Chemical Reactions and Chemical Quantities

Chapter 4: Chemical Reactions and Chemical Quantities

Effects of Global Warming

• The Greenhouse Effect

  • Greenhouse gases in the atmosphere allow sunlight to enter.

  • Sunlight warms Earth's surface.

  • These gases prevent some of the heat generated by the sunlight from escaping.

  • Earth's average temperature is determined by the balance between incoming and outgoing energy from the sun.

• Global Warming Trends

  • Since $1860$, scientists have measured an average of a $0.7^ ext{o}C$ rise in atmospheric temperature.

  • During the same period, atmospheric $\text{CO}_2$ levels have risen by $38\%$.

  • A key question is whether these two trends are causally linked.

• Sources of $\text{CO}_2$

  • One major source of $\text{CO}_2$ is the combustion reactions of fossil fuels used for energy.

  • Another natural source of $\text{CO}_2$ is volcanic action.

  • Distinguishing whether global warming is natural or human-induced (due to fossil fuel combustion) is a critical scientific and societal challenge.

Chemical Reactions

• Definition: A chemical reaction is a process in which one or more substances are converted into one or more different substances.

• Nature of Change: These processes involve chemical changes in matter, resulting in the formation of new chemical substances.

• Combustion Reaction

  • A specific type of chemical reaction where a substance combines with oxygen ($\text{O}_2$) to form one or more oxygen-containing compounds.

  • Combustion reactions also characteristically emit heat.

Chemical Equations

• Purpose: Chemical equations serve as a shorthand way of describing a chemical reaction.

• Information Provided: They offer crucial information about the reaction:

  • The formulas of reactants (starting materials) and products (substances formed).

  • The states of reactants and products.

  • The relative numbers of reactant and product molecules required. This information is critical for determining the weights of reactants used and the amounts of products that can be made.

• Structure: Reactants are listed on the left, products on the right, separated by an arrow ($\rightarrow$), which indicates the direction of the reaction.

  • $\text{Reactants}\rightarrow\text{Products}$

• States of Matter Abbreviations

  • $(g)$: Gas

  • $(l)$: Liquid

  • $(s)$: Solid

  • $(aq)$: Aqueous (dissolved in water solution)

• Balancing Chemical Equations

  • Necessity: Equations must be balanced to obey the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

  • Method: We adjust the coefficients (the numbers in front of the chemical formulas) of the molecules to ensure that there are equal numbers of atoms of each element on both sides of the reaction arrow.

  • Important Note: Subscripts within chemical formulas must never be changed when balancing an equation, as changing them would alter the identity of the substance.

  • Example: Combustion of Methane

    • Unbalanced: $\text{CH}<em>4(g) + \text{O}</em>2(g) \rightarrow \text{CO}<em>2(g) + \text{H}</em>2\text{O}(g)$

    • Problem identified: The left side has four hydrogen atoms, while the right side has only two. Also, oxygen atoms are unbalanced.

    • Balanced: $\text{CH}<em>4(g) + 2\text{O}</em>2(g) \rightarrow \text{CO}<em>2(g) + 2\text{H}</em>2\text{O}(g)$

• Law of Conservation of Mass in Equations

  • The quantity that must always be the same on both sides of a chemical equation is the number of atoms of each kind.

  • The number of molecules or moles of each kind of molecule does not necessarily have to be the same.

• Balancing Procedure (General Conceptual Plan)

  1. Write a skeletal reaction with correct chemical formulas.

  2. Balance atoms in more complex compounds first.

  3. Balance elements that occur as free elements (e.g., $\text{O}<em>2$, $\text{H}</em>2$, $\text{Fe}$) last.

  4. Clear any fractions by multiplying all coefficients by a common denominator.

  5. Check that all atoms are balanced on both sides.

  • Example: Rusting of Iron

    • Skeletal: $\text{Fe}(s) + \text{O}<em>2(g) \rightarrow \text{Fe}</em>2\text{O}_3(s)$

    • Balanced: $4\text{Fe}(s) + 3\text{O}<em>2(g) \rightarrow 2\text{Fe}</em>2\text{O}_3(s)$

    • Check: $4$ Fe atoms and $6$ O atoms on both sides.

  • Example: Ammonia Reaction

    • Skeletal: $\text{NH}<em>3(g) + \text{O}</em>2(g) \rightarrow \text{NO}(g) + \text{H}_2\text{O}(g)$

    • Balanced: $4\text{NH}<em>3(g) + 5\text{O}</em>2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$

    • Check: $4$ N, $12$ H, $10$ O atoms on both sides.

Reaction Stoichiometry

• Definition: Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.

• Coefficients and Mole Ratios: The coefficients in a balanced chemical reaction specify the relative amounts in moles (or molecules) of each substance involved in the reaction.

  • Example: Octane Combustion

    • $2\text{C}<em>8\text{H}</em>{18}(l) + 25\text{O}<em>2(g) \rightarrow 16\text{CO}</em>2(g) + 18\text{H}_2\text{O}(g)$

    • This means: $2$ molecules of $\text{C}<em>8\text{H}</em>{18}$ react with $25$ molecules of $\text{O}<em>2$ to form $16$ molecules of $\text{CO}</em>2$ and $18$ molecules of $\text{H}_2\text{O}$. The same ratio applies to moles.

    • The stoichiometric ratio can be written as: $2\text{mol C}<em>8\text{H}</em>{18} : 25\text{mol O}<em>2 : 16\text{mol CO}</em>2 : 18\text{mol H}_2\text{O}$

• Mole-to-Mole Conversions

  • Stoichiometric ratios derived from balanced equations are used as conversion factors to convert moles of one substance to moles of another.

  • Example: If $22.0$ moles of $\text{C}<em>8\text{H}</em>{18}$ are burned:

    • Moles of $\text{CO}<em>2$ formed: $\left(22.0\text{ mol C}</em>8\text{H}<em>{18}\right) \times \frac{16\text{ mol CO}</em>2}{2\text{ mol C}<em>8\text{H}</em>{18}} = 176\text{ mol CO}_2$

    • Moles of $\text{H}<em>2\text{O}$ produced: $\left(22.0\text{ mol C}</em>8\text{H}<em>{18}\right) \times \frac{18\text{ mol H}</em>2\text{O}}{2\text{ mol C}<em>8\text{H}</em>{18}} = 198\text{ mol H}_2\text{O}$

  • Example: For $2\text{N}<em>2\text{O}</em>5(g) \longrightarrow 4\text{NO}<em>2(g) + \text{O}</em>2(g)$

    • If $6.8$ mol $\text{N}<em>2\text{O}</em>5$ reacts: $\left(6.8\text{ mol N}<em>2\text{O}</em>5\right) \times \frac{4\text{ mol NO}<em>2}{2\text{ mol N}</em>2\text{O}<em>5} = 13.6\text{ mol NO}</em>2$ ($14\text{ mol NO}_2$ to two significant figures)

• Mass-to-Mass Conversions (Making Molecules)

  • To convert between mass of one substance and mass of another in a reaction, a multi-step conversion is typically required:
    $\text{mass of A} \xrightarrow{\text{molar mass A}} \text{moles of A} \xrightarrow{\text{stoichiometric ratio}} \text{moles of B} \xrightarrow{\text{molar mass B}} \text{mass of B}$

  • Example: Global Carbon Dioxide Emission

    • Estimate mass of $\text{CO}<em>2$ produced if $4.0 \times 10^{15} \text{ g}$ of octane ($\text{C}</em>8\text{H}_{18}$) is burned.

    • Molar mass of $\text{C}<em>8\text{H}</em>{18} = 114.22 \text{ g/mol}$

    • Molar mass of $\text{CO}_2 = 44.01 \text{ g/mol}$

    • Balanced equation: $2\text{C}<em>8\text{H}</em>{18}(l) + 25\text{O}<em>2(g) \rightarrow 16\text{CO}</em>2(g) + 18\text{H}<em>2\text{O}(g)$ $4.0 \times 10^{15}\text{ g C}</em>8\text{H}<em>{18} \times \frac{1\text{ mol C}</em>8\text{H}<em>{18}}{114.22\text{ g C}</em>8\text{H}<em>{18}} \times \frac{16\text{ mol CO}</em>2}{2\text{ mol C}<em>8\text{H}</em>{18}} \times \frac{44.01\text{ g CO}<em>2}{1\text{ mol CO}</em>2} = 1.2 \times 10^{16}\text{ g CO}_2$

  • Example: Mass of $\text{N}<em>2\text{O}</em>5$ (in kg) to mol $\text{NO}_2$

    • Given $2.87 \text{ kg N}<em>2\text{O}</em>5$

    • $\text{Molar Mass N}<em>2\text{O}</em>5 = 108.02 \text{ g/mol}$
      $2.87\text{ kg N}<em>2\text{O}</em>5 \times \frac{1000\text{ g}}{1\text{ kg}} \times \frac{1\text{ mol N}<em>2\text{O}</em>5}{108.02\text{ g N}<em>2\text{O}</em>5} \times \frac{4\text{ mol NO}<em>2}{2\text{ mol N}</em>2\text{O}<em>5} = 53.1 \text{ mol NO}</em>2$

  • Example: Mass of KBr from $15.39 \text{ g Br}_2$

    • Reaction: $2\text{ K}(s) + \text{Br}_2(l) \longrightarrow 2\text{ KBr}(s)$

    • $\text{Molar Mass Br}_2 = 159.80 \text{ g/mol}$

    • $\text{Molar Mass KBr} = 119.00 \text{ g/mol}$
      $15.39\text{ g Br}<em>2 \times \frac{1\text{ mol Br}</em>2}{159.80\text{ g Br}<em>2} \times \frac{2\text{ mol KBr}}{1\text{ mol Br}</em>2} \times \frac{119.00\text{ g KBr}}{1\text{ mol KBr}} = 22.91\text{ g KBr}$

Limiting Reactant, Theoretical Yield, and Percent Yield

• Analogy: Making Pizza

  • Imagine a pizza recipe: $1 \text{ crust} + 5 \text{ oz tomato sauce} + 2 \text{ cups cheese} \rightarrow 1 \text{ pizza}$

  • If you have $4$ crusts, $10$ cups of cheese, and $15$ oz tomato sauce:

    • Crusts: $(4 \text{ crusts}) \times \frac{1 \text{ pizza}}{1 \text{ crust}} = 4 \text{ pizzas}$

    • Cheese: $(10 \text{ cups cheese}) \times \frac{1 \text{ pizza}}{2 \text{ cups cheese}} = 5 \text{ pizzas}$

    • Tomato Sauce: $(15 \text{ oz tomato sauce}) \times \frac{1 \text{ pizza}}{5 \text{ oz tomato sauce}} = 3 \text{ pizzas}$

  • The tomato sauce limits production to $3$ pizzas.

• Limiting Reactant (or Limiting Reagent)

  • The reactant that is completely consumed during a chemical reaction.

  • It limits the amount of product that can be formed because once it's used up, the reaction stops.

  • In the pizza analogy, tomato sauce is the limiting reactant.

• Excess Reactant

  • Reactants that are not completely consumed in a reaction are called excess reactants.

• Theoretical Yield

  • The maximum amount of product that can be made in a chemical reaction based on the amount of the limiting reactant.

  • It represents the ideal maximum product under perfect conditions.

  • In the pizza analogy, $3$ pizzas is the theoretical yield.

• Actual Yield

  • The actual amount of product produced by a chemical reaction, measured experimentally.

  • It is often less than the theoretical yield due to various factors (e.g., side reactions, incomplete reactions, loss during purification).

  • If only $2$ pizzas were made due to burning, dropping, etc., then $2$ pizzas would be the actual yield.

• Percent Yield

  • A measure of the efficiency of a chemical reaction.

  • Calculated as the ratio of the actual yield to the theoretical yield, multiplied by $100\%$.

  • $\text{Percent Yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

  • For the pizza example: $\frac{2 \text{ pizzas}}{3 \text{ pizzas}} \times 100\% = 66.7\%$.

• Calculating Limiting Reactant, Theoretical Yield, and Percent Yield from Reactant Masses

  1. Convert the mass of each reactant to moles.

  2. Using the stoichiometric ratios from the balanced equation, calculate the moles of product that could be formed from each reactant.

  3. The reactant that produces the smallest amount of product (in moles or mass) is the limiting reactant. This smallest amount of product is the theoretical yield.

  4. Convert the theoretical yield from moles to mass (if required).

  5. Calculate the percent yield using the actual yield (if given).

  • Example: Magnesium Oxide Formation

    • Reaction: $2\text{Mg}(s) + \text{O}_2(g) \longrightarrow 2\text{MgO}(s)$

    • Given: $10.1 \text{ g Mg}$, $10.5 \text{ g O}_2$ (reactants); $11.9 \text{ g MgO}$ (actual yield).

    • Molar Masses: $\text{Mg} = 24.31 \text{ g/mol}$, $\text{O}_2 = 32.00 \text{ g/mol}$, $\text{MgO} = 40.31 \text{ g/mol}$

    • Moles of MgO from Mg: $(10.1\text{ g Mg}) \times \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \times \frac{2\text{ mol MgO}}{2\text{ mol Mg}} = 0.4155\text{ mol MgO}$

    • Moles of MgO from O$\text{<em>2}$: $(10.5\text{ g O}</em>2) \times \frac{1\text{ mol O}<em>2}{32.00\text{ g O}</em>2} \times \frac{2\text{ mol MgO}}{1\text{ mol O}_2} = 0.6563\text{ mol MgO}$

    • Limiting Reactant: Since $0.4155\text{ mol MgO}$ (from Mg) is less than $0.6563\text{ mol MgO}$ (from O\text{_2}), Magnesium (Mg) is the limiting reactant.

    • Theoretical Yield (in grams): $(0.4155\text{ mol MgO}) \times \frac{40.31\text{ g MgO}}{1\text{ mol MgO}} = 16.75\text{ g MgO} \approx 16.8\text{ g MgO}$

    • Percent Yield: $\frac{11.9\text{ g MgO (actual)}}{16.75\text{ g MgO (theoretical)}} \times 100\% = 71.0\%$

Specific Chemical Reactions

• Alkali Metal Reactions

  • With Nonmetals: Reactions of alkali metals (Group 1) with nonmetals are vigorous.

    • Example: Sodium ($\text{Na}$) and Chlorine ($\text{Cl}<em>2$) form sodium chloride ($\text{NaCl}$): $2\text{Na}(s) + \text{Cl}</em>2(g) \rightarrow 2\text{NaCl}(s)$

  • With Water: Alkali metals also react vigorously with water ($\text{H}<em>2\text{O}$) to form the dissolved alkali metal ion ($\text{M}^+$), the hydroxide ion ($\text{OH}^-$), and hydrogen gas ($\text{H}</em>2$).

    • General equation: $2\text{M}(s) + 2\text{H}<em>2\text{O}(l) \rightarrow 2\text{M}^+(aq) + 2\text{OH}^-(aq) + \text{H}</em>2(g)$

    • The reactions become progressively more vigorous as one moves down the group (Li < Na < K).

• Halogen Reactions

  • Nature: Halogens (Group 17) are among the most active nonmetals in the periodic table.

  • With Metals: Halogens react with many metals to form metal halides.

    • Since metals tend to lose electrons and halogens tend to gain electrons, metal halides typically contain ionic bonds.

    • Example: Iron ($\text{Fe}$) and Chlorine ($\text{Cl}<em>2$) form iron(III) chloride ($\text{FeCl}</em>3$):
      $2\text{Fe}(s) + 3\text{Cl}<em>2(g) \rightarrow 2\text{FeCl}</em>3(s)$

  • With Hydrogen: Halogens react with hydrogen ($\text{H}_2$) to form hydrogen halides ($\text{HX}$).

    • General equation: $\text{H}<em>2(g) + \text{X}</em>2 \rightarrow 2\text{HX}(g)$

  • With Each Other (Interhalogen Compounds): Halogens also react with each other to form interhalogen compounds.

    • Example: Bromine ($\text{Br}<em>2$) and Fluorine ($\text{F}</em>2$) form bromine monofluoride ($\text{BrF}$):
      $\text{Br}<em>2(l) + \text{F}</em>2(g) \rightarrow 2\text{BrF}(g)$