AP · thinka 原創模擬試題

2024 AP AP Chemistry 模擬試題連答案詳解

Thinka May 2024 AP-Style Mock — AP Chemistry

46 分105 分鐘2024

An original Thinka practice paper modelled on the structure and difficulty of the May 2024 AP AP Chemistry paper. Not affiliated with or reproduced from AP.

部分 II: Long Free-Response Questions

Answer Questions 1–3. Each question is worth 10 points. Recommended time: approximately 23 minutes per question.

3 題目 · 30 分

題目 1 · long_free_response

10 分

A student investigates the chemical properties and thermochemistry of glycolic acid, $\text{HOCH}_2\text{COOH}$ (molar mass $76.05\text{ g/mol}$), which is a monoprotic carboxylic acid commonly used in skincare products.

$$\text{HOCH}_2\text{COOH}(aq) + \text{KOH}(aq) \rightarrow \text{KOCH}_2\text{COO}(aq) + \text{H}_2\text{O}(l)$$

(a) The structural formula of glycolic acid is shown below:

```
H O
| ||
H--C---C--O--H
|
O--H
```

Identify the hydrogen atom that is removed when glycolic acid reacts with potassium hydroxide, $\text{KOH}$, by specifying whether it is bonded to the carbon atom, the alcohol oxygen atom, or the carboxyl oxygen atom.

(b) The student prepares a standard solution of potassium hydroxide by dissolving $7.01\text{ g}$ of solid $\text{KOH}$ (molar mass $56.11\text{ g/mol}$) in distilled water to make a total volume of $250.0\text{ mL}$. Calculate the molarity of the $\text{KOH}$ solution.

(c) The student titrates a $20.0\text{ mL}$ sample of a $0.100\text{ M}$ glycolic acid solution with the $\text{KOH}$ solution prepared in part (b). The equivalence point is reached when $4.00\text{ mL}$ of $\text{KOH}(aq)$ is added. At a volume of $2.00\text{ mL}$ of $\text{KOH}(aq)$ added, the measured $\text{pH}$ of the mixture is $3.83$.
(i) State the value of $\text{p}K_a$ for glycolic acid based on this titration.
(ii) Calculate the acid dissociation constant, $K_a$, of glycolic acid.

(d) In another container, the student mixes $30.0\text{ mL}$ of $0.100\text{ M}$ glycolic acid with $10.0\text{ mL}$ of the $0.500\text{ M KOH}$ solution. Does the resulting mixture have a $\text{pH}$ that is less than, equal to, or greater than the $\text{p}K_a$ of glycolic acid? Justify your answer by comparing the concentrations of glycolic acid and glycolate ion present in the solution.

(e) In a separate experiment, the student determines the molar enthalpy of neutralization for the reaction between glycolic acid and potassium hydroxide. The student mixes $50.0\text{ mL}$ of $0.800\text{ M}$ glycolic acid at $21.5^\circ\text{C}$ with $50.0\text{ mL}$ of $0.800\text{ M KOH}$ at $21.5^\circ\text{C}$ in an insulated coffee-cup calorimeter. The maximum temperature reached by the mixture is $26.9^\circ\text{C}$. Assume that the total mass of the mixture is $100.0\text{ g}$ and the specific heat capacity of the mixture is $4.18\text{ J}/(\text{g}\cdot^\circ\text{C})$.
(i) Calculate the quantity of heat, $q$, absorbed by the solution, in joules.
(ii) Calculate the molar enthalpy of neutralization, $\Delta H_{\text{rxn}}$, in $\text{kJ}/\text{mol}_{\text{rxn}}$. Include the algebraic sign in your answer.
(iii) The student did not replace the calorimeter lid quickly enough after mixing, allowing a small amount of heat to escape to the surroundings before the maximum temperature was recorded. Explain how this experimental error affects the experimentally determined value of the magnitude of $\Delta H_{\text{rxn}}$.

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解題

(a) The hydrogen atom bonded to the carboxyl oxygen atom (the $-\text{COOH}$ group) is the most acidic proton due to resonance stabilization of the conjugate base (carboxylate anion).

(b) Moles of $\text{KOH} = \frac{7.01\text{ g}}{56.11\text{ g/mol}} = 0.1249\text{ mol}$
Volume of solution $= 250.0\text{ mL} = 0.2500\text{ L}$
$\text{Molarity} = \frac{0.1249\text{ mol}}{0.2500\text{ L}} = 0.500\text{ M}$

(c)(i) At the half-equivalence point ($2.00\text{ mL}$ added, which is half of $4.00\text{ mL}$), $[\text{HA}] = [\text{A}^-]$, so $\text{pH} = \text{p}K_a$.
Therefore, $\text{p}K_a = 3.83$.

(c)(ii) $K_a = 10^{-\text{p}K_a} = 10^{-3.83} = 1.48 \times 10^{-4}$

(d) Initial moles of glycolic acid $= 0.0300\text{ L} \times 0.100\text{ M} = 0.00300\text{ mol}$
Moles of $\text{OH}^-$ added $= 0.0100\text{ L} \times 0.500\text{ M} = 0.00500\text{ mol}$
Because the moles of strong base ($0.00500\text{ mol}$) exceed the initial moles of weak acid ($0.00300\text{ mol}$), all the acid is converted to conjugate base ($0.00300\text{ mol}$) and excess $\text{OH}^-$ ($0.00200\text{ mol}$) remains. Thus, the solution is basic with a $\text{pH} > 7$, which is significantly greater than $\text{p}K_a = 3.83$.
Alternatively, past the half-equivalence point / equivalence point, $[\text{A}^-] > [\text{HA}]$, so by $\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$, $\text{pH} > \text{p}K_a$.

(e)(i) $\Delta T = 26.9^\circ\text{C} - 21.5^\circ\text{C} = 5.4^\circ\text{C}$
$q = mc\Delta T = (100.0\text{ g})(4.18\text{ J}/(\text{g}\cdot^\circ\text{C}))(5.4^\circ\text{C}) = 2257.2\text{ J} \approx 2.3 \times 10^3\text{ J}$ (or $2260\text{ J}$ to 2 significant figures based on $\Delta T$).

(e)(ii) Moles of glycolic acid $= 0.0500\text{ L} \times 0.800\text{ M} = 0.0400\text{ mol}$
Moles of $\text{KOH} = 0.0500\text{ L} \times 0.800\text{ M} = 0.0400\text{ mol}$
$\text{Moles of reaction} = 0.0400\text{ mol}_{\text{rxn}}$
$q_{\text{rxn}} = -q_{\text{soln}} = -2257.2\text{ J} = -2.2572\text{ kJ}$
$\Delta H_{\text{rxn}} = \frac{-2.2572\text{ kJ}}{0.0400\text{ mol}_{\text{rxn}}} = -56.4\text{ kJ}/\text{mol}_{\text{rxn}}$ (acceptable range $-56$ to $-57\text{ kJ}/\text{mol}_{\text{rxn}}$).

(e)(iii) If heat is lost to the surroundings, the measured maximum temperature will be lower than expected, resulting in an experimentally determined $\Delta T$ that is too small. A smaller $\Delta T$ produces a smaller calculated value of $q_{\text{soln}}$, which makes the calculated magnitude of $\Delta H_{\text{rxn}}$ smaller than the true magnitude.

評分準則

(a) [1 point] For identifying the hydrogen atom attached to the carboxyl group / carboxyl oxygen atom.

(b) [1 point] For the correct calculated molarity with appropriate work:
$\text{Molarity} = \frac{7.01\text{ g} / 56.11\text{ g/mol}}{0.2500\text{ L}} = 0.500\text{ M}$

(c)(i) [1 point] For stating $\text{p}K_a = 3.83$ (the pH at the half-equivalence point).

(c)(ii) [1 point] For the correct calculated value of $K_a$:
$K_a = 10^{-3.83} = 1.5 \times 10^{-4}$ (accept $1.48 \times 10^{-4}$).

(d) [1 point] For predicting $\text{pH} > \text{p}K_a$ with a valid justification based on moles of $\text{OH}^-$ exceeding weak acid or comparing relative amounts of acid and conjugate base.

(e)(i) [1 point] For the correct calculation of heat $q$:
$q = (100.0\text{ g})(4.18\text{ J}/(\text{g}\cdot^\circ\text{C}))(5.4^\circ\text{C}) = 2300\text{ J}$ (or $2260\text{ J}$).

(e)(ii) [2 points]
- 1 point for the correct moles of reaction ($0.0400\text{ mol}$).
- 1 point for the correct numerical value and negative sign of $\Delta H_{\text{rxn}}$ ($-56.4\text{ kJ}/\text{mol}_{\text{rxn}}$).

(e)(iii) [2 points]
- 1 point for stating that the magnitude of $\Delta H_{\text{rxn}}$ will be smaller.
- 1 point for linking the error to a smaller measured $\Delta T$ (or lower peak temperature).

題目 2 · long_free_response

10 分

Solid magnesium carbonate reacts with aqueous hydrochloric acid according to the following balanced chemical equation:

$$\text{MgCO}_3(s) + 2\,\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$$

(a) In an experiment, a student mixes excess $\text{MgCO}_3(s)$ with $50.0\text{ mL}$ of $0.600\text{ M HCl}(aq)$ at $25.0^\circ\text{C}$. The reaction goes to completion.
(i) Calculate the theoretical number of moles of $\text{CO}_2(g)$ produced in this reaction.
(ii) The $\text{CO}_2(g)$ generated is collected over water in a gas collection tube at a barometric pressure of $758\text{ torr}$ and a temperature of $25.0^\circ\text{C}$. The vapor pressure of water at $25.0^\circ\text{C}$ is $23.8\text{ torr}$. Calculate the partial pressure of $\text{CO}_2(g)$, in $\text{atm}$.
(iii) Calculate the volume of dry $\text{CO}_2(g)$, in liters, that the gas would occupy under these conditions ($T = 25.0^\circ\text{C}$ and the partial pressure calculated in part (a)(ii)).

(b) The student conducts two separate trials of the reaction to observe the rate of gas evolution. In Trial 1, large chunks of $\text{MgCO}_3(s)$ are used. In Trial 2, the same mass of $\text{MgCO}_3(s)$ is ground into a fine powder before reacting with an identical sample of $\text{HCl}(aq)$.
(i) State whether the initial rate of $\text{CO}_2(g)$ production in Trial 2 is greater than, less than, or equal to the initial rate in Trial 1.
(ii) Justify your answer to part (b)(i) in terms of collision theory and particle surface area.

(c) For the reaction represented above, $\Delta S^\circ > 0$. Explain in terms of particle arrangements and microstates why the standard entropy change for this reaction is positive.

(d) The reaction is known to be slightly endothermic ($\Delta H^\circ > 0$). A student hypothesizes that the reaction will become non-spontaneous at sufficiently low temperatures. Do you agree or disagree with the student's hypothesis? Justify your answer using the relationship $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

(e) In an aqueous saturated solution of magnesium carbonate, the concentration of $\text{Mg}^{2+}(aq)$ at equilibrium is $1.8 \times 10^{-3}\text{ M}$ at $25^\circ\text{C}$.
$$\text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)$$
Calculate the value of the solubility product constant, $K_{\text{sp}}$, for $\text{MgCO}_3$ at $25^\circ\text{C}$.

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解題

(a)(i) Moles of $\text{HCl} = 0.0500\text{ L} \times 0.600\text{ M} = 0.0300\text{ mol}$
From stoichiometry, $2\text{ mol HCl}$ produce $1\text{ mol CO}_2$:
$\text{Moles of CO}_2 = 0.0300\text{ mol HCl} \times \frac{1\text{ mol CO}_2}{2\text{ mol HCl}} = 0.0150\text{ mol CO}_2$

(a)(ii) By Dalton's law of partial pressures:
$P_{\text{CO}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 758\text{ torr} - 23.8\text{ torr} = 734.2\text{ torr}$
Converting to atmospheres:
$P_{\text{CO}_2} = \frac{734.2\text{ torr}}{760\text{ torr/atm}} = 0.966\text{ atm}$

(a)(iii) Using the ideal gas law $PV = nRT$:
$T = 25.0^\circ\text{C} + 273.15 = 298.15\text{ K}$
$V = \frac{nRT}{P} = \frac{(0.0150\text{ mol})(0.08206\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(298.15\text{ K})}{0.966\text{ atm}} = 0.380\text{ L}$

(b)(i) The initial rate of $\text{CO}_2(g)$ production in Trial 2 is greater than in Trial 1.

(b)(ii) Grinding the solid into a fine powder significantly increases the accessible surface area of the reactant. With more solid surface exposed to the aqueous $\text{H}^+$ ions, the frequency of collisions between reactant particles increases, leading to a higher number of effective collisions per unit time and thus a faster reaction rate.

(c) The reaction converts a solid and aqueous ions into one mole of gas ($\text{CO}_2$). Gas molecules have much higher positional dispersal and translational entropy (occupy far more possible microstates) than particles in the condensed phases (solid and aqueous), resulting in a net increase in disorder and $\Delta S^\circ > 0$.

(d) Agree. Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, with both $\Delta H^\circ > 0$ and $\Delta S^\circ > 0$, the $-T\Delta S^\circ$ term is negative. At high temperatures, $T\Delta S^\circ > \Delta H^\circ$, making $\Delta G^\circ < 0$ (thermodynamically favorable). However, at sufficiently low temperatures, $T\Delta S^\circ < \Delta H^\circ$, which causes $\Delta G^\circ > 0$, making the reaction non-spontaneous / thermodynamically unfavorable.

(e) For $\text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)$:
$[\text{Mg}^{2+}] = [\text{CO}_3^{2-}] = 1.8 \times 10^{-3}\text{ M}$
$K_{\text{sp}} = [\text{Mg}^{2+}][\text{CO}_3^{2-}] = (1.8 \times 10^{-3})^2 = 3.24 \times 10^{-6} \approx 3.2 \times 10^{-6}$

評分準則

(a)(i) [1 point] For the correct calculation of moles of $\text{CO}_2$:
$n = 0.0500\text{ L} \times 0.600\text{ M} \times \frac{1}{2} = 0.0150\text{ mol}$

(a)(ii) [1 point] For the correct partial pressure of $\text{CO}_2$ in atm:
$P = \frac{758 - 23.8}{760} = 0.966\text{ atm}$

(a)(iii) [1 point] For the correct calculation of volume using ideal gas law:
$V = \frac{(0.0150)(0.08206)(298.15)}{0.966} = 0.380\text{ L}$

(b)(i) [1 point] For stating that the rate is greater in Trial 2.

(b)(ii) [1 point] For a valid justification referencing increased surface area leading to increased frequency of collisions between reactant particles.

(c) [1 point] For a valid explanation stating that gas particles are formed, which are more dispersed / occupy more microstates than solid and aqueous reactants.

(d) [2 points]
- 1 point for agreeing with the hypothesis.
- 1 point for justifying that at low temperatures, the positive $\Delta H^\circ$ term outweighs the $T\Delta S^\circ$ term, making $\Delta G^\circ > 0$.

(e) [2 points]
- 1 point for the correct expression $K_{\text{sp}} = [\text{Mg}^{2+}][\text{CO}_3^{2-}]$.
- 1 point for the correct numerical calculation $K_{\text{sp}} = (1.8 \times 10^{-3})^2 = 3.2 \times 10^{-6}$.

題目 3 · long_free_response

10 分

Bronze is an alloy consisting primarily of copper ($\text{Cu}$) with approximately $12\%$ tin ($\text{Sn}$) by mass. An antique bronze artifact is suspected of having a surface corrosion layer of copper(I) sulfide, $\text{Cu}_2\text{S}$.

(a) Determine the oxidation number of copper in:
(i) $\text{Cu}(s)$
(ii) $\text{Cu}_2\text{S}(s)$

(b) The atomic radius of copper is $145\text{ pm}$ and that of tin is $140\text{ pm}$.
(i) Explain why bronze is classified as a substitutional alloy rather than an interstitial alloy.
(ii) Copper (atomic number 29) and zinc (atomic number 30) are adjacent in Period 4 of the periodic table. The atomic radius of copper is $145\text{ pm}$, while that of zinc is $135\text{ pm}$. In terms of atomic structure and Coulomb's law, explain why zinc has a smaller atomic radius than copper.

(c) To analyze the tarnish layer, the $\text{Cu}_2\text{S}$ corrosion (molar mass $159.16\text{ g/mol}$) is removed chemically from the bronze artifact. The mass of the artifact before tarnish removal was $215.48\text{ g}$, and after removal it was $209.12\text{ g}$. Assuming only $\text{Cu}_2\text{S}$ was removed, calculate the number of moles of copper atoms removed from the artifact.

(d) To protect the cleaned bronze artifact from future corrosion, it is coated with a thin layer of nickel by electroplating it in an acidic bath of nickel(II) sulfate, $\text{NiSO}_4(aq)$. Oxygen gas is evolved at the anode.

The standard reduction potentials are given in the table below:

| Half-Reaction | $E^\circ\text{ (V)}$ |
|---|---|
| $\text{Ni}^{2+}(aq) + 2\,e^- \rightarrow \text{Ni}(s)$ | $-0.26$ |
| $\text{O}_2(g) + 4\,\text{H}^+(aq) + 4\,e^- \rightarrow 2\,\text{H}_2\text{O}(l)$ | $+1.23$ |

(i) Write the balanced net ionic equation for the overall electroplating reaction.
(ii) Calculate the standard cell potential, $E^\circ_{\text{cell}}$, for the reaction in part (d)(i).
(iii) Based on your answer to part (d)(ii), explain why an external direct current power supply must be used to carry out this electroplating process.

(e) A constant current of $3.50\text{ A}$ is passed through the electroplating cell to deposit $1.75\text{ g}$ of solid nickel (molar mass $58.69\text{ g/mol}$) onto the bronze artifact. Calculate the time, in seconds, required for this process.

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解題

(a)(i) In elemental copper, $\text{Cu}(s)$, the oxidation number is $0$.
(a)(ii) In $\text{Cu}_2\text{S}(s)$, sulfur has an oxidation number of $-2$, so each copper atom has an oxidation number of $+1$.

(b)(i) Copper and tin have comparable atomic radii ($145\text{ pm}$ and $140\text{ pm}$, a difference of less than $5\%$). In a substitutional alloy, solute atoms of similar size replace host atoms in the crystal lattice. An interstitial alloy requires solute atoms with significantly smaller radii to fit into the spaces (interstices) between host atoms.

(b)(ii) Zinc ($Z = 30$) has one more proton in its nucleus than copper ($Z = 29$), resulting in a greater effective nuclear charge ($Z_{\text{eff}}$). Because the valence electrons of both atoms occupy the same outermost energy level ($n = 4$) and experience similar inner-shell shielding, the valence electrons in zinc experience a stronger Coulombic force of attraction toward the nucleus, drawing the electron cloud inward and resulting in a smaller atomic radius.

(c) Mass of $\text{Cu}_2\text{S}$ removed $= 215.48\text{ g} - 209.12\text{ g} = 6.36\text{ g}$
$\text{Moles of }\text{Cu}_2\text{S} = \frac{6.36\text{ g}}{159.16\text{ g/mol}} = 0.03996\text{ mol}$
Each mole of $\text{Cu}_2\text{S}$ contains $2\text{ mol}$ of $\text{Cu}$ atoms:
$\text{Moles of Cu} = 0.03996\text{ mol} \times 2 = 0.0799\text{ mol Cu}$

(d)(i) Reduction at cathode: $2\,[\text{Ni}^{2+}(aq) + 2\,e^- \rightarrow \text{Ni}(s)]$
Oxidation at anode: $2\,\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\,\text{H}^+(aq) + 4\,e^-$
Overall balanced net ionic equation:
$$2\,\text{Ni}^{2+}(aq) + 2\,\text{H}_2\text{O}(l) \rightarrow 2\,\text{Ni}(s) + \text{O}_2(g) + 4\,\text{H}^+(aq)$$

(d)(ii) $E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{cathode}) - E^\circ_{\text{red}}(\text{anode}) = -0.26\text{ V} - (+1.23\text{ V}) = -1.49\text{ V}$

(d)(iii) The standard cell potential $E^\circ_{\text{cell}}$ is negative ($-1.49\text{ V}$), which means $\Delta G^\circ = -nFE^\circ > 0$. The reaction is thermodynamically unfavorable under standard conditions, so an external power source is necessary to supply electrical energy to drive the non-spontaneous electrolytic process.

(e) Moles of $\text{Ni} = \frac{1.75\text{ g}}{58.69\text{ g/mol}} = 0.02982\text{ mol Ni}$
Moles of electrons needed $= 0.02982\text{ mol Ni} \times \frac{2\text{ mol }e^-}{1\text{ mol Ni}} = 0.05964\text{ mol }e^-$
Total charge $q = 0.05964\text{ mol }e^- \times 96,485\text{ C/mol }e^- = 5754\text{ C}$
Since $I = \frac{q}{t}$:
$t = \frac{q}{I} = \frac{5754\text{ C}}{3.50\text{ C/s}} = 1644\text{ s} \approx 1.64 \times 10^3\text{ s}$ (or $1640\text{ s}$ to 3 significant figures).

評分準則

(a) [1 point] For both correct oxidation numbers: $0$ for $\text{Cu}(s)$ and $+1$ for $\text{Cu}$ in $\text{Cu}_2\text{S}(s)$.

(b)(i) [1 point] For explaining that copper and tin atoms have similar atomic radii, allowing tin atoms to replace/substitute copper atoms in the lattice.

(b)(ii) [1 point] For a valid explanation using Coulomb's law and atomic structure (zinc has greater nuclear charge / more protons with similar shielding in $n = 4$, resulting in stronger attraction on valence electrons).

(c) [2 points]
- 1 point for calculating the mass of $\text{Cu}_2\text{S}$ lost ($6.36\text{ g}$) and converting to moles of $\text{Cu}_2\text{S}$ ($0.0400\text{ mol}$).
- 1 point for multiplying by $2$ to get $0.0799\text{ mol Cu}$.

(d)(i) [1 point] For the correct balanced net ionic equation:
$2\,\text{Ni}^{2+}(aq) + 2\,\text{H}_2\text{O}(l) \rightarrow 2\,\text{Ni}(s) + \text{O}_2(g) + 4\,\text{H}^+(aq)$

(d)(ii) [1 point] For the correct value of $E^\circ_{\text{cell}} = -1.49\text{ V}$.

(d)(iii) [1 point] For explaining that a negative $E^\circ_{\text{cell}}$ corresponds to a thermodynamically unfavorable reaction ($\Delta G^\circ > 0$), requiring continuous input of electrical energy.

(e) [2 points]
- 1 point for calculating the moles of electrons transferred ($0.0596\text{ mol }e^-$).
- 1 point for the correct calculated time with appropriate units and significant figures ($1640\text{ s}$ or $1.64 \times 10^3\text{ s}$).

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部分 II: Short Free-Response Questions

Answer Questions 4–7. Each question is worth 4 points. Recommended time: approximately 9 minutes per question.

4 題目 · 16 分

題目 1 · Free-Response

4 分

A student investigates the dissolution of solid ammonium nitrate, $\text{NH}_4\text{NO}_3(s)$ (molar mass $80.04\text{ g/mol}$), in water using an insulated coffee-cup calorimeter.

A sample of $4.00\text{ g}$ of $\text{NH}_4\text{NO}_3(s)$ at $22.0^\circ\text{C}$ is added to $96.0\text{ g}$ of distilled water at $22.0^\circ\text{C}$. After continuous stirring, all solid dissolves and the lowest temperature recorded by the thermometer is $18.8^\circ\text{C}$. Assume that the specific heat capacity of the resulting solution is $4.18\text{ J}/(\text{g}\cdot^\circ\text{C})$ and that the mass of the solution is the sum of the masses of the solute and water.

(a) Calculate the quantity of heat absorbed by the solution, in Joules.

(b) Calculate the molar enthalpy of dissolution, $\Delta H^\circ_{\text{soln}}$, for $\text{NH}_4\text{NO}_3$ in $\text{kJ/mol}_{\text{rxn}}$. Include the appropriate algebraic sign.

(c) The dissolution of ammonium nitrate in water is thermodynamically favorable at room temperature. Using particle-level reasoning, explain why the standard entropy of the system increases ($\Delta S^\circ > 0$) during the dissolution process.

(d) If the calorimeter used by the student absorbed heat from the warmer surrounding room air during the experiment, would the experimental value of $\Delta H^\circ_{\text{soln}}$ calculated from the data be greater in magnitude, less in magnitude, or equal in magnitude to the actual value? Justify your answer.

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解題

(a)
$m_{\text{total}} = 4.00\text{ g} + 96.0\text{ g} = 100.0\text{ g}$
$\Delta T = T_f - T_i = 18.8^\circ\text{C} - 22.0^\circ\text{C} = -3.2^\circ\text{C}$
$q_{\text{soln}} = m c \Delta T = (100.0\text{ g})(4.18\text{ J}/(\text{g}\cdot^\circ\text{C}))(-3.2^\circ\text{C}) = -1337.6\text{ J} = -1340\text{ J}$ (or heat absorbed by solution / heat of reaction $q_{\text{rxn}} = +1340\text{ J}$).

(b)
$n_{\text{NH}_4\text{NO}_3} = \frac{4.00\text{ g}}{80.04\text{ g/mol}} = 0.04998\text{ mol}$
$q_{\text{rxn}} = -q_{\text{soln}} = +1337.6\text{ J} = +1.3376\text{ kJ}$
$\Delta H^\circ_{\text{soln}} = \frac{+1.3376\text{ kJ}}{0.04998\text{ mol}} = +26.8\text{ kJ/mol}_{\text{rxn}}$

(c)
In solid $\text{NH}_4\text{NO}_3$, the ions are locked into fixed positions within the crystal lattice. When dissolved, the ions ($\text{NH}_4^+$ and $\text{NO}_3^-$) become separated and hydrated, gaining greater freedom of motion and occupying many more possible spatial configurations (microstates), resulting in a positive entropy change ($\Delta S^\circ > 0$).

(d)
Less in magnitude. If heat enters the calorimeter from the surrounding air, the final recorded temperature will be higher than the true minimum temperature. This makes the measured temperature change ($|\Delta T|$) smaller than it should be, resulting in a calculated $q_{\text{rxn}}$ and $|\Delta H^\circ_{\text{soln}}|$ that are smaller (less in magnitude) than the true value.

評分準則

(a) [1 pt] for calculating the correct quantity of heat transferred (1340 J or 1.34 kJ).

(b) [1 pt] for calculating the correct molar enthalpy of dissolution with the correct positive sign (+26.8 kJ/mol_rxn).

(c) [1 pt] for a valid particle-level explanation of the increase in entropy (e.g., transition from an ordered solid lattice to mobile, dispersed aqueous ions with more microstates).

(d) [1 pt] for choosing 'less in magnitude' with a correct justification relating heat gain from the surroundings to a smaller observed temperature drop (smaller |ΔT|).

題目 2 · Free-Response

4 分

Consider the following reversible reaction for the decomposition of dinitrogen tetroxide gas:

\[\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g) \quad \Delta H^\circ = +57.2\text{ kJ/mol}_{\text{rxn}}\]

(a) Write the equilibrium expression, $K_p$, in terms of partial pressures for this reaction.

(b) At a certain temperature $T_1$, a reaction mixture at equilibrium in a closed container has partial pressures $P_{\text{N}_2\text{O}_4} = 0.40\text{ atm}$ and $P_{\text{NO}_2} = 0.60\text{ atm}$. Calculate the value of the equilibrium constant, $K_p$, at this temperature.

(c) In a separate experiment at the same temperature $T_1$, a rigid vessel is filled with gases such that the initial partial pressures are $P_{\text{N}_2\text{O}_4} = 0.20\text{ atm}$ and $P_{\text{NO}_2} = 0.80\text{ atm}$.
(i) Calculate the value of the reaction quotient, $Q_p$.
(ii) Predict whether the partial pressure of $\text{NO}_2(g)$ will increase, decrease, or remain the same as the mixture moves toward equilibrium. Justify your answer by comparing $Q_p$ to $K_p$.

(d) The temperature of the equilibrium mixture is increased while the volume of the rigid vessel is held constant. State whether the value of the equilibrium constant $K_p$ will increase, decrease, or remain the same. Justify your answer.

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解題

(a)
$K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$

(b)
$K_p = \frac{(0.60)^2}{0.40} = \frac{0.36}{0.40} = 0.90$

(c)
(i) $Q_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} = \frac{(0.80)^2}{0.20} = \frac{0.64}{0.20} = 3.2$
(ii) Since $Q_p (3.2) > K_p (0.90)$, the ratio of products to reactants is greater than at equilibrium. The system will shift in the reverse direction (to the left) to reach equilibrium, converting $\text{NO}_2$ to $\text{N}_2\text{O}_4$. Therefore, $P_{\text{NO}_2}$ will decrease.

(d)
Increase. Because the forward reaction is endothermic ($\Delta H^\circ > 0$), adding thermal energy shifts the equilibrium to the right (favoring the forward endothermic reaction), which increases the equilibrium concentration/partial pressure of products and thereby increases $K_p$.

評分準則

(a) [1 pt] for the correct expression for $K_p$ using partial pressure notation.

(b) [1 pt] for the correct calculated value of $K_p$ (0.90).

(c) [1 pt] for calculating $Q_p = 3.2$ AND correctly predicting that $P_{\text{NO}_2}$ will decrease because $Q_p > K_p$.

(d) [1 pt] for stating that $K_p$ increases with a correct justification based on the endothermic nature of the reaction (Le Chatelier's principle / van 't Hoff equation).

題目 3 · Free-Response

4 分

The decomposition of sulfuryl chloride gas in a closed vessel is represented by the following equation:

\[\text{SO}_2\text{Cl}_2(g) \rightarrow \text{SO}_2(g) + \text{Cl}_2(g)\]

A chemist monitors the concentration of $\text{SO}_2\text{Cl}_2(g)$ over time at $600\text{ K}$ and generates three integrated rate plots:
- Plot 1: $[\text{SO}_2\text{Cl}_2]$ versus time yields a downward curved line.
- Plot 2: $\ln[\text{SO}_2\text{Cl}_2]$ versus time yields a straight line with a slope of $-2.2 \times 10^{-4}\text{ s}^{-1}$.
- Plot 3: $\frac{1}{[\text{SO}_2\text{Cl}_2]}$ versus time yields an upward curved line.

(a) State the order of the reaction with respect to $\text{SO}_2\text{Cl}_2$. Justify your answer using the graphical data provided.

(b) Determine the value of the rate constant, $k$, for this reaction at $600\text{ K}$. Include appropriate units.

(c) At a particular instant during the reaction, the rate of disappearance of $\text{SO}_2\text{Cl}_2(g)$ is $4.8 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}$. Determine the rate of appearance, in $\text{M}\cdot\text{s}^{-1}$, of $\text{Cl}_2(g)$ at that same instant.

(d) A student claims that doubling the initial concentration of $\text{SO}_2\text{Cl}_2$ will double the initial reaction rate. Do you agree or disagree with the student's claim? Justify your answer based on the rate law.

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解題

(a)
The reaction is first order with respect to $\text{SO}_2\text{Cl}_2$ because the plot of the natural logarithm of concentration, $\ln[\text{SO}_2\text{Cl}_2]$, versus time is a straight line.

(b)
For a first-order integrated rate law: $\ln[A]_t - \ln[A]_0 = -kt$, which corresponds to $y = mx + b$ where $\text{slope} = -k$.
$k = -\text{slope} = -(-2.2 \times 10^{-4}\text{ s}^{-1}) = 2.2 \times 10^{-4}\text{ s}^{-1}$.

(c)
According to the stoichiometric coefficients in the balanced chemical equation, $1\text{ mol}$ of $\text{SO}_2\text{Cl}_2$ decomposes to produce $1\text{ mol}$ of $\text{Cl}_2$.
$-\frac{\Delta[\text{SO}_2\text{Cl}_2]}{\Delta t} = \frac{\Delta[\text{Cl}_2]}{\Delta t} = 4.8 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}$.

(d)
Agree. The differential rate law is $\text{Rate} = k[\text{SO}_2\text{Cl}_2]$. Because the exponent is $1$, doubling the concentration of $\text{SO}_2\text{Cl}_2$ will double the rate of reaction: $k(2[\text{SO}_2\text{Cl}_2]) = 2(k[\text{SO}_2\text{Cl}_2])$.

評分準則

(a) [1 pt] for identifying the reaction as first order AND explaining that the plot of $\ln[\text{SO}_2\text{Cl}_2]$ versus time is linear.

(b) [1 pt] for the correct numerical value and correct units for the rate constant ($2.2 \times 10^{-4}\text{ s}^{-1}$).

(c) [1 pt] for calculating the correct rate of appearance of $\text{Cl}_2$ ($4.8 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}$) based on the 1:1 stoichiometry.

(d) [1 pt] for agreeing with the claim AND providing a correct justification using the first-order rate law.

題目 4 · Free-Response

4 分

A student uses spectrophotometry to determine the concentration of the iron(III) thiocyanate complex ion, $[\text{Fe(SCN)}]^{2+}(aq)$, in an unknown solution.

(a) An absorbance spectrum of $[\text{Fe(SCN)}]^{2+}(aq)$ shows a maximum absorbance peak at a wavelength of $\lambda_{\max} = 447\text{ nm}$. Explain why quantitative spectrophotometric measurements of unknown solution concentrations should be performed at or near $\lambda_{\max}$.

(b) Using a spectrophotometer set to $447\text{ nm}$ and a standard cuvette with a path length of $1.00\text{ cm}$, the student determines the molar absorptivity, $\varepsilon$, of $[\text{Fe(SCN)}]^{2+}$ to be $4.50 \times 10^3\text{ M}^{-1}\cdot\text{cm}^{-1}$. Calculate the molar concentration of $[\text{Fe(SCN)}]^{2+}$ in an unknown sample that has an absorbance of $0.675$.

(c) To prepare a diluted standard solution, the student uses a pipet to transfer $5.00\text{ mL}$ of a $2.00 \times 10^{-3}\text{ M}$ stock solution into a $50.00\text{ mL}$ volumetric flask and adds distilled water to the calibration mark. Calculate the molarity of the diluted standard solution.

(d) Prior to placing the unknown sample in the spectrophotometer, the student rinses the cuvette with distilled water but fails to dry the inside before filling it with the unknown sample. State whether the calculated concentration of $[\text{Fe(SCN)}]^{2+}$ will be too high, too low, or unaffected. Justify your answer.

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解題

(a)
Measuring at $\lambda_{\max}$ ensures maximum sensitivity because the molar absorptivity is greatest at this wavelength, meaning small differences in concentration produce the largest possible changes in absorbance. This reduces relative experimental error in absorbance measurements.

(b)
Using the Beer-Lambert Law, $A = \varepsilon b c$:
\[c = \frac{A}{\varepsilon b} = \frac{0.675}{(4.50 \times 10^3\text{ M}^{-1}\cdot\text{cm}^{-1})(1.00\text{ cm})} = 1.50 \times 10^{-4}\text{ M}\]

(c)
Using the dilution equation $M_1 V_1 = M_2 V_2$:
\[M_2 = \frac{M_1 V_1}{V_2} = \frac{(2.00 \times 10^{-3}\text{ M})(5.00\text{ mL})}{50.00\text{ mL}} = 2.00 \times 10^{-4}\text{ M}\]

(d)
Too low. The water droplets remaining inside the cuvette dilute the solution when the sample is added. This lower concentration reduces the measured absorbance $A$. Since $c = \frac{A}{\varepsilon b}$, a lower measured absorbance results in an experimentally calculated concentration that is lower (too low) than the actual value.

評分準則

(a) [1 pt] for explaining that measuring at $\lambda_{\max}$ maximizes absorbance/sensitivity and minimizes measurement error.

(b) [1 pt] for calculating the correct concentration ($1.50 \times 10^{-4}\text{ M}$) using Beer's Law.

(c) [1 pt] for calculating the correct diluted concentration ($2.00 \times 10^{-4}\text{ M}$).

(d) [1 pt] for predicting 'too low' AND justifying that residual water dilutes the solution, decreasing measured absorbance and the calculated concentration.

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