In Unit 6, we discussed Hess’s Law and how it can be used to find the value of ΔH for reactions by adding together, flipping, and multiplying reactions. In this section, we’ll apply similar rules to equilibrium! If you did well with Hess’s Law problems then you should do fine with problems involving the properties of the equilibrium constant.
Review of Hess's Law
Let's bring Hess's Law back into memory a little bit, as it'll help you with the content in this study guide. Hess's Law is stated as simply as enthalpy is a state function. No matter what way you go from reactants to products, you will end up with the same enthalpy of reaction for the reaction. This means that if we know the enthalpy of formation for different reactions, we can manipulate them to get a single reaction and find the enthalpy of that single reaction.
There are three major rules of Hess's Law:
- When a reaction is reversed, the enthalpy change stays constant in magnitude but becomes reversed in mathematical sign (Flipping the reaction flips the sign of ΔH).
- If an equation is multiplied by n, ΔH has to also be multiplied by n.
- When two (or more) reactions are added to obtain an overall reaction➕, the individual enthalpy changes of each reaction are added to obtain the net change of enthalpy of the overall reaction. Now, let's see how these rules can be applied to the equilibrium constant and its properties.
Properties of Keq
Flipping Reactions
Flipping reactions is essentially the equivalent of saying, “Let’s start with the products and end with the reactants”. Because of this, we can find out what our equilibrium constant will be by writing out the formula for the equilibrium constant for both equations:
If our original reaction was A ⇌ B, that would make K = [B]/[A]. Let’s think about our K value for the reverse, B ⇌ A. You may think that the equilibrium constant would be the same because A ⇌ B is reversible, but in this case the forward reaction is what the backward reaction really is. You're basically changing your perspective when looking at the chemical equation. Rewriting our equilibrium formula for B ⇌ A we find that K₂ = [A]/[B] = 1/K.
From here we can conclude that flipping a reaction gives it a new K value of 1/K.
Example of Inverting K
Let’s look at the following example and how flipping a reaction quantitatively alters Keq.
Consider the reaction N₂ + 3H₂ ⇌ 2NH₃ that has a K = 0.118. Find the equilibrium constant for the reaction 2NH₃ ⇌ N₂ + 3H₂.
Once you recognize that the new reaction is the reverse of the initial reaction, you can easily find K for 2NH₃ ⇌ N₂ + 3H₂ to be 1/0.118 = 8.47. Think of this as Final K = 1/Initial K.
✨ Fun Fact! This reaction has a special name called the Haber Process and is useful for creating fertilizers among many other useful chemicals!
Multiplying Reactions
Let’s take a look at what happens when we multiply a reaction by a coefficient. When we say multiplying a reaction, we mean taking something like A ⇌ B and multiplying it by a constant n to form nA ⇌ nB.
If n=2, our new chemical equation would be 2A ⇌ 2B. This does not change the reaction besides the number of moles that go in and out (the equation remains balanced either way!), but it does change the equilibrium constant. Let’s explore why:
For the reaction A ⇌ B, K = [B]/[A]. We established this same fact in part 1.
However, for the reaction 2A ⇌ 2B, K = [B]² / [A]² = ([B]/[A])² = K².
Let’s see if we can generalize this to any n coefficient:
For the general multiplied reaction nA ⇌ nB, K = [B]ⁿ / [A]ⁿ = ([B]/[A])ⁿ = Kⁿ.
What does this tell us about how multiplying a reaction impacts the equilibrium constant? It tells us that when we multiply a reaction, our K value gets exponentiated that same amount. For example, multiplying a reaction by 3 cubes K.
Example of Exponentiating K
Let’s take a look at a real example:
Like before, Consider the reaction N₂ + 3H₂ ⇌ 2NH₃ that has a K = 0.118. Find the equilibrium constant for the reaction (1/2)N₂ + (3/2)H₂ ⇌ NH₃.
Comparing the two reactions, you'll quickly notice that only the stoichiometric coefficients have changed, but to what degree? In this case, we've multiplied our reaction by ½. This means that our original K value, 0.118, is raised to the ½ power.
K = (0.118)¹/² = 0.343.
Adding Reactions Together
The crux of this section is what happens when we learn what happens when reactions are added together. Like before, let’s look at a general example without any real chemicals:
A ⇌ B : K = [B]/[A]
C ⇌ D : K = [D]/[C]
A + C ⇌ B + D : K = [B][D] / [A][C].
What does this mean? We have two reactions, A ⇌ B and C ⇌ D that we’re adding together to form A + C ⇌ B + D. However, take a look at what happens to our equilibrium constants. It may seem like nothing happened, but in fact, in adding the reactions together we actually multiplied our equilibrium constants! K1 was [B]/[A] and K2 was [D]/[C] which multiplied to form [B][D] / [A][C]! Therefore, when we add reactions together we multiply the K values together.
Summary of Properties of K
We’ve covered 3 main rules that you may have to combine in problems. Here’s a table of the rules we’ve learned so far and a side-by-side comparison to Hess's Law.
| Manipulation | Properties of K | Properties of ΔH |
|---|---|---|
| Reverse the Reaction | Inverse the value of K (i.e. raise it to the -1 power) | Flip the sign of ΔH |
| Multiplying by a Constant (n) | Raise the equilibrium constant to the power that was multiplied by | Multiply ΔH by n |
| Adding Reactions | Multiply the equilibrium constants by one another | Add each individual ΔH |
Practice Problem
Using the following 2 reactions, find the equilibrium constant for the reaction N₂ + 2O₂ ⇌ 2NO₂.
- Reaction 1: (1/2)N₂ + (1/2)O₂ ⇌ NO …. K₁ = 6.55 * 10⁻¹³
- Reaction 2: 2NO + O₂ ⇌ 2NO₂ …. K₂ = 6.9 * 10⁵
We have two reactions that sorta look like the reaction we want, so let’s think about ways to get these reactions into a form that allows us to form our reaction by adding the two up.
If we multiply reaction 1 by the constant 2, we’ll get N₂ + O₂ ⇌ 2NO and the subsequent K value of (6.55 * 10⁻¹³)² = 4.3 * 10⁻²⁵.
Next, adding together our multiplied reaction 1 with reaction 2, we’ll find that they equal the reaction that we want! The NOs will cancel out, the (1/2)N₂ is now N₂, and our O₂s will add to form 2O₂. Therefore, our final K will be 4.3 * 10⁻²⁵ * 6.9 * 10⁵ = 3.0 * 10⁻¹⁹.
You may be wondering how to know what to do when you’re given one of these problems without any clear steps. There really is no rhyme or reason, you just have to follow the three rules in a way that makes sense to you.
In general, practice makes perfect! Practicing these problems will give you a better chemistry spidey-sense and realize when to flip a reaction, when to multiply a reaction, and when to add them together. You got this! ✨
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| equilibrium constant | A numerical value that expresses the ratio of products to reactants at equilibrium, indicating the extent to which a reaction proceeds. |
| multistep process | A reaction mechanism consisting of two or more elementary steps that combine to produce an overall reaction. |
| overall equilibrium expression | The equilibrium constant expression for the net reaction obtained by adding multiple elementary steps together. |
| reaction quotient | A value calculated using the same expression as the equilibrium constant but using current (non-equilibrium) concentrations or partial pressures. |
| stoichiometric coefficients | The numerical coefficients in a balanced chemical equation that indicate the relative proportions of reactants and products. |
Frequently Asked Questions
What is the equilibrium constant and why does it matter?
The equilibrium constant K is a number that quantifies the ratio of product activities to reactant activities at equilibrium for a given reaction (the law of mass action). It tells you how far a reaction lies to products (large K) or reactants (small K). K matters because it predicts the composition of an equilibrium mixture and how systems respond to changes. Important properties you should know from the CED: reversing a reaction inverts K (K → 1/K); multiplying stoichiometric coefficients by c raises K to the c power (K → K^c); adding reactions multiplies their K’s (overall K = product of constituent K’s). The reaction quotient Q has the same form as K, so the same algebra applies and comparing Q to K predicts direction of shift. Remember K is fixed at a given temperature—changing T changes K. For practice and AP-style guidance on these rules, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and more Unit 7 review (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice questions, use Fiveable’s problem set (https://library.fiveable.me/practice/ap-chemistry).
How do you calculate K for a multistep reaction?
If a process happens in steps, get K for each elementary reaction and combine them using three rules from the CED: reverse a reaction → invert K (Knew = 1/Kold); multiply a balanced equation by c → raise K to the power c (Knew = Kold^c); add reactions → multiply their K values to get the overall K (Koverall = K1 × K2 × ...). So write each step with the stoichiometry you want for the overall equation, adjust Ks with the power rule if you scaled any steps, invert any Ks for reversed steps, then multiply the adjusted Ks. (Remember K is temperature dependent, and the same algebra applies to Q.) For worked examples and quick practice, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and try problems at Fiveable practice (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about what happens to K when you reverse a chemical equation - does it flip or something?
Yes—if you reverse a reaction, the equilibrium constant becomes its reciprocal. That’s CED 7.6.A.1: for A ⇌ B with K = [B]/[A], the reversed reaction B ⇌ A has K' = 1/K. Quick reminders from Topic 7.6: - If you multiply every coefficient by c, Knew = K^c (CED 7.6.A.2). - If you add reactions, Koverall = product of the individual Ks (CED 7.6.A.3). - The same algebra applies to the reaction quotient Q (CED 7.6.A.4), so you can compare Q to K the same way after reversing or scaling. These properties are common on the AP exam when you build overall equilibrium expressions from steps. If you want short worked examples and practice, check the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and more Unit 7 resources (https://library.fiveable.me/ap-chemistry/unit-7) or try practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why does K get inverted when you write the reaction backwards?
K gets inverted when you reverse a reaction because K is defined from the law of mass action as (products)ⁿ/(reactants)ᵐ. If you flip the equation, what used to be products are now reactants and vice versa, so the algebraic form is just the reciprocal. Example: for A ⇌ B, K = [B]/[A]; for B ⇌ A you write K' = [A]/[B] = 1/K. This is just basic algebra applied to equilibrium expressions (CED 7.6.A.1). The same idea explains the other properties: multiplying coefficients raises K to a power, and adding reactions multiplies their K’s (CED 7.6.A.2–A.3). If you want a quick refresher or practice problems tied to this topic, check the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7). For lots of practice, see Fiveable’s AP Chem problems (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between K and Q and do they follow the same math rules?
K is the equilibrium constant—a fixed number (at a given temperature) that comes from the concentrations/partials at equilibrium. Q is the reaction quotient—same algebraic form as K but evaluated with concentrations/pressures at any moment. If Q < K the reaction proceeds forward; if Q > K it goes reverse; if Q = K it’s at equilibrium (this prediction is routinely tested on the AP exam). Yes, they follow the same math rules (CED 7.6.A): reversing a reaction inverts K (and Q); multiplying a reaction by c raises K (and Q) to the c power; adding reactions multiplies K’s (and Q’s) to get the overall constant. The big difference: K depends only on temperature; Q changes with the system’s composition. For more practice and a Topic 7.6 review, see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and try problems at (https://library.fiveable.me/practice/ap-chemistry).
If I multiply all the coefficients in a balanced equation by 2, what happens to the equilibrium constant?
If you multiply every stoichiometric coefficient by 2, the equilibrium constant is raised to the power of 2: Knew = Kold^2. That follows directly from the law of mass action and CED essential knowledge 7.6.A.2—scaling a reaction by factor c scales the exponents in the K expression, so K is raised to c. (Remember this only applies if temperature is constant; K depends on T.) Quick example: for A ⇌ B, K = [B]/[A]. For 2A ⇌ 2B, K' = [B]^2/[A]^2 = ( [B]/[A] )^2 = K^2. For more on these rules (reverse inverts K; add reactions multiply Ks), check the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w). For extra practice, try problems at https://library.fiveable.me/practice/ap-chemistry—these manipulations show up often on the AP exam.
How do I combine equilibrium constants when I add reactions together?
Short answer: use the three K rules from the CED. - If you reverse a reaction, invert K (Krev = 1/Kf). - If you multiply all coefficients by c, raise K to that power (Knew = K^c). - If you add reactions, multiply their K values to get the overall K (Koverall = K1 × K2 × ...). Example: A ⇌ B with K1, and B ⇌ C with K2. Adding them gives A ⇌ C and Koverall = K1 × K2. If the second reaction were written backwards (C ⇌ B), use K2' = 1/K2 before multiplying. If you doubled a step, square its K before combining. Because Q has the same form as K, the same algebra works for Q too (useful for ICE/Q setups on the exam). This is exactly Topic 7.6 in the CED—see the study guide for worked examples (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w). For more practice, try problems at Fiveable’s unit page (https://library.fiveable.me/ap-chemistry/unit-7) or the practice bank (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain why K^c is used when you multiply stoichiometric coefficients by factor c?
When you multiply every stoichiometric coefficient in a balanced reaction by a factor c, every concentration (or activity) term in the equilibrium expression gets raised to that new coefficient. Because K is the product of species’ activities each raised to their stoichiometric power (law of mass action), every exponent in the K expression is multiplied by c—mathematically that makes the whole K value raised to the power c (Knew = Kold^c). Example: A ⇌ B has K = [B]/[A]. If you double the equation (c = 2): 2A ⇌ 2B, equilibrium expression is [B]^2/[A]^2 = ([B]/[A])^2 = K^2. That’s the power rule from the CED (7.6.A.2). This rule follows from identical algebraic form of Q and K (7.6.A.4) and is used with the inversion rule and product rule when reversing or adding reactions (7.6.A.1, 7.6.A.3). For a quick review, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w). For extra practice problems, check Fiveable’s AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how to manipulate equilibrium expressions for complex reactions with multiple steps.
Think of K like a math object you can flip, raise, and multiply. Rules from the CED: reversing a reaction inverts K (Knew = 1/Kold), scaling all coefficients by c raises K to the c power (Knew = Kold^c), and adding steps multiplies their Ks for the overall reaction (Koverall = K1·K2·…). These same manipulations apply to Q because Q has the same form. Example: If step 1: A ⇌ B (K1), step 2: 2B ⇌ C (K2). Overall A + B ⇌ C comes from adding step1 and half of step2 reversed: take step2 reversed so K2' = 1/K2, then scale by 1/2 so K2'' = (1/K2)^(1/2). Multiply K1·K2''. That product is Koverall. On the exam you may need to show each algebraic step and cite which rule you used (CED 7.6.A). For more worked examples and practice, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice problems, check Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).
What are the mathematical properties of equilibrium constants that I need to memorize for the AP exam?
Memorize these core math rules from Topic 7.6—they’re short and show up directly on the exam: - Reverse a reaction → invert K (K_reverse = 1 / K_forward). - Multiply all stoichiometric coefficients by c → K_new = K_old^c. - Add reactions → overall K = product of the individual Ks (K_total = K1 × K2 × …). - K and Q have identical algebraic form, so any algebraic manipulation you do to K applies to Q (useful for comparing Q to K to predict shift). Also remember: K is temperature dependent (don’t change K for pressure/volume changes alone), heterogeneous equilibria omit pure solids/liquids from K expressions, and use the appropriate activity/standard-state forms if given. These are exactly the CED essentials (7.6.A.1–A.4). For a quick Topic 7.6 review see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w). For unit review and lots of practice problems, check the Unit 7 page (https://library.fiveable.me/ap-chemistry/unit-7) and Fiveable practice bank (https://library.fiveable.me/practice/ap-chemistry).
How do you find the overall K when you have like 3 different reactions that add up to one big reaction?
Use the properties in the CED: reversing a step inverts K (K → 1/K), scaling a reaction by c raises K to the c power (K → K^c), and adding reactions multiplies their K’s (overall K = product of constituent K’s)—this is exactly 7.6.A in the CED. So for three steps that add to the overall equation: - If the steps are written exactly as given and you don’t reverse or scale any, Koverall = K1 × K2 × K3. - If you reverse step 2, use K2' = 1/K2. If you double step 3, use K3' = (K3)^2. Then multiply the adjusted K’s. Example: if you reverse reaction 2 and double reaction 3, Koverall = K1 × (1/K2) × (K3)^2. This algebra works the same for Q (reaction quotient). For a quick refresher tied to the AP CED, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and more Unit 7 review (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
Why do all the algebra rules that work for K also work for the reaction quotient Q?
They work the same way because Q and K are the exact same mathematical expression—the law of mass action—just evaluated at different states. K is Q evaluated at equilibrium; Q is evaluated at any moment. So any algebraic manipulation you can do to the equilibrium expression (invert for a reversed reaction, raise to a power when you multiply stoichiometric coefficients, multiply K’s when you add reactions) applies to Q because the underlying products over reactants form is identical. Example: for A ⇌ B, Q = [B]/[A]. If you reverse the reaction, Q_rev = [A]/[B] = 1/Q. If you double the reaction, Q_2 = ([B]^2)/([A]^2) = (Q)^2. These are true whether the system is at equilibrium or not; at equilibrium Q = K. Review the rules in the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and practice with problems on the Unit 7 page (https://library.fiveable.me/ap-chemistry/unit-7) or the large problem bank (https://library.fiveable.me/practice/ap-chemistry). On the AP exam you’ll need to apply these algebraic rules when combining reactions or comparing Q and K to predict direction of reaction.
If K₁ = 5 and K₂ = 10 for two reactions, what's K for the combined reaction when you add them?
If you add the two reactions together, the overall equilibrium constant is the product of the individual K’s. So Koverall = K1 × K2 = 5 × 10 = 50. Remember the related CED rules: if you reverse a reaction you take 1/K, and if you multiply a reaction by a factor c you raise K to the c power. This product rule for K (Topic 7.6.A.3) is exactly what you used here. For more review on these manipulations, check the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
I missed the lecture on equilibrium constant properties - what are the main rules I need to know?
Here are the main rules you need from Topic 7.6 (short and usable for the AP exam): - Reversing a reaction inverts K: Krev = 1/Kfwd. (CED 7.6.A.1) - Multiplying the whole equation by c raises K to that power: Knew = Kold^c. (CED 7.6.A.2) - Adding reactions: the overall K is the product of constituent Ks (multiply the K values). (CED 7.6.A.3) - Any algebraic manipulation valid for K also applies to Q because K and Q have the same form—so you can compare Q to K after manipulating expressions. (CED 7.6.A.4) Extras you should remember for AP problems: K values depend on temperature (don’t change K for pressure/volume or concentrations unless T changes), heterogeneous equilibria omit pure solids/liquids from the K expression, and keep track of stoichiometric scaling when building overall equilibrium expressions. For a quick refresher, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you know when to multiply, divide, or raise K to a power when working with multiple equilibrium expressions?
Use the three CED rules as your checklist: - Reverse a reaction → invert K (Knew = 1/Kold). - Multiply every stoichiometric coefficient by c → raise K to the c power (Knew = Kold^c). - Add reactions → multiply their K values to get the overall K (Koverall = K1·K2·…). Why these work: K expressions come from the law of mass action, so algebraic changes to the reaction change the exponents or direction in the same way. The same rules apply to Q because K and Q share the same form (CED 7.6.A.1–7.6.A.4). Quick example: If A ⇌ B has K1, and 2A ⇌ 2B is needed, K2 = K1^2. If you reverse A ⇌ B, Krev = 1/K1. If you add A ⇌ B (K1) and B ⇌ C (K2), overall A ⇌ C has K = K1·K2. For a focused review, see the Topic 7.6 study guide (https://library.fiveable.me/ap-chemistry/unit-7/properties-equilibrium-constant/study-guide/PGLxCyabBaR6axW5E18w). For lots of practice, check Fiveable’s AP Chem practice problems (https://library.fiveable.me/practice/ap-chemistry).