AP Unit 7 Problems
14 problems for Unit 7. Try the particle picture, chemical setup and explanation before opening the separate worked solution. These are newly authored learning tasks; they are not released College Board exam questions.
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Problem 1
A sealed flask at constant temperature contains A(g) ⇌ B(g). A sensor records [A] = 0.80, 0.62, 0.53, 0.50, 0.50 M at 0, 10, 20, 40, 60 s; [B] = 0.20, 0.38, 0.47, 0.50, 0.50 M. Sketch both concentration curves with labeled axes. Explain why the flat part is consistent with dynamic equilibrium rather than a stopped reaction. Give a second reversible physical process that can reach dynamic equilibrium in a closed container.
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Problem 2
For A(g) ⇌ B(g), the forward and reverse rates at one instant are 0.030 and 0.012 mol L−1 s−1. Determine the signs and magnitudes of the instantaneous changes in [A] and [B]. A student says the reaction is irreversible because the net change is forward. Evaluate the claim.
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Problem 3
- Write Qc for CaCO3(s) ⇌ CaO(s) + CO2(g), and Qp for N2(g) + 3H2(g) ⇌ 2NH3(g). Explain each omitted term and exponent. When can either quotient be called its corresponding equilibrium constant?
- For H2O(l) ⇌ H2O(g), write Qp for the ideal water vapor above a pure-liquid phase. Explain why one water phase appears but the other does not.
- For HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) in a dilute aqueous solution, write Qc using the usual solvent-activity approximation. Explain why the dissolved HA is not omitted with the solvent.
Assume the stated pure phases are present and the gas and solution approximations are appropriate. Use the AP concentration/partial-pressure forms; thermodynamic activities are referenced to their standard states.
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Problem 4
For 2NO2(g) ⇌ N2O4(g), equilibrium samples at one temperature give [NO2] = 0.0400 M and [N2O4] = 0.120 M. Calculate Kc. A second run reports 0.0600 M and 0.270 M, respectively. Does it support the same Kc? Explain why using the initial NO2 concentration would invalidate this determination.
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Problem 5
Two 1:1 isomerizations A ⇌ B begin with only A. At the same temperature Kc is 2.0 × 105 for one and 4.0 × 10−6 for the other. Compare their equilibrium compositions. Can these K values tell you which reaches equilibrium first? Justify both answers.
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Problem 6
At one temperature, A ⇌ B has K1 = 4.0 and B ⇌ C has K2 = 9.0. Determine K for (a) C ⇌ A and (b) 2A ⇌ 2C. For a current mixture with [A] = 0.20 M, [B] = 0.30 M, [C] = 0.60 M, show that the analogous Q expressions multiply for A ⇌ C. Explain why multiplying an equation by two does not merely double K.
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Problem 7
A(g) ⇌ B(g) has Kc = 4.00 at a fixed temperature. A sealed constant-volume flask initially has [A] = 0.300 M and [B] = 0.200 M. Determine the net direction and both equilibrium concentrations. Check the ratio and atom balance.
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Problem 8
- For N2O4(g) ⇌ 2NO2(g), Kc = 0.100 at the stated temperature. Initially [N2O4] = 0.100 M and [NO2] = 0. Calculate both equilibrium concentrations. A student assumes x is negligible compared with 0.100 M. Check that approximation.
- A second sealed, constant-volume flask at the same temperature starts with [N2O4] = 0.0500 M and [NO2] = 0.200 M. Calculate its initial Qc and net reaction direction. Build a stoichiometric change table, solve for both equilibrium concentrations, and explain which root is physically possible. Verify conserved nitrogen units and final Qc = Kc.
Treat both gas mixtures with the same ideal concentration model. The flasks are independent samples.
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Problem 9
In equal-volume boxes for the isomerization A ⇌ B, box I contains 8 A and 2 B particles, box II contains 4 A and 6 B, and box III contains 2 A and 8 B. Each symbol represents the same number of molecules. Kc = 4.0. Identify the equilibrium box, predict the direction for the others, and describe what a second snapshot of the equilibrium box would show. State a limitation of these small-number pictures.
Draw a later equilibrium snapshot with a legend and the same volume/particle scale. Include a forward and a reverse event arrow, and explain what stays constant on average.
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Problem 10
For 2NO2(g) ⇌ N2O4(g), the forward reaction is exothermic; NO2 is brown and N2O4 is colorless. Predict the equilibrium response to (a) heating at fixed volume and (b) decreasing volume at fixed temperature. In (b), distinguish the immediate change in NO2 concentration from its later change during re-equilibration. Explain why the final brown intensity relative to the original cannot be concluded just from the word shift.
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Problem 11
- A(g) ⇌ 2B(g) is at equilibrium with Kp = 0.800. The volume is doubled instantly at constant temperature. Determine the new Qp in terms of its old equilibrium value and predict the net direction. Explain separately the changes to Q and K if reactant A is added at fixed volume, or if temperature changes.
- For a different equilibrium, A(aq) + B(aq) ⇌ AB(aq), Qc initially equals Kc. In separate trials at constant temperature, (i) half the dissolved AB is selectively removed without changing solution volume or other species amounts, and (ii) enough pure water is added to double the volume. For each, obtain the immediate Qc as a multiple of Kc and predict the later net direction. Assume dilute solutions and no competing reactions.
- A saturated solution has MX(s) ⇌ M+(aq) + X−(aq). In separate trials, pure MX solid is added or half the existing solid is removed. A macroscopic amount of the same solid phase remains in both trials, and dissolved concentrations, temperature and solution volume are unchanged by the manipulation. Predict whether either manipulation changes Qsp or the saturated dissolved composition. Why does the condition that solid remains matter?
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Problem 12
At a stated temperature, MX2(s) ⇌ M2+(aq) + 2X−(aq) has Ksp = 3.20 × 10−11. Find its molar solubility in pure water, ignoring side reactions. Then a separate salt NY(s) ⇌ N+ + Y− has molar solubility 3.00 × 10−5 M; calculate its Ksp and compare the two salts’ molar solubilities. Explain why sorting Ksp values alone is insufficient across different formulas.
For a third salt described by the same course model, explain what Ksp > 1 implies for the qualitative soluble/insoluble classification, and identify a limitation of concentration-based predictions when the resulting solution is concentrated.
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Problem 13
Equal 50.0 mL volumes of 2.00 × 10−4 M M+ and 4.00 × 10−4 M X− solutions are mixed. For MX(s) ⇌ M+ + X−, Ksp = 1.00 × 10−8. Assuming additive volumes, predict whether precipitation begins and justify using Qsp. If a student uses the concentrations before mixing, what error does that produce?
A separate saturated pure-water sample of MX at the same temperature contains undissolved solid and measured [M+] = [X−] = 1.00 × 10−4 M. Determine its molar solubility and infer Ksp from this measurement.
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Problem 14
MX(s) ⇌ M+(aq) + X−(aq) has Ksp = 4.00 × 10−10 at 25 °C. Compare its molar solubility in pure water with that in 0.0200 M NaX. Show and check an approximation. Predict how replacing water with NaX solution would change the measured dissolved M+ concentration in a controlled solubility investigation, and whether Ksp itself changes.
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