Thermochemistry Review

LECTURE 12 THERMOCHEMISTRY #2 - OCT. 7, 2025

Section Overview

• Section 6.3: Calorimetry

• Section 6.4: Stoichiometry of Thermochemical Equations

• Section 6.5: Hess’s Law

• Section 6.6: Standard Enthalpies of Reaction

Learning Objectives

Concepts

• Specific Heat Capacity & Heat (#6.3):

  • Relate specific heat capacity to heat transferred in a reaction.

• Calorimetry (#6.3):

  • Describe how constant-pressure (coffee-cup) calorimeters are constructed and how to experimentally determine values of ΔH.

• ΔH & Amount of Substance (#6.4):

  • Relate ΔH to the amount of substance in reactions.

• Hess’s Law (#6.5):

  • Describe Hess’s law and how it applies to thermochemical reactions.

• Standard Enthalpy of Formation (#6.6):

  • Explain the meaning of standard enthalpy of formation.

Skills

• Solve problems involving specific heat capacity and heat transferred in a reaction (SPs 6.4-6.7).

• Calculate heat transferred in physical or chemical processes based on substances involved (SP 6.8).

• Use Hess’s law to find ΔH for an unknown (“target”) reaction (SP 6.9).

• Use ΔHf° values to find ΔH° for a process (SPs 6.10, 6.11).

Ways to Calculate ΔHrxn (Change in Enthalpy)

1. Calorimetry:

   • Use the equation: q = m imes c imes riangle T

   • To find ΔHrxn: ext{ΔHrxn} = rac{q_{rxn}}{mol ext{ LR}}

2. Stoichiometry:

   • Use ΔHrxn provided for the reaction with given coefficients.

3. Hess's Law:

   • Manipulate given reactions and ΔHrxn values to find ΔHrxn for unknown reactions.

4. Using Heats of Formation:

   • Use the relation: ext{ΔH°} = ext{Σ}n ( ext{products}) - ext{Σ}n ( ext{reactants})

Calorimetry Details

• Calorimetry Equation:

  • q = m imes c imes riangle T .

• Coffee-Cup Calorimeter:

  • Measures heat transfer at constant pressure (qP).

  • Used for reactions in solutions.

  • For the reaction: q{solution} + q{reaction} = 0

  • Rearranged as: q{solution} = -q{reaction} .

Example: Specific Heat Capacity of a Solid

• Given:

  • Mass of solid: 22.05 g,

  • Water mass: 50.00 g,

  • Initial water temperature: 25.10 °C,

  • Final water temperature: 28.49 °C,

  • Specific heat of water: c_{water} = 4.184 ext{ J/g⋅K} .

  • Find the specific heat capacity of the solid.

Heat vs. Enthalpy Change

• The heat absorbed or released (q) is expressed in J or kJ.

  • Enthalpy change (ΔH) is expressed on a molar basis in ext{kJ/mol} .

Stoichiometry of Thermochemical Equations

• Thermochemical Equation:

  • A balanced equation that includes ΔHrxn.

  • Example reaction: ext{CH}4(g) + 2 ext{O}2(g) ightarrow ext{CO}2(g) + 2 ext{H}2 ext{O}(l)

    • Where ext{ΔHrxn} = -891 ext{ kJ} .

• Exothermic vs. Endothermic:

  • The sign of ΔH indicates the nature of the reaction.

  • Example: Combustion of methane is exothermic, releasing 891 kJ of energy.

Example: Heat from Combustion

• Given:

  • Combustion reaction for methane:

  • ext{CH}4(g) + 2 ext{O}2(g)
    ightarrow ext{CO}2(g) + 2 ext{H}2 ext{O}(l)

  • ext{ΔH} = -891 ext{ kJ} .

• Tasks:
  a. Calculate heat from combustion of 3.30 mol CH4.
  b. Calculate change in enthalpy for modified equation:
  2 ext{CH}4(g) + 4 ext{O}2(g)
  ightarrow 2 ext{CO}2(g) + 4 ext{H}2 ext{O}(l) .

Hess’s Law Concepts

• Enthalpy changes for difficult-to-measure reactions can be calculated by manipulating known reactions.

• Hess's Law Principle:

  • When chemical equations are added to yield a different equation, corresponding ΔH values are added to get ΔH for the desired equation.

Rules for Manipulating Chemical Equations

• Rule 1:

  • When an equation is reversed, the sign of its ΔH changes.

  • Example:
    ext{N}2(g) + ext{O}2(g)
    ightarrow 2 ext{NO}(g); ext{ΔH} = 180.6 ext{ kJ}

  • Reversed:
    2 ext{NO}(g)
    ightarrow ext{N}2(g) + ext{O}2(g); ext{ΔH} = -180.6 ext{ kJ} .

• Rule 2:

  • When coefficients are multiplied by a factor, the ΔH value is multiplied by that factor.

  • Example: ext{Ca}(s) + rac{1}{2} ext{O}_2(g) ightarrow ext{CaO}(s); ext{ΔH} = -635.1 ext{ kJ}

    • When doubled:
      2 ext{Ca}(s) + ext{O}_2(g)
      ightarrow 2 ext{CaO}(s); ext{ΔH} = 2 imes (-635.1 ext{ kJ}) = -1270 ext{ kJ} .

• Rule 3:

  • When reactions are summed, the overall reaction's enthalpy is the sum of component reaction enthalpies.

  • Example: ext{C(diamond)} + ext{O}2(g) ightarrow ext{CO}2(g); ext{ΔH} = -395.4 ext{ kJ} and ext{CO}2(g) ightarrow ext{C(graphite)} + ext{O}2(g); ext{ΔH} = 393.5 ext{ kJ}

    • Thus,
      ext{C(diamond)}
      ightarrow ext{C(graphite)}; ext{ΔH} = -395.4 ext{ kJ} + 393.5 ext{ kJ} = -1.9 ext{ kJ} .

Example of Using Hess’s Law

• Pollution Mitigation Reaction:

  • ext{CO(g)} + ext{NO(g)}
    ightarrow ext{CO}2(g) + rac{1}{2} ext{N}2(g); ΔH = ? .

  • Given equations:

  • Equation A: ext{CO(g)} + rac{1}{2} ext{O}2(g) ightarrow ext{CO}2(g); ext{ΔH}_A = -283.0 ext{ kJ} .

  • Equation B: ext{N}2(g) + ext{O}2(g)
    ightarrow 2 ext{NO}(g); ext{ΔH}_B = 180.6 ext{ kJ} .

Standard Enthalpies of Formation

• Standard State:

  • The state of a substance at 25 °C (298 K) and 1 atm pressure, which is stable.

• Standard Enthalpy of Formation, ΔH°f:

  • The enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states.

  • Example:
    ext{H}2(g) + rac{1}{2} ext{O}2(g)
    ightarrow ext{H}_2 ext{O}(l); ΔH°f = -285.8 ext{ kJ/mol} .

  • Note: For any element in its standard state, ΔH°f = 0 .

Practice Example: Formation Reactions

• Which of the following is NOT a correct formation reaction?

  • Options are given:

  • A) ext{H}2(g) + ext{O(g)} ightarrow ext{H}2 ext{O}(l)

  • B) rac{1}{2} ext{H}2(g) + rac{1}{2} ext{Cl}2(g)
    ightarrow ext{HCl}(g)

  • C) 6 ext{C(graphite)} + 3 ext{H}2(g) ightarrow ext{C}6 ext{H}_6(l)

  • D) ext{C(graphite)}
    ightarrow ext{C(diamond)}

  • E) 6 ext{C(graphite)} + 6 ext{H}2(g) + 3 ext{O}2(g)
    ightarrow ext{C}6 ext{H}{12} ext{O}_6(s) .

Calculating ΔH°f Using Heats of Formation

• Equation for calculating the reaction's ΔH°:

  • Δ_{rxn} = ext{Σ}[ ext{ΔH}°(products)] - ext{Σ}[ ext{ΔH}°(reactants)] .

  • where m and n represent moles of products and reactants, respectively.

• Example: Complete combustion of propane at standard conditions:

  • Reaction:
    ext{C}3 ext{H}8(g) + 5 ext{O}2(g) ightarrow 3 ext{CO}2(g) + 4 ext{H}_2 ext{O}(l) .

Practice Calculations

1. For combustion of 1.0 mol ext{C}3 ext{H}8 at 25°C and 1 atm conditions.

   • Setup:

   • [(-285.8) + (-393.5)] - (-74.6)

   • or variations as needed to calculate.

Additional Practice Problems

• Past Exam Questions:

  • Calculate the quantity of heat released when 7.40 g of NaOH dissolves in 100.0 g water in a calorimeter.

  • Heat required to raise the temperature of copper in a skillet.

• More Practice:

  • Given thermochemical equations for finding ΔH rxn for elemental lead extraction from galena, etc.