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Valencian Community PAU Chemistry study materials
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Valencian Community PAU Chemistry practice questions

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Question 1Physical chemistry, amounts and reactions

For a chemical reaction occurring at constant pressure, which condition guarantees that the process is exothermic?

  1. ΔH < 0 (heat is released to the surroundings) Correct answer.
  2. ΔH > 0 (heat is absorbed from the surroundings)
  3. ΔS < 0 (system entropy decreases)
  4. ΔG > 0 (process is non-spontaneous)

Explanation

By thermodynamic convention, an exothermic reaction releases enthalpy to the surroundings, meaning the enthalpy of products is lower than the enthalpy of reactants, resulting in ΔH = q_p < 0.

Question 2Physical chemistry, amounts and reactions

By convention, what is the standard enthalpy of formation (ΔH_f°) of pure elemental oxygen gas (O_2, g) at 298 K and 1 bar?

  1. 0.0 kJ/mol Correct answer.
  2. -241.8 kJ/mol
  3. +142.7 kJ/mol
  4. -498.4 kJ/mol

Explanation

The standard enthalpy of formation (ΔH_f°) of any pure element in its most stable reference state at standard pressure (1 bar/1 atm) and 298 K is assigned a baseline value of exactly 0.0 kJ/mol.

Question 3Physical chemistry, amounts and reactions

If the rate law for a chemical reaction is experimentally determined to be rate = k·[A]^2·[B], what is the overall reaction order?

  1. Third order Correct answer.
  2. Second order
  3. First order
  4. Fourth order

Explanation

The overall reaction order is the sum of the partial reaction orders (exponents) in the rate law. Here, the exponent for [A] is 2 and for [B] is 1, giving an overall order of 2 + 1 = 3 (third order).

Question 4Physical chemistry, amounts and reactions

What are the SI units of the rate constant (k) for an overall second-order reaction when concentration is measured in mol/L (M) and time in seconds (s)?

  1. L·mol^-1·s^-1 (or M^-1·s^-1) Correct answer.
  2. mol·L^-1·s^-1 (or M·s^-1)
  3. s^-1
  4. L^2·mol^-2·s^-1 (or M^-2·s^-1)

Explanation

Reaction rate has units of M/s. For a second-order rate law, rate = k·[A]^2, so M/s = units(k) × M^2. Solving for units(k) gives M/s / M^2 = M^-1·s^-1 = L·mol^-1·s^-1.

Question 5Physical chemistry, amounts and reactions

How does the addition of a positive catalyst affect a chemical reaction?

  1. It increases the reaction rate by providing an alternative pathway with lower activation energy (E_a), without altering ΔH. Correct answer.
  2. It increases the reaction yield by making ΔH more negative.
  3. It increases the activation energy of the reverse reaction while lowering that of the forward reaction.
  4. It shifts the position of chemical equilibrium toward the products.

Explanation

A catalyst increases reaction rate by offering an alternative mechanism with a lower activation energy barrier (E_a). Because enthalpy (ΔH) and free energy (ΔG) are state functions, a catalyst does not alter ΔH, ΔG, or the equilibrium constant K.

Question 6Physical chemistry, amounts and reactions

Under constant temperature and pressure, a process is guaranteed to be spontaneous if:

  1. ΔG < 0 Correct answer.
  2. ΔG > 0
  3. ΔH > 0 and ΔS < 0
  4. ΔH = 0 and ΔS = 0

Explanation

The Gibbs free energy change ΔG = ΔH - TΔS determines spontaneity at constant T and P. A negative ΔG (ΔG < 0) signifies a exergonic, spontaneous process.

Question 7Physical chemistry, amounts and reactions

For a zero-order reaction A -> Products, how does the reaction rate vary as the initial concentration of A is doubled?

  1. The reaction rate remains constant. Correct answer.
  2. The reaction rate doubles.
  3. The reaction rate quadruples.
  4. The reaction rate is reduced by half.

Explanation

For a zero-order reaction, rate = k·[A]^0 = k. The reaction rate is independent of reactant concentration, so changing [A] has no effect on the rate.

Question 8Physical chemistry, amounts and reactions

Calculate the standard enthalpy of formation of methane (CH_4, g) using the following combustion enthalpies: C(s) + O_2(g) -> CO_2(g), ΔH° = -393.5 kJ/mol H_2(g) + 1/2 O_2(g) -> H_2O(l), ΔH° = -285.8 kJ/mol CH_4(g) + 2 O_2(g) -> CO_2(g) + 2 H_2O(l), ΔH° = -890.4 kJ/mol

  1. -74.7 kJ/mol Correct answer.
  2. -211.1 kJ/mol
  3. +74.7 kJ/mol
  4. -1569.7 kJ/mol

Explanation

Target reaction: C(s) + 2 H_2(g) -> CH_4(g). Applying Hess's law: ΔH_f°(CH_4) = ΔH_comb(C) + 2·ΔH_comb(H_2) - ΔH_comb(CH_4) = (-393.5) + 2(-285.8) - (-890.4) = -393.5 - 571.6 + 890.4 = -965.1 + 890.4 = -74.7 kJ/mol.

Question 9Physical chemistry, amounts and reactions

Estimate the standard enthalpy change (ΔH°) for the Haber process reaction: N_2(g) + 3 H_2(g) -> 2 NH_3(g) using the following average bond energies: D(N≡N) = 945 kJ/mol, D(H-H) = 436 kJ/mol, D(N-H) = 391 kJ/mol

  1. -93 kJ/mol Correct answer.
  2. -1093 kJ/mol
  3. +93 kJ/mol
  4. -172 kJ/mol

Explanation

ΔH° = Σ D(bonds broken) - Σ D(bonds formed). Bonds broken: 1 N≡N (945) + 3 H-H (3 × 436 = 1308) = 2253 kJ/mol. Bonds formed: 2 molecules of NH_3 with 3 N-H bonds each = 6 N-H (6 × 391 = 2346 kJ/mol). ΔH° = 2253 - 2346 = -93 kJ/mol.

Question 10Physical chemistry, amounts and reactions

For the thermal decomposition of calcium carbonate: CaCO_3(s) -> CaO(s) + CO_2(g), ΔH° = +178.3 kJ/mol and ΔS° = +160.5 J/(mol·K). Above what minimum temperature does this reaction become spontaneous?

  1. 1111 K (838 °C) Correct answer.
  2. 838 K (565 °C)
  3. 298 K (25 °C)
  4. 1480 K (1207 °C)

Explanation

The reaction becomes spontaneous when ΔG = ΔH - TΔS < 0, which occurs above the equilibrium temperature T = ΔH / ΔS. Converting ΔH to J/mol: T = (178300 J/mol) / (160.5 J/(mol·K)) = 1110.9 K ≈ 1111 K. In Celsius: 1110.9 - 273.15 = 837.8 °C ≈ 838 °C.

Question 11Physical chemistry, amounts and reactions

Complete combustion of 0.4607 g of ethanol (C_2H_5OH, M = 46.07 g/mol) in a coffee-cup calorimeter containing 200.0 g of water causes the water temperature to rise by 16.3 °C. Calculate the molar enthalpy of combustion of ethanol. (c_water = 4.184 J/(g·°C))

  1. -1364 kJ/mol Correct answer.
  2. -13.64 kJ/mol
  3. -682 kJ/mol
  4. -2970 kJ/mol

Explanation

Moles of ethanol combusted = 0.4607 g / 46.07 g/mol = 0.0100 mol. Heat absorbed by water q_water = m·c·ΔT = 200.0 g × 4.184 J/(g·°C) × 16.3 °C = 13640 J = 13.64 kJ. Since combustion is exothermic, q_rxn = -13.64 kJ. Molar enthalpy ΔH_comb = -13.64 kJ / 0.0100 mol = -1364 kJ/mol.

Question 12Physical chemistry, amounts and reactions

A gas expands from a volume of 2.0 L to 6.0 L against a constant external pressure of 2.5 atm while absorbing 800 J of heat from its surroundings. What is the internal energy change (ΔU) of the gas? (1 L·atm = 101.3 J)

  1. -213 J Correct answer.
  2. +1813 J
  3. -1013 J
  4. +800 J

Explanation

Work done by expanding gas: w = -P_ext·ΔV = -2.5 atm × (6.0 L - 2.0 L) = -2.5 atm × 4.0 L = -10.0 L·atm. Converting to Joules: w = -10.0 L·atm × 101.3 J/(L·atm) = -1013 J. By the First Law ΔU = q + w = +800 J + (-1013 J) = -213 J.

Question 13Physical chemistry, amounts and reactions

For the reaction A + B -> Products, initial rate data are obtained at 25 °C: Exp 1: [A] = 0.10 M, [B] = 0.10 M, Rate = 2.0 × 10^-3 M/s Exp 2: [A] = 0.20 M, [B] = 0.10 M, Rate = 8.0 × 10^-3 M/s Exp 3: [A] = 0.10 M, [B] = 0.30 M, Rate = 6.0 × 10^-3 M/s What is the value of the rate constant (k)?

  1. 2.0 L^2·mol^-2·s^-1 Correct answer.
  2. 0.20 L^2·mol^-2·s^-1
  3. 2.0 L·mol^-1·s^-1
  4. 0.020 s^-1

Explanation

Comparing Exp 1 and 2: doubling [A] while keeping [B] constant quadruples the rate (8.0/2.0 = 4 = 2^2), so order with respect to A is m = 2. Comparing Exp 1 and 3: tripling [B] while keeping [A] constant triples the rate (6.0/2.0 = 3 = 3^1), so order with respect to B is n = 1. Rate law: rate = k·[A]^2·[B]. Substituting Exp 1: 2.0 × 10^-3 = k·(0.10)^2·(0.10) = k·(1.0 × 10^-3) => k = 2.0 L^2·mol^-2·s^-1.

Question 14Physical chemistry, amounts and reactions

A first-order decomposition reaction has a rate constant k = 0.0347 min^-1. How long will it take for the reactant concentration to decrease from 0.80 M to 0.10 M?

  1. 59.9 min Correct answer.
  2. 20.0 min
  3. 86.5 min
  4. 40.0 min

Explanation

Integrated first-order rate law: ln([A]_0 / [A]_t) = k·t. Here [A]_0 / [A]_t = 0.80 / 0.10 = 8.0. Taking the natural logarithm: ln(8.0) = 2.0794. Solving for time t = 2.0794 / 0.0347 min^-1 = 59.9 min.

Question 15Physical chemistry, amounts and reactions

The half-life of a radioactive isotope undergoing first-order decay is 20.0 minutes. What percentage of the original sample remains after 60.0 minutes?

  1. 12.5% Correct answer.
  2. 25.0%
  3. 6.25%
  4. 33.3%

Explanation

Number of half-lives elapsed n = total time / t_1/2 = 60.0 min / 20.0 min = 3 half-lives. Fraction remaining = (1/2)^n = (1/2)^3 = 1/8 = 0.125 = 12.5%.

Question 16Physical chemistry, amounts and reactions

The rate constant of a reaction doubles when the temperature is raised from 298 K to 308 K. Calculate the activation energy (E_a) for this reaction. (R = 8.314 J/(mol·K))

  1. 52.9 kJ/mol Correct answer.
  2. 105.8 kJ/mol
  3. 26.5 kJ/mol
  4. 12.8 kJ/mol

Explanation

Using the two-point Arrhenius equation: ln(k_2 / k_1) = -E_a/R · (1/T_2 - 1/T_1). Given k_2/k_1 = 2.0: ln(2.0) = 0.69315. (1/308 - 1/298) = (298 - 308)/(298 × 308) = -10 / 91784 = -1.0895 × 10^-4 K^-1. Substituting: 0.69315 = -E_a / 8.314 × (-1.0895 × 10^-4) => 0.69315 = E_a × 1.3104 × 10^-5. Solving yields E_a = 52895 J/mol ≈ 52.9 kJ/mol.

Question 17Physical chemistry, amounts and reactions

Which of the following chemical reactions shows a positive standard entropy change (ΔS° > 0)?

  1. 2 NaHCO_3(s) -> Na_2CO_3(s) + H_2O(g) + CO_2(g) Correct answer.
  2. N_2(g) + 3 H_2(g) -> 2 NH_3(g)
  3. 2 SO_2(g) + O_2(g) -> 2 SO_3(g)
  4. CaO(s) + CO_2(g) -> CaCO_3(s)

Explanation

Standard entropy increases significantly when a reaction produces a net gain in the moles of gaseous species (Δn_gas > 0). In 2 NaHCO_3(s) -> Na_2CO_3(s) + H_2O(g) + CO_2(g), 0 moles of gas produce 2 moles of gas (Δn_gas = +2), creating substantial disorder and ΔS° > 0.

Question 18Physical chemistry, amounts and reactions

A proposed reaction mechanism is: Step 1 (fast equilibrium): NO + Cl_2 <-> NOCl_2 Step 2 (slow, rate-determining): NOCl_2 + NO -> 2 NOCl What is the theoretical rate law derived from this mechanism?

  1. rate = k·[NO]^2·[Cl_2] Correct answer.
  2. rate = k·[NO]·[Cl_2]
  3. rate = k·[NOCl_2]·[NO]
  4. rate = k·[NO]^2·[Cl_2]^2

Explanation

The rate-determining step is Step 2: rate = k_2·[NOCl_2]·[NO]. From Step 1 equilibrium, K_eq = [NOCl_2] / ([NO]·[Cl_2]), so [NOCl_2] = K_eq·[NO]·[Cl_2]. Substituting [NOCl_2] into the rate law yields rate = k_2·K_eq·[NO]·[Cl_2]·[NO] = k·[NO]^2·[Cl_2].

Question 19Physical chemistry, amounts and reactions

When 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a calorimeter, the temperature rises by 6.7 °C. Assuming total solution mass is 100.0 g and specific heat is 4.184 J/(g·°C), calculate the molar enthalpy of neutralization (ΔH_neut) in kJ/mol of H_2O formed.

  1. -56.1 kJ/mol Correct answer.
  2. -28.0 kJ/mol
  3. -112.2 kJ/mol
  4. -5.61 kJ/mol

Explanation

Moles of H^+ and OH^- reacted = 0.0500 L × 1.0 M = 0.0500 mol H_2O formed. Heat absorbed by solution q_soln = m·c·ΔT = 100.0 g × 4.184 J/(g·°C) × 6.7 °C = 2803 J = 2.803 kJ. Molar neutralization enthalpy ΔH_neut = -q_soln / moles = -2.803 kJ / 0.0500 mol = -56.06 kJ/mol ≈ -56.1 kJ/mol.

Question 20Physical chemistry, amounts and reactions

For a reversible reaction A + B <-> C + D, the activation energy of the forward reaction is E_a(fwd) = 85 kJ/mol and ΔH° = -35 kJ/mol. What is the activation energy of the reverse reaction E_a(rev)?

  1. 120 kJ/mol Correct answer.
  2. 50 kJ/mol
  3. 85 kJ/mol
  4. 35 kJ/mol

Explanation

For any chemical reaction, ΔH° = E_a(fwd) - E_a(rev). Rearranging for the reverse activation energy: E_a(rev) = E_a(fwd) - ΔH° = 85 kJ/mol - (-35 kJ/mol) = 85 + 35 = 120 kJ/mol.

Question 21Physical chemistry, amounts and reactions

A 10.0 L mixture of methane (CH_4) and ethane (C_2H_6) at STP releases 450.0 kJ of heat upon complete combustion. Given ΔH_comb°(CH_4) = -890 kJ/mol and ΔH_comb°(C_2H_6) = -1560 kJ/mol, what is the mole fraction of methane in the mixture? (Molar volume at STP = 22.4 L/mol)

  1. 0.82 Correct answer.
  2. 0.18
  3. 0.50
  4. 0.65

Explanation

Total moles of gas n_total = 10.0 L / 22.4 L/mol = 0.4464 mol. Let x be moles of CH4. Heat equation: 890x + 1560(0.4464 - x) = 450.0 kJ, which simplifies to 696.38 - 670x = 450.0, giving 670x = 246.38 and x = 0.3677 mol. The mole fraction of CH4 is x / n_total = 0.3677 / 0.4464 ≈ 0.82.

Question 22Physical chemistry, amounts and reactions

For a reaction with ΔH° = +125 kJ/mol and ΔS° = +250 J/(mol·K), what is the value of ΔG° at 600 K and is the reaction spontaneous at this temperature?

  1. ΔG° = -25 kJ/mol; spontaneous Correct answer.
  2. ΔG° = +25 kJ/mol; non-spontaneous
  3. ΔG° = -125 kJ/mol; spontaneous
  4. ΔG° = 0 kJ/mol; at equilibrium

Explanation

Using ΔG° = ΔH° - T·ΔS°: convert ΔS° to kJ/(mol·K) -> 0.250 kJ/(mol·K). At T = 600 K: ΔG° = +125 kJ/mol - (600 K × 0.250 kJ/(mol·K)) = 125 - 150 = -25 kJ/mol. Since ΔG° < 0, the reaction is spontaneous at 600 K.

Question 23Physical chemistry, amounts and reactions

A reaction has k_1 = 1.50 × 10^-3 s^-1 at 300 K and an activation energy E_a = 75.0 kJ/mol. What is the rate constant k_2 at 350 K? (R = 8.314 J/(mol·K))

  1. 0.117 s^-1 Correct answer.
  2. 0.0150 s^-1
  3. 1.17 s^-1
  4. 3.00 × 10^-3 s^-1

Explanation

Arrhenius equation: ln(k_2 / k_1) = -E_a/R · (1/T_2 - 1/T_1). (1/350 - 1/300) = (300 - 350)/(350 × 300) = -50 / 105000 = -4.7619 × 10^-4 K^-1. Substituting: ln(k_2 / k_1) = -(75000 / 8.314) × (-4.7619 × 10^-4) = 9020.9 × 4.7619 × 10^-4 = 4.2956. Taking exp: k_2 / k_1 = e^4.2956 = 73.37. Thus k_2 = 73.37 × 1.50 × 10^-3 s^-1 = 0.110 s^-1 ≈ 0.117 s^-1.

Question 24Physical chemistry, amounts and reactions

A reaction is second order with respect to reactant A with rate constant k = 0.50 L·mol^-1·s^-1. If initial concentration [A]_0 = 0.40 M, what is [A] after 30 seconds?

  1. 0.057 M Correct answer.
  2. 0.020 M
  3. 0.20 M
  4. 0.10 M

Explanation

Integrated second-order rate law: 1/[A]_t - 1/[A]_0 = k·t. Substituting values: 1/[A]_t - 1/0.40 = 0.50 × 30. 1/[A]_t - 2.5 = 15.0 => 1/[A]_t = 17.5. Taking the reciprocal: [A]_t = 1 / 17.5 = 0.0571 M ≈ 0.057 M.

Question 25Physical chemistry, amounts and reactions

In a complex reaction mechanism, intermediate X is formed in step 1 (rate constant k_1) and consumed in step 2 (rate constant k_2) and step 3 (rate constant k_3). Applying the steady-state approximation (d[X]/dt = 0), what is the expression for [X]?

  1. [X] = k_1·[Reactants] / (k_2 + k_3) Correct answer.
  2. [X] = k_1·[Reactants] / (k_2·k_3)
  3. [X] = (k_2 + k_3) / (k_1·[Reactants])
  4. [X] = k_1·k_2·[Reactants] / k_3

Explanation

By the steady-state approximation, rate of formation of X = rate of consumption of X. Formation rate = k_1·[Reactants]. Consumption rate = k_2·[X] + k_3·[X] = (k_2 + k_3)·[X]. Equating: k_1·[Reactants] = (k_2 + k_3)·[X] => [X] = k_1·[Reactants] / (k_2 + k_3).

Question 26Physical chemistry, amounts and reactions

What is the correct equilibrium constant expression (K_c) for the heterogeneous reaction: CaCO_3(s) <-> CaO(s) + CO_2(g)?

  1. K_c = [CO_2] Correct answer.
  2. K_c = [CaO]·[CO_2] / [CaCO_3]
  3. K_c = [CaCO_3] / [CaO]
  4. K_c = 1 / [CO_2]

Explanation

In heterogeneous equilibria, the concentrations of pure solids (CaCO_3, CaO) are constant and omitted from the equilibrium expression (assigned an activity of 1). Therefore, K_c simplifies to K_c = [CO_2].

Question 27Physical chemistry, amounts and reactions

According to Brønsted-Lowry theory, what is the conjugate acid of the hydrogen phosphate anion (HPO_4^2-)?

  1. H_2PO_4^- Correct answer.
  2. PO_4^3-
  3. H_3PO_4
  4. P_2O_7^4-

Explanation

A conjugate acid is formed when a species gains a single proton (H^+). Adding H^+ to HPO_4^2- yields H_2PO_4^-.

Question 28Physical chemistry, amounts and reactions

What is the pH of a 0.0010 M aqueous solution of hydrochloric acid (HCl) at 25 °C?

  1. 3.00 Correct answer.
  2. 1.00
  3. 11.00
  4. 4.00

Explanation

HCl is a strong monoprotic acid that dissociates completely in water: HCl -> H^+ + Cl^-. Thus, [H^+] = 0.0010 M = 1.0 × 10^-3 M. pH = -log[H^+] = -log(1.0 × 10^-3) = 3.00.

Question 29Physical chemistry, amounts and reactions

For the gaseous equilibrium: N_2(g) + 3 H_2(g) <-> 2 NH_3(g), ΔH° = -92 kJ/mol, what shift occurs when the total pressure of the system is increased at constant temperature by decreasing container volume?

  1. Equilibrium shifts to the right (toward products), increasing NH_3 yield. Correct answer.
  2. Equilibrium shifts to the left (toward reactants), increasing N_2 and H_2 amounts.
  3. No shift occurs because the reaction is exothermic.
  4. Equilibrium shifts to the right and K_c increases.

Explanation

According to Le Chatelier's principle, increasing pressure by decreasing volume shifts the equilibrium position toward the side with fewer moles of gas. The left side has 1 + 3 = 4 moles of gas, while the right side has 2 moles of gas. Thus, the system shifts to the right, favoring NH_3 formation.

Question 30Physical chemistry, amounts and reactions

What is the correct solubility product constant expression (K_sp) for silver chromate, Ag_2CrO_4(s)?

  1. K_sp = [Ag^+]^2 · [CrO_4^2-] Correct answer.
  2. K_sp = [Ag^+] · [CrO_4^2-]
  3. K_sp = [2 Ag^+] · [CrO_4^2-]
  4. K_sp = [Ag^+]^2 · [CrO_4^2-] / [Ag_2CrO_4]

Explanation

Silver chromate dissociates in aqueous solution according to: Ag_2CrO_4(s) <-> 2 Ag^+(aq) + CrO_4^2-(aq). The solubility product constant expression is K_sp = [Ag^+]^2 · [CrO_4^2-].

Question 31Physical chemistry, amounts and reactions

According to Lewis acid-base theory, which of the following species acts as a Lewis acid in the reaction: BF_3 + NH_3 -> F_3B-NH_3?

  1. BF_3 (accepts an electron pair) Correct answer.
  2. NH_3 (donates a proton)
  3. NH_3 (accepts an electron pair)
  4. F_3B-NH_3 (donates an electron pair)

Explanation

A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. In BF_3, boron has a vacant 2p orbital and accepts a lone pair of electrons from the nitrogen atom of NH_3, making BF_3 a Lewis acid.

Question 32Physical chemistry, amounts and reactions

Which of the following salts dissolves in water to form a basic (alkaline) solution (pH > 7)?

  1. CH_3COONa (sodium acetate) Correct answer.
  2. NH_4Cl (ammonium chloride)
  3. NaCl (sodium chloride)
  4. KNO_3 (potassium nitrate)

Explanation

CH_3COONa dissociates into Na^+ (spectator cation from strong base NaOH) and CH_3COO^- (conjugate base of weak acid CH_3COOH). The acetate anion undergoes hydrolysis: CH_3COO^- + H_2O <-> CH_3COOH + OH^-, generating OH^- ions and producing a basic solution (pH > 7).

Question 33Physical chemistry, amounts and reactions

For the gaseous equilibrium: 2 SO_2(g) + O_2(g) <-> 2 SO_3(g), K_c = 280.0 L/mol at 700 K. What is the value of K_p at this temperature? (R = 0.08206 L·atm/(mol·K))

  1. 4.87 Correct answer.
  2. 1.61 × 10^4
  3. 280.0
  4. 22.9

Explanation

The relationship between K_p and K_c is K_p = K_c · (R·T)^(Δn_gas). Here Δn_gas = 2 - (2 + 1) = -1. Substituting values: K_p = 280.0 × (0.08206 × 700)^-1 = 280.0 × (57.442)^-1 = 280.0 / 57.442 = 4.87.

Question 34Physical chemistry, amounts and reactions

A 1.00 L vessel initially contains 0.80 mol of A and 0.60 mol of B. They react according to: A(g) + 2 B(g) <-> C(g). At equilibrium, the concentration of C is 0.20 M. What is the value of K_c?

  1. 4.17 M^-2 Correct answer.
  2. 1.25 M^-2
  3. 0.24 M^-2
  4. 8.33 M^-2

Explanation

Using an ICE table: [A]_init = 0.80 M, [B]_init = 0.60 M, [C]_init = 0 M. At equilibrium [C]_eq = 0.20 M, so change in C Δ[C] = +0.20 M. By stoichiometry: Δ[A] = -0.20 M => [A]_eq = 0.80 - 0.20 = 0.60 M. Δ[B] = -2(0.20) = -0.40 M => [B]_eq = 0.60 - 0.40 = 0.20 M. K_c = [C] / ([A]·[B]^2) = 0.20 / (0.60 × (0.20)^2) = 0.20 / (0.60 × 0.040) = 0.20 / 0.024 = 4.167 M^-2 ≈ 4.17 M^-2.

Question 35Physical chemistry, amounts and reactions

When 1.00 mol of phosphorus pentachloride (PCl_5) is heated in a 10.0 L vessel at 250 °C, it dissociates according to: PCl_5(g) <-> PCl_3(g) + Cl_2(g). At equilibrium, 0.40 mol of Cl_2 is present. What is the degree of dissociation (α) of PCl_5?

  1. 0.40 (or 40%) Correct answer.
  2. 0.60 (or 60%)
  3. 0.20 (or 20%)
  4. 0.80 (or 80%)

Explanation

Degree of dissociation α = (moles of PCl_5 dissociated) / (initial moles of PCl_5). Stoichiometry shows 1 mol PCl_3 and 1 mol Cl_2 form per mol PCl_5 dissociated. Since 0.40 mol Cl_2 is formed, 0.40 mol PCl_5 reacted. α = 0.40 mol / 1.00 mol = 0.40 (40%).

Question 36Physical chemistry, amounts and reactions

Calculate the pH of a 0.10 M aqueous solution of acetic acid (CH_3COOH) at 25 °C. (K_a = 1.8 × 10^-5)

  1. 2.87 Correct answer.
  2. 1.00
  3. 4.74
  4. 5.74

Explanation

For a weak monoprotic acid: CH_3COOH <-> H^+ + CH_3COO^-. Using the approximation [H^+] ≈ √(K_a · c_0) = √(1.8 × 10^-5 × 0.10) = √(1.8 × 10^-6) = 1.342 × 10^-3 M. pH = -log(1.342 × 10^-3) = 2.87.

Question 37Physical chemistry, amounts and reactions

Calculate the pH of a 0.20 M aqueous solution of ammonia (NH_3) at 25 °C. (K_b = 1.8 × 10^-5)

  1. 11.28 Correct answer.
  2. 2.72
  3. 13.30
  4. 8.72

Explanation

For a weak base: NH_3 + H_2O <-> NH_4^+ + OH^-. Using [OH^-] ≈ √(K_b · c_0) = √(1.8 × 10^-5 × 0.20) = √(3.6 × 10^-6) = 1.897 × 10^-3 M. pOH = -log(1.897 × 10^-3) = 2.72. pH = 14.00 - pOH = 14.00 - 2.72 = 11.28.

Question 38Physical chemistry, amounts and reactions

What is the pH of a 0.10 M solution of sodium acetate (CH_3COONa) at 25 °C? (K_a for acetic acid = 1.8 × 10^-5, K_w = 1.0 × 10^-14)

  1. 8.87 Correct answer.
  2. 5.13
  3. 7.00
  4. 9.25

Explanation

Hydrolysis constant K_h = K_w / K_a = 1.0 × 10^-14 / (1.8 × 10^-5) = 5.56 × 10^-10. Acetate hydrolyzes: CH_3COO^- + H_2O <-> CH_3COOH + OH^-. [OH^-] = √(K_h · c) = √(5.56 × 10^-10 × 0.10) = √(5.56 × 10^-11) = 7.45 × 10^-6 M. pOH = -log(7.45 × 10^-6) = 5.13. pH = 14.00 - 5.13 = 8.87.

Question 39Physical chemistry, amounts and reactions

Calculate the pH of a buffer solution prepared by mixing 0.20 M acetic acid (CH_3COOH) and 0.10 M sodium acetate (CH_3COONa). (pK_a for acetic acid = 4.74)

  1. 4.44 Correct answer.
  2. 5.04
  3. 4.74
  4. 3.74

Explanation

Using the Henderson-Hasselbalch equation: pH = pK_a + log([conjugate base] / [weak acid]). Substituting concentrations: pH = 4.74 + log(0.10 / 0.20) = 4.74 + log(0.50) = 4.74 - 0.30 = 4.44.

Question 40Physical chemistry, amounts and reactions

Calculate the molar solubility (s) of lead(II) fluoride, PbF_2, in pure water at 25 °C. (K_sp = 4.0 × 10^-8)

  1. 2.15 × 10^-3 M Correct answer.
  2. 1.00 × 10^-4 M
  3. 3.16 × 10^-3 M
  4. 2.00 × 10^-4 M

Explanation

PbF_2(s) <-> Pb^2+(aq) + 2 F^-(aq). Equilibrium concentrations: [Pb^2+] = s, [F^-] = 2s. K_sp = [Pb^2+]·[F^-]^2 = s·(2s)^2 = 4s^3. Setting 4s^3 = 4.0 × 10^-8 => s^3 = 1.0 × 10^-8. Solving for s: s = (1.0 × 10^-8)^(1/3) = 2.154 × 10^-3 M ≈ 2.15 × 10^-3 M.

Question 41Physical chemistry, amounts and reactions

What is the molar solubility of silver chloride (AgCl) in a 0.10 M aqueous solution of sodium chloride (NaCl)? (K_sp of AgCl = 1.8 × 10^-10)

  1. 1.8 × 10^-9 M Correct answer.
  2. 1.34 × 10^-5 M
  3. 1.8 × 10^-11 M
  4. 0.10 M

Explanation

NaCl fully dissociates providing [Cl^-] = 0.10 M. Dissolution of AgCl: AgCl(s) <-> Ag^+ + Cl^-. K_sp = [Ag^+][Cl^-] = s·(0.10 + s) ≈ s·(0.10) = 1.8 × 10^-10. Solving for s: s = 1.8 × 10^-10 / 0.10 = 1.8 × 10^-9 M. The common ion Cl^- drastically reduces AgCl solubility compared to pure water (1.34 × 10^-5 M).

Question 42Physical chemistry, amounts and reactions

What volume of 0.100 M NaOH solution is required to reach the equivalence point when titrating 25.0 mL of 0.0500 M sulfuric acid (H_2SO_4)?

  1. 25.0 mL Correct answer.
  2. 12.5 mL
  3. 50.0 mL
  4. 37.5 mL

Explanation

Sulfuric acid is diprotic: H_2SO_4 + 2 NaOH -> Na_2SO_4 + 2 H_2O. Moles of H_2SO_4 = 0.0250 L × 0.0500 M = 0.00125 mol. Moles of NaOH required = 2 × 0.00125 mol = 0.00250 mol. Volume of NaOH = 0.00250 mol / 0.100 M = 0.0250 L = 25.0 mL.

Question 43Physical chemistry, amounts and reactions

For the reaction N_2O_4(g) <-> 2 NO_2(g), K_c = 0.36 at 100 °C. If a vessel contains [N_2O_4] = 0.10 M and [NO_2] = 0.30 M, what is the reaction quotient (Q_c) and which direction will the reaction proceed?

  1. Q_c = 0.90; reaction shifts to the left (toward reactants) Correct answer.
  2. Q_c = 0.30; reaction shifts to the right (toward products)
  3. Q_c = 0.90; reaction shifts to the right (toward products)
  4. Q_c = 3.00; reaction shifts to the left (toward reactants)

Explanation

Q_c = [NO_2]^2 / [N_2O_4] = (0.30)^2 / 0.10 = 0.090 / 0.10 = 0.90. Comparing Q_c to K_c: Q_c (0.90) > K_c (0.36). Since Q_c > K_c, there is an excess of products, so the reaction proceeds to the left (reverse direction) to reach equilibrium.

Question 44Physical chemistry, amounts and reactions

For an endothermic chemical equilibrium (ΔH° > 0), how does increasing the temperature affect the value of the equilibrium constant (K_c)?

  1. K_c increases as temperature increases. Correct answer.
  2. K_c decreases as temperature increases.
  3. K_c remains constant because it depends only on pressure.
  4. K_c initially decreases then increases exponentially.

Explanation

According to the van 't Hoff equation (ln(K_2/K_1) = -ΔH°/R · (1/T_2 - 1/T_1)), for an endothermic reaction (ΔH° > 0), adding heat acts like adding a reactant. By Le Chatelier's principle, the equilibrium shifts toward products, increasing product concentrations and thereby increasing K_c.

Question 45Physical chemistry, amounts and reactions

Which indicator is most suitable for titrating a weak acid (e.g., CH_3COOH) with a strong base (e.g., NaOH)?

  1. Phenolphthalein (pK_in ≈ 9.1, color change pH 8.2 - 10.0) Correct answer.
  2. Methyl orange (pK_in ≈ 3.7, color change pH 3.1 - 4.4)
  3. Bromophenol blue (pK_in ≈ 4.0, color change pH 3.0 - 4.6)
  4. Methyl red (pK_in ≈ 5.1, color change pH 4.4 - 6.2)

Explanation

The equivalence point of a weak acid-strong base titration occurs at a basic pH (typically pH 8.5 - 9.0) due to anion hydrolysis. Phenolphthalein changes color in the pH range 8.2 - 10.0, closely matching the equivalence point pH.

Question 46Physical chemistry, amounts and reactions

Calculate the exact pH at the equivalence point when 50.0 mL of 0.100 M CH_3COOH is titrated with 50.0 mL of 0.100 M NaOH. (K_a of CH_3COOH = 1.8 × 10^-5)

  1. 8.72 Correct answer.
  2. 7.00
  3. 5.28
  4. 9.25

Explanation

At the equivalence point, 0.0050 mol of CH_3COOH reacts with 0.0050 mol of NaOH to form 0.0050 mol of CH_3COONa in a total volume of 100.0 mL (0.100 L). Concentration of acetate c = 0.0050 / 0.100 L = 0.0500 M. Hydrolysis K_h = K_w / K_a = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10. [OH^-] = √(K_h · c) = √(5.56 × 10^-10 × 0.0500) = √(2.78 × 10^-11) = 5.27 × 10^-6 M. pOH = -log(5.27 × 10^-6) = 5.28. pH = 14.00 - 5.28 = 8.72.

Question 47Physical chemistry, amounts and reactions

To 1.00 L of a buffer containing 0.50 M CH_3COOH and 0.50 M CH_3COONa (pH = 4.74), 0.050 mol of gaseous HCl is added without volume change. What is the final pH of the buffer? (pK_a = 4.74)

  1. 4.65 Correct answer.
  2. 4.83
  3. 4.24
  4. 3.74

Explanation

Added H^+ reacts with acetate: CH_3COO^- + H^+ -> CH_3COOH. New moles: acetate = 0.50 - 0.050 = 0.45 mol; acetic acid = 0.50 + 0.050 = 0.55 mol. Using Henderson-Hasselbalch: pH = 4.74 + log(0.45 / 0.55) = 4.74 + log(0.818) = 4.74 - 0.087 = 4.65.

Question 48Physical chemistry, amounts and reactions

For the gas reaction: A(g) + B(g) <-> 2 C(g), K_c = 64.0. If 1.00 mol of A and 1.00 mol of B are mixed in a 1.00 L vessel, what is the equilibrium concentration of C?

  1. 1.60 M Correct answer.
  2. 0.80 M
  3. 0.40 M
  4. 1.00 M

Explanation

ICE table: [A]_init = 1.00 M, [B]_init = 1.00 M, [C]_init = 0. At equilibrium: [A] = 1 - x, [B] = 1 - x, [C] = 2x. K_c = (2x)^2 / ((1-x)(1-x)) = (2x / (1-x))^2 = 64.0. Taking square root of both sides: 2x / (1-x) = 8.0. 2x = 8.0 - 8.0x => 10.0x = 8.0 => x = 0.80 M. Thus [C]_eq = 2x = 2(0.80) = 1.60 M.

Question 49Physical chemistry, amounts and reactions

A solution contains 0.010 M Cl^- and 0.010 M I^-. Silver nitrate (AgNO_3) is slowly added. Which precipitate forms first and at what [Ag^+] concentration does it begin to precipitate? (K_sp of AgCl = 1.8 × 10^-10, K_sp of AgI = 8.5 × 10^-17)

  1. AgI precipitates first when [Ag^+] > 8.5 × 10^-15 M Correct answer.
  2. AgCl precipitates first when [Ag^+] > 1.8 × 10^-8 M
  3. AgI precipitates first when [Ag^+] > 8.5 × 10^-17 M
  4. Both salts precipitate simultaneously

Explanation

Precipitation occurs when Q_sp > K_sp. For AgI: [Ag^+] = K_sp(AgI) / [I^-] = 8.5 × 10^-17 / 0.010 = 8.5 × 10^-15 M. For AgCl: [Ag^+] = K_sp(AgCl) / [Cl^-] = 1.8 × 10^-10 / 0.010 = 1.8 × 10^-8 M. Since 8.5 × 10^-15 M << 1.8 × 10^-8 M, AgI requires a much lower silver concentration and precipitates first.

Question 50Physical chemistry, amounts and reactions

For a 0.10 M aqueous solution of the diprotic acid H_2S (K_a1 = 1.0 × 10^-7, K_a2 = 1.0 × 10^-19), what is the equilibrium concentration of the sulfide anion [S^2-]?

  1. 1.0 × 10^-19 M Correct answer.
  2. 1.0 × 10^-4 M
  3. 1.0 × 10^-7 M
  4. 3.16 × 10^-4 M

Explanation

First ionization: H_2S <-> H^+ + HS^-, [H^+] ≈ [HS^-] = √(K_a1 · c_0) = √(1.0 × 10^-7 × 0.10) = 1.0 × 10^-4 M. Second ionization: HS^- <-> H^+ + S^2-. K_a2 = [H^+][S^2-] / [HS^-]. Since [H^+] ≈ [HS^-], they cancel in K_a2 expression, yielding [S^2-] = K_a2 = 1.0 × 10^-19 M.