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.
• $CaCl<em>2 + Na</em>2SO<em>4 \rightarrow CaSO</em>4 + NaCl$ (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: $N<em>2 + O</em>2 \rightarrow 2NO$
• 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 $N<em>2$. Oxygen atoms (red spheres) form $O</em>2$.
• 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: $C<em>3H</em>8 + 5O<em>2 \rightarrow 3CO</em>2 + 4H_2O$

• Reactants: Propane ($C<em>3H</em>8$) and oxygen ($O_2$).

• Products: Carbon dioxide ($CO<em>2$) and water ($H</em>2O$).

• 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.

  • $C<em>3H</em>8$: 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: $H<em>2 + O</em>2 \rightarrow H_2O$
• 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

• $CH<em>4 + 2O</em>2 \rightarrow CO<em>2 + 2H</em>2O$
• 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 $C<em>6H</em>{14} + O<em>2 \rightarrow CO</em>2 + H_2O$

• Balance carbon first.
• $C<em>6H</em>{14} + O<em>2 \rightarrow 6CO</em>2 + H_2O$ (6 carbon atoms on both sides).
• Balance hydrogen next.
• $C<em>6H</em>{14} + O<em>2 \rightarrow 6CO</em>2 + 7H_2O$ (14 hydrogen atoms on both sides).

Balancing Oxygen

• Reactant side: 2 oxygen atoms.
• Product side: 6 ($CO<em>2$) * 2 + 7 ($H</em>2O$) * 1 = 19 oxygen atoms.
• Make reactant side equal to 19 by multiplying by $19/2$.

Adjusting for Whole Numbers

• $C<em>6H</em>{14} + \frac{19}{2}O<em>2 \rightarrow 6CO</em>2 + 7H_2O$
• Coefficients must be whole numbers.
• Multiply the entire equation by 2 to eliminate the fraction.
• $2C<em>6H</em>{14} + 19O<em>2 \rightarrow 12CO</em>2 + 14H_2O$

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: $C<em>6H</em>{12} + O<em>2 \rightarrow CO</em>2 + H_2O$
• Balance carbon: $C<em>6H</em>{12} + O<em>2 \rightarrow 6CO</em>2 + H_2O$
• Balance hydrogen: $C<em>6H</em>{12} + O<em>2 \rightarrow 6CO</em>2 + 6H_2O$
• Balance oxygen: $C<em>6H</em>{12} + 8O<em>2 \rightarrow 6CO</em>2 + 6H_2O$
• 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: $C<em>2H</em>6O + 3O<em>2 \rightarrow 2CO</em>2 + 3H_2O$

• 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

• $\frac{1 \text{ mol } C<em>2H</em>6O}{3 \text{ mol } O<em>2} \text{ or } \frac{3 \text{ mol } O</em>2}{1 \text{ mol } C<em>2H</em>6O}$
• Use dimensional analysis.

Calculation

• $5 \text{ mol } C<em>2H</em>6O \times \frac{3 \text{ mol } O<em>2}{1 \text{ mol } C</em>2H<em>6O} = 15 \text{ mol } O</em>2$

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 ($N<em>2$) is produced when 130 grams of sodium azide ($NaN</em>3$) 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

$2NaN<em>3 -> 3N</em>2$

Molar mass to Nitrogen to 28 g

Solution Summary

• $130 \text{ g } NaN<em>3 \times \frac{1 \text{ mol } NaN</em>3}{65 \text{ g } NaN<em>3} \times \frac{3 \text{ mol } N</em>2}{2 \text{ mol } NaN<em>3} \times \frac{ 28 \text{ g } N</em>2}{1 \text{ mol } N_2}$

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