Ap Chemistry 2차 - Unit 7

Unit 7: Equilibrium — Multiple Choice Questions

K Expression Q vs K Le Chatelier Manipulating K Ksp Common Ion Effect Ka · Kb · Kw

1. Chemical equilibrium is the state where:

(A) Reactant and product concentrations are equal
(B) Forward and reverse rates are equal, concentrations are constant
(C) Only the forward reaction occurs
(D) ΔG° = 0 for all temperatures

View Answer

Correct Answer: (B) Explanation: At equilibrium, forward rate = reverse rate; concentrations are constant (not necessarily equal).

2. Which species are omitted from an equilibrium expression?

(A) Gases only
(B) Aqueous species only
(C) Pure solids and pure liquids
(D) All species are included

View Answer

Correct Answer: (C) Explanation: Only gases and aqueous species appear; pure solids/liquids have activity ≈ 1 and are omitted.

3. For \( \text{CaCO}_3(s)\rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)\), the correct \(K_c\) is:

(A) \(\dfrac{[\text{CaO}][\text{CO}_2]}{[\text{CaCO}_3]}\)
(B) \([\text{CO}_2]\)
(C) \(\dfrac{1}{[\text{CO}_2]}\)
(D) \([\text{CaO}][\text{CO}_2]\)

View Answer

Correct Answer: (B) Explanation: Solids are omitted; \(K_c=[\text{CO}_2]\).

4. For \( \text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons 2\text{NH}_3(g)\), which is \(K_c\)?

(A) \(\dfrac{[\text{NH}_3]}{[\text{N}_2][\text{H}_2]}\)
(B) \(\dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\)
(C) \(\dfrac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}\)
(D) \(\dfrac{[\text{NH}_3]^3}{[\text{N}_2]^2[\text{H}_2]}\)

View Answer

Correct Answer: (B) Explanation: Coefficients become exponents in the law of mass action.

5. For the reaction in Q4, \(K_c=0.50\) at a certain T. If \([\text{N}_2]=1.0\), \([\text{H}_2]=1.0\), \([\text{NH}_3]=1.0\) (all M), the reaction will:

(A) Shift right (Q < K)
(B) Shift left (Q > K)
(C) Be at equilibrium
(D) Not proceed

View Answer

Correct Answer: (B) Explanation: \(Q=\dfrac{1^2}{1\cdot1^3}=1.0>K\Rightarrow\) shifts left.

6. Which change alters the numerical value of \(K\) for a reaction?

(A) Adding a catalyst
(B) Changing concentrations
(C) Changing temperature
(D) Adding inert gas at constant volume

View Answer

Correct Answer: (C) Explanation: Only temperature changes \(K\); catalysts and concentration changes do not.

7. For \(2\text{SO}_2(g)+\text{O}_2(g)\rightleftharpoons 2\text{SO}_3(g)\), decreasing volume (at constant T) shifts equilibrium:

(A) Left, toward more moles of gas
(B) Right, toward fewer moles of gas
(C) No change
(D) Depends on catalyst presence

View Answer

Correct Answer: (B) Explanation: 3 moles ⇄ 2 moles; higher pressure favors fewer moles (right).

8. If the forward reaction is exothermic, increasing temperature will:

(A) Increase \(K\) and shift right
(B) Decrease \(K\) and shift left
(C) Leave \(K\) unchanged but shift right
(D) Leave both \(K\) and position unchanged

View Answer

Correct Answer: (B) Explanation: Heat is a product; adding heat shifts left, and \(K\) decreases.

9. Which statement about catalysts and equilibrium is TRUE?

(A) Catalysts change \(K\)
(B) Catalysts shift equilibrium to products
(C) Catalysts speed both directions equally; \(K\) and equilibrium composition unchanged
(D) Catalysts consume reactants

View Answer

Correct Answer: (C) Explanation: They lower activation barriers, reducing time to reach equilibrium without altering \(K\).

10. If \(A\rightleftharpoons B\) has \(K=4.0\), then for \(2A\rightleftharpoons 2B\), the equilibrium constant is:

(A) 2.0
(B) 4.0
(C) 8.0
(D) 16

View Answer

Correct Answer: (D) Explanation: Multiplying coefficients by 2 squares \(K\): \(K' = K^2 = 16\).

11. If \(A\rightleftharpoons B\) has \(K=0.25\), then \(B\rightleftharpoons A\) has:

(A) \(K=0.25\)
(B) \(K=4.0\)
(C) \(K=0.50\)
(D) \(K=2.0\)

View Answer

Correct Answer: (B) Explanation: Reversing inverts \(K\): \(K' = 1/K = 4.0\).

12. If Reaction 3 = Reaction 1 + Reaction 2, then \(K_3\) equals:

(A) \(K_1+K_2\)
(B) \(K_1/K_2\)
(C) \(K_1\cdot K_2\)
(D) \(\sqrt{K_1K_2}\)

View Answer

Correct Answer: (C) Explanation: Adding reactions multiplies equilibrium constants.

13. For \( \text{AB}(s)\rightleftharpoons \text{A}^+(aq)+\text{B}^-(aq)\), \(K_{sp}\) is:

(A) \([\text{AB}]\)
(B) \([\text{A}^+][\text{B}^-]\)
(C) \(\dfrac{[\text{A}^+]}{[\text{B}^-]}\)
(D) \([\text{A}^+]^2[\text{B}^-]^2\)

View Answer

Correct Answer: (B) Explanation: Solids omitted; product of ion molarities at equilibrium.

14. If \(K_{sp}(\text{AB})=1.0\times10^{-8}\) for \( \text{AB}(s)\rightleftharpoons \text{A}^+ + \text{B}^- \), the molar solubility \(s\) in pure water is:

(A) \(1.0\times10^{-8}\ \text{M}\)
(B) \(1.0\times10^{-4}\ \text{M}\)
(C) \(1.0\times10^{-2}\ \text{M}\)
(D) \(1.0\times10^{-16}\ \text{M}\)

View Answer

Correct Answer: (B) Explanation: \(K_{sp}=s^2\Rightarrow s=\sqrt{10^{-8}}=10^{-4}\ \text{M}\).

15. For \( \text{AB}_2(s)\rightleftharpoons \text{A}^{2+}+2\text{B}^- \) with \(K_{sp}=3.2\times10^{-9}\), the molar solubility \(s\) is (assume pure water):

(A) \(9.3\times10^{-4}\ \text{M}\)
(B) \(1.6\times10^{-3}\ \text{M}\)
(C) \(3.2\times10^{-3}\ \text{M}\)
(D) \(3.2\times10^{-4}\ \text{M}\)

View Answer

Correct Answer: (A) Explanation: \(K_{sp}=4s^3\Rightarrow s=(K_{sp}/4)^{1/3}=(8.0\times10^{-10})^{1/3}\approx 9.3\times10^{-4}\ \text{M}\).

16. \( \text{AgCl}(s)\rightleftharpoons \text{Ag}^+ + \text{Cl}^- \), \(K_{sp}=1.8\times10^{-10}\). In 0.10 M NaCl, the approximate solubility of AgCl is:

(A) \(1.8\times10^{-5}\ \text{M}\)
(B) \(1.8\times10^{-9}\ \text{M}\)
(C) \(1.8\times10^{-10}\ \text{M}\)
(D) \(1.8\times10^{-8}\ \text{M}\)

View Answer

Correct Answer: (B) Explanation: \(K_{sp}\approx s(0.10)\Rightarrow s\approx 1.8\times10^{-9}\ \text{M}\) (since \(s\ll0.10\)).

17. Equal volumes of 0.010 M AgNO₃ and 0.010 M NaCl are mixed. Will AgCl(s) precipitate? \(K_{sp}=1.8\times10^{-10}\).

(A) No, because \(Q_{sp}=1.8\times10^{-10}\)
(B) Yes, because \(Q_{sp} > K_{sp}\)
(C) No, because \(Q_{sp} < K_{sp}\)
(D) Undetermined without volumes

View Answer

Correct Answer: (B) Explanation: After mixing, \([\text{Ag}^+]=[\text{Cl}^-]=0.0050\ \text{M}\). \(Q_{sp}=2.5\times10^{-5}\gg K_{sp}\) ⇒ precipitate forms.

18. For conjugate pair HA/A⁻ at 25 °C, \(K_a(\text{HA})=1.8\times10^{-5}\). What is \(K_b(\text{A}^-)\)? (\(K_w=1.0\times10^{-14}\))

(A) \(5.6\times10^{-10}\)
(B) \(1.8\times10^{-9}\)
(C) \(5.6\times10^{-14}\)
(D) \(1.8\times10^{-19}\)

View Answer

Correct Answer: (A) Explanation: \(K_aK_b=K_w\Rightarrow K_b=K_w/K_a=10^{-14}/1.8\times10^{-5}=5.6\times10^{-10}\).

19. For \(2\text{NO}_2(g)\rightleftharpoons \text{N}_2\text{O}_4(g)\), \(K_c=0.15\). If \([\text{NO}_2]=0.60\) M and \([\text{N}_2\text{O}_4]=0.10\) M, the system will:

(A) Shift right (Q < K)
(B) Shift left (Q > K)
(C) Be at equilibrium
(D) Be undefined

View Answer

Correct Answer: (B) Explanation: \(Q=\dfrac{0.10}{(0.60)^2}=0.278>0.15\Rightarrow\) shift left.

20. For \(\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g)\), decreasing volume at constant T will:

(A) Shift right (fewer moles)
(B) Shift left (fewer moles)
(C) Not change equilibrium position (same total moles of gas)
(D) Make \(K\) increase

View Answer

Correct Answer: (C) Explanation: \(\Delta n_{\text{gas}}=0\); pressure/volume changes do not shift equilibrium.

21. For \( \text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons 2\text{NH}_3(g) \) at 700 K, \(K_c=0.500\). What is \(K_p\)? (Use \(R=0.08206\ \text{L·atm·mol}^{-1}\text{·K}^{-1}\))

(A) \(1.5\times10^{-3}\)
(B) \(1.5\times10^{-4}\)
(C) \(3.0\times10^{-2}\)
(D) \(0.500\)

View Answer

Correct Answer: (B) Explanation: \(\Delta n_{\text{gas}}=2-4=-2\). \(K_p=K_c(RT)^{\Delta n}=0.500\,(0.08206\times700)^{-2}\approx 0.500/3302\approx 1.5\times10^{-4}\).

22. For \( \text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) \), which statement is TRUE?

(A) \(K_p = K_c(RT)^{-1}\)
(B) \(K_p = K_c\)
(C) \(K_p = K_c(RT)^{+1}\)
(D) \(K_p\) and \(K_c\) are unrelated

View Answer

Correct Answer: (B) Explanation: \(\Delta n_{\text{gas}}=2-2=0\Rightarrow K_p=K_c(RT)^0=K_c\).

23. For \( \text{A}\rightleftharpoons \text{B} \) with \(K_c=4.0\) and initial \([\text{A}]_0=1.00\ \text{M}, [\text{B}]_0=0\), the equilibrium \([\text{B}]\) is:

(A) 0.20 M
(B) 0.40 M
(C) 0.67 M
(D) 0.80 M

View Answer

Correct Answer: (D) Explanation: Let \(x\) form B: \(K=\dfrac{x}{1-x}=4\Rightarrow x=0.80\ \text{M}\).

24. For \( 2\text{A}\rightleftharpoons \text{B} \) with \(K_c=1.0\), initial \([\text{A}]_0=0.50\ \text{M}, [\text{B}]_0=0\). What is \([\text{B}]_{eq}\)?

(A) 0.050 M
(B) 0.096 M
(C) 0.25 M
(D) 0.40 M

View Answer

Correct Answer: (B) Explanation: Change \(-2x,+x\). \(K=\dfrac{x}{(0.50-2x)^2}=1\Rightarrow x\approx0.096\ \text{M}\).

25. Mix 25.0 mL of 0.100 M Ca(NO₃)₂ with 75.0 mL of 0.100 M NaF. Will CaF₂(s) precipitate? \(K_{sp}(\text{CaF}_2)=3.9\times10^{-11}\).

(A) No, \(Q_{sp} < K_{sp}\)
(B) Yes, \(Q_{sp} > K_{sp}\)
(C) No change
(D) Cannot decide without density

View Answer

Correct Answer: (B) Explanation: After mixing, \([\text{Ca}^{2+}]=0.025\ \text{M}, [\text{F}^-]=0.075\ \text{M}\). \(Q_{sp}=0.025(0.075)^2\approx1.4\times10^{-4}\gg K_{sp}\).

26. \( \text{PbF}_2(s)\rightleftharpoons \text{Pb}^{2+}+2\text{F}^- \), \(K_{sp}=3.3\times10^{-8}\). In 0.10 M NaF, \([\text{Pb}^{2+}]\) at equilibrium is approximately:

(A) \(3.3\times10^{-5}\ \text{M}\)
(B) \(3.3\times10^{-6}\ \text{M}\)
(C) \(3.3\times10^{-7}\ \text{M}\)
(D) \(6.6\times10^{-8}\ \text{M}\)

View Answer

Correct Answer: (B) Explanation: \(K_{sp}\approx[\text{Pb}^{2+}](0.10)^2\Rightarrow [\text{Pb}^{2+}]\approx3.3\times10^{-6}\ \text{M}\).

27. For \( \text{A(OH)}_3(s)\rightleftharpoons \text{A}^{3+}+3\text{OH}^- \) with \(K_{sp}=2.7\times10^{-23}\), the molar solubility \(s\) in pure water is:

(A) \(1.0\times10^{-8}\ \text{M}\)
(B) \(1.0\times10^{-7}\ \text{M}\)
(C) \(1.0\times10^{-6}\ \text{M}\)
(D) \(1.0\times10^{-5}\ \text{M}\)

View Answer

Correct Answer: (C) Explanation: \(K_{sp}=s(3s)^3=27s^4\Rightarrow s=(K_{sp}/27)^{1/4}=(10^{-24})^{1/4}=10^{-6}\ \text{M}\).

28. If \(K_a(\text{HA})=4.0\times10^{-4}\) at 25 °C, then \(K_b(\text{A}^-)\) equals:

(A) \(2.5\times10^{-11}\)
(B) \(4.0\times10^{-10}\)
(C) \(4.0\times10^{-18}\)
(D) \(2.5\times10^{-7}\)

View Answer

Correct Answer: (A) Explanation: \(K_b=K_w/K_a=10^{-14}/4.0\times10^{-4}=2.5\times10^{-11}\).

29. A buffer contains 0.10 M HA (weak) and 0.30 M A⁻. If \(K_a=1.8\times10^{-5}\), the pH is closest to:

(A) 4.74
(B) 5.22
(C) 5.74
(D) 6.22

View Answer

Correct Answer: (B) Explanation: \( \mathrm{p}K_a=4.74\). \( \mathrm{pH}=\mathrm{p}K_a+\log\!\frac{0.30}{0.10}=4.74+0.477\approx5.22\).

30. The correct relation between \(K_p\) and \(K_c\) for a gas reaction is:

(A) \(K_p=K_c(RT)^{\Delta n_{\text{gas}}}\)
(B) \(K_p=K_c/(RT)\)
(C) \(K_p=K_c\cdot RT\)
(D) \(K_p=K_c^{\Delta n_{\text{gas}}}\)

View Answer

Correct Answer: (A) Explanation: \(\Delta n_{\text{gas}}=\sum\nu_{\text{gas,prod}}-\sum\nu_{\text{gas,react}}\).

31. In \( \text{CaCO}_3(s)\rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)\), adding more solid CaCO₃ will:

(A) Increase \([\text{CO}_2]\) at equilibrium
(B) Decrease \([\text{CO}_2]\) at equilibrium
(C) Not change equilibrium \([\text{CO}_2]\)
(D) Double \(K_c\)

View Answer

Correct Answer: (C) Explanation: Activities of pure solids are constant; equilibrium \([\text{CO}_2]\) (and \(K\)) unchanged.

32. For \( 2\text{NO}_2(g)\rightleftharpoons \text{N}_2\text{O}_4(g) \), adding helium at constant volume will:

(A) Shift left
(B) Shift right
(C) Not shift equilibrium
(D) Make \(K_p\) larger

View Answer

Correct Answer: (C) Explanation: Partial pressures of reacting species unchanged at constant volume; no shift.

33. \( \text{N}_2\text{O}_4(g)\rightleftharpoons 2\text{NO}_2(g) \), \(K_c=0.15\). Starting with \([\text{N}_2\text{O}_4]_0=1.00\ \text{M}\), \([\text{NO}_2]_0=0\). What is \([\text{NO}_2]_{eq}\)?

(A) 0.20 M
(B) 0.28 M
(C) 0.35 M
(D) 0.50 M

View Answer

Correct Answer: (C) Explanation: Let \(x\) dissociate: \(K=\dfrac{(2x)^2}{1-x}=0.15\Rightarrow x\approx0.176\). Then \([\text{NO}_2]=2x\approx0.35\ \text{M}\).

34. For an endothermic forward reaction, increasing temperature will:

(A) Decrease \(K\)
(B) Increase \(K\)
(C) Leave \(K\) unchanged
(D) Make the reaction stop

View Answer

Correct Answer: (B) Explanation: Heat treated as reactant; higher \(T\) shifts right and increases \(K\).

35. Adding NaA (source of A⁻) to a solution of weak acid HA will:

(A) Increase HA ionization
(B) Decrease HA ionization (shift left)
(C) Not affect HA equilibrium
(D) Change \(K_a\)

View Answer

Correct Answer: (B) Explanation: Common ion (A⁻) raises \(Q\); system shifts left. \(K_a\) is constant at fixed \(T\).

36. For \( \text{H}^+ + \text{A}^- \rightleftharpoons \text{HA} \) in aqueous solution, evaporation of water (volume ↓, \(T\) constant) will:

(A) Shift left (toward ions)
(B) Shift right (toward HA)
(C) Not shift
(D) Change \(K_c\)

View Answer

Correct Answer: (B) Explanation: All concentrations scale up; \(Q'=\dfrac{\alpha[\text{HA}]}{\alpha[\text{H}^+]\cdot \alpha[\text{A}^-]}=\dfrac{Q}{\alpha}<Q\). With \(K\) unchanged, \(Q'<K\) ⇒ shift right.

37. An AB₃ salt has molar solubility \(s=1.0\times10^{-6}\ \text{M}\) in pure water. What is \(K_{sp}\)?

(A) \(1.0\times10^{-18}\)
(B) \(1.0\times10^{-24}\)
(C) \(2.7\times10^{-23}\)
(D) \(9.0\times10^{-18}\)

View Answer

Correct Answer: (C) Explanation: \( \text{AB}_3\to \text{A}^{3+}+3\text{B}^- \Rightarrow K_{sp}=s(3s)^3=27s^4=27\times10^{-24}=2.7\times10^{-23}\).

38. Mix 50.0 mL of 0.020 M Pb(NO₃)₂ with 100.0 mL of 0.030 M KI. Will PbI₂(s) form? \(K_{sp}(\text{PbI}_2)=7.9\times10^{-9}\).

(A) No, \(Q_{sp} < K_{sp}\)
(B) Yes, \(Q_{sp} > K_{sp}\)
(C) At equilibrium only
(D) Cannot decide without temperature

View Answer

Correct Answer: (B) Explanation: After mixing (150 mL): \([\text{Pb}^{2+}]=0.00667\ \text{M}\), \([\text{I}^-]=0.020\ \text{M}\). \(Q_{sp}=6.67\times10^{-3}\times(0.020)^2\approx2.7\times10^{-6}\gg K_{sp}\).

39. In Q38, assuming \([\text{I}^-]\approx0.020\ \text{M}\) remains large, the approximate equilibrium \([\text{Pb}^{2+}]\) is:

(A) \(2.0\times10^{-3}\ \text{M}\)
(B) \(2.0\times10^{-4}\ \text{M}\)
(C) \(2.0\times10^{-5}\ \text{M}\)
(D) \(2.0\times10^{-6}\ \text{M}\)

View Answer

Correct Answer: (C) Explanation: \(K_{sp}=[\text{Pb}^{2+}][\text{I}^-]^2\Rightarrow [\text{Pb}^{2+}]\approx 7.9\times10^{-9}/(0.020)^2\approx2.0\times10^{-5}\ \text{M}\).

40. If \(K_a(\text{NH}_4^+)=5.6\times10^{-10}\) at 25 °C, then \(K_b(\text{NH}_3)\) equals:

(A) \(5.6\times10^{-10}\)
(B) \(1.8\times10^{-5}\)
(C) \(5.6\times10^{-5}\)
(D) \(1.8\times10^{-10}\)

View Answer

Correct Answer: (B) Explanation: \(K_b=K_w/K_a=10^{-14}/5.6\times10^{-10}\approx1.8\times10^{-5}\).

41. For \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \) at a given T, \(K_p=0.15\). If initially \(P_{\text{NO}_2}=0.20\ \text{atm}\) and \(P_{\text{N}_2\text{O}_4}=0.10\ \text{atm}\), the system will:

(A) Shift right (since \(Q_p<K_p\))
(B) Shift left (since \(Q_p>K_p\))
(C) Be at equilibrium
(D) Not enough information

View Answer

Correct Answer: (B) Explanation: \(Q_p=\frac{(0.20)^2}{0.10}=0.40>0.15\Rightarrow\) shifts left.

42. For \( \text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) \), at equilibrium \([\text{H}_2]=0.10\ \text{M}\), \([\text{I}_2]=0.10\ \text{M}\), \([\text{HI}]=0.60\ \text{M}\). What is \(K_c\)?

(A) 3.6
(B) 6.0
(C) 36
(D) 60

View Answer

Correct Answer: (C) Explanation: \(K_c=\dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}=\dfrac{0.36}{0.01}=36.\)

43. For \( \text{A}\rightleftharpoons 2\text{B} \) with \(K_c=1.0\), start with \([\text{A}]_0=0.50\ \text{M}\), \([\text{B}]_0=0\). What is \([\text{A}]_{eq}\)?

(A) 0.10 M
(B) 0.25 M
(C) 0.33 M
(D) 0.40 M

View Answer

Correct Answer: (B) Explanation: Let \(x\) react: \([\text{A}]=0.50-x\), \([\text{B}]=2x\). \(K=\frac{(2x)^2}{0.50-x}=1\Rightarrow 4x^2+x-0.50=0\Rightarrow x=0.25\). Thus \([\text{A}]_{eq}=0.25\ \text{M}.\)

44. For \( \text{A}(g)+\text{B}(g)\rightleftharpoons \text{C}(g) \) at equilibrium, some A is removed (T, V constant). The system will:

(A) Shift right to replace A
(B) Shift left to form more A
(C) Not shift (A is a solid)
(D) Increase \(K\)

View Answer

Correct Answer: (B) Explanation: Removing a reactant makes \(Q>K\), shifting left to produce more A.

45. For \( \text{Ca(OH)}_2(s)\rightleftharpoons \text{Ca}^{2+}+2\text{OH}^- \), \(K_{sp}=5.5\times10^{-6}\) at 25 °C. In a solution where \([\text{OH}^-]=1.0\times10^{-1}\ \text{M}\), the equilibrium \([\text{Ca}^{2+}]\) is:

(A) \(5.5\times10^{-6}\ \text{M}\)
(B) \(5.5\times10^{-5}\ \text{M}\)
(C) \(5.5\times10^{-4}\ \text{M}\)
(D) \(5.5\times10^{-3}\ \text{M}\)

View Answer

Correct Answer: (C) Explanation: \(K_{sp}=[\text{Ca}^{2+}][\text{OH}^-]^2\Rightarrow [\text{Ca}^{2+}]=5.5\times10^{-6}/(10^{-1})^2=5.5\times10^{-4}\ \text{M}.\)

46. A solution has \([\text{Ag}^+]=0.010\ \text{M}\) and \([\text{Pb}^{2+}]=0.010\ \text{M}\). Cl⁻ is slowly added. Which chloride precipitates first?
Data: \(K_{sp}(\text{AgCl})=1.8\times10^{-10}\), \(K_{sp}(\text{PbCl}_2)=1.6\times10^{-5}\).

(A) AgCl
(B) PbCl₂
(C) Both simultaneously
(D) Neither

View Answer

Correct Answer: (A) Explanation: Threshold \([\text{Cl}^-]_{\text{crit}}:\) AgCl: \(K_{sp}/[\text{Ag}^+]=1.8\times10^{-8}\ \text{M}\) (very small); PbCl₂: \([\text{Cl}^-]=\sqrt{K_{sp}/[\text{Pb}^{2+}]}\approx0.040\ \text{M}\). AgCl forms first.

47. \( \text{AB}(s)\rightleftharpoons \text{A}^+ + \text{B}^- \) with \(K_{sp}=1.0\times10^{-12}\). In 0.010 M NaB, the molar solubility of AB is approximately:

(A) \(1.0\times10^{-12}\ \text{M}\)
(B) \(1.0\times10^{-10}\ \text{M}\)
(C) \(1.0\times10^{-8}\ \text{M}\)
(D) \(1.0\times10^{-6}\ \text{M}\)

View Answer

Correct Answer: (B) Explanation: \(K_{sp}\approx s(0.010)\Rightarrow s\approx 10^{-12}/10^{-2}=10^{-10}\ \text{M}.\)

48. For a certain reaction, \(K\) increases when temperature is raised. The forward reaction is most likely:

(A) Endothermic
(B) Exothermic
(C) Athermal (ΔH = 0)
(D) Impossible to tell

View Answer

Correct Answer: (A) Explanation: Higher \(T\) increases \(K\) for endothermic forward reactions (heat acts as a reactant).

49. For \( \text{N}_2\text{O}_4(g)\rightleftharpoons 2\text{NO}_2(g) \) (\(\Delta n_{\text{gas}}=+1\)), adding an inert gas at constant pressure (isothermal) will:

(A) Shift left (toward fewer moles)
(B) Shift right (toward more moles)
(C) Not shift
(D) Decrease \(K\)

View Answer

Correct Answer: (B) Explanation: At constant pressure, adding inert gas increases volume ⇒ shift toward more moles of gas.

50. \( \text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) \) is at equilibrium. The volume is suddenly halved (isothermal). Immediately after compression, the reaction quotient \(Q_p\):

(A) Becomes > \(K_p\) (shift left)
(B) Becomes < \(K_p\) (shift right)
(C) Remains = \(K_p\) (no net shift due to \(\Delta n=0\))
(D) Becomes undefined

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Correct Answer: (C) Explanation: All partial pressures scale by the same factor; with exponents summing equally on both sides, \(Q_p\) remains unchanged (= \(K_p\)).

Ap Chemistry 2차 - Unit 7

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