Calculating the Equilibrium Constant - AP Chemistry

  1. What is Δngas\Delta n_{\text{gas}} for 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g)+\text{O}_2(g)\rightleftharpoons^2\text{SO}_3(g)?
    Δngas=−1\Delta n_{\text{gas}}=-1. 22 moles gas products minus 33 moles gas reactants equals −1-1.
  2. What is KpK_p from Kc=0.50K_c=0.50 at T=300 KT=300\,K for N2O4(g)⇌2NO2(g)N_2O_4(g)\rightleftharpoons 2NO_2(g)? Use R=0.0821R=0.0821.
    Kp=0.50 (0.0821⋅300)1=12K_p=0.50\,(0.0821\cdot 300)^1=12. Δn=2−1=1\Delta n=2-1=1; multiply KcK_c by (RT)1(RT)^1.
  3. Calculate KpK_p for 2A(g)⇌B(g)2A(g)\rightleftharpoons B(g) if PA=0.50 atmP_A=0.50\,\text{atm} and PB=2.0 atmP_B=2.0\,\text{atm} at equilibrium.
    Kp=8.0K_p=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 = 8.0
  4. What is KK for the reverse reaction if K=4.0×10−3K=4.0\times10^{-3} for the forward reaction?
    Kreverse=2.5×102K_{\text{reverse}}=2.5\times10^2. Kreverse=1÷(4.0×10−3)=2.5×102K_{reverse} = 1 ÷ (4.0×10^{-3}) = 2.5×10^2
  5. State the relationship between KpK_p and KcK_c using Δngas\Delta n_{\text{gas}}.
    Kp=Kc(RT)ΔngasK_p=K_c(RT)^{\Delta n_{\text{gas}}}. Relates pressure and concentration constants via ideal gas law.
  6. What is the relationship between ΔG∘\Delta G^\circ, KK, RR, and TT?
    ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K. Links thermodynamic favorability to equilibrium position.
  7. Identify the correct KcK_c for N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons^2\text{NH}_3(g).
    Kc=[NH3]2[N2][H2]3K_c=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}. Products over reactants with coefficients: 22 for NH₃, 33 for H₂.
  8. What is KK for the reaction if the given reaction has K=3.0K=3.0 and all coefficients are doubled?
    Knew=9.0K_{\text{new}}=9.0. Doubling coefficients squares KK: 3.02=9.03.0^2 = 9.0
  9. What is Δngas\Delta n_{\text{gas}} for Kp=Kc(RT)ΔngasK_p=K_c(RT)^{\Delta n_{\text{gas}}}?
    Δngas=∑νprod,gas−∑νreact,gas\Delta n_{\text{gas}}=\sum \nu_{\text{prod,gas}}-\sum \nu_{\text{react,gas}}. Change in moles of gas: product coefficients minus reactant coefficients.
  10. What is the numerical value of KK when ΔG∘=0\Delta G^\circ=0 at that temperature?
    K=1K=1. At equilibrium, forward and reverse rates are equal.
  11. What is the value of KK for any reaction written as X⇌XX\rightleftharpoons X?
    K=1K=1. Same species on both sides cancel out, leaving unity.
  12. What is the equilibrium constant expression KcK_c for 2NO2(g)⇌N2O4(g)2NO_2(g)\rightleftharpoons N_2O_4(g)?
    Kc=[N2O4][NO2]2K_c=\frac{[N_2O_4]}{[NO_2]^2}. Products over reactants with stoichiometric exponents.
  13. Which species are omitted from KK expressions: pure solids, pure liquids, gases, or aqueous solutes?
    Pure solids and pure liquids are omitted. Their activities equal 1, so they don't affect the equilibrium constant.
  14. What is the relationship between KcK_c and KpK_p using Δn\Delta n and R,TR,T?
    Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}. Relates concentration and pressure equilibrium constants.
  15. Which temperature unit must be used in Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n} calculations?
    TT must be in kelvin (K)(\text{K}). Absolute temperature required for ideal gas calculations.
  16. What is KcK_c for CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)?
    Kc=[CO2]K_c=[CO_2]. Solids omitted; only gas concentration appears.
  17. Find KcK_c if Kp=2.0K_p=2.0, Δngas=−2\Delta n_{\text{gas}}=-2, T=400 KT=400\,\text{K}, and R=0.0821R=0.0821.
    Kc=2.15×103K_c=2.15\times10^3. Kc=2.0÷(0.0821×400)−2=2.0×1079=2.15×103K_c = 2.0 ÷ (0.0821 × 400)^{-2} = 2.0 × 1079 = 2.15×10^3
  18. Identify the correct KcK_c for CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s)\rightleftharpoons\text{CaO}(s)+\text{CO}_2(g).
    Kc=[CO2]K_c=[\text{CO}_2]. Solids omitted; only gaseous CO₂ appears in the expression.
  19. What is the general expression for KpK_p for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD (gases)?
    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 gas-phase equilibria.
  20. Find KpK_p if Kc=0.50K_c=0.50, Δngas=1\Delta n_{\text{gas}}=1, T=300 KT=300\,\text{K}, and R=0.0821R=0.0821.
    Kp=12.3K_p=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 = 12.3
  21. Which species are omitted from KK for a heterogeneous equilibrium involving pure solids and liquids?
    Pure solids and pure liquids are omitted (activity =1=1). Their concentrations remain constant during reaction.
  22. What is KcK_c for CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)CH_3COOH(aq)\rightleftharpoons H^+(aq)+CH_3COO^-(aq)?
    Kc=[H+][CH3COO−][CH3COOH]K_c=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}. Weak acid dissociation with products over reactant.
  23. Find KoverallK_\text{overall} if reaction 2 is the reverse of a step with constant K2K_2 and reactions are added.
    Koverall=K1(1K2)K_\text{overall}=K_1\left(\frac{1}{K_2}\right). Reversing step 2 inverts its KK before multiplying.
  24. What happens to KK when all coefficients in the reaction are multiplied by nn?
    Knew=KnK_{\text{new}}=K^n. Each concentration term gets raised to nn, so KK is raised to nn.
  25. What is the general expression for KpK_p for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD?
    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.
  26. Identify KcK_c for 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g).
    Kc=[SO3]2[SO2]2[O2]K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}. Products over reactants with stoichiometric powers.
  27. What is Δn\Delta n (used in Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}) for gas-phase equilibria?
    Δn=mol gas products−mol gas reactants\Delta n=\text{mol gas products}-\text{mol gas reactants}. Change in moles of gas from reactants to products.
  28. What is the general expression for KcK_c for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD?
    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 stoichiometric coefficients.
  29. What is the general expression for KcK_c for aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD?
    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 their stoichiometric coefficients.
  30. What is the equilibrium constant expression KpK_p for N2(g)+3H2(g)⇌2NH3(g)N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)?
    Kp=(PNH3)2PN2(PH2)3K_p=\frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}. Partial pressures raised to stoichiometric coefficients.
  31. What is KpK_p for CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)?
    Kp=PCO2K_p=P_{CO_2}. Solids omitted; only gas pressure appears.
  32. What is KK for an overall reaction made by adding reactions with constants K1K_1 and K2K_2?
    Koverall=K1K2K_{\text{overall}}=K_1K_2. When reactions add, their equilibrium constants multiply.
  33. Calculate KcK_c if at equilibrium [A]=0.20 M[A]=0.20\,M, [B]=0.30 M[B]=0.30\,M, [C]=0.40 M[C]=0.40\,M for A+B⇌CA+B\rightleftharpoons C.
    Kc=0.40(0.20)(0.30)=6.7K_c=\frac{0.40}{(0.20)(0.30)}=6.7. Substitute equilibrium concentrations into KcK_c expression.
  34. Calculate KpK_p if PCO=2.0 atmP_{CO}=2.0\,atm, PH2=3.0 atmP_{H_2}=3.0\,atm, PCH3OH=1.5 atmP_{CH_3OH}=1.5\,atm for CO+2H2⇌CH3OHCO+2H_2\rightleftharpoons CH_3OH.
    Kp=1.5(2.0)(3.0)2=0.083K_p=\frac{1.5}{(2.0)(3.0)^2}=0.083. Substitute equilibrium pressures into KpK_p expression.
  35. What is the effect on KK when a reaction is reversed?
    Krev=1KK_\text{rev}=\frac{1}{K}. Reversing flips the fraction, giving the reciprocal.
  36. What happens to KK when the reaction is reversed?
    Kreverse=1KforwardK_{\text{reverse}}=\frac{1}{K_{\text{forward}}}. Reversing swaps products and reactants, inverting the fraction.
  37. What is the effect on KK when all coefficients are multiplied by nn?
    Knew=KnK_\text{new}=K^n. Multiplying coefficients raises KK to that power.
  38. Identify the correct KcK_c for CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\text{CH}_3\text{COOH}(aq)\rightleftharpoons\text{H}^+(aq)+\text{CH}_3\text{COO}^-(aq).
    Kc=[H+][CH3COO−][CH3COOH]K_c=\frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}. Weak acid dissociation: products over undissociated acid.
  39. Calculate KcK_c for A⇌2BA\rightleftharpoons^2B if [A]=0.50 M[A]=0.50\,\text{M} and [B]=0.20 M[B]=0.20\,\text{M} at equilibrium.
    Kc=0.080K_c=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 = 0.080
  40. What is KK for the overall reaction if two steps have constants K1K_1 and K2K_2 and are added?
    Koverall=K1K2K_\text{overall}=K_1K_2. When reactions add, equilibrium constants multiply.

Study these flashcards