Chemical Reactions and Equations Notes

Chemical Reactions

  • Chemical reactions are processes where elements and compounds interact to form new compounds. Key aspects include:
    • Formation of new compounds.
    • Breaking of existing bonds.
    • Formation of new bonds.

Evidences of Chemical Reactions

  • Due to the small size of atoms and bonds, we look for indirect evidence of chemical reactions:
    • Change in color.
    • Formation of a solid (e.g., a yellow solid forming when mixing two liquids).
    • Production of heat (e.g., fireworks).
    • Evolution of a gas (bubbling).
    • Noticeable aroma.

Biochemical Reactions (Metabolism)

  • Millions occur every second in biological cells.
  • Produce compounds and energy for living processes.
  • Enable breathing, growth, and reproduction.
  • Involve the breakdown of food compounds, forming waste material, and releasing energy.
  • Energy produced is vital for biological functions like speech, breathing, growth, and reproduction.

Molecular Behavior During Chemical Reactions

  • Atoms, ions, and molecules are in constant motion.
  • When chemicals mix (e.g., green and blue in a container), their molecules move randomly, colliding with each other and the walls of the container.
  • Collisions are random and numerous.
  • Most collisions result in particles bouncing off each other.
  • Occasionally, collisions have enough energy to break existing bonds and form new ones, leading to the creation of new products.

Example of Nitrogen Monoxide Formation

  • Oxygen molecules collide with nitrogen molecules.
  • Some collisions are strong enough to break bonds between oxygen and nitrogen atoms.
  • New bonds form, resulting in nitrogen monoxide (NO).

Chemical Equations

  • Represent chemical changes using a reaction equation.
  • Reactants are on the left-hand side (e.g., nitrogen gas and oxygen gas). The (g) indicates that these molecules are gases.
  • Products are on the right-hand side (e.g., nitrogen monoxide).
  • Reactants and products contain the same atoms but are arranged differently.

Analogy with LEGOs

  • A chemical reaction is like rearranging LEGOs to create new designs.
  • The number of LEGOs (atoms) remains the same before and after the reaction.
  • Only the bonds between atoms change; atoms are neither created nor destroyed.
  • Reactants and products differ but contain the same atoms.

Law of Conservation of Mass

  • Atoms are neither created nor destroyed in chemical reactions.

Example: Sodium Sulfate and Calcium Chloride

  • Sodium sulfate solution + calcium chloride solution = 300.23 grams.
  • After mixing, calcium sulfate (white precipitate) and NaCl form.
  • The mass remains the same (300.23 grams), demonstrating that atoms are conserved.
  • CaCl2 + Na2SO4 \rightarrow CaSO4 + NaCl (The coefficients to balance the equation are discussed later).

Expressing Chemical Reactions

  • A chemical equation describes a chemical reaction.
  • Identifies reactants, products, relative proportions, and physical states.
  • Physical states: solid (s), liquid (l), gas (g).
  • (aq) indicates aqueous, meaning dissolved in water.
  • Reactants and products are separated by a plus sign (+).
  • Reactants are on the left, products on the right, separated by an arrow ($\rightarrow$).
  • Whole number coefficients indicate relative proportions.

Reading a Chemical Equation

  • Example: N2 + O2 \rightarrow 2NO
  • Read as: One nitrogen molecule reacts with one oxygen molecule to produce two molecules of nitrogen monoxide.

Explanation of the Process

  • Nitrogen atoms (purple spheres) form N2. Oxygen atoms (red spheres) form O2.
  • Bonds between nitrogen atoms and oxygen atoms break.
  • One nitrogen atom combines with one oxygen atom to form NO.
  • For every one nitrogen and one oxygen, two NO molecules are formed.
  • Coefficients show that the law of conservation of mass is followed.

Analyzing a Chemical Reaction

Example: C3H8 + 5O2 \rightarrow 3CO2 + 4H_2O

  • Reactants: Propane (C3H8) and oxygen (O_2).

  • Products: Carbon dioxide (CO2) and water (H2O).

  • Coefficients:

    • 1 propane molecule combines with 5 oxygen molecules.
    • Produces 3 carbon dioxide molecules and 4 water molecules.

  • Molar Ratio: Reactants are 1:5 (one propane to five oxygen). Products are 3:4 (three carbon dioxide to four water).

  • Physical State: All reactants and products are gases (g).

  • Atom Count: Analyzing each side of the arrow confirms that the number of atoms is the same.

    • C3H8: 3 carbon atoms, 8 hydrogen atoms.
    • 5O_2: 10 oxygen atoms (5 molecules, each with 2 oxygen atoms).

Product Side

  • 3CO_2: 3 carbon atoms (3 molecules, each with 1 carbon atom), 6 oxygen atoms (3 molecules, each with 2 oxygen atoms).
  • 4H_2O: 8 hydrogen atoms (4 molecules, each with 2 hydrogen atoms), 4 oxygen atoms (4 molecules, each with 1 oxygen atom).
  • Total Atoms: 3 carbon, 8 hydrogen, and 10 oxygen atoms are on both sides.
    • Carbon: 3 on each side.
    • Hydrogen: 8 on each side.
    • Oxygen: 10 on each side.
  • The law of conservation of mass is followed.

Balanced Chemical Equations

  • Total number of each atom type must be equal on both sides of the equation.
  • Displays the lowest whole number ratio of coefficients.

Example: Hydrogen and Oxygen Forming Water

  • Unbalanced equation: H2 + O2 \rightarrow H_2O
  • Reactant side: 2 hydrogen atoms, 2 oxygen atoms.
  • Product side: 2 hydrogen atoms, 1 oxygen atom.
  • This equation is incorrect because it loses one oxygen atom.
  • Unbalanced equations do not follow the law of conservation of mass.

Balancing Chemical Equations

  • A balanced equation demonstrates that the total number of atoms on the reactant side equals the total number on the product side.

Example: Combustion of Methane

  • CH4 + 2O2 \rightarrow CO2 + 2H2O
  • 1 carbon atom (black sphere).
  • 4 hydrogen atoms (white spheres).
  • 4 oxygen atoms (red spheres).
  • Product side:
    • 1 carbon atom.
    • 4 hydrogen atoms.
    • 4 oxygen atoms.
  • Atoms are balanced on both sides.

Steps for Balancing Chemical Equations

  1. Balance one atom at a time.
  2. Start with an atom found in only one compound on each side.
  3. Balance pure elements (e.g., oxygen) at the end.

Example: Balancing C6H{14} + O2 \rightarrow CO2 + H_2O

  • Balance carbon first.
  • C6H{14} + O2 \rightarrow 6CO2 + H_2O (6 carbon atoms on both sides).
  • Balance hydrogen next.
  • C6H{14} + O2 \rightarrow 6CO2 + 7H_2O (14 hydrogen atoms on both sides).

Balancing Oxygen

  • Reactant side: 2 oxygen atoms.
  • Product side: 6 (CO2) * 2 + 7 (H2O) * 1 = 19 oxygen atoms.
  • Make reactant side equal to 19 by multiplying by 19/2.

Adjusting for Whole Numbers

  • C6H{14} + \frac{19}{2}O2 \rightarrow 6CO2 + 7H_2O
  • Coefficients must be whole numbers.
  • Multiply the entire equation by 2 to eliminate the fraction.
  • 2C6H{14} + 19O2 \rightarrow 12CO2 + 14H_2O

Verification

  • Double-check the number of atoms on each side.

    • 2 * 6 = 12 carbon atoms.
    • 2 * 14 = 28 hydrogen atoms.
    • 19 * 2 = 38 oxygen atoms.

  • Product side:

    • 12 * 1 = 12 carbon atoms.
    • 14 * 2 = 28 hydrogen atoms.
    • (12 * 2) + (14 * 1) = 38 oxygen atoms.

  • The equation is balanced.

Practice Problem

  • Balance: C6H{12} + O2 \rightarrow CO2 + H_2O
  • Balance carbon: C6H{12} + O2 \rightarrow 6CO2 + H_2O
  • Balance hydrogen: C6H{12} + O2 \rightarrow 6CO2 + 6H_2O
  • Balance oxygen: C6H{12} + 8O2 \rightarrow 6CO2 + 6H_2O
  • Verification: Carbon (6), Hydrogen (12), Oxygen (18) are balanced.

Stoichiometry Calculations

  • Calculate the maximum amount of product from a given amount of reactant.
  • Determine the mass of product formed from the mass of reactant.

Sandwich Example

  • Analogy: Making a sandwich is like a chemical reaction.
  • 3 bread + 2 turkey + 4 bacon = 1 sandwich
  • If you start with any reactant, you can calculate how much product you can obtain.
  • If two turkey slices give one sandwich, four turkey slices give two sandwiches.
  • Mole ratio or the ratio between reactants and products is key.
  • From the balance equation to make the sandwich, I know that three bread slices give me one sandwich

Expansion

  • If I have 12 breads, how many sandwiches can I use so I can use this as a conversion factor?

Applying Stoichiometry to Chemical Reactions

  • Balance Equation: C2H6O + 3O2 \rightarrow 2CO2 + 3H_2O

  • From the equation, we can say that:

    • The coefficients are moles. (so these are called mole to mole speculators)
    • One mole of ethanol reacts with three moles of oxygen to form two moles of carbon dioxide and three moles of water.

  • Balance equation tells us the mole to mole relationships between any reactant or any product (mole to mole conversion).

    • Example: One mole ethanol gives how many moles of water? Three moles of water.

Mole-to-Mole Relationships

  • How many moles of oxygen are needed to react with five moles of ethanol?
  • Given: 5 moles of ethanol. Find: Moles of oxygen.
  • Relate ethanol and oxygen using the balanced equation.
  • Mole to mole relationship: 1 mole ethanol requires 3 moles oxygen.

Conversion Factor

  • \frac{1 \text{ mol } C2H6O}{3 \text{ mol } O2} \text{ or } \frac{3 \text{ mol } O2}{1 \text{ mol } C2H6O}
  • Use dimensional analysis.

Calculation

  • 5 \text{ mol } C2H6O \times \frac{3 \text{ mol } O2}{1 \text{ mol } C2H6O} = 15 \text{ mol } O2

Mass-to-Mass Stoichiometry

  • Converting mass of any compound into moles learned already.
  • So once you know how to convert that, then you can convert back after doing geometric calculation.

Key steps

Convert mass of a given to moles of the substance using the molar mass concept.

  • Convert moles of reactants to moles of products.
  • Convert moles of the known to moles of the unknown using the coefficients in the balanced equation (mole ratio).
  • Convert moles of the product B into mass of B again using the molar mass concept.

Sample Problem

  • What mass of nitrogen (N2) is produced when 130 grams of sodium azide (NaN3) decomposes in an airbag?

  • From balance equation we can see two moles of sodium azide will give us three moles of nitrogen.

  • First, convert mass to mole by applying the concept of molar mass.

  • the molar mass of sodium azide = 23 + 3(14) = 65grams

  • Second, you convert moles of NaN3 to mole of your product: N2
    From here we will use STOICHIOMETRY

Calculations

  • Starting point - > converting mass of azide to moles right away.
  • The molar mass of azide = 23 + 3(14) = 65grams (one mole divided by number of grams)
    Next step conversion will come from the coefficients

2NaN3 -> 3N2

Molar mass to Nitrogen to 28 g

Solution Summary

  • 130 \text{ g } NaN3 \times \frac{1 \text{ mol } NaN3}{65 \text{ g } NaN3} \times \frac{3 \text{ mol } N2}{2 \text{ mol } NaN3} \times \frac{ 28 \text{ g } N2}{1 \text{ mol } N_2}

=\frac{130 \times 1 \times 3 \times 28}{ 65 \times 2 \times 1} =\frac{10920}{130}= 84 g N_2