CHM 101.pdf

# Chemical reactions and stoichiometry This topic explores the fundamental principles of chemical reactions, including their identification, the process of balancing chemical equations, and the quantitative relationships governing reactants and products through stoichiometry. ### 1.1 Understanding chemical reactions A chemical reaction is defined as a process where at least one new substance is produced as a result of a chemical change [3](#page=3). #### 1.1.1 Indications of a chemical reaction Several observable signs can indicate that a chemical reaction has occurred: * Evolution of heat, light, and/or sound [3](#page=3). * Production of a gas or vapor [3](#page=3). * Formation of a precipitate (a solid formed from a solution) [3](#page=3). * Color change [3](#page=3). ### 1.2 Balancing chemical equations The principle of the conservation of matter dictates that a chemical equation must be balanced, meaning it must contain the same number of atoms of each element on both sides of the reaction [4](#page=4). #### 1.2.1 Key terms in chemical equations * **Reactants:** The substances that exist before a chemical change takes place [4](#page=4). * **Products:** The new substances formed during the chemical change [4](#page=4). * **Chemical Equation:** A representation that indicates the reactants and products of a reaction [4](#page=4). #### 1.2.2 Steps for balancing chemical equations 1. Write a word equation for the reaction. 2. Write the correct chemical formulas for all reactants and products. 3. Determine the appropriate coefficients to ensure the number of atoms of each element is equal on both sides of the equation [5](#page=5). #### 1.2.3 Subscripts vs. Coefficients * **Subscripts:** Indicate the number of atoms of each element within a molecule or compound. For example, in $H_2O$, the subscript '2' indicates two hydrogen atoms [6](#page=6). * **Coefficients:** Indicate the number of molecules or formula units of a compound involved in the reaction. For example, in $2H_2O$, the coefficient '2' indicates two molecules of water [6](#page=6). #### 1.2.4 Example of balancing an equation To balance the reaction between aluminum sulfate and calcium chloride to form calcium sulfate, follow these steps: 1. **Word equation:** aluminum sulfate + calcium chloride → calcium sulfate + aluminum chloride [7](#page=7). 2. **Chemical formulas:** $Al_2(SO_4)_3 + CaCl_2 \rightarrow CaSO_4 + AlCl_3$ [7](#page=7). 3. **Balancing coefficients:** $Al_2(SO_4)_3 + 3CaCl_2 \rightarrow 3CaSO_4 + 2AlCl_3$ [7](#page=7). ### 1.3 Types of chemical reactions Chemical reactions can be categorized into several types based on the rearrangement of atoms and molecules: * **Combination reactions:** Two or more substances combine to form a single product. The general form is $A + B \rightarrow AB$ [8](#page=8). * **Example:** Nitrogen gas reacts with hydrogen gas to form ammonia gas: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ [8](#page=8). * **Decomposition reactions:** A single substance breaks down into two or more simpler substances. The general form is $AB \rightarrow A + B$ [8](#page=8). * **Example:** Calcium carbonate decomposes into calcium oxide and carbon dioxide gas: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ [8](#page=8). * **Combustion reactions:** Rapid reactions involving oxygen as a reactant, often producing a flame. Hydrocarbons typically react with oxygen in the air to produce carbon dioxide and water [9](#page=9). * **Example:** Propane reacts with oxygen to form carbon dioxide and water: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$ [9](#page=9). * **Neutralization reactions:** An acid reacts with a base to form a salt and water through an exchange of ions [9](#page=9). * **Example:** Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water: $NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$ [9](#page=9). * **Displacement (Substitution or Single Replacement) reactions:** A more reactive element replaces a less reactive element in a compound. The general form is $A + BC \rightarrow AC + B$, where A is more reactive than B [10](#page=10). * **Example:** Zinc metal reacts with hydrochloric acid to form zinc chloride and hydrogen gas: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$ [10](#page=10). * **Double displacement reactions:** Ions are exchanged between two reactant compounds to form new compounds. The general form is $AB + CD \rightarrow AD + CB$ [10](#page=10). * **Example:** Barium chloride reacts with sodium sulfate to form barium sulfate (a precipitate) and sodium chloride: $BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$ [10](#page=10). * **Oxidation and Reduction (Redox) reactions:** These reactions involve the transfer of electrons. * **Oxidation:** Characterized by the addition of oxygen or a non-metallic element, or the removal of hydrogen or a metallic element from a compound [11](#page=11). * **Reduction:** Characterized by the addition of hydrogen or a metallic element, or the removal of oxygen or a non-metallic element from a compound [11](#page=11). ### 1.4 Stoichiometry: Quantitative relationships in reactions Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions [12](#page=12). #### 1.4.1 Mass calculations * **Formula Mass (FM):** The sum of the masses of the atoms in the empirical formula of a compound [12](#page=12). * **Example:** For hydrogen peroxide ($HO$), FM = mass of H (1.0 amu) + mass of O (16.0 amu) = 17.0 amu [12](#page=12). * **Molecular Mass (MM):** The sum of the masses of the atoms in the molecular formula of a compound [12](#page=12). * **Example:** For hydrogen peroxide ($H_2O_2$), MM = 2(mass of H) + 2(mass of O) = 2(1.0 amu) + 2(16.0 amu) = 34.0 amu [12](#page=12). #### 1.4.2 Atomic mass unit and the mole * **Atomic Mass Unit (amu):** $1 \text{ amu} = 1.6605 \times 10^{-24} \text{ g}$ [13](#page=13). * **Avogadro's Constant ($N_A$):** $6.022 \times 10^{23}$ amu = 1 g. This is the number of elementary entities (atoms, molecules, etc.) in one mole of a substance [13](#page=13). * **Mole:** The amount of any substance containing as many elementary entities as there are atoms in exactly 12 grams of carbon-12 [13](#page=13). #### 1.4.3 Molar Mass Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol) [15](#page=15). #### 1.4.4 Conversions between grams and moles * **Grams to Moles:** Divide the mass in grams by the molar mass of the substance. * **Example:** To find the moles in 15.32 g of sodium chloride (NaCl, molar mass ≈ 23 + 35.45 = 58.45 g/mol): $15.32 \text{ g NaCl} / 58.45 \text{ g/mol} = 0.2621 \text{ mol}$ [14](#page=14). * **Moles to Grams:** Multiply the number of moles by the molar mass of the substance. * **Example:** To find the mass of 2.50 mol of ethanol ($C_2H_5OH$, molar mass ≈ 2(12.01) + 6(1.008) + 16.00 = 46.07 g/mol): $2.50 \text{ mol} \times 46.07 \text{ g/mol} = 115.2 \text{ g}$ [14](#page=14). #### 1.4.5 Percentage composition The percentage composition by mass of a compound shows the proportion of each element's mass within the compound. It can be calculated using the formula: $$ \% \text{ element} = \frac{(\text{number of atoms})(\text{atomic weight})}{\text{molecular mass of the compound}} \times 100\% $$ [15](#page=15). * **Example:** For ethane ($C_2H_6$, molecular mass = 2(12.0 amu) + 6(1.0 amu) = 30.0 amu): * $\% C = \frac{2(12.0 \text{ amu})}{30.0 \text{ amu}} \times 100\% = 80\%$ [16](#page=16). * $\% H = \frac{6(1.0 \text{ amu})}{30.0 \text{ amu}} \times 100\% = 20\%$ [16](#page=16). * **Example:** For ammonia ($NH_3$, molar mass = 14.0 + 3(1.0) = 17.0 g/mol): * Mass percent N = $\frac{14.0 \text{ g}}{17.0 \text{ g}} \times 100\% = 82.35\%$ [16](#page=16). * Mass percent H = $\frac{3.0 \text{ g}}{17.0 \text{ g}} \times 100\% = 17.65\%$ [16](#page=16). #### 1.4.6 Calculating empirical formulas from percent composition The empirical formula represents the simplest whole-number ratio of atoms in a compound. * **Steps:** 1. Assume a 100 g sample of the compound, so percentages directly convert to grams. 2. Convert the mass of each element to moles by dividing by its atomic mass. 3. Divide the mole values by the smallest number of moles calculated to obtain a ratio. 4. If the ratios are not whole numbers, multiply all ratios by a small integer (e.g., 2, 3, 4) to obtain whole numbers. * **Example:** Para-aminobenzoic acid is composed of C (61.31%), H (5.14%), N (10.21%), and O (23.33%) [17](#page=17). 1. **Convert mass to moles:** * C: $61.31 \text{ g} / 12.01 \text{ g/mol} = 5.105 \text{ mol C}$ [17](#page=17). * H: $5.14 \text{ g} / 1.01 \text{ g/mol} = 5.09 \text{ mol H}$ [17](#page=17). * N: $10.21 \text{ g} / 14.01 \text{ g/mol} = 0.7288 \text{ mol N}$ [17](#page=17). * O: $23.33 \text{ g} / 16.00 \text{ g/mol} = 1.456 \text{ mol O}$ [17](#page=17). 2. **Divide by the smallest number of moles (0.7288 mol N):** * C: $5.105 / 0.7288 \approx 7$ [18](#page=18). * H: $5.09 / 0.7288 \approx 7$ [18](#page=18). * N: $0.7288 / 0.7288 = 1$ [18](#page=18). * O: $1.458 / 0.7288 \approx 2$ [18](#page=18). 3. **Empirical Formula:** $C_7H_7NO_2$ [18](#page=18). #### 1.4.7 Empirical Formula by Synthesis This method determines the empirical formula of a compound formed by reacting elements. * **Example:** 2.435 g of antimony reacts with sulfur to form 3.397 g of an antimony-sulfur compound [19](#page=19). 1. **Calculate the mass of sulfur:** Mass sulfur = 3.397 g - 2.435 g = 0.962 g [19](#page=19). 2. **Convert masses to moles:** * Sb: $2.435 \text{ g} / 121.8 \text{ g/mol} = 0.02 \text{ mol Sb}$ [19](#page=19). * S: $0.962 \text{ g} / 32.06 \text{ g/mol} = 0.03 \text{ mol S}$ [19](#page=19). 3. **Divide by the smallest mole value (0.02 mol Sb):** * Sb: $0.02 / 0.02 = 1$ [19](#page=19). * S: $0.03 / 0.02 = 1.5$ [19](#page=19). 4. **Multiply to get whole numbers:** Multiply by 2 to clear the 1.5: * Sb: $1 \times 2 = 2$ [19](#page=19). * S: $1.5 \times 2 = 3$ [19](#page=19). 5. **Empirical Formula:** $Sb_2S_3$ [19](#page=19). #### 1.4.8 Stoichiometric calculations Stoichiometric calculations use balanced chemical equations to determine the amounts of reactants and products involved in a reaction. The general pathway is: Mass of substance A $\rightarrow$ Moles of substance A $\rightarrow$ Moles of substance B $\rightarrow$ Mass of substance B. * **Example:** 10 g of glucose ($C_6H_{12}O_6$) undergoes combustion. Calculate the grams of each product ($CO_2$ and $H_2O$) [20](#page=20). * Balanced equation: $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$ [20](#page=20). * Molar masses: $C_6H_{12}O_6$ (180 g/mol), $CO_2$ (44 g/mol), $H_2O$ (18 g/mol) [20](#page=20). 1. **Convert grams of glucose to moles:** $10 \text{ g} / 180 \text{ g/mol} = 0.055 \text{ mol glucose}$ [20](#page=20). 2. **Use mole ratios from the balanced equation to find moles of products:** * Moles of $CO_2 = 0.055 \text{ mol glucose} \times \frac{6 \text{ mol } CO_2}{1 \text{ mol glucose}} = 0.33 \text{ mol } CO_2$ (Note: The calculation in the document shows $6 \times 0.055 = 0.33 \text{ mol } CO_2$, but the output shows 15g for CO2 which corresponds to $0.33 \times 44 \approx 14.5$g. The direct calculation $6 \times 0.055 = 0.33 \text{ mol }$ implies $0.33 \times 44 = 14.52$ g. However, the document's final numbers are 15g and 5.9g, which are derived from a slightly different intermediate calculation of moles or rounding. Let's follow the document's result pathway for consistency.) * Moles of $H_2O = 0.055 \text{ mol glucose} \times \frac{6 \text{ mol } H_2O}{1 \text{ mol glucose}} = 0.33 \text{ mol } H_2O$. (Revisiting document's math: $10 \text{ g} / 180 \text{ g/mol} \approx 0.05555... \text{ mol}$. Using this more precise value: $6 \times 0.05555... = 0.3333... \text{ mol } CO_2$. Then $0.3333... \text{ mol } \times 44 \text{ g/mol} = 14.66... \text{ g } CO_2$. The document states 15 g for $CO_2$. For $H_2O$, $0.3333... \text{ mol } \times 18 \text{ g/mol} = 6 \text{ g}$. The document states 5.9 g for $H_2O$. The document's provided numbers for grams (15g and 5.9g) seem to arise from a calculation where the moles of glucose might have been rounded differently, or there's a slight discrepancy in the document's numbers. Sticking to the document's final provided grams: 15 g $CO_2$ and 5.9 g $H_2O$.) [20](#page=20). * **Example:** Magnesium metal ignites in oxygen to form MgO. What mass of Mg reacts completely to give 1.000 g of MgO? [21](#page=21). * Balanced equation: $2Mg + O_2 \rightarrow 2MgO$ [21](#page=21). * Molar masses: Mg (24 g/mol), O (16 g/mol), MgO (24 + 16 = 40 g/mol) [21](#page=21). * From the balanced equation: 2 moles of Mg (2 $\times$ 24 g = 48 g) produce 2 moles of MgO (2 $\times$ 40 g = 80 g) [21](#page=21). * Using proportions: $$ \frac{X \text{ g Mg}}{1.000 \text{ g MgO}} = \frac{48 \text{ g Mg}}{80 \text{ g MgO}} $$ $$ X = 1.000 \text{ g} \times \frac{48 \text{ g Mg}}{80 \text{ g MgO}} = 0.6 \text{ g of Mg} $$ [21](#page=21). #### 1.4.9 Limiting reactants The limiting reactant (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed [22](#page=22). * **Determining the limiting reactant:** 1. Convert the masses of all reactants to moles. 2. For each reactant, calculate how much of the other reactant(s) would be needed to react completely using the mole ratios from the balanced equation. 3. The reactant that is "used up" first (i.e., for which there is insufficient amount of the other reactant) is the limiting reactant. Alternatively, calculate the moles of a specific product that can be formed from each reactant. The reactant that produces the smallest amount of product is the limiting reactant. * **Example:** Reacting 1.0 g of sodium bicarbonate ($NaHCO_3$) with 1.0 g of citric acid ($H_3C_6H_5O_7$) to produce $CO_2$ [22](#page=22). * Balanced equation: $3NaHCO_3(aq) + H_3C_6H_5O_7(aq) \rightarrow 3CO_2(g) + 3H_2O(l) + Na_3C_6H_5O_7(aq)$ [23](#page=23). * Molar masses: $NaHCO_3$ (84 g/mol), $H_3C_6H_5O_7$ (192 g/mol), $CO_2$ (44 g/mol) [23](#page=23). 1. **Convert masses to moles:** * Moles $NaHCO_3 = 1.0 \text{ g} / 84 \text{ g/mol} = 0.012 \text{ mol}$ [23](#page=23). * Moles $H_3C_6H_5O_7 = 1.0 \text{ g} / 192 \text{ g/mol} = 0.0052 \text{ mol}$ [23](#page=23). 2. **Determine limiting reactant by calculating moles of $CO_2$ from each:** * From $NaHCO_3$: $0.012 \text{ mol } NaHCO_3 \times \frac{3 \text{ mol } CO_2}{3 \text{ mol } NaHCO_3} = 0.012 \text{ mol } CO_2$ [24](#page=24). * From $H_3C_6H_5O_7$: $0.0052 \text{ mol } H_3C_6H_5O_7 \times \frac{3 \text{ mol } CO_2}{1 \text{ mol } H_3C_6H_5O_7} = 0.0156 \text{ mol } CO_2$ [24](#page=24). * Since $NaHCO_3$ produces fewer moles of $CO_2$ (0.012 mol vs. 0.0156 mol), sodium bicarbonate is the limiting reactant [24](#page=24). 3. **Calculate the amount of $CO_2$ produced:** Use the moles of $CO_2$ determined by the limiting reactant. * Amount of $CO_2 = 0.012 \text{ mol} \times 44 \text{ g/mol} = 0.53 \text{ g}$ [24](#page=24). #### 1.4.10 Theoretical yield, actual yield, and percent yield * **Theoretical Yield:** The maximum amount of product that can be formed from a given amount of reactants, as calculated by stoichiometry [25](#page=25). * **Actual Yield:** The amount of product that is experimentally obtained and measured [25](#page=25). * **Percent Yield:** A measure of the efficiency of a reaction, calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100%. $$ \% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% $$ [25](#page=25). * **Example:** 5.393 g of 1-bromo-2-methylpropane is formed when 6.034 g of 2-methyl-1-propanol reacts with excess PBr3. Calculate the percent yield [26](#page=26). * We need to determine the theoretical yield of 1-bromo-2-methylpropane ($C_4H_9Br$) from 6.034 g of 2-methyl-1-propanol ($C_4H_9OH$) [27](#page=27). * Assume the reaction is: $C_4H_9OH \rightarrow C_4H_9Br$ (simplified representation, actual reaction involves PBr3) [26](#page=26). * Molar masses: $C_4H_9OH$ (74 g/mol), $C_4H_9Br$ (137 g/mol) [27](#page=27). 1. **Calculate the theoretical yield of $C_4H_9Br$:** $$ \text{Theoretical Yield} = 6.034 \text{ g } C_4H_9OH \times \frac{1 \text{ mol } C_4H_9OH}{74 \text{ g } C_4H_9OH} \times \frac{1 \text{ mol } C_4H_9Br}{1 \text{ mol } C_4H_9OH} \times \frac{137 \text{ g } C_4H_9Br}{1 \text{ mol } C_4H_9Br} $$ The document simplifies this calculation: $x = (3 \times 137 \text{ g } C_4H_9Br) \times 6.034 \text{ g } C_4H_9OH / (3 \times 74 \text{ g } C_4H_9OH) = 11.17 \text{ g } C_4H_9Br$. (Note: The factor of 3 appears to be a placeholder or error from a specific reaction pathway not fully detailed, but we follow the document's calculation) [27](#page=27). 2. **Calculate the percent yield:** $$ \% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{5.393 \text{ g}}{11.17 \text{ g}} \times 100\% = 48.28\% $$ [27](#page=27). ### 1.5 Practice Problems and Applications The document includes several practice problems that apply these stoichiometric principles: 1. **Molarity Calculation:** Calculating the molarity of a sodium phosphate solution [28](#page=28). 2. **Neutralization Reaction:** Determining the volume of sodium hydroxide needed to neutralize phosphoric acid [28](#page=28). 3. **Precipitation Reaction:** Calculating the maximum mass of silver chloride that will precipitate from mixing solutions of magnesium chloride and silver nitrate [29](#page=29). 4. **Industrial Process (Steam Reforming):** Involves writing a balanced equation for the reaction of methane with water vapor and calculating moles of hydrogen formed and moles of water vapor reacted [30](#page=30). * **Balanced Equation:** $CH_4(g) + H_2O(g) \rightarrow CO(g) + 3H_2(g)$ [30](#page=30). * **Moles of hydrogen formed from 4 moles of methane:** Using the 1:3 mole ratio ($CH_4: H_2$), 4 moles of $CH_4$ will produce $4 \times 3 = 12$ moles of $H_2$ [30](#page=30). * **Moles of water vapor reacted to yield 174.82 moles of hydrogen:** Using the 1:3 mole ratio ($H_2O: H_2$), moles of $H_2O$ = $174.82 \text{ moles } H_2 \times \frac{1 \text{ mol } H_2O}{3 \text{ mol } H_2} \approx 58.27 \text{ moles } H_2O$ [30](#page=30). --- # Bonding and intermolecular forces Bonding and intermolecular forces describe the attractive and repulsive forces that hold substances together, determining their physical properties. Intramolecular forces, like covalent bonds, exist within molecules, while intermolecular forces operate between molecules [31](#page=31). ### 2.1 States of matter and intermolecular forces The state of a substance (solid, liquid, or gas) is determined by the balance between the kinetic energy of its molecules and the strength of the intermolecular forces between them [31](#page=31). * **Gases:** The kinetic energy of molecules significantly exceeds the intermolecular attractive forces, allowing molecules to move freely and independently. Deviations from ideal gas behavior are described by Van der Waals [33](#page=33). * **Liquids:** Intermolecular forces are strong enough to keep molecules close together, but not so strong as to lock them into fixed positions, resulting in a lack of long-range order [33](#page=33). * **Solids:** Intermolecular forces are sufficiently strong to hold molecules in fixed positions, leading to a highly ordered structure [33](#page=33). ### 2.2 Strength of intermolecular forces The strengths of intermolecular forces are generally considerably weaker than intramolecular forces like ionic or covalent bonds. For instance, separating molecules might require around 16 kJ/mol, while breaking a bond can demand up to 431 kJ/mol [34](#page=34). ### 2.3 Van der Waals forces Van der Waals forces encompass the relatively weak attractive forces that arise between neutral atoms and molecules due to temporary or permanent electric polarizations. These include London dispersion forces, dipole-dipole interactions, and hydrogen bonding [35](#page=35). #### 2.3.1 Dipole-dipole forces Dipole-dipole forces are attractive forces that occur between polar molecules. Polar molecules possess permanent dipoles arising from differences in electronegativity between bonded atoms. The partially positive end of one molecule is attracted to the partially negative end of another. Substances with dipole-dipole attractions typically exhibit higher melting and boiling points compared to nonpolar molecules which are only influenced by London dispersion forces [36](#page=36). #### 2.3.2 Hydrogen bonding Hydrogen bonding is a particularly strong type of dipole-dipole interaction. It occurs when a hydrogen atom covalently bonded to a highly electronegative and small atom (such as fluorine, oxygen, or nitrogen) is attracted to a lone pair of electrons on another nearby small, electronegative atom (F, O, or N) [37](#page=37). #### 2.3.3 London dispersion forces (LDF) London dispersion forces arise from the transient, induced polarities that form within molecules or atoms due to the random movement and distribution of electrons. These temporary dipoles can then induce similar temporary dipoles in adjacent particles, leading to a weak attraction. LDFs are present in all substances, regardless of whether they are polar or nonpolar [38](#page=38). * **Polarizability:** This refers to the ease with which the electron cloud of an atom or molecule can be distorted to create an instantaneous dipole. Larger molecules tend to be more polarizable than smaller ones [39](#page=39). > **Tip:** The strength of intermolecular forces directly correlates with physical properties like boiling point. Stronger intermolecular forces lead to higher boiling points because more energy is required to overcome these attractions and transition to the gaseous state [39](#page=39). #### 2.3.4 Identifying intermolecular forces A systematic approach helps identify the dominant intermolecular forces present in a substance or mixture [40](#page=40). * **London Dispersion Force:** Present in all molecular mixtures. It is the strongest force in nonpolar molecules [40](#page=40). * **Dipole-Dipole Force:** Present in mixtures containing molecules with permanent dipoles [40](#page=40). * **Hydrogen Bonding:** The strongest type of dipole-dipole interaction. Present in mixtures containing molecules where hydrogen is covalently bonded to nitrogen, oxygen, or fluorine [40](#page=40). ### 2.4 Other types of forces holding solids together Beyond intermolecular forces between discrete molecules, other bonding types exist, particularly in solid structures where individual molecules may not be the fundamental units. #### 2.4.1 Ionic bonding Ionic bonding involves the electrostatic attraction between oppositely charged ions, which are held together by their charges. In ionic solids, there are no discrete molecules; the entire structure is an extended lattice of ions [41](#page=41). #### 2.4.2 Metallic bonding Metallic bonding is characterized by a "sea of electrons". In this type of bonding, valence electrons are delocalized and shared among a large number of metal atoms within a crystal lattice. This occurs because metals have valence electrons that are not tightly held by the nucleus, allowing them to contribute to a collective electron pool [42](#page=42). --- # Energy changes in phase transitions The energy required to change the phase of a substance is directly related to the strength of its intermolecular forces [44](#page=44). ### 3.1 Intermolecular forces and phase transitions Phase transitions, such as melting (solid to liquid) and vaporization (liquid to gas), involve overcoming the attractive forces between molecules. Stronger intermolecular forces mean that more energy, typically in the form of heat, must be supplied to break these attractions and facilitate the phase change [44](#page=44). ### 3.2 Endothermic and exothermic processes Phase transitions can be classified as either endothermic or exothermic based on the direction of heat flow. * **Endothermic processes:** These transitions require the input of energy (heat) from the surroundings. Examples include melting, vaporization, and sublimation (solid to gas). Energy is absorbed by the substance to overcome intermolecular forces [44](#page=44). * **Exothermic processes:** These transitions release energy (heat) into the surroundings. Examples include freezing (liquid to solid), condensation (gas to liquid), and deposition (gas to solid). Energy is released as molecules move into a more ordered state with weaker intermolecular attractions [44](#page=44). > **Tip:** Think of it this way: to break bonds (like in melting or boiling), you need to add energy (endothermic). When bonds form (like in freezing or condensing), energy is released (exothermic). ### 3.3 Relative energy requirements The energy required for a phase transition increases with the strength of the intermolecular forces. Therefore, substances with stronger intermolecular forces will require more energy to melt or vaporize compared to substances with weaker forces. The progression from solid to liquid to gas generally requires increasing amounts of energy because the intermolecular attractions become progressively weaker with each phase change [44](#page=44). --- ## Common mistakes to avoid - Review all topics thoroughly before exams - Pay attention to formulas and key definitions - Practice with examples provided in each section - Don't memorize without understanding the underlying concepts

| Term | Definition | |------|------------| | Chemical Reaction | A process where at least one new substance is produced as a result of a chemical change, often indicated by changes in heat, light, sound, gas production, precipitate formation, or color. | | Stoichiometry | The study of the quantitative relationships between reactants and products in chemical reactions, based on the law of conservation of mass. | | Balanced Equation | A chemical equation where the number of atoms of each element is the same on both the reactant and product sides, adhering to the principle of conservation of matter. | | Reactants | The substances that are present at the beginning of a chemical reaction and undergo chemical change. | | Products | The new substances that are formed as a result of a chemical reaction. | | Combination Reaction | A type of reaction where two or more substances combine to form a single product. | | Decomposition Reaction | A reaction in which a single compound breaks down into two or more simpler substances. | | Combustion Reaction | A rapid reaction involving oxygen, often producing heat and light, and typically involving hydrocarbons reacting to form carbon dioxide and water. | | Neutralization Reaction | A reaction between an acid and a base that produces a salt and water, involving an exchange of ions. | | Displacement Reaction | A reaction where a more reactive element replaces a less reactive element in a compound; also known as substitution or single replacement reactions. | | Double Displacement Reaction | A reaction where ions are exchanged between two reactants, leading to the formation of new compounds. | | Oxidation | The process involving the addition of oxygen or a non-metallic element, or the removal of hydrogen or a metallic element from a compound. | | Reduction | The process involving the addition of hydrogen or a metallic element, or the removal of oxygen or a non-metallic element from a compound. | | Formula Mass | The sum of the masses of the atoms in the empirical formula of a compound. | | Molecular Mass | The sum of the masses of the atoms in the molecular formula of a compound. | | Atomic Mass Unit (amu) | A unit of mass used to express atomic and molecular masses, defined as 1/12th the mass of a carbon-12 atom. | | Mole | The amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. It is a key unit for chemical calculations. | | Avogadro’s Constant (NA) | The number of elementary entities (such as atoms or molecules) in one mole of a substance, approximately equal to $6.022 \times 10^{23}$ entities per mole. | | Molar Mass | The mass of one mole of a substance, typically expressed in grams per mole (g/mol). | | Percentage Composition | The percentage by mass of each element present in a compound, calculated using atomic weights and the molecular mass of the compound. | | Empirical Formula | The simplest whole-number ratio of atoms of each element in a compound. | | Intermolecular Forces | Attractive and repulsive forces that exist between molecules, influencing the physical properties of substances such as their state (solid, liquid, gas) and boiling points. | | Intramolecular Forces | Forces that hold atoms together within a molecule, such as covalent bonds. | | London Dispersion Forces (LDF) | Weak attractive forces that arise from temporary, induced dipoles in molecules due to the random movement of electrons, present in all substances. | | Dipole-Dipole Forces | Attractive forces that occur between polar molecules, where the partially positive end of one molecule is attracted to the partially negative end of another. | | Hydrogen Bonding | A particularly strong type of dipole-dipole interaction that occurs when a hydrogen atom is bonded to a highly electronegative atom (N, O, or F) and is attracted to a lone pair of electrons on another electronegative atom. | | Polarizability | The ease with which the electron cloud of an atom or molecule can be distorted to create a temporary dipole. | | Ionic Bonding | An electrostatic attraction between oppositely charged ions, forming a crystal lattice structure in ionic compounds. | | Metallic Bonding | A type of chemical bonding that arises from the electrostatic attractive force between positively charged metal ions and delocalized valence electrons, forming a "sea of electrons". | | Theoretical Yield | The maximum amount of product that can be produced from a given amount of reactants in a chemical reaction, calculated using stoichiometry. | | Actual Yield | The amount of product that is experimentally obtained and measured in a chemical reaction. | | Percent Yield | A measure of the efficiency of a chemical reaction, calculated as the ratio of the actual yield to the theoretical yield, expressed as a percentage. |