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

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.
Density:
- The solid phase is denser than the liquid phase since the melting point curve slopes to the right with increasing pressure.
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.
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.
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.
\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.