask at that temperature contains 0.500 M

\mathrm{N_2O_4}

and 0.300 M

\mathrm{NO_2}

. Which way will it shift? Answer with the equation written forward, then reversed.

Forward.

Q_c = \frac{(0.300)^2}{0.500} = 0.180 < K_c = 0.360

Q < K

, so the net reaction runs forward (Topic 7.7):

\mathrm{N_2O_4}

breaks apart and more

\mathrm{NO_2}

forms.

Reversed. For

\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}

, both values invert:

Q_c' = \frac{1}{0.180} = 5.56 > K_c' = \frac{1}{0.360} = 2.78

Now

Q > K

, so the reversed equation runs in reverse, which means

\mathrm{N_2O_4}

turns into

\mathrm{NO_2}

. Same prediction. Flipping the equation flips both Q and K, so their comparison flips too, and the chemistry stays the same. Halving the equation gives the same result:

Q^{1/2} = 0.424

is still less than

K^{1/2} = 0.600

Common mistakes

Multiplying K by $c$ instead of raising it to the power $c$ . Doubling an equation squares K.

Adding K values when you add reactions. K values multiply; it's $\Delta H$ values that add.

Reversing a reaction but keeping the same K. A reversed reaction always gets $1/K$ .

Forgetting that a fractional coefficient means a root: multiplying by $\tfrac{1}{2}$ gives $K^{1/2}$ , not $K/2$ .

Mixing $K_c$ and $K_p$ values in one combination.

Using a K value without checking which equation (and which temperature) it belongs to.

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