AP Chemistry Review

Thermochemistry

100

216 J of energy is required to raise the temperature of a piece of aluminum from 15 degrees Celsius to 35 degrees. Calculate the mass of aluminum used. (Specific heat capacity of aluminum= .90 J/(Celsius x g) ) A. 12.0g B. 15.0g C. 18.0g D. 21.0g E. 24.0g

A. q= m x c x (final temperature-initial temperature) 216 J = m x .90 J x (35-15) m= 12.0 g

200

Two solutions, both at the same temperature, are mixed. After mixing, the temperature drops dramatically. Which of the following are true? I. Products have stronger bonds (lower enthalpy) than the reactants. II. Products have weaker bonds (higher enthalpy) than the reactants. III. Products are more stable than reactants. IV. Products are less stable than reactants. A. I B. III C. I and III D. II and IV E. IV

D. This is an endothermic reaction (positive delta H) because it absorbs heat. In endothermic reactions, products have higher enthalpy and are less stable than reactants.

300

A certain non-spontaneous reaction has a delta H of 12.3 kJ and a delta S of 42 J/K. At approximately what temperature will the reaction become spontaneous? A. -20 °C B. 0 °C C. 20 °C D. 100 °C E. 200 °C

C. T= (delta H/ delta S) = (12,300J / 42) = 293 K 293K-273= 20 °C

400

The standard molar heat of vaporization of methanol is 35.3 kJ/mol and the boiling point at 1 atm is 64.6 degrees Celsius. What is ΔS° for the vaporization of .5 mol of ethanol? A. 17.4 J x mol / K B. 52.3 J / K C. 76.3 J x K /mol D. 88.4 J/k E. 332 J x K

B. 64.6 + 273= 337.6 K ΔS° = (ΔH/T) = (.5 mol/1) x (35.3 kJ/ mol x 337.6kJ) x (10^3 J/ kJ)= 52.3 J/K

500

.30 kg of ice at 0.0°C is added to 1.0 kg of water at 45°C. What is the final temperature, assuming there is no heat exchange with the surroundings? Specific heat capacity od water is 4200 J/ kg x K and the specific latent heat of fusion of ice is 3.4 x 10^5 J/kg. A. 0°C B. 8°C C. 16°C D. 22°C E. 44°C

C. heat lost by water = heat gained in melting ice + heat gained in warming the ice water. -(1kg H2O)x(4200J)x(?-45°C)=(.30kg ice) x(3.4x10^5 J) + (.30 kg ice) x (4200J) x (?-0.0°C) ?=16.0°C

Equilibrium

100

The reaction below has an equlibrium constant of 1000 at a given temperature. 2NO2(g) <==> N2O4(g) What is the Kc for the reaction 2N2O4(g) <==> 4NO2(g)? (A) 1000 (B) 1.0 x 10-3 (C) 1.0 x 10-6 (D) 5.0 x 10-4 (E) 2000

C. Kc of the reverse reaction is equal to the reciprocal of the forward reaction. When the reaction is doubled, the new Kc is equal to the original Kc squared.

200

For the reaction A(g) + B(g) --> C(g) + D(g), in order to increase the amount of D, you could: A) add more C for D to react with. B) remove some C. C) decrease the volume of the container. D) increase the volume of the container. E) add a catalyst to speed up the reaction and make more D.

B) Removing C causes the equilibrium to shift right, producing more D. Changing the volume of the container would change the concentrations of all species, but no shift would occur because there are equal numbers of moles of gas on both sides of the equation.

300

Given the following reaction: 2A(g) <---> B₂(g) + 3C(g). ΔHᵒ = +78.4 kJ To achieve the maximum amount of products, the reaction should be carried out at: A) low T, low P. B) low T, high P. C) high T, high P. D) high T, low P. E) none of the above.

D. Heat + 2A (g)<---> B₂(g) + 3C(g). More heat shifts the equilibrium to the right (products). Low pressure favors the formation of the larger number of moles of gas.

400

If 300 mL of .0500 M barium fluoride is added to 400 mL of .0750 M sodium sulfate, which of the following is true? (Ksp BaSO4 = 1.1 x 10^-10) A) Barium sulfate will precipitate out, Q>Ksp. B) Barium sulfate will not precipitate out, Q>Ksp. C) Barium sulfate will precipitate out, Q

Ba(2+) + SO4(2-) ---> BaSO4 number of moles of Ba2+ initially present: 300mL x (1L/1000mL) x (.05 mol/L) = .015 mol Ba2+. Total volume= 300mL+400mL=700mL=0.7L. concentration of Ba2+ = .015mol/.7L=.0214M. Number of moles of SO4 initially present: 400mL x (1L/1000mL) x (.0750mol/L)=.03 mol SO4. concentration of SO4=.03mol/.7L=.0429M. Q=(concentration of Ba) x (concentration of SO4). Q=9.18x10^-4. Q>Ksp, so a precipitate will form.

500

8.What is the ionization constant, Ka, for a weak monoprotic acid if a 0.30-molar solution has a pH of 4.0? a.9.7 × 10–10 b.4.7 × 10–2 c.1.7 × 10–6 d.3.0 × 10–4 e.3.3 × 10–8

E. If pH = 4.0, then [H+] = 1 × 10–4 = [A–], and [HA] = 0.30 – 1 × 10–4. The generic Ka is [H+][A–]/[HA], and when the values are entered into this equation: (1 × 10–4)2/0.30 = 3.3 × 10–8. Since you can estimate the answer, no actual calculations need be done.

Kinetics

100

A reaction follows the rate law: Rate = k[A]2. Which of the following plots will give a straight line? a.1/[A] versus 1/time b.[A]2 versus time c.1/[A] versus time d.ln[A] versus time e.[A] versus time

C—The "2" exponent means this is a second-order rate law. Second-order rate laws give a straight-line plot for 1/[A] versus t.

200

2.For the following reaction: NO2(g) + CO(g) → NO(g) + CO2(g), the rate law is: Rate = k[NO2]2. If a small amount of gaseous carbon monoxide (CO) is added to a reaction mixture that was 0.10 molar in NO2 and 0.20 molar in CO, which of the following statements is true? a.Both k and the reaction rate remain the same. b.Both k and the reaction rate increase. c.Both k and the reaction rate decrease. d.Only k increases, the reaction rate remains the same. e.Only the reaction rate increases; k remains the same.

A. The value of k remains the same unless the temperature is changed or a catalyst is added. Only materials that appear in the rate law, in this case NO2, will affect the rate. Adding NO2 would increase the rate, and removing NO2 would decrease the rate. CO has no effect on the rate.

300

3.The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds? a.0.250 mg b.0.125 mg c.0.0625 mg d.0.375 mg e.0.500 mg

B—The half-life is 0.693/k = 0.693/0.049 s–1 = 14 s. The time given, 28 s, represents two half-lives. The first half-life uses one-half of the isotope, and the second half-life uses one-half of the remaining material, so only one-fourth of the original material remains.

400

The equation C --> A + B which was found to second order, had a rate of .40 M/s and with an initial concentration of C of .25 M. If the concentration of C was increased to .45 M, then the rate would change to: A) .028 M/s. B) .125 M/s C) .50 M/s D) .75 M/s E) 3.6 M/s

E) The concentration of C increases from .25 to .75 (a three-fold increase). Being second order, the rate would increase 3^2= 9 times. 9(.40 M/s)=3.6M/s

500

The thermal decomposition of PH3 obeys first-order kinetics. At 680 degrees Celsius, a plot of ln[PH3] versus t gives a slope of -0.0198/s. What is the half-life of the reaction? A) 35.0 min B) 70.0 min C) 140.0 min D) 175.0 min E) 210.0 min

A) 4PH3 ---> P4 + 6H2 For first order reactions, the slope of the line is equal to -k. Therefore, k=0.0198/s. t^1/2=(.693/k)=(.693/.0198)= 35.0 min

Electrochemistry

100

S2O32– + OH– → SO42– + H2O + e– After the above half-reaction is balanced, which of the following are the respective coefficients of OH– and SO42–in the balanced half-reaction? a.8 and 3 b.6 and 2 c.10 and 2 d.5 and 2 e.5 and 1

C—The balanced equation is: S2O32– + 10 OH– → 2 SO42– + 5 H2O + 8 e–

200

All of the following may serve as reducing agents, EXCEPT: a.Mg b.Cs c.Fe2+ d.MnO4– e.Br

D—For a substance to serve as a reducing agent, it must be capable of being oxidized. The manganese, in the MnO4–, is already in its highest oxidation state, so it could not be oxidized. All other answers contain a substance that may be oxidized.

300

A sample of silver is to be purified by electrorefining. This will separate the silver from an impurity of gold. The impure silver is made into an electrode. Which of the following is the best way to set up the electrolytic cell? a.an impure silver cathode and an inert anode b.an impure silver cathode and a pure gold anode c.a pure silver cathode with an impure silver anode d.a pure gold cathode with an impure silver anode e.an impure silver cathode with a pure silver anode

C—The impure silver must be oxidized so it will go into solution. Oxidation occurs at the anode. Reduction is required to convert the silver ions to pure silver. Reduction occurs at the cathode. The cathode must be pure silver, otherwise it could be contaminated with the cathode material.

400

2 Fe3+ + Zn → Zn2+ + 2 Fe2+ The reaction shown above was used in an electrolytic cell. The voltage measured for the cell was not equal to the calculated E ° for the cell. This discrepancy could be caused by which of the following? a.The anion in the anode compartment was chloride, instead of nitrate as in the cathode compartment. b.One or more of the ion concentrations was not 1 M. c.Both of the solutions were at 25°C instead of 0°C. d.The solution in the salt bridge was Na2SO4 instead of KNO3. e.The anode and cathode were different sizes.

B—If the voltage was not equal to E °, then the cell was not standard. Standard cells have 1 M concentrations, and operate at 25°C with a partial pressure of each gas equal to 1 atm. No gases are involved in this reaction, so the cell must be operating at a different temperature or a different concentration (or both).

500

When a basic solution of KMnO4 is added to an SnCl2 solution, a brown precipitate of MnO2 forms and Sn4+ remains in solution. When the same basic solution of KMnO4 is added to an NaF solution, no reaction occurs. Which of the substances involved in these reactions serves as the best reducing agent? a.SnCl2 b.KMnO4 c.NaF d.MnO2 e.Sn4+

A—The Sn2+, from SnCl2, reduces the manganese from +7 to +4. This makes SnCl2 a reducing agent. The tin is oxidized to Sn4+, so KMnO4 is an oxidizing agent. NaF did nothing, so it behaves as neither an oxidizing nor as a reducing agent.

Acids and Bases

100

A Bronsted-Lowry base is defined as: A) a hydroxide donor B) a proton acceptor C) a proton donor D) a hydroxide acceptor E) an electron pair acceptor

B) A Bronsted-Lowry base is defined as a proton acceptor.

200

What is the pH of a 0.00001 molar HCl solution? A) 1 B) 9 C) 5 D) 4 E) 2

C) By definition pH is the negative log (logarithm) of the hydronium ion concentration. A 0.00001 molar solution has a H+ concentration of 10-5 M (move the decimal point 5 places to the right). The value of the negative exponent (-5) gives a pH of 5.

300

What is the conjugate base of C5H5NH+? A) C5H5N- B) C5H5NH C) C5H5NH2+ D) C5H5N+ E) C5H5N

E) Removal of a proton from an acid forms its conjugate base (C5H5NH+ ---> C5H5N + H+). Addition of a proton to a base forms its conjugate acid.

400

The following data were collected at the endpoint of a titration performed to find the molarity of an HCl solution. Volume of acid (HCl) used = 14.4 mL Volume of base (NaOH) used = 22.4 mL Molarity of standard base (NaOH) = 0.20 M What is the molarity of the acid solution? A) 1.6 M B) 0.64 M C) 0.31 M D) 0.13 M E) none of the above

C- Using the titration method, the volume of acid x the molarity of the acid neutralizes an equal volume of base x the molarity of the base. Or VAcid x MAcid = VBase x MBase where V = volume and M = molarity of HCl and NaOH. You are looking for the molarity of HCL or (14.4 mL) (M of HCl) = (22.4 mL) ( 0.20M). Rearranging terms, M of NaOH = (22.4 mL) (0.20 M) / 14.4 mL = 0.31 M.

500

Which of the following would not make an effective buffer system? A) NaH2PO4 and H3PO4 B) KClO4 and HClO4 C) KCH3COO and CH3COOH D) NaHCO3 and Na2CO3 E) Na2HPO4 and NaH2PO4

B) Buffers consist of mixtures of a weak acid and its conjugate base. In (B), HClO4 is a strong acid and its conjugate base (ClO4-) is a very weak base; therefore, ClO4- cannot combine with H+ to produce HClO4.