Chemistry Study Notes: Thermochemistry

Chapter 6: Thermochemistry: Energy Flow and Chemical Change

6.1 Forms of Energy and Their Interconversion

  • Thermodynamics is the study of energy and its transformations.

  • Thermochemistry is a branch of thermodynamics that deals with the heat involved in chemical and physical changes.

  • Energy transfer occurs from one object to another manifesting as work and/or heat.

6.2 A Chemical System and its Surroundings

  • The system refers to the contents of the reaction flask.

  • The surroundings comprise everything else including the flask itself.

6.3 Defining System and Surroundings

  • A meaningful study of energy transfer requires clear definitions of both the system and surroundings.

  • Equation: System + Surroundings = Universe.

  • Internal energy, $E$, of a system is the sum of the potential and kinetic energies of all particles present:
    E=E<em>potential+E</em>kineticE = E<em>{potential} + E</em>{kinetic}

  • The total energy of the universe remains constant.

  • A change in the energy of the system must result in an equal and opposite change in the energy of the surroundings.

6.4 Transfer of Internal Energy

  • Change in internal energy ($ riangle E$) is defined as:
    riangleE=E<em>finalE</em>initialriangle E = E<em>{final} - E</em>{initial}

  • This reflects the difference in energy between the products and the reactants.

6.5 Heat and Work: Two Forms of Energy Transfer

  • Energy transfer occurs in two forms:

    • Heat ($q$): energy transferred due to temperature difference between system and surroundings.

    • Work ($w$): energy transferred when an object is moved by a force.

  • Total change in a system's internal energy:
    riangleE=q+wriangle E = q + w

6.6 Sign Conventions for q, w, and ΔE

  • For $q$ (heat):

    • $+$ indicates heat absorbed by the system from surroundings.

    • $-$ indicates heat released by the system to the surroundings.

  • For $w$ (work):

    • $+$ indicates work done on the system by surroundings.

    • $-$ indicates work done by the system on surroundings.

6.7 The Law of Energy Conservation

  • First law of thermodynamics states: the total energy of the universe is constant.

  • Energy can neither be created nor destroyed, only transferred as heat or work.

  • Equation:
    riangleE<em>universe=riangleE</em>system+riangleEsurroundings=0riangle E<em>{universe} = riangle E</em>{system} + riangle E_{surroundings} = 0

6.8 Units of Energy

  • The SI unit of energy is the joule (J):
    1J=1extkgm2/exts21 J = 1 ext{ kg m}^2/ ext{s}^2

  • The calorie is defined as the quantity of energy needed to raise the temperature of 1 g of water by 1°C:

    • Conversion:
      1extcal=4.184J1 ext{ cal} = 4.184 J

  • The nutritional Calorie (Kilocalorie):

    • Conversion:
      1extCal=1000extcal=1extkcal=4184J1 ext{ Cal} = 1000 ext{ cal} = 1 ext{ kcal} = 4184 J

  • British thermal unit (Btu):

    • Conversion:
      1extBtu=1055J1 ext{ Btu} = 1055 J

6.9 Some Quantities of Energy

  • Daily solar energy falling on Earth: 1024J10^{24} J.

  • Energy of a strong earthquake: 1018J10^{18} J.

  • Daily electrical output of Hoover Dam: 1015J10^{15} J.

  • Combustion of 1 mol of glucose: 103J10^{3} J.

  • 1 kilowatt-hour of electrical energy: 106J10^{6} J.

  • Energy from fission of one 235U^{235}U atom: 1015J10^{-15} J.

6.10 Two Different Paths for the Energy Change of a System

  • The total $ riangle E$ remains the same although $q$ and $w$ differ for different paths taken through the system's states.

6.11 Pressure-Volume Work

  • Expanding gas performs PV work on surroundings:
    w=PriangleVw = -P riangle V

6.12 Enthalpy: Chemical Change at Constant Pressure

  • Enthalpy ($H$) defined as:
    H=E+PVH = E + PV
    Thus:
    riangleH=riangleE+PriangleVriangle H = riangle E + P riangle V

  • Under constant pressure conditions, if the volume change is minimal,
    riangleHextisapproximatelyequaltoriangleE.riangle H ext{ is approximately equal to } riangle E.

6.13 ΔH as a Measure of ΔE

  • $ riangle H$ represents the change in heat for a system at constant pressure:
    qp=riangleE+PriangleVq_p = riangle E + P riangle V

  • For reactions that involve gases, when the amount of gas does not change, $ riangle H ext{ approximates } riangle E$.

6.14 Enthalpy of Exothermic and Endothermic Processes

  • Exothermic: releases heat;

    • $ riangle H < 0$.

  • Endothermic: absorbs heat;

    • $ riangle H > 0$.

6.15 Calorimetry

  • Heat exchanged is represented by: q=cimesmimesriangleTq = c imes m imes riangle T where:

    • $q$: heat lost or gained

    • $c$: specific heat capacity

    • $m$: mass in grams

    • $ riangle T = T{final} - T{initial}$.

6.16 Specific Heat Capacities of Elements and Compounds

  • Common specific heat capacities (c) values at 298 K (25°C):

    • Aluminum (Al): 0.897 J/g K

    • Water (H2O(l)): 4.18 J/g K

    • Ethyl alcohol (C2H5OH(l)): 2.438 J/g K

    • Iron (Fe): 0.449 J/g K

6.17 Sample Problem 6.4: Heat and Temperature Change

  • Problem: Calculate heat needed to raise temperature of 125 g of copper from 25°C to 90°C.

    • extc(Cu)=0.385J/gKext{c (Cu)} = 0.385 J/g K

    • T<em>final=90°C,T</em>initial=25°C,riangleT=65°CT<em>{final} = 90°C, T</em>{initial} = 25°C, riangle T = 65°C

  • Solution:
    qCu=cimesmimesriangleT=0.385imes125imes65=3.13imes103Jq_{Cu} = c imes m imes riangle T = 0.385 imes 125 imes 65 = 3.13 imes 10^{3} J

  • Repeat calculation for water:

    • extc(H2O)=4.18J/gKext{c (H2O)} = 4.18 J/g K

    • qH2O=4.18imes125imes65=4.30imes104Jq_{H2O} = 4.18 imes 125 imes 65 = 4.30 imes 10^{4} J

  • Reason for difference: heat required for water is 10.9 times that of copper due to specific heat capacity differences.

6.18 Calorimetry Techniques

  • Constant-Pressure Calorimetry: Illustrated Colloquial setup includes coffee-cup calorimeter, which consists of:

    • Stirrer

    • Thermometer

    • Cork stopper

    • Nested Styrofoam cups for insulation

  • A Bomb Calorimeter measures heat at constant volume.

6.19 Stoichiometry of Thermochemical Equations

  • A thermochemical equation integrates enthalpy $ riangle H$.

  • Sign of $ riangle H$: Indicates exothermic or endothermic reactions.

  • Magnitude of $ riangle H$: Directly proportional to the amount of substance.

6.20 Sample Problem 6.8: Thermochemical Equations

  • Problem: Decompose bauxite ($Al2O3$).

  • Analyze heat transferred to grams of aluminum produced when 3.0 x 10² kJ is absorbed:
    2Al<em>2O</em>3<br>ightarrow4Al+3O2+1676kJ2 Al<em>2O</em>3 <br>ightarrow 4 Al + 3 O_2 + 1676 kJ

  • Plan: Use thermal relations with reactions to convert between moles and heat absorbed.

6.21 Sample Problem 6.8: Solutions

  • Steps indicating conversion of heat kJ to grams:
    extmass(Al)=rac3.0imes103kJrac1676kJ2extmolAlimes26.98g=32.20gext{mass}(Al) = rac{3.0 imes 10^{3} kJ}{ rac{1676 kJ}{2 ext{ mol } Al}} imes 26.98 g = 32.20 g

  • Conversion heat absorbed for $Al2O3$ reaction:

    • Compute resultant from known $ riangle H$ values.

6.22 Hess’s Law

  • Hess’s law: Enthalpy change of an overall process equals sum of changes of individual steps.
    riangleH<em>overall=riangleH</em>1+riangleH<em>2++riangleH</em>nriangle H<em>{overall} = riangle H</em>1 + riangle H<em>2 + … + riangle H</em>n

  • $ riangle H$ values for individual steps can recompute overall $ riangle H$.

6.23 Calculating ΔH for Overall Process

  • Identify target equation and analyze reactants and products.

  • Manipulate known equations and their respective $ riangle H$ values.

  • Reverse signs for negative changes and multiply accordingly.

6.24 Standard Enthalpy of Formation

  • Standard enthalpy of formation, $f^ullet riangle H$, refers to the formation energy of 1 mol compound from elements in standard states.

  • For elements:
    friangleH=0f^\bullet riangle H = 0

  • Common compounds release heat when forming from elements, most have negative $f^ullet riangle H$ values.

6.25 Selected Standard Enthalpies of Formation

  • At 25°C (298 K), common enthalpies include:

    • $Br_2 (l)
      ightarrow 0$

    • $CaCO_3(s)
      ightarrow -1207.6$ kJ

    • $H_2O(g)
      ightarrow -241.8$ kJ

6.26 Sample Problem 6.11: Using ΔH Values

  • Problem: Calculate enthalpy change for the reaction of nitrogen dioxide with oxygen to form dinitrogen pentoxide.

  • Utilize $f^ullet riangle H$ values to assess product and reactant conversions:
    friangleHreaction=extproductsextreactantsf^\bullet riangle H_{reaction} = ext{products} - ext{reactants}

6.27 Sample Problem 6.11: Solution

  • Compute reaction enthalpy changes based on established equations and known value manipulations leading to resultant $H_{rxn} = -219 kJ$.