General Chemistry Study Guide: Chemical Reactions, Stoichiometry, Solutions, and Gas Laws

Chemical Reactions and Stoichiometry

Types of Chemical Reactions

Chemical reactions can be classified based on how reactants are transformed into products. Understanding reaction types helps predict products and balance equations.

• Double Replacement: Two compounds exchange ions to form two new compounds. Example: PbCl2 forms from Pb(NO3)2 and CuCl2.

• Decomposition: A single compound breaks down into two or more simpler substances. Example: CuCO3 decomposes into CuO and CO2.

Balanced Chemical Equations: Equations must be balanced to obey the law of conservation of mass. Coefficients are adjusted so the number of atoms of each element is the same on both sides.

• Example: CuCO3 → CuO + CO2

Identifying Reaction Types: Look for patterns in reactants and products to classify reactions (e.g., synthesis, decomposition, single/double replacement, combustion).

Stoichiometry

Stoichiometry involves quantitative relationships between reactants and products in chemical reactions.

• Mole-to-mole conversions: Use balanced equations to convert between moles of different substances.

• Example: For Mg + 2HCl → MgCl2 + H2, 1 mol Mg produces 1 mol H2.

General Steps:

1. Write the balanced chemical equation.

2. Convert given quantities to moles.

3. Use mole ratios from the equation to find moles of desired substance.

4. Convert moles to grams or other units as needed.

Solutions and Chemical Quantities

Molarity and Solution Calculations

Molarity (M) is the concentration of a solution, defined as moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Example: 3.2 g glucose (C6H12O6, molar mass = 180 g/mol) in 250 mL:

  • Moles: $\frac{3.2\,g}{180\,g/mol} = 0.018\,mol$

  • Molarity: $\frac{0.018\,mol}{0.250\,L} = 0.071\,M$

Mass-Volume Calculations: Use molarity to find mass or volume of solute/solution.

• Example: 12 mL of 5.5 M solution contains $12\,mL \times \frac{1\,L}{1000\,mL} \times 5.5\,mol/L \times 180\,g/mol = 12\,g$ glucose.

Titration Calculations

Titration is used to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

• Formula: $M_1V_1 = M_2V_2$ (for monoprotic acids/bases)

• Example: 1.25 M × 20.0 mL = 2.5 M × V2 → $V_2 = \frac{1.25 \times 20.0}{2.5} = 10.0\,mL$

Acid-Base Chemistry

Brønsted-Lowry Acids and Bases

The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors.

• Example: NH3 is a base; its conjugate acid is NH4+.

• Conjugate Acid-Base Pairs: Differ by one proton (H+).

pH Calculations: pH is a measure of hydrogen ion concentration.

• Formula: $\text{pH} = -\log[\text{H}^+]$

• Example: If $[\text{H}^+] = 1.5 \times 10^{-6}$, $\text{pH} = -\log(1.5 \times 10^{-6}) = 5.82$

Gas Laws and Properties

Ideal Gas Law

The ideal gas law relates pressure, volume, temperature, and amount of gas.

• Formula: $PV = nRT$

• Variables:

  • P = pressure (atm)

  • V = volume (L)

  • n = moles of gas

  • R = gas constant ($0.08206\,L\cdot atm\cdot mol^{-1}\cdot K^{-1}$)

  • T = temperature (K)

• Standard Temperature and Pressure (STP): 0°C (273.15 K) and 1 atm; 1 mole of gas occupies 22.4 L.

Example Calculation: How many moles of gas in 25 mL at 19°C and 1.45 atm?

• Convert 19°C to K: $19 + 273.15 = 292.15\,K$

• Convert 25 mL to L: $0.025\,L$

• $n=\frac{PV}{RT}=\frac{1.45\times0.025}{0.08206\times292.15}=0.0015\,mol$

Gas Stoichiometry

Gas stoichiometry combines chemical reaction stoichiometry with gas law calculations.

• Example: 0.32 g Mg reacts with HCl to produce H2 gas. Find volume at 20°C and 1.2 atm.

• Calculate moles Mg: $0.32\,g \times \frac{1\,mol}{24.305\,g} = 0.013\,mol$

• Use $PV = nRT$ to find volume: $V = \frac{nRT}{P} = \frac{0.013 \times 0.08206 \times 293.15}{1.2} = 0.26\,L$

Other Gas Laws

Several other gas laws describe relationships between two variables while others are held constant.

• Boyle's Law: $P_1V_1 = P_2V_2$ (at constant T and n)

• Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (at constant P and n)

• Avogadro's Law: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$ (at constant P and T)

Thermochemistry

Heat and Calorimetry

Calorimetry measures heat changes in chemical reactions or physical processes.

• Specific Heat Capacity (c): Amount of heat required to raise the temperature of 1 g of a substance by 1°C.

• Formula: $q = m c \Delta T$

• Example: 18.5 g copper heated from 22°C to 26.1°C:

  • $\Delta T = 26.1 - 22 = 4.1\,°C$

  • $q = 18.5\,g \times 0.374\,J/g\cdot°C \times 4.1\,°C = 28.3\,J$

Calorimeter Calculations: For solutions, $q_{\text{solution}} = m_{\text{solution}} c_{\text{solution}} \Delta T$

• Example: $q_{\text{water}} = 58.5\,mL \times 1.00\,g/mL \times 4.18\,J/g\cdot°C \times 3.8\,°C = 922\,J$

Acid-Base Equilibria

Writing Equations for Acid-Base Reactions

Acid-base reactions involve the transfer of protons between reactants. The products are the conjugate acid and base.

• Example: NH3 + H2O → NH4+ + OH-

Identifying Conjugate Acid-Base Pairs

Conjugate pairs differ by one proton. The acid donates a proton, forming its conjugate base; the base accepts a proton, forming its conjugate acid.

• Example: NH3 (base) → NH4+ (conjugate acid)

Summary Table: Key Equations and Constants

Equation	Purpose	Variables
$PV = nRT$	Ideal Gas Law	P = pressure, V = volume, n = moles, R = gas constant, T = temperature
$M = \frac{\text{mol}}{\text{L}}$	Molarity	mol = moles of solute, L = liters of solution
$q = m c \Delta T$	Heat (calorimetry)	q = heat, m = mass, c = specific heat, $\Delta T$ = temperature change
$\text{pH} = -\log[\text{H}^+]$	Acidity	[H+] = hydrogen ion concentration

Additional info: Some explanations and examples have been expanded for clarity and completeness. All equations are provided in LaTeX format for reference.