Theory Exercises

Reaction Rates and Stoichiometry

Part 1: Reaction Rates (Quantitative)

Defining Reaction Rate

Reaction rate is the change in concentration of a substance per unit time.

\[\text{Reaction Rate} = \frac{\text{Change in Concentration}}{\text{Change in Time}} = \frac{\Delta C}{\Delta t}\]

Units: mol/(L·s) or mol·L⁻¹·s⁻¹

Factors Affecting Reaction Rate (Quantitative)

1. Concentration Dependence - Rate Laws

For a simple reaction: A → Products

The rate law relates the reaction rate to reactant concentration:

\[\text{Rate} = k[A]^n\]

Where:

  • k = rate constant (depends on temperature)
  • [A] = concentration of reactant A
  • n = order of reaction (0, 1, 2, or more)

First-order reaction (n = 1):
\[\text{Rate} = k[A]\]
  • Doubling [A] doubles the rate
Second-order reaction (n = 2):
\[\text{Rate} = k[A]^2\]
  • Doubling [A] quadruples the rate

> [Ejemplo: For the reaction: 2NO + O₂ → 2NO₂ > The rate law is: Rate = k[NO]²[O₂] > This shows that the rate depends on the square of NO concentration.]

2. Temperature Dependence - Arrhenius Equation

The relationship between temperature and rate constant:

\[k = Ae^{-\frac{E_a}{RT}}\]

Where:

  • A = pre-exponential factor (frequency of collisions)
  • Ea = activation energy (J/mol)
  • R = gas constant (8.314 J/mol·K)
  • T = temperature (K)
  • e = mathematical constant ≈ 2.718

Key insight: A 10°C increase typically doubles or triples the reaction rate.

Part 2: Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions.

Balanced Chemical Equations

A balanced equation shows the correct ratio of reactants to products.

Example:
\[2H_2 + O_2 \rightarrow 2H_2O\]
Meaning:
  • 2 molecules of hydrogen react with 1 molecule of oxygen
  • to produce 2 molecules of water
  • The ratio is 2 : 1 : 2

How to Balance Equations

  1. Count atoms on each side of the equation
  2. Adjust coefficients (not subscripts) to balance
  3. Check that all elements are balanced

> [Ejemplo: Balance the equation: C + O₂ → CO₂ > > Left side: 1 C, 2 O > Right side: 1 C, 2 O > Already balanced! Answer: C + O₂ → CO₂]

> [Ejemplo: Balance: Fe + O₂ → Fe₂O₃ > > Try: 2Fe + O₂ → Fe₂O₃ > Left: 2 Fe, 2 O > Right: 2 Fe, 3 O (not balanced) > > Try: 4Fe + 3O₂ → 2Fe₂O₃ > Left: 4 Fe, 6 O > Right: 4 Fe, 6 O > Balanced! Answer: 4Fe + 3O₂ → 2Fe₂O₃]

The Mole Concept

A mole is a unit that represents 6.022 × 10²³ particles (Avogadro's number).

Key Conversions:
  • Moles to particles: n × (6.022 × 10²³)
  • Particles to moles: N ÷ (6.022 × 10²³)
  • Moles to grams: n × M (where M = molar mass)
  • Grams to moles: m ÷ M

> [Ejemplo: How many moles are in 18 g of water (H₂O)? > Molar mass of H₂O = 2(1) + 16 = 18 g/mol > Moles = 18 g ÷ 18 g/mol = 1 mol]

Stoichiometric Calculations

Use the balanced equation to convert between quantities of reactants and products.

General steps:
  1. Write the balanced equation
  2. Convert given amount to moles
  3. Use mole ratio from coefficients
  4. Convert to desired units

> [Ejemplo: Given the reaction: 2H₂ + O₂ → 2H₂O > How many grams of water are produced from 4 mol of hydrogen? > > Step 1: From equation, 2 mol H₂ produces 2 mol H₂O > Step 2: 4 mol H₂ × (2 mol H₂O / 2 mol H₂) = 4 mol H₂O > Step 3: 4 mol H₂O × 18 g/mol = 72 g H₂O]

Limiting Reactant

The limiting reactant (or limiting reagent) is the substance that is completely consumed first and determines how much product can be formed.

How to find:
  1. Convert each reactant amount to moles
  2. Divide by its stoichiometric coefficient
  3. The reactant with the smallest value is limiting
  4. Use the limiting reactant to calculate products

> [Ejemplo: Given: 2H₂ + O₂ → 2H₂O > You have 4 mol H₂ and 2 mol O₂ > > H₂: 4 mol ÷ 2 = 2 > O₂: 2 mol ÷ 1 = 2 > > Both give 2, so both are limiting reactants (or in perfect stoichiometric ratio). > Product: 2 × 2 mol H₂O = 4 mol H₂O]

> [Ejemplo: Given: N₂ + 3H₂ → 2NH₃ > You have 1 mol N₂ and 4 mol H₂ > > N₂: 1 mol ÷ 1 = 1 > H₂: 4 mol ÷ 3 = 1.33 > > N₂ has the smallest value (1 < 1.33), so N₂ is limiting > Product: 1 × 2 mol NH₃ = 2 mol NH₃ > H₂ left over: 4 - (1 × 3) = 1 mol H₂]

Percent Yield

Theoretical yield is the maximum amount of product that could form. Actual yield is the amount actually obtained from the experiment.

\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]

> [Ejemplo: A reaction theoretically produces 50 g of product, but only 40 g was actually collected. > Percent Yield = (40 ÷ 50) × 100% = 80%]

Integrated Problem Example

Given: 2Fe + 3Cl₂ → 2FeCl₃

Problem: If 5 mol Fe reacts with 6 mol Cl₂, find:
  1. Limiting reactant
  2. Moles of FeCl₃ produced
  3. Moles of excess reactant remaining
Solution:

Step 1 - Find limiting reactant:

  • Fe: 5 mol ÷ 2 = 2.5
  • Cl₂: 6 mol ÷ 3 = 2.0

Cl₂ has smallest value → Cl₂ is limiting

Step 2 - Calculate FeCl₃:

  • 6 mol Cl₂ × (2 mol FeCl₃ / 3 mol Cl₂) = 4 mol FeCl₃

Step 3 - Calculate excess Fe:

  • Fe consumed: 6 mol Cl₂ × (2 mol Fe / 3 mol Cl₂) = 4 mol Fe
  • Fe remaining: 5 - 4 = 1 mol Fe excess