Calorimetry and Enthalpy Concepts

Review and Introduction to Calorimetry

  • Last session: Started Chapter 5, progressed significantly through it.

  • Stopped at BOM calorimetry, which is the final section of calorimetry covered in this course.

Key Review Questions in Calorimetry

Conceptual Question on Heat Transfer

  • Question: If 25 kilojoules of heat enter a system and the system does 10 kilojoules of work, what is dealt to you?

    • Understanding the question: Looking for implications of energy transfer involving heat and work,

    • Internal Energy: Symbolized as 'E'.

    • Equation from prior lesson: ext{U} = Q + W

      • Where:

      • Q = Heat entered into the system

      • W = Work done by the system

    • Positive vs. negative signs:

      • Anything entering the system (i.e., heat) is positive, while anything leaving (i.e., work) is negative.

      • Heat entering: 25 kilojoules (positive)

      • Work done by the system: 10 kilojoules (as volume increases, the sign is negative)

    • Solution:

      • Total change in energy (E): E = 25 ext{ kJ} - 10 ext{ kJ} = 15 ext{ kJ}

Heat Capacity and Energy Requirements

  • Question: Which of the following requires the least energy to raise the temperature by one degree Celsius?

    • Set:

      • 20 grams (Heat capacity = 0.5 J/g°C)

      • 20 grams (Heat capacity = 0.25 J/g°C)

      • 10 grams (Heat capacity = 0.5 J/g°C)

      • 10 grams (Heat capacity = 1 J/g°C)

      • 5 grams (Heat capacity = 0.2 J/g°C)

    • Equation: Q = C imes m imes ext{ΔT} ( where)

      • C = heat capacity

      • m = mass of the substance

      • ext{ΔT} = change in temperature

    • Calculation for energy requirements with ext{ΔT} = 1

      • Energy required for each:

      • 20 g at 0.5 J/g°C: 20 imes 0.5 imes 1 = 10 ext{ J}

      • 20 g at 0.25 J/g°C: 20 imes 0.25 imes 1 = 5 ext{ J}

      • 10 g at 0.5 J/g°C: 10 imes 0.5 imes 1 = 5 ext{ J}

      • 10 g at 1 J/g°C: 10 imes 1 imes 1 = 10 ext{ J}

      • 5 g at 0.2 J/g°C: 5 imes 0.2 imes 1 = 1 ext{ J}

    • Conclusion: The one that requires the least energy is for the 5 g at 0.2 J/g°C.

Conceptual Question About Specific Heat

  • Question: Which of the following metals has the largest change in temperature if 10,000 joules were added?

    • Metals included: Aluminum, Gold, Silver, Platinum

    • Specific Heat values (hypothetical):

      • Aluminum: 0.900 J/g°C

      • Gold: 0.129 J/g°C

      • Silver: 0.240 J/g°C

      • Platinum: 0.133 J/g°C

    • Concept: Higher specific heat means less temperature change for the same amount of energy.

    • Behavior of water vs. metals: Water heats up slower than metals.

Calculation of Temperature Change in Context of Water

  • Question: How many degrees does the temperature of water change if 1,050 joules of heat is lost from 5 grams of water?

    • Specific heat of water: 4.184 J/g°C

    • Using the formula: Q = mC ext{ΔT}

      • Transitioning to find ext{ΔT}

      • Rearranged: ext{ΔT} = - rac{Q}{mC}

    • Inserting values:

      • - rac{1050 ext{ J}}{(5 ext{ g}) imes (4.184 ext{ J/g°C})}

      • Result: Calculate ext{ΔT}, determining it should yield a negative number reflecting a drop in temperature, specifically 50°C.

Introduction to BOM Calorimetry

  • Definition: BOM calorimetry is often regarded as the simplest form of calorimetry, focused on combustion reactions.

  • Key Features:

    • Measures heat released or absorbed during combustion reactions.

    • Uses a specific heat value ( ext{C}_{sample}) for the calorimeter.

  • Equation: To calculate heat change in bom calorimetry, the primary equation is:

    • Q = C imes m imes ext{ΔT}

    • Measure how much heat upsurges the temperature in the calorimeter.

Using BOM Calorimetry in Problems

  • Example problem setup: Using 0.1 moles, 8.2 kJ/°C. What is the temperature change?

    • Pay attention to input units: moles serve to add complexity.

    • Result calculated through formula manipulation, resulting in temperature changes that can be tied to energy provided during combustion.

Transition to Enthalpy

  • What is Enthalpy?: A state function representing the total heat content of a system under constant pressure (denoted as H).

    • Formula relating heat and enthalpy: ext{ΔH} = ext{ΔE} + P ext{ΔV}

    • If $ ext{ΔV} = 0$, then ext{ΔH} = Q_P.

    • Implies that enthalpy is pivotal because it's typically constant pressure.

Endothermic vs Exothermic Reactions

  • Definition of reactions:

    • If ext{ΔH} is negative, the reaction is exothermic (releases heat).

    • If ext{ΔH} is positive, the reaction is endothermic (absorbs heat).

  • Examples of reactions:

    • Ice melting is endothermic (requires energy input).

    • Combustion of gasoline is exothermic (energy is released).

Problems with Enthalpy and Standard Reactions

  • Comprehension of how to connect available data for standard reactions with enthalpy changes.

    • Reaction: Combustion of propane $ ext{C}3 ext{H}8 + 5O2 ightarrow 3CO2 + 4H_2O, ext{ΔH} = -2043 kJ per mole}.$

  • The importance of using ΔH values originating from literature to determine feasibility in practical applications.

Hess’s Law Overview

  • Hess’s Law: Addresses multi-step reactions, relying on the fact that enthalpy is a state function.

    • Procedure in Hess's Law problems: Identify how reactions can be manipulated, reversed, and combined to find overall $ ext{ΔH}$.

    • For each manipulation:

      • Multiplication of reactions changes $ ext{ΔH}$ based on magnitudes.

      • Flipping reactions inversely changes the sign of $ ext{ΔH}$.

      • Sum of multiple steps gives the total change in enthalpy.

  • Illustrative Example:

    • Given specific data, derive through relational equations tying together known enthalpies and balancing chemical equations for desired outcomes.

  • Final Note: Expect problems from calorimetry topics and public application of $ ext{ΔH}$ ranges in exam questions, preparation with illustrative data will enhance grasp of concepts.