When calculating an equilibrium constant for AP Chem 7.4, you write the equilibrium expression (products over reactants, each raised to its coefficient) and plug in concentrations or partial pressures measured at equilibrium. Leave out pure solids and liquids, and remember that K is reported as a unitless number.

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Why This Matters for the AP Chemistry Exam

Calculating Kc or Kp from experimental data is a core equilibrium skill that shows up across the exam. You need it to find K from measured concentrations or pressures, to compare a system to equilibrium, and to set up later problems like finding equilibrium concentrations and predicting shifts.

On free-response questions, you may be asked to write a correct equilibrium expression and then calculate K from given data. Writing the expression in the wrong form (for example, using Kc form when the question gives pressures) can cost points, so match your expression to the data you are given.

Key Takeaways

Write K as products over reactants, each concentration or partial pressure raised to its stoichiometric coefficient.
Use Kc with molar concentrations and Kp with partial pressures, and keep the two formats separate.
Only plug in values measured at equilibrium. Using values from any other point gives Q, not K.
Leave pure solids and pure liquids out of the equilibrium expression.
Report K as a unitless number.
Convert to the right units first (grams to moles, moles to molarity, or moles and total pressure to partial pressures) before plugging in.

Calculate Kc Step by Step

Calculate the value of Kc for the system below if 0.1908 moles of CO₂, 0.0908 moles of H₂, 0.0092 moles of CO, and 0.0092 moles of H₂O vapor are present in a 2.00 L reaction vessel at equilibrium.

$\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)$

Write the Kc expression as products over reactants:

$K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}$

Convert each mole amount to molarity by dividing by 2.00 L:

CO: 0.0092 / 2.00 = 0.0046 M
H₂O: 0.0092 / 2.00 = 0.0046 M
CO₂: 0.1908 / 2.00 = 0.0954 M
H₂: 0.0908 / 2.00 = 0.0454 M

Plug in and solve:

$K_c = \frac{(0.0046)(0.0046)}{(0.0954)(0.0454)} = 4.9 \times 10^{-3}$

Every species here is a gas with a coefficient of one, so there are no solids or liquids to leave out and no exponents to add.

Calculate Kp Step by Step

Calculate Kp for the reaction below using the given equilibrium partial pressures:

$2\text{N}_2\text{O}_5(g) \rightleftharpoons \text{O}_2(g) + 4\text{NO}_2(g)$

P(N₂O₅) = 2.00
P(O₂) = 0.296
P(NO₂) = 1.70

Write the Kp expression using partial pressures, and raise each one to its coefficient:

$K_p = \frac{(P_{\text{O}_2})(P_{\text{NO}_2})^4}{(P_{\text{N}_2\text{O}_5})^2}$

Plug in and solve:

$K_p = \frac{(0.296)(1.70)^4}{(2.00)^2} = 0.618$

The coefficients of 4 and 2 become exponents, so do not skip them.

Why the Equilibrium Expression Works

For a general reaction A + B ⇌ C + D, the equilibrium constant is the ratio of product concentrations to reactant concentrations, each raised to its coefficient:

$K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

These are equilibrium concentrations, so the ratio answers a simple question: how do the amounts of product compare to the amounts of reactant once the system stops changing? When K is greater than 1, the products outweigh the reactants. When K is less than 1, reactants are still present in larger amounts. K can be extremely small but is never negative.

How to Use This on the AP Chemistry Exam

Free Response

Expect to write an equilibrium expression and then calculate K from data. Match the expression to the data type: use concentrations for Kc and partial pressures for Kp. A correct expression often earns its own point, so write it carefully before you start plugging in numbers.

Problem Solving

Confirm the values are measured at equilibrium. Otherwise you are calculating Q, not K. For more on that comparison, see the Reaction Quotient and Equilibrium Constant guide.
Check your units first. If you are given moles and a volume, divide to get molarity. If you are given grams, convert to moles first.
For partial pressures from a total pressure, use the mole fraction relationship $P_A = X_A \times P_{\text{total}}$ , which connects to the Ideal Gas Law guide.
Turn every stoichiometric coefficient into an exponent in the expression.

Common Trap

Writing the wrong form of the expression is the most common slip. If a question gives partial pressures, set up Kp, not Kc. Also remember to leave out pure solids and pure liquids, since their amounts do not change the value of K.

Worked AP-Style Example

This setup is based on a released free-response problem.

$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$

At high temperatures, N₂(g) and O₂(g) react to produce nitrogen monoxide, NO(g).

Write the expression for the equilibrium constant, Kp, for the forward reaction:

$K_p = \frac{(P_{\text{NO}})^2}{(P_{\text{N}_2})(P_{\text{O}_2})}$

The coefficient of 2 on NO becomes an exponent, while N₂ and O₂ each have a coefficient of 1. Even though pressures carry units individually, the reported K value is unitless.

Common Misconceptions

"K has units." Report K as a unitless number, even though the concentrations or pressures you plug in have units.
"Any measured concentration works." Only equilibrium values give K. Values from any other moment give Q.
"Solids and liquids belong in the expression." Pure solids and pure liquids are left out. Only gases and aqueous species appear.
"Kc and Kp are interchangeable." They use different inputs. Use concentrations for Kc and partial pressures for Kp, and match the expression to the data given.
"Coefficients can be ignored." Each stoichiometric coefficient becomes an exponent in the expression, so a coefficient of 4 means raising that term to the fourth power.
"A negative K is possible." K can be very small but never negative, since concentrations and pressures are not negative.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
concentrations at equilibrium	The molar amounts of reactants and products per unit volume when a reversible reaction reaches equilibrium and no net change occurs.
equilibrium constant expression	A mathematical expression that relates the concentrations or partial pressures of products and reactants at equilibrium, with each raised to the power of its stoichiometric coefficient.
Kc	The equilibrium constant expressed in terms of molar concentrations of reactants and products at equilibrium.
Kp	The equilibrium constant expressed in terms of partial pressures of gaseous reactants and products at equilibrium.
partial pressures	The individual pressure exerted by each gas in a mixture of gases at equilibrium.

Frequently Asked Questions

How do you calculate the equilibrium constant in AP Chemistry?

Write the equilibrium expression as products over reactants, raise each term to its stoichiometric coefficient, and plug in equilibrium concentrations for Kc or equilibrium partial pressures for Kp. Pure solids and pure liquids are omitted.

What is Kc in AP Chemistry?

Kc is the equilibrium constant written with molar concentrations. Use Kc when the problem gives equilibrium concentrations or gives moles and volume so you can calculate molarity.

What is Kp in AP Chemistry?

Kp is the equilibrium constant written with equilibrium partial pressures of gases. Use Kp when the problem gives partial pressures or enough gas data to calculate them.

Do equilibrium constants have units?

On the AP Chemistry exam, report K as a unitless number. The concentrations or pressures you substitute may have units, but the equilibrium constant itself is reported without units.

What does K depend on in AP Chemistry?

For a given reaction, K depends on temperature. It does not depend on starting amounts, volume changes, or catalysts, though those changes can affect concentrations or how fast equilibrium is reached.

What is the difference between K and Q?

K uses concentrations or pressures measured at equilibrium. Q uses the same expression at any moment that may not be equilibrium. If the values are not equilibrium values, you are calculating Q, not K.