2023 AP AP Chemistry Practice Paper with Answers

(a)(i) Ground-state electron configuration of neutral cobalt (\(Z = 27\)):
\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\) or \([\text{Ar}]\,4s^2 3d^7\) (or \([\text{Ar}]\,3d^7 4s^2\)).

(a)(ii) Transition metals lose their outermost valence electrons first upon ionization. Therefore, electrons are removed from the \(4s\) subshell first.

(b) Mass of \(\text{I}\) in product = (Mass of crucible and product) \(-\) (Mass of crucible and \(\text{Co}\))
\[ \text{Mass of I} = 34.704\text{ g} - 29.628\text{ g} = 5.076\text{ g I} \]

(c) Moles of \(\text{I}\) atoms:
\[ n_{\text{I}} = 5.076\text{ g I} \times \frac{1\text{ mol I}}{126.90\text{ g I}} = 0.04000\text{ mol I} \]

(d) Mole ratio of \(\text{I}\) to \(\text{Co}\):
\[ \frac{0.04000\text{ mol I}}{0.0200\text{ mol Co}} = \frac{2.00\text{ mol I}}{1.00\text{ mol Co}} \]
Empirical formula: \(\text{CoI}_2\).

(e) Less than. If solid product splatters out of the crucible, the final recorded mass of crucible and product will be too low. Since the mass of iodine is calculated by subtracting the initial mass of cobalt and crucible from the final mass, the calculated mass (and therefore calculated moles) of iodine will be less than the actual value.

(f)(i) The spontaneous reaction pairs the oxidation of \(\text{Co}\) with the reduction of \(\text{Ag}^+\):
Anode (oxidation): \(\text{Co}(s) \rightarrow \text{Co}^{2+}(aq) + 2e^-\)
Cathode (reduction): \(2\,\text{Ag}^+(aq) + 2e^- \rightarrow 2\,\text{Ag}(s)\)
Net ionic equation: \(\text{Co}(s) + 2\,\text{Ag}^+(aq) \rightarrow \text{Co}^{2+}(aq) + 2\,\text{Ag}(s)\)

(f)(ii) Standard cell potential:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V} \]

(f)(iii) Gibbs free energy calculation:
\[ \Delta G^\circ = -nFE^\circ = -(2\text{ mol } e^-)\left(96{,}485\text{ C/mol } e^-\right)(1.08\text{ J/C}) = -208{,}408\text{ J/mol}_{\text{rxn}} = -208\text{ kJ/mol}_{\text{rxn}} \]

(f)(iv) Disagree. The battery is a closed system. Although the cobalt anode dissolves into solution as \(\text{Co}^{2+}\), silver ions in solution are simultaneously reduced and deposited onto the cathode as solid \(\text{Ag}\). No matter enters or leaves the system, so the total mass of the sealed battery remains constant according to the law of conservation of mass.

(a)(i) [1 point] For the correct ground-state electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\) or \([\text{Ar}]\,4s^2 3d^7\).

(a)(ii) [1 point] For identifying the \(4s\) subshell (must include both principal quantum number 4 and letter s).

(b) [1 point] For the correct calculated mass of iodine with appropriate units: \(34.704\text{ g} - 29.628\text{ g} = 5.076\text{ g}\).

(c) [1 point] For the correct calculated moles of iodine atoms: \(0.04000\text{ mol I}\) (accept \(0.0400\text{ mol}\)).

(d) [1 point] For the correct empirical formula consistent with part (c): \(\text{CoI}_2\) with supporting whole-number mole ratio calculation shown.

(e) [1 point] For choosing 'less than' AND providing a valid justification connecting the loss of mass from splattering to a decreased calculated mass/moles of iodine.

(f)(i) [1 point] For the correct balanced net ionic equation: \(\text{Co}(s) + 2\,\text{Ag}^+(aq) \rightarrow \text{Co}^{2+}(aq) + 2\,\text{Ag}(s)\) (states of matter not required).

(f)(ii) [1 point] For the correct calculated value: \(E^\circ_{\text{cell}} = +1.08\text{ V}\) (or consistent with half-reactions chosen in (f)(i)).

(f)(iii) [1 point] For the correct calculated value of \(\Delta G^\circ\) in \(\text{kJ/mol}_{\text{rxn}}\) including correct algebraic negative sign: \(-208\text{ kJ/mol}_{\text{rxn}}\) (using \(n = 2\)).

(f)(iv) [1 point] For disagreeing and providing a valid justification (e.g., stating that the battery is a closed/sealed system so total mass is conserved, or noting that mass lost at the anode is offset by mass gained at the cathode).

(a) Stoichiometric conversion from moles of \(\text{PCl}_3\) to grams of \(\text{Cl}\):
\[ 0.850\text{ mol PCl}_3 \times \frac{3\text{ mol Cl}}{1\text{ mol PCl}_3} \times \frac{35.45\text{ g Cl}}{1\text{ mol Cl}} = 90.3975\text{ g Cl} \approx 90.4\text{ g Cl} \]

(b) Applying Hess's law to obtain Reaction 1:
1. Reverse Reaction 2 and divide by 4:
\[ \text{PCl}_3(g) \rightarrow \frac{1}{4}\,\text{P}_4(s) + \frac{3}{2}\,\text{Cl}_2(g) \quad \Delta H = -\frac{1}{4}(-1148\text{ kJ/mol}) = +287\text{ kJ/mol} \]
2. Add Reaction 3:
\[ \frac{1}{4}\,\text{P}_4(s) \rightarrow \text{P}(g) \quad \Delta H = +316\text{ kJ/mol} \]
3. Multiply Reaction 4 by \(\frac{3}{2}\):
\[ \frac{3}{2}\,\text{Cl}_2(g) \rightarrow 3\,\text{Cl}(g) \quad \Delta H = \frac{3}{2}(+243\text{ kJ/mol}) = +364.5\text{ kJ/mol} \]

Adding these three manipulated equations gives Reaction 1:
\[ \Delta H_1^\circ = +287 + 316 + 364.5 = +967.5\text{ kJ/mol}_{\text{rxn}} \approx +968\text{ kJ/mol}_{\text{rxn}} \]

(c)(i)
- Internuclear distance (x-axis): The minimum of the \(\text{P}-\text{Cl}\) potential energy curve is located further to the right (at \(204\text{ pm}\) compared to \(200\text{ pm}\) for \(\text{Cl}_2\)).
- Depth of well (y-axis): The minimum of the \(\text{P}-\text{Cl}\) curve is deeper / lower on the y-axis (at \(-326\text{ kJ/mol}\) compared to \(-243\text{ kJ/mol}\) for \(\text{Cl}_2\)), reflecting a stronger bond.

(d)(i) Structure B can be eliminated. In Structure B, the central phosphorus atom has 3 single bonds and 0 lone pairs, giving 3 electron domains. According to VSEPR theory, 3 bonding domains adopt a trigonal planar geometry with \(120^\circ\) bond angles, which is inconsistent with the observed trigonal pyramidal geometry.

(d)(ii) Structure A is the best representation. Formal charges in Structure A:
- \(\text{FC}(\text{P}) = 5 - 2 - \frac{1}{2}(6) = 0\)
- \(\text{FC}(\text{Cl}) = 7 - 6 - \frac{1}{2}(2) = 0\) for each chlorine atom.
All formal charges are zero, which is the most stable arrangement.

(e) Equilibrium expression:
\[ K_p = \frac{P_{\text{PCl}_5}}{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}} \]

(f) Total moles of gas = \(3.0 + 3.0 + 6.0 = 12.0\text{ mol}\).
Mole fractions:
\[ X_{\text{PCl}_3} = \frac{3.0}{12.0} = 0.250 \]
\[ X_{\text{Cl}_2} = \frac{3.0}{12.0} = 0.250 \]
\[ X_{\text{PCl}_5} = \frac{6.0}{12.0} = 0.500 \]
Partial pressures (\(P_i = X_i \cdot P_{\text{total}}\)):
\[ P_{\text{PCl}_3} = 0.250 \times 8.00\text{ atm} = 2.00\text{ atm} \]
\[ P_{\text{Cl}_2} = 0.250 \times 8.00\text{ atm} = 2.00\text{ atm} \]
\[ P_{\text{PCl}_5} = 0.500 \times 8.00\text{ atm} = 4.00\text{ atm} \]

Calculating \(K_p\):
\[ K_p = \frac{4.00\text{ atm}}{(2.00\text{ atm})(2.00\text{ atm})} = 1.00\text{ atm}^{-1} \quad (\text{or } 1.00) \]

(a) [1 point] For the correct calculated mass of \(\text{Cl}(g)\) reported to 3 significant figures: \(90.4\text{ g}\).

(b) [2 points]
- [1 point] For correctly reversing and dividing Reaction 2 by 4 (\(+287\text{ kJ/mol}\)) OR multiplying Reaction 4 by 1.5 (\(+364.5\text{ kJ/mol}\)).
- [1 point] For the correct overall calculated value of \(\Delta H_1^\circ = +968\text{ kJ/mol}_{\text{rxn}}\) (accept 967.5 to 968).

(c)(i) [2 points]
- [1 point] For correctly stating that the minimum of the curve is located at a larger internuclear distance (\(x = 204\text{ pm}\) vs \(200\text{ pm}\)).
- [1 point] For correctly stating that the well is deeper / potential energy minimum is lower (\(y = -326\text{ kJ/mol}\) vs \(-243\text{ kJ/mol}\)).

(d)(i) [1 point] For identifying Structure B and justifying with VSEPR theory (3 electron domains / 0 lone pairs produce trigonal planar geometry).

(d)(ii) [1 point] For identifying Structure A and showing that formal charges on all atoms (P and Cl) equal zero.

(e) [1 point] For the correct \(K_p\) expression using partial pressures: \(K_p = \frac{P_{\text{PCl}_5}}{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}}\).

(f) [1 point] For the correct calculated value of \(K_p\) consistent with part (e): \(K_p = 1.00\) (or \(1.0\)).

(a) Solid \(\text{MgCO}_3\) is insoluble / undissociated, while \(\text{HNO}_3\) is a strong acid that fully dissociates into \(\text{H}^+\) and \(\text{NO}_3^-\). Nitrate is a spectator ion:
\[ \text{MgCO}_3(s) + 2\,\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \]
(or using hydronium: \(\text{MgCO}_3(s) + 2\,\text{H}_3\text{O}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{CO}_2(g) + 3\,\text{H}_2\text{O}(l)\))

(b) The data for Trial 5 are anomalous because:
- Comparing Trial 5 (small pieces, \(2.00\text{ M}\)) to Trial 2 (small pieces, \(1.00\text{ M}\)): increasing \([\text{HNO}_3]\) should decrease the reaction time (as seen from Trial 1 to Trial 4), but Trial 5 (\(195\text{ s}\)) is much longer than Trial 2 (\(88\text{ s}\)).
- Alternatively, for \(2.00\text{ M HNO}_3\), small pieces (Trial 5, \(195\text{ s}\)) should react faster than a large chunk (Trial 6, \(131\text{ s}\)), but Trial 5 has a longer reaction time than Trial 6.

(c) The fine powder in Trial 1 has a significantly greater surface area than the large chunk in Trial 3. A greater surface area exposes a larger number of \(\text{MgCO}_3\) particles at the solid-liquid interface, leading to a higher frequency of collisions between \(\text{H}^+\) ions and \(\text{MgCO}_3\) per unit time, thereby increasing the reaction rate and decreasing the time required for completion.

(d) Disagree. If the reaction were zeroth-order with respect to \(\text{HNO}_3\), changing the acid concentration would have no effect on the reaction rate or completion time for trials with identical surface area. However, comparing Trial 1 (\(1.00\text{ M}\), \(45\text{ s}\)) and Trial 4 (\(2.00\text{ M}\), \(23\text{ s}\)), doubling the concentration cuts the reaction time approximately in half (doubling the rate), proving that the rate depends on \([\text{HNO}_3]\).

(e) Initial moles of \(\text{HNO}_3\):
\[ n_{\text{initial}} = 0.0500\text{ L} \times 1.00\text{ M} = 0.0500\text{ mol HNO}_3 \]
Moles of \(\text{MgCO}_3\) reacted:
\[ n_{\text{MgCO}_3} = 1.00\text{ g} \times \frac{1\text{ mol}}{84.31\text{ g}} = 0.01186\text{ mol} \]
Moles of \(\text{HNO}_3\) consumed:
\[ n_{\text{consumed}} = 0.01186\text{ mol MgCO}_3 \times \frac{2\text{ mol HNO}_3}{1\text{ mol MgCO}_3} = 0.02372\text{ mol HNO}_3 \]
Moles of \(\text{HNO}_3\) remaining:
\[ n_{\text{remaining}} = 0.0500\text{ mol} - 0.02372\text{ mol} = 0.02628\text{ mol} \]
Molarity of remaining \(\text{HNO}_3\):
\[ [\text{HNO}_3]_{\text{remaining}} = \frac{0.02628\text{ mol}}{0.0500\text{ L}} = 0.526\text{ M} \]

(f) Exothermic. The temperature of the solution increased (from \(22.00^\circ\text{C}\) to \(24.40^\circ\text{C}\)), indicating that heat was released by the chemical reaction into the aqueous solution (the surroundings).

(g)(i) Magnitude of heat absorbed by the solution:
\[ \Delta T = 24.40^\circ\text{C} - 22.00^\circ\text{C} = 2.40^\circ\text{C} \]
\[ q_{\text{solution}} = mc\Delta T = (51.0\text{ g})\left(4.18\frac{\text{J}}{\text{g}\cdot^\circ\text{C}}\right)(2.40^\circ\text{C}) = 511.63\text{ J} \approx 512\text{ J} \]

(g)(ii) Enthalpy of reaction per mole of \(\text{MgCO}_3\):
\[ q_{\text{rxn}} = -q_{\text{solution}} = -511.63\text{ J} = -0.51163\text{ kJ} \]
\[ \Delta H_{\text{rxn}}^\circ = \frac{-0.51163\text{ kJ}}{0.01186\text{ mol}} = -43.14\text{ kJ/mol}_{\text{rxn}} \approx -43.1\text{ kJ/mol}_{\text{rxn}} \text{ (or } -43.2\text{ kJ/mol}_{\text{rxn}}\text{)} \]

(a) [1 point] For the correct balanced net ionic equation: \(\text{MgCO}_3(s) + 2\,\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\) (state symbols not required; \(\text{H}_3\text{O}^+\) is acceptable).

(b) [1 point] For a correct explanation citing specific data from the table (e.g., pointing out that Trial 5 has a longer time than Trial 2 despite higher concentration, or longer than Trial 6 despite smaller particle size).

(c) [2 points]
- [1 point] For correctly relating particle size to surface area (fine powder has greater surface area / more exposed reactant particles than a large chunk).
- [1 point] For connecting greater surface area to a higher frequency of collisions between reacting particles.

(d) [1 point] For disagreeing and citing specific data comparing Trial 1 to Trial 4 (or 3 to 6) showing that increasing \([\text{HNO}_3]\) decreases reaction time / increases rate.

(e) [2 points]
- [1 point] For correctly calculating the moles of \(\text{HNO}_3\) consumed: \(0.0237\text{ mol}\).
- [1 point] For correctly calculating the remaining molarity: \([\text{HNO}_3] = 0.526\text{ M}\) (accept \(0.525\text{ M}\) to \(0.53\text{ M}\)).

(f) [1 point] For stating 'exothermic' and justifying by noting the increase in solution temperature.

(g)(i) [1 point] For the correct calculated value: \(q = 512\text{ J}\) (or \(0.512\text{ kJ}\)).

(g)(ii) [1 point] For calculating \(\Delta H_{\text{rxn}}^\circ\) by dividing \(q\) in kJ by moles of \(\text{MgCO}_3\) reacted and including a negative sign: \(-43.1\text{ kJ/mol}_{\text{rxn}}\) to \(-43.2\text{ kJ/mol}_{\text{rxn}}\).

(a) \(\text{mass} = 0.0100\text{ mol} \times 96.06\text{ g/mol} = 0.961\text{ g}\)

(b)
- Step 1: Place a weighing dish on the electronic balance (tare it), and use the spatula to measure out \(0.961\text{ g}\) of solid \(\text{NaC}_2\text{H}_5\text{COO}\).
- Step 4: Rinse the buret with a small amount of the \(0.200\text{ M } \text{C}_2\text{H}_5\text{COOH}(aq)\) solution, drain it through the tip, and then fill the buret with the \(0.200\text{ M } \text{C}_2\text{H}_5\text{COOH}(aq)\) solution.

(c) Equal to. According to the Henderson-Hasselbalch equation, \(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = \text{p}K_a + \log\left(\frac{n_{\text{A}^-}}{n_{\text{HA}}}\right)\). Since both the moles of \(\text{C}_2\text{H}_5\text{COOH}\) and the moles of \(\text{C}_2\text{H}_5\text{COO}^-\text{ (from }\text{NaC}_2\text{H}_5\text{COO})\) are reduced by half (to \(0.00500\text{ mol}\) each), the ratio of conjugate base to weak acid remains \(1:1\), keeping the \(\text{pH}\) unchanged.

- (a) 1 point for the correct calculated mass of \(\text{NaC}_2\text{H}_5\text{COO}\) with appropriate significant figures (0.961 g).
- (b) 2 points total:
- 1 point for a correct description of Step 1 specifying the use of the balance, weighing dish/paper, spatula, and the calculated mass.
- 1 point for a correct description of Step 4 specifying rinsing the buret with the acid solution (to prevent dilution from residual water) prior to filling.
- (c) 1 point for stating 'equal to' with a valid justification based on the unchanged conjugate base-to-acid ratio.

(a)(i) Using the ideal gas law \(PV = nRT\):
\(n = \frac{PV}{RT} = \frac{(5.20\text{ atm})(4.50\text{ L})}{(0.08206\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(310.\text{ K})} = 0.920\text{ mol}\)

(a)(ii) Since volume and moles are constant, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\):
\(P_2 = P_1 \times \frac{T_2}{T_1} = 5.20\text{ atm} \times \frac{385\text{ K}}{310.\text{ K}} = 6.46\text{ atm}\)

(b)(i) \(\text{HF}\). The diagram depicts a predominance of non-ionized acid molecules (8 out of 9 parent molecules remain intact), demonstrating partial ionization characteristic of a weak acid (\(K_a \ll 1\)). The only weak acid listed is \(\text{HF}\) (\(K_a = 6.8 \times 10^{-4}\)), as \(\text{HBr}\) and \(\text{HClO}_4\) are strong acids with \(K_a \gg 1\) that would dissociate almost completely.

(b)(ii) The partially positive hydrogen atoms of polar water molecules are oriented toward the negatively charged \(\text{Br}^-\) ion to maximize attractive ion-dipole interactions.

- (a)(i) 1 point for the correct calculated number of moles (0.920 mol) with appropriate work.
- (a)(ii) 1 point for the correct calculated new pressure (6.46 atm).
- (b)(i) 1 point for selecting \(\text{HF}\) and explaining that the particulate diagram displays incomplete ionization consistent with a weak acid having \(K_a < 1\).
- (b)(ii) 1 point for correctly explaining/identifying that the hydrogen ends (positive dipole) of water molecules are oriented toward the \(\text{Br}^-\) anion.

(a)(i) In \(\text{CH}_3\text{SH}(l)\): London dispersion forces and dipole-dipole forces.
(a)(ii) In \(\text{CH}_3\text{OH}(l)\): London dispersion forces, dipole-dipole forces, and hydrogen bonding.

(b)(i) While \(\text{CH}_3\text{SH}\) has stronger London dispersion forces due to having more electrons, \(\text{CH}_3\text{OH}\) exhibits strong hydrogen bonding between the \(-\text{OH}\) groups. These hydrogen bonding interactions are significantly stronger than the dipole-dipole and dispersion forces in \(\text{CH}_3\text{SH}\), so more thermal energy is required to overcome intermolecular attractions and vaporize \(\text{CH}_3\text{OH}\).

(b)(ii)
\(n_{\text{CH}_3\text{OH}} = \frac{12.8\text{ g}}{32.04\text{ g/mol}} = 0.3995\text{ mol}\)
\(q = 0.3995\text{ mol} \times 35.2\text{ kJ/mol} = 14.1\text{ kJ}\)

(c) A sulfur atom has three occupied electron shells (valence electrons in \(n=3\)), whereas an oxygen atom has only two occupied electron shells (valence electrons in \(n=2\)). Because the valence electrons of sulfur are in a higher energy level further from the nucleus, sulfur has a larger atomic radius than oxygen, leading to a greater internuclear distance and a longer \(\text{C}-\text{S}\) bond compared to the \(\text{C}-\text{O}\) bond.

- (a) 1 point for correctly listing all intermolecular forces for both substances (identifying dispersion and dipole-dipole for both, and hydrogen bonding uniquely for \(\text{CH}_3\text{OH}\)).
- (b)(i) 1 point for attributing the higher enthalpy of vaporization of \(\text{CH}_3\text{OH}\) to hydrogen bonding, which is stronger than the IMFs in \(\text{CH}_3\text{SH}\).
- (b)(ii) 1 point for the correct calculation of thermal energy (14.1 kJ) with units and appropriate significant figures.
- (c) 1 point for explaining that sulfur has an additional occupied electron shell (or valence electrons in \(n = 3\) vs. \(n = 2\) for oxygen), resulting in a larger atomic radius and greater bond length.

(a) The student's drawing shows an equal number of \(\text{Pb}^{2+}\) and \(\text{Cl}^-\) ions (a \(1:1\) ratio), which is incorrect because the chemical formula \(\text{PbCl}_2\) requires twice as many chloride ions as lead(II) ions (a \(1:2\) mole ratio) to maintain electrical neutrality.

(b)(i) Based on the stoichiometry of dissolution, \([\text{Cl}^-] = 2[\text{Pb}^{2+}] = 2(1.6 \times 10^{-2}\text{ M}) = 3.2 \times 10^{-2}\text{ M} = 0.032\text{ M}\).

(b)(ii) \(K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (1.6 \times 10^{-2})(3.2 \times 10^{-2})^2 = (1.6 \times 10^{-2})(1.024 \times 10^{-3}) = 1.6 \times 10^{-5}\).

(c) Less than. The \(0.10\text{ M } \text{NaCl}\) solution contains \(\text{Cl}^-(aq)\), which acts as a common ion in the \(\text{PbCl}_2\) dissolution equilibrium. According to Le Chatelier's principle / the common-ion effect, the presence of initial \(\text{Cl}^-\) suppresses the dissolution of \(\text{PbCl}_2(s)\), reducing its molar solubility and yielding a lower equilibrium \([\text{Pb}^{2+}]\) than in pure water.

- (a) 1 point for identifying that the ratio of \(\text{Pb}^{2+}\) to \(\text{Cl}^-\) is incorrectly shown as 1:1 instead of 1:2 (or that the solution lacks charge neutrality).
- (b)(i) 1 point for calculating \([\text{Cl}^-] = 0.032\text{ M}\).
- (b)(ii) 1 point for correctly calculating \(K_{sp} = 1.6 \times 10^{-5}\) consistent with (b)(i).
- (c) 1 point for stating 'less than' and providing a valid justification based on the common-ion effect (or Le Chatelier's principle reducing the solubility of \(\text{PbCl}_2\)).