University College Chemistry Semester 2 Final Review Study Guide

Comprehensive Overview of Semester 2 College Chemistry Final Review

The final examination consists of 30 multiple-choice questions spanning five major units: Unit 6 (Intermolecular Forces and Properties), Unit 7 (Gases), Unit 8 (Matter and Energy), Unit 9 (Stoichiometry), and Unit 10 (Reactions). Permitted materials during the exam include a regular periodic table, a periodic table of ions, a molecular geometry chart, a stoichiometry map, and a scientific calculator. All necessary equations will be provided on the exam. A specific review of previous problems confirms that while a periodic table of ions (polyatomic ions) is provided, it is not strictly required for questions like the stoichiometry of $H_2S$ ($S = 32.06\,g\,mol^{-1}$), as standard periodic tables provide necessary molar masses.

Unit 6: Intermolecular Forces and Molecular Properties

Intermolecular forces (IMFs) are determined by the polarity and structure of covalent molecules. For ammonia ($NH_3$), the molecule is polar and exhibits London Dispersion Forces (LDFs), Dipole-Dipole interactions, and Hydrogen bonding. Methane ($CH_4$) is non-polar and only possesses LDFs. Hydrogen sulfide ($H_2S$) is polar with LDFs and Dipole-Dipole interactions. Methanol ($CH_3OH$) is polar and demonstrates LDFs, Dipole-Dipole, and Hydrogen bonding. Ethene ($C_2H_4$) is non-polar with only LDFs. Trifluoromethane ($CHF_3$) is polar, exhibiting LDFs and Dipole-Dipole forces, but specifically lacks Hydrogen bonding because the Hydrogen is not bonded directly to Nitrogen, Oxygen, or Fluorine.

Boiling point predictions are based on the strength of these IMFs. Methanol ($CH_3OH$) is predicted to have the highest boiling point among these examples because it can Hydrogen bond and has a larger molecular size than $NH_3$. Furthermore, Oxygen is more electronegative than Nitrogen, making the $O-H$ bond more polar. Conversely, ethene ($C_2H_4$) is predicted to have the lowest boiling point because it only possesses weak LDFs. In a comparison between $CH_4$, $CF_4$, $CH_3F$, $O_2$, and $NH_3$, $NH_3$ has the highest boiling point due to Hydrogen bonding.

Bond polarity is ranked by the difference in electronegativity ($\Delta EN$). For the bonds $S-H$, $C-Br$, $F-N$, and $O-H$, the calculated differences are $\Delta EN = 0.4$ for $S-H$ ($2.6 - 2.2$), $\Delta EN = 0.4$ for $C-Br$ ($3.0 - 2.6$), $\Delta EN = 1.0$ for $F-N$ ($4.0 - 3.0$), and $\Delta EN = 1.5$ for $O-H$ ($3.4 - 1.9$). Therefore, the bonds ranked from least polar to most polar are $S-H$ and $C-Br$ (tie), followed by $F-N$, and finally $O-H$.

Unit 7: Kinetic Molecular Theory and Gas Laws

A gas exerts pressure on the walls of its container due to the force of collisions of the gas particles with the container walls. Several gas laws describe the relationship between pressure ($P$), volume ($V$), and temperature ($T$). Boyle's Law states that as Volume increases, Pressure decreases; mathematically, this is an inverse relationship where $P = \frac{1}{V}$ or $P_1V_1 = P_2V_2$. Charles's Law states that as Temperature increases, Volume increases; this is a direct relationship where $V = T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. Gay-Lussac's Law states that as Temperature increases, Pressure increases; this is a direct relationship where $P = T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$. If the volume of a gas is decreased at a constant temperature, its pressure will increase.

Dalton's Law of Partial Pressure describes the pressure of individual gases in a mixture. The total pressure is the sum of partial pressures ($P_{total} = P_1 + P_2 + ... + P_n$). For a mixture of $4.3\,mol$ Argon, $2.8\,mol\,CO_2$, and $6.2\,mol\,N_2$ with a total pressure of $8.7\,atm$, the total moles are $13.3\,mol$. The partial pressures are calculated as $P_{Ar} = \frac{4.3}{13.3} \times 8.7 = 2.8\,atm$, $P_{CO_2} = \frac{2.8}{13.3} \times 8.7 = 1.8\,atm$, and $P_{N_2} = \frac{6.2}{13.3} \times 8.7 = 4.1\,atm$. In another example with a total pressure of $833\,torr$, the mole fractions ($X_i$) for $N_2$ ($204\,torr$), $Ne$ ($172\,torr$), and $H_2$ ($457\,torr$) are $X_{N_2} = 0.245$, $X_{Ne} = 0.206$, and $X_{H_2} = 0.549$.

The Ideal Gas Law is defined as $PV = nRT$. For a gas with $6.82\,mol$, a volume of $53.5\,L$, and a pressure of $2.65\,atm$, using $R = 0.0821\,L \cdot atm \cdot mol^{-1} \cdot K^{-1}$, the temperature is calculated as $T = \frac{2.65 \times 53.5}{6.82 \times 0.0821} = 253\,K$. For $8.73\,moles$ of Helium at $344\,kPa$ and $58.1^{\circ}C$ ($331.1\,K$), using $R = 8.31\,L \cdot kPa \cdot mol^{-1} \cdot K^{-1}$, the volume is $V = \frac{8.73 \times 8.31 \times 331.1}{344} = 69.8\,L$. The Combined Gas Law ($\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$) is used for changing conditions. A balloon with $5.25\,L$ at $136\,kPa$ and $28.2^{\circ}C$ ($301.2\,K$) moved to a pressure of $189\,kPa$ and $17.3^{\circ}C$ ($290.3\,K$) will have a new volume of $V_2 = \frac{136 \times 5.25 \times 290.3}{301.2 \times 189} = 3.64\,L$.

Unit 8: Matter and Energy

Phase properties distinguish solids, liquids, and gases. Solids have a definite volume and a definite shape, with no space between particles and no translational motion. Liquids have a definite volume but no definite shape; they also lack significant space between particles and translational motion. Gases have no definite volume or shape, are compressible, have significant space between particles, and exhibit translational motion.

Definitions of energy include: Heat, which is the total energy due to particle movement, and Temperature, which is the average kinetic energy of a substance. During a phase change, the potential energy changes while temperature remains constant. Chemical bonds and intermolecular forces are considered forms of potential energy. Specific heat represents the kinetic energy change for $1\,g$ of a substance per $1^{\circ}C$. Calculations for heat transfer use $q = mc\Delta T$, where $c$ for gaseous water is $1.84\,J \cdot g^{-1} \cdot ^{\circ}C^{-1}$ or $2.08\,J \cdot g^{-1} \cdot ^{\circ}C^{-1}$ depending on the reference used. For $15.0\,g$ of water going from $T_i = 150.8^{\circ}C$ to $T_f = 105.9^{\circ}C$, the heat transfer is $q = 15.0 \times 1.84 \times (105.9 - 150.8) = -1239\,J$.

Phase changes require enthalpy of fusion ($\Delta H_{fus}$) or vaporization ($\Delta H_{vap}$). Calculating the energy to condense $156\,g$ of gaseous water ($H_2O$) to liquid involves $\Delta H_{vap} = 40.67\,kJ\,mol^{-1}$. The calculation is $\Delta H = 156\,g \times \frac{1\,mol}{18.02\,g} \times (-40.67\,kJ) = -352\,kJ$. Melting $150.0\,g$ of ice ($\Delta H_{fus} = 6.01\,kJ\,mol^{-1}$) requires $\Delta H = 150.0\,g \times \frac{1\,mol}{18.02\,g} \times 6.01\,kJ = 50.0\,kJ$. Calorimetry calculation for enthalpy of solution ($\Delta H_{soln}$) uses $\Delta H_{soln} = -m_{total} \cdot 4.18 \cdot (T_f - T_i)$. For example, adding $10\,g$ of $CaCl_2$ to $100\,g$ of $H_2O$ ($T_i = 24.5^{\circ}C$, $T_f = 39.4^{\circ}C$) results in $\Delta H_{soln} = -110 \times 4.18 \times (39.4 - 24.5) = -6851\,J$, which is an exothermic process.

Unit 9: Stoichiometry and Yields

Stoichiometry requires balanced chemical equations to determine relationships between reactants and products. In the reaction $2\,Al(s) + 6\,HCl(aq) \rightarrow 2\,AlCl_{3}(aq) + 3\,H_{2}(g)$, reacting $5.2\,mol\,Al$ requires $15.6\,mol\,HCl$ ($5.2 \times \frac{6}{2}$). Reacting $7.5\,mol\,HCl$ produces $3.75\,mol\,H_2$ ($7.5 \times \frac{3}{6}$). Making $3.25\,mol\,AlCl_3$ requires the reaction of $3.25\,mol\,Al$ ($3.25 \times \frac{2}{2}$). At STP (Standard Temperature and Pressure), $1\,mol$ of any gas occupies $22.4\,L$. For the reaction $2\,H_2S(l) + 3\,O_2(g) \rightarrow 2\,SO_2(g) + 2\,H_2O(l)$, reacting $103.5\,g\,H_2S$ requires $102\,L\,O_2$ calculated as: $103.5\,g \times \frac{1\,mol}{34.08\,g} \times \frac{3\,O_2}{2\,H_2S} \times 22.4\,L = 102\,L$.

Limiting reactants determine the theoretical yield. If $175\,g\,CuCl_2$ reacts with $35.1\,g\,Al$ in the reaction $2\,Al + 3\,CuCl_2 \rightarrow 2\,AlCl_3 + 3\,Cu$, the conversion from $CuCl_2$ yields $0.868\,mol\,AlCl_3$ ($116\,g$), while $Al$ yields $1.30\,mol\,AlCl_3$. Thus, $CuCl_2$ is the limiting reactant and the theoretical yield is $116\,g$. Percent yield is defined as $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$. If $4.28\,L\,H_2$ is actually produced from a theoretical yield of $5.04\,L$, the percent yield is $84.9\%$. In another reaction producing $8.07\,g\,NaOH$ from a theoretical $12.9\,g$, the percent yield is $62.6\%$. For magnesium chloride reacting with sodium ($2\,Na + MgCl_{2} \rightarrow 2\,NaCl + Mg$), reacting $0.255\,L$ of $1.75\,M\,MgCl_2$ requires $20.5\,g\,Na$ calculated as: $0.255\,L \times 1.75\,mol\,L^{-1} \times \frac{2\,Na}{1\,MgCl_2} \times 22.99\,g\,mol^{-1} = 20.5\,g$.

Unit 10: Chemical Reactions, Kinetics, and Equilibrium

Collision theory states that for a reaction to occur, a collision must have two things: the Activation Energy ($E_a$) and the proper orientation. Five factors affect the rate of reaction: temperature, surface area of solids, concentration of reactants, catalysts, and the pressure of gaseous reactants. Increasing the concentration of products has no effect on the forward reaction rate. Enthalpy of reaction ($\Delta H$) is the energy difference between reactants and products. In an energy profile, activation energy ($E_a$) is the energy needed to start bond breaking.

Energy changes can be calculated from bond enthalpies using $\Delta H = \text{bonds broken} - \text{bonds formed}$. For the reaction $N_2H_4 + O_2 \rightarrow N_2 + 2\,H_2O$, given bonds $N-N = 163$, $N-H = 391$, $O=O = 495$, $N \equiv N = 941$, and $O-H = 463\,kJ\,mol^{-1}$, the calculation is $\Delta H = (163 + 4 \times 391 + 495) - (941 + 4 \times 463) = -571\,kJ$. Since $\Delta H$ is negative, the reaction is exothermic. For the combustion of methane ($CH_4 + 2\,O_2 \rightarrow CO_2 + 2\,H_2O$), given $C-H = 413$, $O=O = 495$, $C=O = 799$, and $O-H = 463\,kJ\,mol^{-1}$, $\Delta H = (4 \times 413 + 2 \times 495) - (2 \times 799 + 4 \times 463) = -808\,kJ$.

Equilibrium is described by the equilibrium constant ($K_c$ or $K_p$). For the reaction $C(s) + 2\,H_2(g) \rightleftharpoons CH_4(g)$, the expressions are $K_c = \frac{[CH_4]}{[H_2]^2}$ and $K_p = \frac{P_{CH_4}}{P_{H_2}^2}$, noting that solids like Carbon are excluded. If $K_p = 0.277$ and $P_{H_2} = 1.21\,atm$, then $P_{CH_4} = 1.21^2 \times 0.277 = 0.406\,atm$. If an equilibrium constant $K_{eq}$ is much less than 1 (e.g., $0.001$), the equilibrium mixture contains mostly reactants. Le Chatelier's Principle predicts reaction shifts: for $2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g) + \text{heat}$ ($\Delta H = 39\,kJ$ - wait, if heat is a product, it is exothermic, but the transcript notes $\Delta H = 39\,kJ$. Based on the provided answer key: adding $SO_2$ shifts right, removing $SO_2$ shifts left, adding $O_2$ shifts right, decreasing pressure shifts left, and increasing temperature shifts left for exothermic reactions).

Questions & Discussion

Question: Does any question from the beginning need to use the periodic table of ions?

Answer: Looking back at the questions we have done, no, you do not need a periodic table of ions (polyatomic ions) to solve them. For example, in Question 3 (Stoichiometry), you only needed a regular periodic table to find the molar mass of $H_2S$ ($S = 32.06\,g\,mol^{-1}$).

Question: Why does a gas exert pressure on the walls of its container?

Answer: The force of the collisions of the gas particles with the container walls creates the pressure.

Question: What are the two things a collision must have in order to produce a reaction?

Answer: It must have sufficient Activation Energy ($E_a$) and the proper orientation.

Question: What is the temperature of $15.0\,g$ of water moving between $150.8^{\circ}C$ and $105.9^{\circ}C$?

Answer (from ChatGPT margin note): The water is in a gaseous state because the temperature is above $100^{\circ}C$. Use the formula $q = mc\Delta T$. Calculation: $15.0\,g \times 1.84\,J \cdot g^{-1} \cdot ^{\circ}C^{-1} \times (105.9 - 150.8) = -1239\,J$.