7.3 Reaction Quotient and Equilibrium Constant

Q is the reaction quotient—the same algebraic form as the equilibrium constant but using concentrations (or partial pressures) at any moment: for aA + bB ⇌ cC + dD, Qc = [C]^c[D]^d / [A]^a[B]^b (and Qp uses partial pressures). K is the value of that ratio specifically at equilibrium (Kc or Kp). Key differences for AP Chem: (1) Q tells you where the reaction currently is; K tells you where it will be at equilibrium. If Q < K the reaction shifts toward products; if Q > K it shifts toward reactants; if Q = K the system is at equilibrium. (2) Neither Q nor K includes pure solids or liquids (their activities are constant). Remember the exam won’t test converting Kc↔Kp—just know conceptual differences. For a focused review check Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat), the unit overview (https://library.fiveable.me/ap-chemistry/unit-7), and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Brackets [ ] are the shorthand chemists use for molar concentration (mol·L⁻¹), so Kc and Qc are built from [A], [B], etc. Parentheses ( ) around a symbol mean the partial pressure of a gas (usually in atm), so Kp and Qp use (PA), (PB), etc. Practically it’s just a notation convention that tells you which physical measure you’re using in the law of mass action: concentrations for Kc/Qc, partial pressures for Kp/Qp. Behind that is the idea of “activity”—for ideal gases activity ≈ partial pressure, for solutes activity ≈ concentration—but on the AP exam you only need to recognize and use Kc vs Kp and omit pure solids/liquids from expressions (CED Topic 7.3). For more review, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and the unit overview (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice, check the AP Chem practice problems (https://library.fiveable.me/practice/ap-chemistry).

Write the equilibrium expression from the balanced equation using the law of mass action. For a general reaction a A + b B ⇌ c C + d D - Kc (or Qc at any moment) = [C]^c [D]^d / ([A]^a [B]^b—concentrations in mol·L⁻¹). - For gases use Kp (or Qp) with partial pressures: Kp = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b. - Use the stoichiometric coefficients as exponents. - Omit pure solids and pure liquids from Q or K because their concentrations (activities) are essentially constant (e.g., CaCO3(s) or H2O(l) are left out). To write one quickly: balance the equation, put products over reactants, raise each concentration/pressure to its coefficient, and decide Kc vs Kp based on whether you’re given concentrations or partial pressures. Remember Q has the same form as K; compare Q to K to predict the reaction’s shift. For a refresher and examples see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and extra practice problems (https://library.fiveable.me/practice/ap-chemistry).

Use Kc when you write the equilibrium expression with concentrations [ ] (molarity); use Kp when you write it with partial pressures P (for gases). For a general reaction aA + bB ⇌ cC + dD: - Kc = [C]^c[D]^d / [A]^a[B]^b (Qc is the same form at any time) - Kp = (PC)^c(PD)^d / (PA)^a(PB)^b (Qp is the reaction quotient in pressures) If species are pure solids or pure liquids, you leave them out of K expressions because their concentrations/activities don’t change. At equilibrium Q → K, so at equilibrium Qc = Kc and Qp = Kp. On the AP exam, pay attention to which form the question gives (concentrations vs. partial pressures). The CED says you don’t need to convert Kc ⇄ Kp for the exam, but you should understand the conceptual difference (Topic 7.3). For a quick refresher, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and the unit overview (https://library.fiveable.me/ap-chemistry/unit-7). For more practice problems, check (https://library.fiveable.me/practice/ap-chemistry).

The law of mass action says that for a reaction a A + b B ⇌ c C + d D, the equilibrium state is described by a constant K whose expression is [C]^c[D]^d / [A]^a[B]^b (or with partial pressures for Kp). It’s important because K (and the reaction quotient Q, the same expression at any time) tells you the equilibrium position: if Q < K the reaction goes forward, if Q > K it goes reverse, and at equilibrium Q = K. Practically, you use it to predict direction of shift, set up ICE tables, and solve equilibrium concentrations—key AP skills in Topic 7.3 and 7.4. Remember pure solids and liquids don’t appear in K or Q. For more review and worked examples see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and practice questions (https://library.fiveable.me/practice/ap-chemistry).

Short answer: solids and pure liquids aren’t in K or Q because their concentrations (more correctly, their activities) don’t change during the reaction. For a pure solid or pure liquid, the “concentration” depends only on density and molar mass, not on how much of that phase you have, so it’s effectively constant. The law of mass action (K = [products]^c/[reactants]^a…) therefore omits species whose amounts don’t affect the equilibrium position—these are heterogeneous equilibria. This is exactly what the CED states in 7.3.A.2: don’t include substances whose concentrations or partial pressures are independent of the amount, such as solids and pure liquids. (If you need the formal treatment, chemists use activities to be precise; for AP you can just say “constant so omitted.”) For more practice and examples, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) or the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7).

Write the reaction’s Q expression from the balanced equation (use concentrations for Qc or partial pressures for Qp). For aA + bB ⇌ cC + dD: Qc = [C]^c[D]^d / [A]^a[B]^b Then plug in the concentrations at that time (raise each to its stoichiometric exponent) and ignore pure solids or liquids (they don’t appear in Q). Example: if at some time [A] = 0.20 M, [B] = 0.10 M, [C] = 0.50 M for A + 2B ⇌ C, then Qc = [C] / ([A][B]^2) = 0.50 / (0.20 × 0.10^2) = 0.50 / (0.20 × 0.01) = 0.50 / 0.002 = 250. Compare Q to K: Q < K (shifts right), Q = K (at equilibrium), Q > K (shifts left). For more practice and the AP-aligned summary, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).

If Q = K, the system is at equilibrium—meaning the relative concentrations (or partial pressures) of reactants and products match the equilibrium constant, so there’s no net change in composition over time. Microscopically it’s dynamic: forward and reverse reaction rates are equal, so molecules keep reacting but the macroscopic amounts stay constant. Use Qc/Kc for concentrations and Qp/Kp for gases; pure solids and liquids aren’t included in the Q expression. If Q ≠ K, the reaction will shift (via more forward or reverse reaction) until Q moves to equal K (ICE tables are a common way to show that mathematically). Note: on the AP exam you won’t be asked to convert Kc ↔ Kp, but you should be comfortable writing Q and K expressions and predicting shift directions (Topic 7.3 study guide: https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat; unit overview: https://library.fiveable.me/ap-chemistry/unit-7). For extra practice, try problems at https://library.fiveable.me/practice/ap-chemistry.

Good question—the exponents equal the coefficients because of the law of mass action and how equilibrium is defined. Two linked ways to think about it: - Kinetics view (easy intuition): for an elementary step, the rate of the forward reaction depends on how often the reactant particles collide. If the balanced elementary reaction is aA + bB → ..., the collision frequency—and so the rate—is proportional to [A]^a[B]^b. At equilibrium the forward and reverse rates balance, so the concentrations enter the equilibrium expression with those powers. - Thermodynamics view (more general): equilibrium comes from equal chemical potentials. When you express that equality using activities (or concentrations for AP work), the math yields a product of activities raised to the stoichiometric coefficients. That’s why Kc = [C]^c[D]^d / [A]^a[B]^b. Pure solids and liquids drop out because their activity is ~1 (so they don’t appear). This is exactly what the CED calls the law of mass action for Kc/Qc (Topic 7.3). If you want a quick refresher or practice, check the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and try some practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Start with the balanced chemical equation. Then follow these steps: 1. Identify phases: include only gases and aqueous species in the expression; exclude pure solids and pure liquids (CED 7.3.A.2). 2. Write product concentrations (or partial pressures) in the numerator and reactant ones in the denominator. 3. Use stoichiometric coefficients as exponents. For aA + bB ⇌ cC + dD: Kc = [C]^c[D]^d / [A]^a[B]^b (or Kp using partial pressures: Kp = (PC)^c(PD)^d/(PA)^a(PB)^b). 4. For a snapshot at any time use Q (Qc or Qp) with the same form; at equilibrium Q = K. Compare Q and K to predict shift (Q < K → shift right; Q > K → shift left). Quick AP notes: show the balanced equation, use correct phases, and label whether you’re using Kc or Kp (AP won’t test conversions between them). For a targeted review, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

If you reverse a chemical equation, the equilibrium expression becomes the reciprocal. For example, if for aA + bB ⇌ cC + dD you have K = [C]^c[D]^d / [A]^a[B]^b, then for the reversed reaction (cC + dD ⇌ aA + bB) Krev = 1/K = [A]^a[B]^b / [C]^c[D]^d. More generally, if you multiply the entire balanced equation by a factor n, the new K equals the original K raised to the n power (Knew = K^n). Remember to keep Kc vs Kp consistent (use concentrations for Kc, partial pressures for Kp) and omit pure solids and liquids from the expression because their activities are constant (CED Topic 7.3). For a quick refresher, see the Topic 7.3 study guide on Fiveable (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat), and practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Because Kc and Kp (or Q) come from the law of mass action, they use the concentrations (or partial pressures) of species that change as the reaction shifts. Pure solids and pure liquids have fixed densities and therefore constant concentrations (or activities) at a given temperature—their “effective concentration” doesn’t change when you add or remove some of that phase unless you remove it entirely. So including them would just multiply K by a constant; AP practice omits them and either treats their activity as 1 or absorbs that constant into K. That’s why heterogeneous equilibrium expressions only list gases and aqueous species (CED 7.3.A.2). If you want a deeper look or examples (ICE tables, activities), check the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) or the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice, use Fiveable’s AP Chem problems (https://library.fiveable.me/practice/ap-chemistry).

Look at what the problem gives or asks for: if it uses concentrations (mol/L or [X]) you need Kc (or Qc); if it uses partial pressures (atm or PX) you need Kp (or Qp). For gas-phase reactions you can write either form, but the AP exam will tell you which one by the units it uses. Remember pure solids and pure liquids don’t appear in the equilibrium expression (they’re omitted from Kc or Kp). Also watch wording: “partial pressure” or “atm” → Kp; “[ ]”, “molarity”, or “M” → Kc. Don’t try to convert Kc↔Kp on the exam (that conversion won’t be tested). If you’re unsure, scan for what the reaction mixture is described in—concentrations or pressures—and use that form. For more practice and examples see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and Unit 7 resources (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice try the AP problems set (https://library.fiveable.me/practice/ap-chemistry).

The reaction quotient Q (or Qc/Qp) is the same formula as the equilibrium expression but evaluated at any moment—it uses the current concentrations (or partial pressures) of reactants and products. K (Kc or Kp) is that expression evaluated at equilibrium and is constant for a given reaction at a fixed temperature. Key points you need for AP Chem: - If Q < K, the net reaction shifts right (forms more products). If Q > K, it shifts left (forms more reactants). If Q = K, the system is at dynamic equilibrium. - Q is time-dependent; K is time-independent (at constant T). - Don’t include pure solids or pure liquids in Q or K expressions (they’re omitted). - For gases use Qp/Kp; for solutions use Qc/Kc. For a clear walkthrough and examples, see the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat). More unit review (https://library.fiveable.me/ap-chemistry/unit-7) and practice Qs (https://library.fiveable.me/practice/ap-chemistry) are great for drilling this.

Short answer: No—you don’t have to memorize converting Kc ↔ Kp for the AP exam. The CED explicitly excludes conversion between Kc and Kp from assessment, so you won’t be asked to do those conversions on the test. That said, you should understand the conceptual difference: Kc uses concentrations, Kp uses partial pressures, and for gas-phase reactions they both express the same equilibrium idea (Kc = Qc, Kp = Qp). It’s helpful to know qualitatively that Kp and Kc can differ when the number of moles of gas changes (Δn ≠ 0), and to pay attention to which form a question gives you. If you want a quick refresher or practice on these ideas (not conversion problems), check the Topic 7.3 study guide (https://library.fiveable.me/ap-chemistry/unit-7/reaction-quotient-equilibrium-constant/study-guide/Y12LZEr3OWIAqV3gErat) and the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7). For lots of practice problems, see (https://library.fiveable.me/practice/ap-chemistry).