AP Chemistry Flashcards: Properties Of The Equilibrium Constant

Study Properties Of 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.

AP Chemistry

Properties Of The Equilibrium Constant

0 mastered0 still learning

0% Complete

QUESTION
1/ 40

What does Q<KQ<K indicate about the direction the system will shift to reach equilibrium?

Tap card or press Space to flip

ANSWER

Shifts right (toward products). System needs more products to reach equilibrium.

How well did you know it?

Card 1 / 40

What this deck covers

This deck focuses on Properties Of 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.

All flashcards

Flashcard 1: What does Q<KQ<K indicate about the direction the system will shift to reach equilibrium?

Answer: Shifts right (toward products). System needs more products to reach equilibrium.

Flashcard 2: What is Δn\Delta n in the formula Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n} for a gas-phase reaction?

Answer: Δn=mol gas productsmol gas reactants\Delta n=\text{mol gas products}-\text{mol gas reactants}. Counts net change in moles of gas from reactants to products.

Flashcard 3: What is the general equilibrium constant expression KpK_p for aA+bBcC+dDaA+bB\rightleftharpoons 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}. Uses partial pressures instead of concentrations for gases.

Flashcard 4: Which species are omitted from KK expressions for heterogeneous equilibria: pure solids, pure liquids, or both?

Answer: Both pure solids and pure liquids are omitted. Their activities are constant at 1, so they don't affect K.

Flashcard 5: If Δngas=0\Delta n_{\text{gas}}=0, what is the relationship between KpK_p and KcK_c?

Answer: Kp=KcK_p=K_c. When gas moles don't change, (RT)0=1(RT)^0=1.

Flashcard 6: What is the general rule for how KK changes when a reaction is reversed?

Answer: Krev=1KK_{\text{rev}}=\frac{1}{K}. Reversing flips products and reactants, so K becomes its reciprocal.

Flashcard 7: What is the equilibrium-constant expression for aA+bBcC+dDaA+bB\rightleftharpoons cC+dD using partial pressures?

Answer: Kp=(PC)c(PD)d(PA)a(PB)bK_p=\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}. Same form as KcK_c but uses partial pressures instead of concentrations.

Flashcard 8: What is the general rule for how KK changes when all coefficients are multiplied by nn?

Answer: Knew=KnK_{\text{new}}=K^n. Multiplying coefficients by n raises K to the nth power.

Flashcard 9: What is the general rule for how KK changes when two reactions are added to give an overall reaction?

Answer: Koverall=K1K2K_{\text{overall}}=K_1K_2. When reactions add, their equilibrium constants multiply.

Flashcard 10: What is the general equilibrium constant expression KcK_c for aA+bBcC+dDaA+bB\rightleftharpoons cC+dD?

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

Flashcard 11: Identify the correct KK when K=3.0K=3.0 and the balanced equation coefficients are doubled.

Answer: 9.09.0. Knew=3.02=9.0K_{\text{new}} = 3.0^2 = 9.0 when coefficients are doubled.

Flashcard 12: What happens to KK when the initial concentrations or partial pressures are changed at constant temperature?

Answer: KK is unchanged. K depends only on temperature, not on initial conditions.

Flashcard 13: Find KK for 2×2\times the reaction if the original reaction has K=3.0×102K=3.0\times10^2.

Answer: Knew=9.0×104K_{\text{new}}=9.0\times10^4. Knew=(3.0×102)2=9.0×104K_{\text{new}}=(3.0\times10^2)^2=9.0\times10^4.

Flashcard 14: How are equilibrium constants combined when reactions are added to give an overall reaction?

Answer: Koverall=K1×K2×K_{\text{overall}}=K_1\times K_2\times\cdots. Adding reactions multiplies their equilibrium constants.

Flashcard 15: What is the activity value used for any pure solid or pure liquid in an equilibrium expression?

Answer: 11. Pure solids and liquids have unit activity by definition.

Flashcard 16: Identify the equilibrium direction if K=2.0×103K=2.0\times10^3 for a reaction written as products over reactants.

Answer: Product-favored (equilibrium lies to the right). K>>1K>>1 means products predominate at equilibrium.

Flashcard 17: What is the relationship between KpK_p and KcK_c for gases using Δngas\Delta n_{\text{gas}}?

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

Flashcard 18: How does KK change if all coefficients in the balanced reaction are multiplied by nn?

Answer: Knew=KnK_{\text{new}}=K^n. Each species' exponent multiplies by nn, so KK raises to nn.

Flashcard 19: Which species are omitted from KK for a heterogeneous equilibrium: pure solids, pure liquids, or aqueous solutes?

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

Flashcard 20: Identify the correct KpK_p if Kc=0.50K_c=0.50, Δn=2\Delta n=2, and RT=25RT=25 for an ideal-gas reaction.

Answer: 312.5312.5. Kp=0.50×252=312.5K_p = 0.50 × 25^2 = 312.5 using Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}

Flashcard 21: Identify the correct KK when K=4.0K=4.0 for ABA\rightleftharpoons B and the reaction is reversed.

Answer: 0.250.25. Krev=14.0=0.25K_{\text{rev}} = \frac{1}{4.0} = 0.25 by the reversal rule.

Flashcard 22: What is Δngas\Delta n_{\text{gas}} for Kp=Kc(RT)ΔngasK_p=K_c(RT)^{\Delta n_{\text{gas}}}?

Answer: Δngas=νprod,gasνreact,gas\Delta n_{\text{gas}}=\sum\nu_{\text{prod,gas}}-\sum\nu_{\text{react,gas}}. Moles of gas products minus moles of gas reactants.

Flashcard 23: What is the sign of ΔG\Delta G^\circ when K<1K<1 at a given temperature?

Answer: ΔG>0\Delta G^\circ>0. Since lnK<0\ln K < 0 when K<1K < 1, ΔG\Delta G^\circ must be positive.

Flashcard 24: What happens to KK when a catalyst is added to a system at constant temperature?

Answer: KK is unchanged. Catalysts speed up both forward and reverse reactions equally.

Flashcard 25: Find KrevK_{\text{rev}} if K=5.0×104K=5.0\times10^{-4} for the forward reaction.

Answer: Krev=2.0×103K_{\text{rev}}=2.0\times10^3. Krev=15.0×104=2.0×103K_{\text{rev}}=\frac{1}{5.0\times10^{-4}}=2.0\times10^3.

Flashcard 26: What is the sign of ΔG\Delta G^\circ when K>1K>1 at a given temperature?

Answer: ΔG<0\Delta G^\circ<0. Since lnK>0\ln K > 0 when K>1K > 1, ΔG\Delta G^\circ must be negative.

Flashcard 27: Identify the correct KK for the overall reaction if K1=2.0×103K_1=2.0\times10^3 and K2=5.0×102K_2=5.0\times10^{-2} are added.

Answer: 1.0×1021.0\times10^2. Koverall=(2.0×103)(5.0×102)=1.0×102K_{\text{overall}} = (2.0×10^3)(5.0×10^{-2}) = 1.0×10^2

Flashcard 28: How does KK change if the balanced reaction is reversed?

Answer: Krev=1KK_{\text{rev}}=\frac{1}{K}. Reversing swaps products and reactants, inverting KK.

Flashcard 29: Identify the equilibrium direction if K=4.0×106K=4.0\times10^{-6} for a reaction written as products over reactants.

Answer: Reactant-favored (equilibrium lies to the left). K<<1K<<1 means reactants predominate at equilibrium.

Flashcard 30: What is the only common experimental change that can change KK for a given reaction?

Answer: Changing temperature. K is a function of temperature only for a given reaction.

Flashcard 31: What does Q>KQ>K indicate about the direction the system will shift to reach equilibrium?

Answer: Shifts left (toward reactants). System has too many products, needs more reactants.

Flashcard 32: Find KK for 12×\frac{1}{2}\times the reaction if the original reaction has K=1.6×105K=1.6\times10^{-5}.

Answer: Knew=4.0×103K_{\text{new}}=4.0\times10^{-3}. Knew=(1.6×105)1/2=4.0×103K_{\text{new}}=(1.6\times10^{-5})^{1/2}=4.0\times10^{-3}.

Flashcard 33: What is the relationship between KpK_p and KcK_c for ideal gases in terms of Δn\Delta n?

Answer: Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}. Relates pressure and concentration equilibrium constants via ideal gas law.

Flashcard 34: How does KK change if all coefficients in the balanced reaction are divided by nn?

Answer: Knew=K1nK_{\text{new}}=K^{\frac{1}{n}}. Each exponent divides by nn, so KK raises to 1n\frac{1}{n}.

Flashcard 35: What is the relationship between KK and ΔG\Delta G^\circ at a given temperature?

Answer: ΔG=RTlnK\Delta G^\circ=-RT\ln K. Links thermodynamic favorability to equilibrium position.

Flashcard 36: What does Q=KQ=K indicate about the system?

Answer: The system is at equilibrium. No net change; forward and reverse rates are equal.

Flashcard 37: What is the reaction quotient expression QcQ_c for aA+bBcC+dDaA+bB\rightleftharpoons cC+dD?

Answer: Qc=[C]c[D]d[A]a[B]bQ_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}. Same form as KcK_c but uses actual concentrations.

Flashcard 38: What is the equilibrium-constant expression for aA+bBcC+dDaA+bB\rightleftharpoons cC+dD using concentrations?

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

Flashcard 39: What is the value of the activity used for any pure solid or pure liquid in an equilibrium expression?

Answer: a=1a=1. By definition, pure solids/liquids have unit activity.

Flashcard 40: Identify the correct KK if ΔG=0\Delta G^\circ=0 at a given temperature.

Answer: 11. When ΔG=0\Delta G^\circ = 0, then lnK=0\ln K = 0, so K=1K = 1.