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AP Chemistry Flashcards: Calculating The Equilibrium Constant

Study Calculating The Equilibrium Constant in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Calculating The Equilibrium Constant, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Chemistry Flashcards: Calculating The Equilibrium Constant

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QUESTION

Identify the correct KcK_cKc​ for CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s)\rightleftharpoons\text{CaO}(s)+\text{CO}_2(g)CaCO3​(s)⇌CaO(s)+CO2​(g).

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ANSWER

Kc=[CO2]K_c=[\text{CO}_2]Kc​=[CO2​]. Solids omitted; only gaseous CO₂ appears in the expression.

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Flashcard 1: Identify the correct KcK_cKc​ for CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s)\rightleftharpoons\text{CaO}(s)+\text{CO}_2(g)CaCO3​(s)⇌CaO(s)+CO2​(g).

Answer: Kc=[CO2]K_c=[\text{CO}_2]Kc​=[CO2​]. Solids omitted; only gaseous CO₂ appears in the expression.

Flashcard 2: Identify the correct KcK_cKc​ for N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons^2\text{NH}_3(g)N2​(g)+3H2​(g)⇌2NH3​(g).

Answer: Kc=[NH3]2[N2][H2]3K_c=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}Kc​=[N2​][H2​]3[NH3​]2​. Products over reactants with coefficients: 222 for NH₃, 333 for H₂.

Flashcard 3: Which temperature unit must be used in Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}Kp​=Kc​(RT)Δn calculations?

Answer: TTT must be in kelvin (K)(\text{K})(K). Absolute temperature required for ideal gas calculations.

Flashcard 4: What is Δngas\Delta n_{\text{gas}}Δngas​ for Kp=Kc(RT)ΔngasK_p=K_c(RT)^{\Delta n_{\text{gas}}}Kp​=Kc​(RT)Δngas​?

Answer: Δngas=∑νprod,gas−∑νreact,gas\Delta n_{\text{gas}}=\sum \nu_{\text{prod,gas}}-\sum \nu_{\text{react,gas}}Δngas​=∑νprod,gas​−∑νreact,gas​. Change in moles of gas: product coefficients minus reactant coefficients.

Flashcard 5: What is the value of KKK for any reaction written as X⇌XX\rightleftharpoons XX⇌X?

Answer: K=1K=1K=1. Same species on both sides cancel out, leaving unity.

Flashcard 6: Which species are omitted from KKK expressions: pure solids, pure liquids, gases, or aqueous solutes?

Answer: Pure solids and pure liquids are omitted. Their activities equal 1, so they don't affect the equilibrium constant.

Flashcard 7: What is the general expression for KpK_pKp​ for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dDaA+bB⇌cC+dD (gases)?

Answer: Kp=(PC)c(PD)d(PA)a(PB)bK_p=\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}Kp​=(PA​)a(PB​)b(PC​)c(PD​)d​. Uses partial pressures instead of concentrations for gas-phase equilibria.

Flashcard 8: What is the general expression for KcK_cKc​ for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dDaA+bB⇌cC+dD?

Answer: Kc=[C]c[D]d[A]a[B]bK_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}Kc​=[A]a[B]b[C]c[D]d​. Products over reactants, each raised to their stoichiometric coefficients.

Flashcard 9: Calculate KcK_cKc​ for A⇌2BA\rightleftharpoons^2BA⇌2B if [A]=0.50 M[A]=0.50\,\text{M}[A]=0.50M and [B]=0.20 M[B]=0.20\,\text{M}[B]=0.20M at equilibrium.

Answer: Kc=0.080K_c=0.080Kc​=0.080. Kc=[B]2/[A]=(0.20)2/0.50=0.040/0.50=0.080K_c = [B]^2/[A] = (0.20)^2/0.50 = 0.040/0.50 = 0.080Kc​=[B]2/[A]=(0.20)2/0.50=0.040/0.50=0.080

Flashcard 10: What is KKK for the reaction if the given reaction has K=3.0K=3.0K=3.0 and all coefficients are doubled?

Answer: Knew=9.0K_{\text{new}}=9.0Knew​=9.0. Doubling coefficients squares KKK: 3.02=9.03.0^2 = 9.03.02=9.0

Flashcard 11: What is KKK for the reverse reaction if K=4.0×10−3K=4.0\times10^{-3}K=4.0×10−3 for the forward reaction?

Answer: Kreverse=2.5×102K_{\text{reverse}}=2.5\times10^2Kreverse​=2.5×102. Kreverse=1÷(4.0×10−3)=2.5×102K_{reverse} = 1 ÷ (4.0×10^{-3}) = 2.5×10^2Kreverse​=1÷(4.0×10−3)=2.5×102

Flashcard 12: Find KcK_cKc​ if Kp=2.0K_p=2.0Kp​=2.0, Δngas=−2\Delta n_{\text{gas}}=-2Δngas​=−2, T=400 KT=400\,\text{K}T=400K, and R=0.0821R=0.0821R=0.0821.

Answer: Kc=2.15×103K_c=2.15\times10^3Kc​=2.15×103. Kc=2.0÷(0.0821×400)−2=2.0×1079=2.15×103K_c = 2.0 ÷ (0.0821 × 400)^{-2} = 2.0 × 1079 = 2.15×10^3Kc​=2.0÷(0.0821×400)−2=2.0×1079=2.15×103

Flashcard 13: Find KpK_pKp​ if Kc=0.50K_c=0.50Kc​=0.50, Δngas=1\Delta n_{\text{gas}}=1Δngas​=1, T=300 KT=300\,\text{K}T=300K, and R=0.0821R=0.0821R=0.0821.

Answer: Kp=12.3K_p=12.3Kp​=12.3. Kp=0.50×(0.0821×300)1=0.50×24.63=12.3K_p = 0.50 × (0.0821 × 300)^1 = 0.50 × 24.63 = 12.3Kp​=0.50×(0.0821×300)1=0.50×24.63=12.3

Flashcard 14: State the relationship between KpK_pKp​ and KcK_cKc​ using Δngas\Delta n_{\text{gas}}Δngas​.

Answer: Kp=Kc(RT)ΔngasK_p=K_c(RT)^{\Delta n_{\text{gas}}}Kp​=Kc​(RT)Δngas​. Relates pressure and concentration constants via ideal gas law.

Flashcard 15: What is KKK for an overall reaction made by adding reactions with constants K1K_1K1​ and K2K_2K2​?

Answer: Koverall=K1K2K_{\text{overall}}=K_1K_2Koverall​=K1​K2​. When reactions add, their equilibrium constants multiply.

Flashcard 16: What happens to KKK when all coefficients in the reaction are multiplied by nnn?

Answer: Knew=KnK_{\text{new}}=K^nKnew​=Kn. Each concentration term gets raised to nnn, so KKK is raised to nnn.

Flashcard 17: What happens to KKK when the reaction is reversed?

Answer: Kreverse=1KforwardK_{\text{reverse}}=\frac{1}{K_{\text{forward}}}Kreverse​=Kforward​1​. Reversing swaps products and reactants, inverting the fraction.

Flashcard 18: Calculate KpK_pKp​ for 2A(g)⇌B(g)2A(g)\rightleftharpoons B(g)2A(g)⇌B(g) if PA=0.50 atmP_A=0.50\,\text{atm}PA​=0.50atm and PB=2.0 atmP_B=2.0\,\text{atm}PB​=2.0atm at equilibrium.

Answer: Kp=8.0K_p=8.0Kp​=8.0. Kp=PB/(PA)2=2.0/(0.50)2=2.0/0.25=8.0K_p = P_B/(P_A)^2 = 2.0/(0.50)^2 = 2.0/0.25 = 8.0Kp​=PB​/(PA​)2=2.0/(0.50)2=2.0/0.25=8.0

Flashcard 19: What is Δngas\Delta n_{\text{gas}}Δngas​ for 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g)+\text{O}_2(g)\rightleftharpoons^2\text{SO}_3(g)2SO2​(g)+O2​(g)⇌2SO3​(g)?

Answer: Δngas=−1\Delta n_{\text{gas}}=-1Δngas​=−1. 222 moles gas products minus 333 moles gas reactants equals −1-1−1.

Flashcard 20: Identify the correct KcK_cKc​ for CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\text{CH}_3\text{COOH}(aq)\rightleftharpoons\text{H}^+(aq)+\text{CH}_3\text{COO}^-(aq)CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq).

Answer: Kc=[H+][CH3COO−][CH3COOH]K_c=\frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}Kc​=[CH3​COOH][H+][CH3​COO−]​. Weak acid dissociation: products over undissociated acid.

Flashcard 21: What is KKK for the overall reaction if two steps have constants K1K_1K1​ and K2K_2K2​ and are added?

Answer: Koverall=K1K2K_\text{overall}=K_1K_2Koverall​=K1​K2​. When reactions add, equilibrium constants multiply.

Flashcard 22: Calculate KcK_cKc​ if at equilibrium [A]=0.20 M[A]=0.20\,M[A]=0.20M, [B]=0.30 M[B]=0.30\,M[B]=0.30M, [C]=0.40 M[C]=0.40\,M[C]=0.40M for A+B⇌CA+B\rightleftharpoons CA+B⇌C.

Answer: Kc=0.40(0.20)(0.30)=6.7K_c=\frac{0.40}{(0.20)(0.30)}=6.7Kc​=(0.20)(0.30)0.40​=6.7. Substitute equilibrium concentrations into KcK_cKc​ expression.

Flashcard 23: Which species are omitted from KKK for a heterogeneous equilibrium involving pure solids and liquids?

Answer: Pure solids and pure liquids are omitted (activity =1=1=1). Their concentrations remain constant during reaction.

Flashcard 24: What is the effect on KKK when a reaction is reversed?

Answer: Krev=1KK_\text{rev}=\frac{1}{K}Krev​=K1​. Reversing flips the fraction, giving the reciprocal.

Flashcard 25: What is the relationship between KcK_cKc​ and KpK_pKp​ using Δn\Delta nΔn and R,TR,TR,T?

Answer: Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}Kp​=Kc​(RT)Δn. Relates concentration and pressure equilibrium constants.

Flashcard 26: What is Δn\Delta nΔn (used in Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}Kp​=Kc​(RT)Δn) for gas-phase equilibria?

Answer: Δn=mol gas products−mol gas reactants\Delta n=\text{mol gas products}-\text{mol gas reactants}Δn=mol gas products−mol gas reactants. Change in moles of gas from reactants to products.

Flashcard 27: What is the equilibrium constant expression KcK_cKc​ for 2NO2(g)⇌N2O4(g)2NO_2(g)\rightleftharpoons N_2O_4(g)2NO2​(g)⇌N2​O4​(g)?

Answer: Kc=[N2O4][NO2]2K_c=\frac{[N_2O_4]}{[NO_2]^2}Kc​=[NO2​]2[N2​O4​]​. Products over reactants with stoichiometric exponents.

Flashcard 28: What is the equilibrium constant expression KpK_pKp​ for N2(g)+3H2(g)⇌2NH3(g)N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)N2​(g)+3H2​(g)⇌2NH3​(g)?

Answer: Kp=(PNH3)2PN2(PH2)3K_p=\frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}Kp​=PN2​​(PH2​​)3(PNH3​​)2​. Partial pressures raised to stoichiometric coefficients.

Flashcard 29: What is KcK_cKc​ for CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)CaCO3​(s)⇌CaO(s)+CO2​(g)?

Answer: Kc=[CO2]K_c=[CO_2]Kc​=[CO2​]. Solids omitted; only gas concentration appears.

Flashcard 30: What is KpK_pKp​ for CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)CaCO3​(s)⇌CaO(s)+CO2​(g)?

Answer: Kp=PCO2K_p=P_{CO_2}Kp​=PCO2​​. Solids omitted; only gas pressure appears.