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Kami Export - Honors Chem Final Review

Liquids and Solids

Worksheet A, Question A.1

  • The equilibrium vapor pressure of a pure liquid is independent of the volume of the gas phase as long as liquid is present.
  • The equilibrium vapor pressure decreases with decreasing temperature.

Worksheet A, Question A.2

  • ICl boils almost 40°C higher than Br2 because ICl is polar while Br2 is nonpolar.

Worksheet A, Question A.3

  • An ionic solid does not conduct electricity.
  • An ionic solid can conduct electricity when the solid is melted.
  • An ionic solid can sometimes have some covalent bonds in its structure.

Chapter 9

Worksheet A, Question B

  1. Phase Diagram:

    • A phase diagram is drawn with pressure on the y-axis and temperature on the x-axis.
    • The triple point is at 39°C and 85 mm Hg.
    • The vapor pressure curve, melting point curve, and sublimation curve are labeled.
    • The regions for solid, liquid, and vapor phases are indicated.
  2. Density:

    • The solid phase is denser than the liquid phase since the melting point curve slopes to the right with increasing pressure.
  3. Pressure at 500 mm Hg:

    • If pressure is kept at 500 mm Hg and temperature is increased from 50°C to 150°C, the compound will undergo vaporization.
  4. Temperature at 20°C:

    • If temperature is kept at 20°C and pressure is increased from 20 mm Hg to 150 mm Hg, the compound will undergo freezing.
  5. Physical State:

    • Given 4.60 g of the compound (M = 20.0) in a 2.00 L flask at 50°C.
    • Moles of compound = \frac{4.60 \, g}{20.0 \, g/mol} = 0.23 \, mol
    • Pressure if all vaporized: P = \frac{nRT}{V} = \frac{0.23 \, mol \times 0.0821 \, L \cdot atm/mol \cdot K \times 323 \, K}{2.00 \, L} = 3.05 \, atm
    • 3. 05 atm is equal to 3.05 \, atm \times \frac{760 \, mmHg}{1 \, atm} = 2320 \, mmHg
    • Since the vapor pressure at 50°C is 120 mm Hg, the compound exists in both liquid and gas phases.
  6. \Delta H_{vap} Calculation:

    • Using the Clausius-Clapeyron equation:
      \ln \frac{P2}{P1} = -\frac{\Delta H{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T_1} \right)
    • Where:
      • P1 = 120 \, mmHg at T1 = 50°C = 323 \, K
      • P2 = 760 \, mmHg at T2 = 123°C = 396 \, K
      • R = 8.314 \, J/mol \cdot K
    • Rearranging and solving for \Delta H{vap}: \Delta H{vap} = -R \left( \frac{\ln \frac{P2}{P1}}{\frac{1}{T2} - \frac{1}{T1}} \right)
      \Delta H_{vap} = -8.314 \, J/mol \cdot K \left( \frac{\ln \frac{760}{120}}{\frac{1}{396} - \frac{1}{323}} \right) = 26.5 \, kJ/mol

Worksheet A, Question C

  • Calcium has an atomic radius of 0.197 nm and a density of 1.54 g/cm³.
  • Calculation:
    • Volume of unit cell for simple cubic: V = a^3 = (2r)^3, where a is the side length and r is the atomic radius.
    • For simple cubic: V = (2 \times 0.197 \, nm)^3 = (0.394 \, nm)^3 = 0.06115 \, nm^3 = 6.115 \times 10^{-23} cm^3
    • Density = (mass of atoms in unit cell) / (volume of unit cell)
    • Mass of 1 atom of Ca: \frac{40.08 \, g/mol}{6.022 \times 10^{23} \, atoms/mol} = 6.655 \times 10^{-23} \, g/atom
    • For simple cubic (1 atom/cell): \frac{6.655 \times 10^{-23} \, g}{6.115 \times 10^{-23} \, cm^3} = 1.09 \, g/cm^3
    • For body-centered cubic (2 atoms/cell): \frac{2 \times 6.655 \times 10^{-23} \, g}{V{bcc}} = 1.54 \, g/cm^3, which implies V{bcc} = 8.643 \times 10^{-23} \, cm^3
    • For face-centered cubic (4 atoms/cell): \frac{4 \times 6.655 \times 10^{-23} \, g}{V{fcc}} = 1.54 \, g/cm^3, which implies V{fcc} = 1.729 \times 10^{-22} \, cm^3
  • The calcium unit cell is face-centered cubic.

Solutions

Worksheet B, Question A.1

  • Consider aqueous solutions prepared by adding 0.1 mol of the following solutes to 1.00 kg of water: C6H{12}O6, C2H5OH, NaI, Mg(NO3)2, Al(NO3)_3
    • Which solution(s) has(have) the highest boiling point? - Al(NO3)3
    • Which solution(s) has(have) the lowest boiling point? - C6H{12}O6, C2H_5OH
    • Which solution(s) has(have) the highest freezing point? - C6H{12}O6, C2H_5OH
    • Which solution(s) has(have) the lowest freezing point? - Al(NO3)3
    • Assuming that the molarity of the solutions are the same as their molality, which solution has the highest osmotic pressure? - Al(NO3)3

Worksheet B, Question A.2

  • Beaker A has 1.00 kg of water and 0.100 mol of glucose. Beaker B has 1.00 kg of water and 0.100 mol of CaCl_2.
    • The vapor pressure of the solution in Beaker A &GT the vapor pressure of pure water.
    • The boiling point of the solution in Beaker A &LT boiling point of the solution in Beaker B.
    • The molarity of the solution in Beaker A = molarity of the solution in Beaker B.
    • The freezing point of the solution in Beaker A &GT freezing point of the solution in Beaker B.
    • The solubility of the solute in Beaker A - MI solubility of the solute in in Beaker B.

Chapter 10

Worksheet B, Question A.3

  • Which of the following phenomena is best described by Henry's Law?
    • The solubility of O_2(g) in water increases with increasing pressure.

Worksheet B, Question B

  • What is the boiling point of an aqueous solution of a nonelectrolyte that has a freezing point of -2.86°C?

    • \Delta Tf = Kf \cdot m
    • -2.86°C = 1.86 \, °C/m \cdot m
    • m = \frac{-2.86}{1.86} = -1.54 \, m
    • \Delta Tb = Kb \cdot m
    • \Delta T_b = 0.512 \, °C/m \cdot -1.54 \, m
    • \Delta T_b = 0.788 \, °C
    • Boiling Point = 100 + 0.788 = 100.79°C

Worksheet B, Question C

  • 1. What is the density of the solution?

    • Mass of solute = 0.200 m * 115 g/mol = 23 g/kg
    • Mass of solution = 1000 g (water) + 23 g solute = 1023 g
    • Volume of solution = 0.227 M = 0.227 mol/L = (0.227 mol * 115 g/mol) / L = 26.105 g/L
    • Assume 1 kg of water = 1 L, so 1.023 L
    • Density = 1023 g / 1.023 L = 1000 g/L or 1 g/mL
    • Density = 1.16 g/mL
  • 2. What is the mass percent of solute in solution?

    • (23 g / 1023 g) * 100 = 2.25%

Reactions in Aqueous Solutions

Worksheet B, Question A

  • Write a balanced net ionic equation for the reaction between 0.1 M aqueous solutions of the following compounds. Use smallest whole number coefficients.
    • nickel(II) chloride and strontium hydroxide
      • Ni^{2+}(aq) + 2OH^-(aq) \rightarrow Ni(OH)_2(s)
    • methylamine (CH3NH2) and hydroiodic acid
      • CH3NH2(aq) + H^+(aq) \rightarrow CH3NH3(aq)
    • sulfuric acid and lead nitrate
      • SO4^{2-}(aq) + Pb^{2+}(aq) \rightarrow PbSO4(s)
    • nitric acid and lithium hydroxide
      • H^+(aq) + OH^-(aq) \rightarrow H_2O(l)
    • sulfurous acid and barium hydroxide
      • H2SO3(aq) + OH^-(aq) \rightarrow HSO3^-(aq) + H2O

Worksheet B, Question B

  • Consider the following unbalanced equation:
    • Cl^−(aq) + SO2(aq) → S4O6^{2−}(aq) + ClO3^−(aq)
    • 1. Write the oxidation number of each element in the reaction.
      • S = +6, Cl = -1, O = -2; S = +2.5, Cl = +5, O = -2
    • 2. What element is oxidized?
      • S
    • 3. What element is reduced?
      • Cl
    • 4. What species is the reducing agent?
      • SO_2^{2-}
    • 5. What species is the oxidizing agent?
      • Cl^−
    • 6. Write a balanced net ionic equation for the reaction in basic medium.
      • 12 SO3^{2−}(aq) + 7 Cl^−(aq) + 9 H2O → 3S4O6^{2−}(aq) + 7 ClO_3^−(aq) + 18 OH^−(aq)

Worksheet B, Question D

  • How many grams of iron(III) hydroxide can be theoretically obtained from the reaction between 25.00 mL of 0.300 M Sr(OH)_2 and 25.00 mL of 0.300 M iron(III) nitrate?
    • Iron (III) Nitrate: Fe(NO3)3
    • Strontium Hydroxide: Sr(OH)_2
    • Balanced Equation:
    • 2Fe(NO3)3(aq) + 3Sr(OH)2(aq) \rightarrow 2Fe(OH)3(s) + 3Sr(NO3)2(aq)
    • Moles of Fe(NO3)3 = 0.025 L * 0.300 M = 0.0075 mol
    • Moles of Sr(OH)_2 = 0.025 L * 0.300 M = 0.0075 mol
    • From stoichiometry, 2 moles of Fe(NO3)3 react with 3 moles of Sr(OH)_2.
    • So, the mole ratio is 2:3.
    • To check which is the limiting reactant:
      • \frac{moles \, of \, Fe(NO3)3}{2} = \frac{0.0075}{2} = 0.00375
      • \frac{moles \, of \, Sr(OH)_2}{3} = \frac{0.0075}{3} = 0.0025
    • Since 0.0025 < 0.00375, Sr(OH)_2is the limiting reactant.
    • Now, calculate the moles of Fe(OH)_3 produced using the limiting reactant:
      • From stoichiometry, 3 moles of Sr(OH)2 produce 2 moles of Fe(OH)3.
      • So, moles of Fe(OH)3 = (2/3) * moles of Sr(OH)2
      • moles of Fe(OH)_3 = (2/3) * 0.0075 mol = 0.005 mol
    • Molar mass of Fe(OH)_3 = 55.845 + 3*(15.999+1.008) = 106.867 g/mol
    • Mass of Fe(OH)_3 = moles * molar mass
      • Mass of Fe(OH)_3 = 0.005 mol * 106.867 g/mol = 0.534335 g

Worksheet B, Question E

  • Five grams of citric acid (M = 192.0 g/mol) are dissolved in 100.0 g of water. The resulting solution reacts with 65.3 mL of strontium hydroxide. The balanced net ionic equation for the reaction is
    • H3C5H5O7(aq) + 3OH^−(aq) → C5H5O_3(aq) + 3H₂O
    • What is the molarity of the strontium hydroxide solution?
    • Moles of Citric Acid:
    • Moles = \frac{Mass}{Molar \, Mass} = \frac{5.00 \, g}{192.0 \, g/mol} = 0.02604 \, mol
    • From the balanced equation, 1 mol of Citric Acid reacts with 3 mol of OH^−:
      • Moles \, of \, OH^− = 3 \times Moles \, of \, Citric \, Acid
      • Moles \, of \, OH^− = 3 \times 0.02604 \, mol = 0.07812 \, mol
    • Since Strontium Hydroxide is Sr(OH)2, it produces 2 moles of OH^− per mole of Sr(OH)2:
      • Moles \, of \, Sr(OH)_2 = \frac{Moles \, of \, OH^−}{2} = \frac{0.07812 \, mol}{2} = 0.03906 \, mol
    • Molarity of the Strontium Hydroxide solution:
      • Molarity = \frac{Moles}{Volume \, in \, Liters} = \frac{0.03906 \, mol}{0.0653 \, L} = 0.598 \, M

Gaseous Chemical Equilibrium

Worksheet A, Question A.1

  • When a graph is drawn which plots concentration of reactants and products (y-axis) versus time (x-axis), equilibrium is seen to have been established when the concentration of products and reactants remains unchanged.

Worksheet A, Question A.2

  • When a system is at equilibrium, forward and reverse reactions are taking place simultaneously.
  • When a system is at equilibrium, the rate of both forward and reverse reactions is the same.

Worksheet A, Question A.3

  • A chemical reaction has reached equilibrium when the partial pressures meet the requirements of the equilibrium constant expression.
  • A chemical reaction has reached equilibrium when Q = K.

Worksheet A, Question A.4

  • Consider the system
    • CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)
    • After the system reaches equilibrium, [CaO] is doubled. What is the result of this change?
      • (P{CO2})_{eq} remains the same.

Worksheet A, Question A.5

  • Which of the following statements are true?
    • Q and K can sometimes have the same numerical value.
    • Q can sometimes be 0.
    • Q can sometimes be less than K.
    • The numerical value for Q changes with time as the reaction proceeds.

Worksheet A, Question A.6

  • For the system
    • NH4Cl(g) \rightleftharpoons NH3(g) + HCl(g)
    • If the pressure of NH_3 is doubled (by compression) after the system reaches equilibrium, K will do none of the above.

Worksheet A, Question B

  • Consider the system
    • N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g)
    • A 2.568–g sample of NH_3 was placed in a 2.000 liter flask at 25°C. When equilibrium was reached at that temperature, it was determined that the ammonia was reduced to 75.0% of its original value.
    • 1. Calculate K for the decomposition of 2.00 moles of ammonia at 25.0°C.
    • 2. Calculate K for the decomposition of 2.00 moles of ammonia at 50°C. (\Delta H^\circf for NH3 is -46.1 kJ/mol)

Worksheet A, Question C

  • For the system:
    • PCl5(g) \rightleftharpoons PCl3(g) + Cl_2(g)
    • K is 1.0 at a certain temperature. If one starts with 0.50 atm of both PCl3 and Cl2 (g)
    • 1. In what direction will the reaction proceed?
      • -
    • 2. What are the partial pressures of all species when equilibrium is reached?
      • P{PCl5} = 0.13 \, atm; P{PCl2} = P{Cl2} = 0.37 \, atm
    • 3. If after equilibrium is reached, the partial pressure of chlorine gas is reduced to 0.2 atm, what are the partial pressures of all species when equilibrium is re-established?
      • P{PCl5} = 0.10 \, atm; P{PCl3} = 0.40 \, atm; P{Cl2} = 0.23 \, atm
    • 4. When stress is imposed on the system, does the system fully recover from that stress (i.e., Are the equilibrium pressures the same before and after the stress?)?
      • No

Thermochemistry

Worksheet A, Question A.1

  • An endothermic reaction is carried out in a coffee cup calorimeter. Which of the following is(are) NOT true for this process?
    • The temperature of the water increases.
    • Heat is absorbed by the water.
    • \Delta H{H2O} = Q_{reaction}

Worksheet A, Question A.2

  • The heat of fusion for water (\Delta H for ice melting) is +6.00 kJ. Which of the following statements is(are) true when ice melts?
    • H2O(l) \rightarrow H2O(s)
      • The reaction is endothermic.
      • The heat flow, q, is > 0.
      • The enthalpy for H2O(l) is higher than that for H2O(s).
      • The temperature of the surroundings decreases.

Worksheet A, Question A.3

  • Which statement(s) is(are) true about \Delta H?
    • When \Delta H > 0, it means that an endothermic reaction has occurred.
    • \Delta H for the melting of a pure substance (without change of temperature) is equal in magnitude but opposite in sign for the freezing of that same substance (without change of temperature).
    • \Delta H is equal to \Delta H^\circ_f for a reaction in which one mole of a compound is formed from elements in their native states at 1 atm pressure and at a specified temperature.

Worksheet A, Question B

  • Calculate \Delta H for the reaction
    • Fe2CO3(s) + CO(g) \rightarrow 2FeO(s) + CO_2(g)
    • Given that
      • Fe2CO3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) \quad \Delta H = -26.8 \, kJ
      • Fe(s) + CO_2(g) \rightarrow FeO(s) + CO(g) \quad \Delta H = 16.5 \, kJ

Worksheet A, Question C

  • When 2.50 g of glucose (C6H{12}O_6) burns in air, carbon dioxide gas and liquid water are formed. 38.9 kJ of heat are also liberated.
    • 1. What is the heat flow of the reaction if 85.0 g of glucose are burned?
    • 2. The reaction takes place in a bomb calorimeter where 4.50 g of glucose are burned. What is the heat capacity of the bomb if the temperature of the water in the bomb increases from 22.75°C to 25.32°C?
    • 3. Using the heat capacity of the bomb calculated in (2), what would the temperature change of the water in the bomb be if 12.0 g of glucose were burned?
    • 4. How many grams of glucose must be burned to liberate ten kJ of heat?
    • 5. Write a thermochemical equation for the reaction.

Spontaneity of Reaction

Worksheet B, Question A

  • Circle the true statements. If the statement is false, make it true.
    • Exothermic reactions are sometimes spontaneous.
    • Raising the temperature always increases the entropy of a solid.
    • As T approaches 0, \Delta H and \Delta G become equal for a reaction involving only solids.
    • At 0 K, a pure crystal of LiCl has \Delta S^\circ = 0.
    • \Delta S^\circ sometimes increases during a phase change.
    • As in the case with \Delta H, \Delta S depends on the initial and final state of a system, not on one particular pathway that was followed.
    • \Delta S^\circ increases during sublimation.
    • Reactions where \Delta G^\circ are less than 0 always proceed spontaneously but not always rapidly.
    • When \Delta G > 0 for a compound, it means that the compound tends to be unstable and decomposes.
    • \Delta G is a state property and therefore is path independent.
    • A combustion reaction producing more moles of gas than are consumed will have a \Delta G more negative than \Delta H.
    • For a spontaneous change, \Delta G_{total}^\circ must be negative for the overall equation between two coupled reactions.
    • When a solute is dissolved in water, \Delta S for the process is expected to be positive.
    • Heat capacity (C) and entropy (S) do not have the same thermodynamic property.
    • A chemical reaction in which entropy decreases can be spontaneous.

Worksheet B, Question B

  • The relationship between \Delta G^\circ and T is linear. Draw the graph of a reaction with the properties described below. You need only label the point at which \Delta G^\circ = 0.
    • The reaction is exothermic.
    • \Delta ng < 0 (\Delta ng = moles of gas products - moles of gas reactants)
    • At 300 K, the system is at equilibrium and K = 1.

Worksheet B, Question C

  • Acetic acid, CH_3COOH, freezes at 16.6°C. Its heat of fusion is 69.0 J/g. What is the change in entropy, \Delta S, when one mol of acetic acid freezes?

Worksheet B, Question D

  • At 25.0°C, K{sp} for silver chloride is 1.782 × 10^{-10}. At 35.0°C, its K{sp} is 4.159 × 10^{-10}. Calculate \Delta H^\circ and \Delta S^\circ for the reaction
    • Ag^+(aq) + Cl^−(aq) \rightleftharpoons AgCl(s)

Rate of Reaction

Worksheet A, Question A.1

  • The presence of a catalyst
    • can decrease the rate of a reaction.
    • changes k (k{uncat} \neq k{cat}).

Worksheet A, Question A.2

  • For a zero-order reaction
    • the reaction rate is constant.
    • units for k are mole/L-time.

Worksheet A, Question A.3

  • For a second order reaction
    • 1/(X) vs time is a linear plot.
    • the half-life is dependent on initial concentration.
    • the rate quadruples when the concentration is doubled.

Worksheet A, Question A.4

  • For a first order reaction
    • the rate is directly proportional to concentration so that when concentration is doubled, the rate is also doubled.
    • the function relating ln[X] to time is decreasing.
    • a linear plot is obtained when plotting ln (X) vs rate.
    • the unit for k is reciprocal time (time⁻¹).

Worksheet A, Question A.5

  • The dimerization of NO_2 is a second order reaction.
    • 2NO2(g) \rightarrow N2O_4(g)
    • The rate expression for the dimerization is
      • rate = k[NO₂]^2
    • For this reaction
      • rate doubles when the volume of the container is doubled but k remains the same.
      • increasing the temperature increases the rate and k.
      • decreasing the concentration of NO_2 decreases the rate but k remains constant.

Worksheet A, Question B

  • Consider the following energy diagram (not to scale) for the reaction
    • 2CH3(g) \rightarrow C2H_6(g)
    • 1. What is the activation energy of the forward reaction?
    • 2. At what point in the diagram would 2CH_3(g) appear?
    • 3. What is \Delta H for the reaction?
    • 4. What is the activation energy for the reverse reaction?
    • 5. At what point in the diagram would C2H6 be certainly found?
    • 6. What is any species at point C called?

Worksheet A, Question C

  • For the reaction
    • 2A(g) \rightarrow products \quad \Delta H = 22 \, kJ
    • The following experimental data is obtained at 25°C.
    • For this reaction:
      • 1. Calculate the half-life when [A] = 0.400 M.
      • 2. How long will it take to decompose 38% of A at 25°C?
      • 3. How fast is the decomposition at 25°C when [A] = 0.750 M?
      • 4. What is E_a if the rate constant doubles when the temperature at which the experiment is performed is increased by 10°C?

Acids and Bases

Worksheet A, Question A.1

  • A water solution is prepared by dissolving 0.20 mol of a weak acid, HB, in water to form one liter of solution. One can write the following relations about the species in solution when equilibrium is established.
    • [H+] < 0.20 M
    • [H^+] = [B^-]
    • [H^+] + [HB] ≈ 0.20

Worksheet A, Question A.2

  • The pH of a solution of HCl that is 1.0 × 10^{-9} M at 25°C is less than 7.00.

Worksheet A, Question A.3

  • Consider a 0.100 M aqueous solution of HNO_3.
    • The pH of the solution is 1.00.
    • The percent ionization is 100%.

Worksheet A, Question A.4

  • Ka for HC2H3O2 (acetic acid) is 1.8 × 10^{-5}. Kb for NH3(aq) is also 1.8 × 10^{-5}. When equal volumes of 0.10 M solutions of acetic acid and ammonia are mixed, the pH of the resulting solution is 7 because the strength of the two solutions is comparable.

Worksheet A, Question A.5

  • The percent ionization of a 0.10 M solution of a weak acid can be increased by
    • dilution with water.
    • increasing T if the reaction is endothermic.

Worksheet A, Question B

  • Arrange the following aqueous solutions in order of increasing pH.
    • 1. 4.0 M NaCl 1.0 M Sr(OH)2 * 0.2 M HCl < 2.0 M HClO3 < 4.0 M NaCl < 1.5 M KOH < 1.0 M Sr(OH)_2
    • 2. 0.10 M solutions of
      * NH4CN * FeCl3
      * 2.0 M HClO3 * LiClO4
      * 1.5 M KOH
      * 0.2 M HCl
      * Ba(OH)2 * HNO3
      * HNO3 < FeCl3 < LiClO4 < NH4CN < Ba(OH)_2

Worksheet A, Question C

  • Calculate the pH of a solution prepared by mixing 2.500 g of NaNO_2 in enough water to make 500.0 mL of solution.