Thermochemistry: Energy and Enthalpy Changes

Thermochemistry (CHY 102 - Chapter 6)

6.2 - The Nature of Energy: Key Definitions

• Thermodynamics: The general study of energy and its interconversions.

• Energy (E): The capacity to do work.

• Work (w): The result of a force acting through distance.

• Heat (q): The flow of energy caused by a temperature difference.

• Kinetic energy: Energy associated with motion.

• Thermal energy: Energy associated with temperature.

• Potential energy: Energy associated with position or composition.

• Chemical energy: A form of potential energy associated with the positions of electrons and nuclei within a system.

• System: The specific portion of the universe singled out for investigation.

• Surroundings: Everything with which the system can exchange energy.

• Law of conservation of energy: Energy can neither be created nor destroyed; it can only be transferred from one object or system to another.

Units of Energy

• Units of energy are derived from the definition of kinetic energy ($KE = rac{1}{2}mv^2$ ): $ext{kg m}^2 ext{ s}^{-2} = ext{J}$ (Joule, the SI unit of energy).

• Other common units:

  • $1 ext{ calorie (cal)} = 4.184 ext{ joules (J)}$

  • $1 ext{ Calorie (Cal) or kilocalorie (kcal)} = 1000 ext{ cal} = 4184 ext{ J}$

  • $1 ext{ kilowatt-hour (kWh)} = 3.60 imes 10^6 ext{ J}$

6.3 - The First Law of Thermodynamics: There Is No Free Lunch

• First Law of Thermodynamics: The total energy of the universe is constant.

• Internal energy (U): The sum of the kinetic and potential energies of all the particles that compose a system.

  • $U = U<em>{ ext{kinetic}} + U</em>{ ext{potential}}$

• The change in internal energy ($ext{ extDelta}U$) is measurable, not the absolute internal energy.

  • $ext{ extDelta}U = q + w$

Sign Conventions for $q$, $w$, and $ext{ extDelta}U$

Example Problem: Sign Conventions

• Scenario: A reaction in a flask releases $890 ext{ J}$ of heat to the surroundings and the gas produced performs $450 ext{ J}$ of work on the surroundings by pushing a piston upward.

• Analysis:

  • Heat is released by the system to the surroundings, so $q = -890 ext{ J}$ .

  • Work is done by the system on the surroundings, so $w = -450 ext{ J}$.

  • $ext{ extDelta}U = q + w = (-890 ext{ J}) + (-450 ext{ J}) = -1340 ext{ J}$.

• Conclusion: $q ext{ is } (-), w ext{ is } (-), ext{ extDelta}U ext{ is } (-)$.

• State function: A property of a system that is invariably the same if a system is in the same state. It is independent of any previous history of the system (path-independent).

• Relationship between reactants, products, and internal energy:

  • If reactants have higher internal energy compared to products, energy will be lost from the system over the course of the reaction.

  • If products have higher internal energy compared to reactants, energy will be effectively gained by the system.

6.4 - Quantifying Heat and Work

• Temperature is a measure of the thermal energy within a sample of matter.

• Heat is the transfer of thermal energy.

• Heat transfers thermal energy from a hotter substance to a colder substance until they reach the same temperature, a state known as thermal equilibrium.

• The amount of heat transferred depends on the temperature change and the substance's capacity to hold heat.

Specific Heat Capacity

• Specific heat capacity ($C_s$): The quantity of heat required to change the temperature of $1.0 ext{ g}$ of a substance by $1^ ext{oC}$.

  • Units: $ext{J g}^{-1} {}^ ext{oC}^{-1}$.

• Equation for heat transfer (in the absence of a phase change):

  • $q = m C_s ext{ extDelta}T$

  • Where: $m$ = mass of the substance (g), $C_s$ = specific heat capacity, $ext{ extDelta}T$ = change in temperature ($^ ext{oC}$ or K).

Selected Specific Heat Capacities (at 298 K)

Example 6.1: Calculating Heat Absorption

• Problem: How much heat is absorbed by a $25.175 ext{ g}$ silver dollar ($C_s = 0.235 ext{ J g}^{-1} {}^ ext{oC}^{-1}$) as it warms from $-10.0^ ext{oC}$ (snow temperature) to $37.0^ ext{oC}$ (body temperature)?

• Given: $m = 25.175 ext{ g}$, $T<em>i = -10.0^ ext{oC}$, $T</em>f = 37.0^ ext{oC}$, $C_s = 0.235 ext{ J g}^{-1} {}^ ext{oC}^{-1}$.

• Calculation:

  • $ext{ extDelta}T = T<em>f - T</em>i = 37.0^ ext{oC} - (-10.0^ ext{oC}) = 47.0^ ext{oC}$

  • $q = m C_s ext{ extDelta}T = (25.175 ext{ g}) (0.235 ext{ J g}^{-1} {}^ ext{oC}^{-1}) (47.0^ ext{oC})$

  • $q = 278 ext{ J}$

Types of Systems

• Open system: Both matter and energy can move between the system and the surroundings.

• Closed system: Energy but not matter can move between the system and the surroundings.

• Isolated system: Neither matter nor energy can leave or enter the system.

Heat Transfer in an Isolated System

• In an isolated system, if there's a temperature difference between the system and surroundings, heat will flow until thermal equilibrium is achieved, with no heat lost to the universe.

• Example: A piece of metal at $55^ ext{oC}$ placed in $25^ ext{oC}$ water.

  • Heat transfers from the metal to the water until they reach a common final temperature ($T_f$).

  • The amount of heat gained by the water equals the amount lost by the metal.

  • Equation for heat transfer: $-m<em>{ ext{metal}} C</em>{ ext{metal}} ext{ extDelta}T<em>{ ext{metal}} = m</em>{ ext{H2O}} C<em>{ ext{H2O}} ext{ extDelta}T</em>{ ext{H2O}}$

Example 6.2: Calculating Final Temperature at Thermal Equilibrium

• Problem: A $32.5 ext{ g}$ cube of aluminum ($C<em>s = 0.903 ext{ J g}^{-1} {}^ ext{oC}^{-1}$) initially at $45.8^ ext{oC}$ is submerged into $105.3 ext{ g}$ of water ($C</em>s = 4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1}$) at $15.4^ ext{oC}$. What is the final temperature ($T_f$) of both substances at thermal equilibrium? (Assume an isolated system).

• Given:

  • Al: $m<em>{ ext{Al}} = 32.5 ext{ g}$, $C</em>{ ext{Al}} = 0.903 ext{ J g}^{-1} {}^ ext{oC}^{-1}$, $T_{i, ext{Al}} = 45.8^ ext{oC}$

  • Water: $m<em>{ ext{H2O}} = 105.3 ext{ g}$, $C</em>{ ext{H2O}} = 4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1}$, $T_{i, ext{H2O}} = 15.4^ ext{oC}$

• Calculation:

  • $-m<em>{ ext{Al}} C</em>{ ext{Al}} (T<em>f - T</em>{i, ext{Al}}) = m<em>{ ext{H2O}} C</em>{ ext{H2O}} (T<em>f - T</em>{i, ext{H2O}})$

  • $-(32.5 ext{ g})(0.903 ext{ J g}^{-1} {}^ ext{oC}^{-1})(T<em>f - 45.8^ ext{oC}) = (105.3 ext{ g})(4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1})(T</em>f - 15.4^ ext{oC})$

  • $-29.3475 (T<em>f - 45.8) = 440.6352 (T</em>f - 15.4)$

  • $-29.3475 T<em>f + 1344.02 = 440.6352 T</em>f - 6784.80$

  • $1344.02 + 6784.80 = 440.6352 T<em>f + 29.3475 T</em>f$

  • $8128.82 = 469.9827 T_f$

  • $T_f = rac{8128.82}{469.9827} = 17.3^ ext{oC}$

Pressure-Volume Work

• Pressure-Volume Work: The work that occurs when a volume change takes place against an external pressure.

  • Example: Combustion of gasoline in an engine causes the piston to move, doing P-V work.

• When a chemical reaction occurs, work may be done on the atmospheric substances (expansion) or by them (compression) to accommodate products or reactants.

  • The volume of liquids or solids is often negligible compared to gases, so only the change in moles of gas is typically considered.

• Equation for P-V work (assuming ideal gas):

  • $w = -P ext{ extDelta}V$

  • Using the ideal gas law ($PV=nRT$), if pressure is constant: $P ext{ extDelta}V = ext{ extDelta}n_{ ext{gas}}RT$

  • Therefore, $w = - ext{ extDelta}n_{ ext{gas}}RT$

  • Where: $ext{ extDelta}n_{ ext{gas}}$ = (moles of product gas) - (moles of reactant gas), $R = 8.314 ext{ J mol}^{-1} ext{K}^{-1}$ (ideal gas constant), $T$ = temperature in Kelvin.

Example: P-V Work Calculation

• Reaction at 298K: $C<em>3H</em>8(g) + 5O<em>2(g) ightarrow 3CO</em>2(g) + 4H_2O(l)$

• Determine $ext{ extDelta}n_{ ext{gas}}$:

  • Moles of reactant gas: $1 ext{ mol } C<em>3H</em>8 + 5 ext{ mol } O_2 = 6 ext{ mol}$

  • Moles of product gas: $3 ext{ mol } CO_2$

  • $ext{ extDelta}n_{ ext{gas}} = 3 - 6 = -3 ext{ mol}$

• Calculate work ($w$):

  • $w = -(-3 ext{ mol})(8.314 ext{ J mol}^{-1} ext{K}^{-1})(298 ext{ K})$

  • $w = 7432.8 ext{ J} = 7.43 imes 10^3 ext{ J mol}^{-1}$

Example Problem: P-V Work with Methanoic Acid Combustion

• Problem: Methanoic acid ($CH<em>2O</em>2(l)$) burns in oxygen to produce gaseous carbon dioxide and liquid water. What is the P-V work associated with the combustion of methanoic acid at $298 ext{ K}$? Is work being done by the system or on it?

• Reaction: $CH<em>2O</em>2(l) + rac{1}{2}O<em>2(g) ightarrow CO</em>2(g) + H_2O(l)$

• Determine $ext{ extDelta}n_{ ext{gas}}$:

  • Moles of reactant gas: $rac{1}{2} ext{ mol } O_2$

  • Moles of product gas: $1 ext{ mol } CO_2$

  • $ext{ extDelta}n_{ ext{gas}} = 1 - rac{1}{2} = rac{1}{2} ext{ mol}$

• Calculate work ($w$):

  • $w = -( rac{1}{2} ext{ mol})(8.314 ext{ J mol}^{-1} ext{K}^{-1})(298 ext{ K})$

  • $w = -1238.786 ext{ J mol}^{-1} ext{ or } -1.24 ext{ kJ mol}^{-1}$

• Conclusion: The negative value indicates that work is being done by the system.

6.5 - Measuring $ext{ extDelta}_rU$ for Chemical Reactions: Constant-Volume Calorimetry

• Recall $ext{ extDelta}U = q + w$. By measuring temperature and volume changes, we can determine $ext{ extDelta}U$.

• Calorimetry: The experimental procedure used to measure the heat evolved in a chemical reaction.

• If a reaction is carried out at constant volume ($ext{ extDelta}V = 0$):

  • $w = -P ext{ extDelta}V = 0$

  • Therefore, $ext{ extDelta}U = q_V$ (the heat exchanged at constant volume).

• Constant-volume calorimetry (e.g., using a bomb calorimeter) allows the measurement of $ext{ extDelta}U$ for a chemical reaction by observing the temperature change in the surroundings.

Bomb Calorimeter

• A bomb calorimeter is designed to measure $ext{ extDelta}U$ for combustion reactions.

• It functions as an isolated system.

• The energy released by the combustion reaction is absorbed by the surrounding water and the calorimeter itself.

• The heat of the reaction ($q<em>r$) is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter ($q</em>{ ext{cal}}$):

  • $q<em>r = -q</em>{ ext{cal}}$

• The heat absorbed by the calorimeter is calculated using its heat capacity ($C_{ ext{cal}}$) and the observed temperature change:

  • $q<em>{ ext{cal}} = C</em>{ ext{cal}} imes ext{ extDelta}T$

Example 6.3: Bomb Calorimetry Calculation

• Problem: $1.010 ext{ g}$ of sucrose ($C<em>{12}H</em>{22}O<em>{11}$) undergoes combustion in a bomb calorimeter, causing a temperature increase from $24.922^ ext{oC}$ to $28.331^ ext{oC}$. Given the heat capacity of the bomb calorimeter is $4.901 ext{ kJ }^ ext{oC}^{-1}$, find $ext{ extDelta}</em>rU$ for sucrose combustion in $ext{kJ mol}^{-1}$.

• Given: $m<em>{ ext{sucrose}} = 1.010 ext{ g}$, $T</em>i = 24.922^ ext{oC}$, $T<em>f = 28.331^ ext{oC}$, $C</em>{ ext{cal}} = 4.901 ext{ kJ }^ ext{oC}^{-1}$.

• Calculations:

  • $ext{ extDelta}T = T<em>f - T</em>i = 28.331^ ext{oC} - 24.922^ ext{oC} = 3.409^ ext{oC}$

  • $q<em>{ ext{cal}} = C</em>{ ext{cal}} imes ext{ extDelta}T = (4.901 ext{ kJ }^ ext{oC}^{-1})(3.409^ ext{oC}) = 16.708 ext{ kJ}$

  • $q<em>r = - ext{q}</em>{ ext{cal}} = -16.708 ext{ kJ}$ (This is the heat for the combustion of 1.010 g of sucrose).

  • Molar mass of sucrose ($C<em>{12}H</em>{22}O_{11}$) = $12(12.01) + 22(1.008) + 11(16.00) = 342.30 ext{ g/mol}$.

  • Convert to $ext{kJ mol}^{-1}$: $ext{ extDelta}U = rac{-16.708 ext{ kJ}}{1.010 ext{ g}} imes rac{342.30 ext{ g}}{1 ext{ mol}} = -5663 ext{ kJ mol}^{-1}$.

6.6 - Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure

• Enthalpy (H): A thermodynamic property which is the sum of the internal energy ($U$) of a system and the product of its pressure ($P$) and volume ($V$).

  • $H = U + PV$

• Enthalpy represents the energy associated with the breaking and forming of bonds in a chemical reaction.

• Enthalpy is a state function.

• At constant pressure (e.g., atmospheric pressure), the change in enthalpy ($ ext{ extDelta}H$) is equal to the heat evolved or absorbed ($q_P$):</p><ul><li><p>$ ext{ extDelta}H = ext{ extDelta}U + P ext{ extDelta}V $(from definition)</p></li><li><p>$ ext{ extDelta}U = q + w $(from First Law)</p></li><li><p>$ w = -P ext{ extDelta}V $(for P-V work)</p></li><li><p>Substituting:$ ext{ extDelta}H = (q + w) + P ext{ extDelta}V = (q - P ext{ extDelta}V) + P ext{ extDelta}V = q_P $</p></li><li><p>Therefore,$ ext{ extDelta}H = qP$. For ideal gases,$P ext{ extDelta}V$can be approximated by$ ext{ extDelta}n{ ext{gas}}RT$.</p></li></ul></li></ul><h5 id="69a642ff-a999-4d7c-aafd-ba766e94e934" data-toc-id="69a642ff-a999-4d7c-aafd-ba766e94e934" collapsed="false" seolevelmigrated="true">Exothermic and Endothermic Reactions</h5><ul><li><p><strong>Exothermic reaction:</strong> A chemical reaction that releases heat to its surroundings.</p><ul><li><p>For an exothermic reaction,$ ext{ extDelta}H < 0$. The system loses potential energy, which is converted to thermal energy.</p></li></ul></li><li><p><strong>Endothermic reaction:</strong> A chemical reaction that absorbs heat from its surroundings.</p><ul><li><p>For an endothermic reaction,$ ext{ extDelta}H > 0$. The system gains potential energy by absorbing thermal energy.</p></li></ul></li></ul><h5 id="32b7ec18-84b4-48d7-bd7b-1c2d65c1b8aa" data-toc-id="32b7ec18-84b4-48d7-bd7b-1c2d65c1b8aa" collapsed="false" seolevelmigrated="true">Molecular Perspective of Enthalpy Changes</h5><ul><li><p>The total energy of a system is the sum of its kinetic and potential energies ($E{ ext{total}} = E{ ext{kinetic}} + E_{ ext{potential}}$).</p></li><li><p>Thermal energy is a form of kinetic energy, but the primary source of energy change in chemical reactions is the <strong>potential energy</strong> stored in chemical bonds.</p></li><li><p>Potential energy arises from the electrostatic forces between electrons and protons within atoms and molecules.</p></li><li><p>During a reaction, bonds break and reform, leading to a reorganization of electrons and nuclei. This process changes the potential energy, which is then converted into or absorbed as thermal energy.</p></li></ul><h5 id="72ffce26-a7d5-4bc9-950c-242cde1969b9" data-toc-id="72ffce26-a7d5-4bc9-950c-242cde1969b9" collapsed="false" seolevelmigrated="true">Enthalpy of Reaction ($ ext{ extDelta}_rH$)</h5><ul><li><p><strong>Enthalpy of reaction ($ ext{ extDelta}_rH$):</strong> The enthalpy change for a chemical reaction. Also called heat of reaction. The value typically corresponds to the stoichiometric amounts of reactants and products as written in the balanced chemical equation.</p></li><li><p><strong>Thermochemical reaction:</strong> A chemical equation that includes the associated$ ext{ extDelta}_rH$value.</p><ul><li><p>Example:$C3H8(g) + 5O2(g) ightarrow 3CO2(g) + 4H_2O(l) ext{ } ext{ extDelta}H = -2044 ext{ kJ mol}^{-1}$</p></li><li><p>This means that for every$1 ext{ mole}$of$C3H8$reacted (or$5 ext{ moles}$of$O_2$),$2044 ext{ kJ}$of heat is released.</p></li></ul></li></ul><h5 id="b394969d-bc79-4a31-a65e-0b8088b551e6" data-toc-id="b394969d-bc79-4a31-a65e-0b8088b551e6" collapsed="false" seolevelmigrated="true">Example 6.4: Calculating Heat of Reaction from Stoichiometry</h5><ul><li><p><strong>Problem:</strong> What is the heat of reaction associated with the complete consumption of$47.9 ext{ kg}$of$O_2$?</p></li><li><p><strong>Reaction:</strong>$C3H8(g) + 5O2(g) ightarrow 3CO2(g) + 4H_2O(l) ext{ } ext{ extDelta}H = -2044 ext{ kJ mol}^{-1}$</p></li><li><p><strong>Approach:</strong> Convert mass of$O_2$to moles, then use the stoichiometric ratio with$ ext{ extDelta}H$.</p></li><li><p><strong>Calculations:</strong></p><ul><li><p>Molar mass of$O_2 = 32.00 ext{ g/mol}$.</p></li><li><p>Convert mass to grams:$47.9 ext{ kg } O2 = 47.9 imes 10^3 ext{ g } O2$.</p></li><li><p>Moles of$O2 = rac{47.9 imes 10^3 ext{ g}}{32.00 ext{ g/mol}} = 1496.875 ext{ mol } O2$.</p></li><li><p>Heat ($q$) =$1496.875 ext{ mol } O2 imes rac{-2044 ext{ kJ}}{5 ext{ mol } O2} = -611762.5 ext{ kJ}$.</p></li><li><p>Rounding to relevant significant figures:$q = -6.12 imes 10^5 ext{ kJ}$.</p></li></ul></li></ul><h5 id="52a67f8e-a0fb-4aa5-bb13-76a06ba4dc14" data-toc-id="52a67f8e-a0fb-4aa5-bb13-76a06ba4dc14" collapsed="false" seolevelmigrated="true">Example Problem: Heat of Reaction with Ammonia</h5><ul><li><p><strong>Problem:</strong> What is the heat (in kJ) associated with the complete reaction of$155.0 ext{ g}$of$NH_3$in the following reaction?</p></li><li><p><strong>Reaction:</strong>$4NH3(g) + 5O2(g)
  ightarrow 4NO(g) + 6H_2O(g) ext{ } ext{ extDelta}H = -902.0 ext{ kJ}$</p></li><li><p><strong>Calculations:</strong></p><ul><li><p>Molar mass of$NH_3 = 14.01 ext{ g/mol} + 3(1.008 ext{ g/mol}) = 17.034 ext{ g/mol}$.</p></li><li><p>Moles of$NH3 = rac{155.0 ext{ g}}{17.034 ext{ g/mol}} = 9.0994 ext{ mol } NH3$.</p></li><li><p>Heat =$9.0994 ext{ mol } NH3 imes rac{-902.0 ext{ kJ}}{4 ext{ mol } NH3} = -2052 ext{ kJ}$.</p></li></ul></li></ul><h4 id="e0c2dbbb-7a2c-4616-a9bd-6c0427fd4abf" data-toc-id="e0c2dbbb-7a2c-4616-a9bd-6c0427fd4abf" collapsed="false" seolevelmigrated="true">6.7 - Constant-Pressure Calorimetry: Measuring$ ext{ extDelta}_rH$</h4><ul><li><p>For many aqueous reactions, enthalpy changes ($ ext{ extDelta}_rH$) can be measured using a <strong>coffee-cup calorimeter</strong>.</p></li><li><p>These reactions typically occur under <strong>constant pressure</strong> (atmospheric pressure).</p></li><li><p>Under these conditions,</p><ul><li><p>$qP = ext{ extDelta}rH$</p></li><li><p>The heat energy evolved/absorbed ($q{ ext{soln}}$) by the solution is calculated as:$q{ ext{soln}} = m{ ext{soln}} C{s, ext{soln}} ext{ extDelta}T$</p></li><li><p>The heat of reaction ($qr$) is the negative of the heat absorbed by the solution:$qr = -q_{ ext{soln}}$.</p></li><li><p>To get$ ext{ extDelta}_rH$per mole of reactant, divide by the number of moles of the limiting reactant.</p></li></ul></li></ul><h5 id="a01ee909-c0f3-4b17-8868-22a7c00884d1" data-toc-id="a01ee909-c0f3-4b17-8868-22a7c00884d1" collapsed="false" seolevelmigrated="true">Summary of Calorimetry Types</h5><ul><li><p><strong>Bomb calorimetry:</strong> Occurs at constant volume and measures$q_V = ext{ extDelta}U$for a reaction.</p></li><li><p><strong>Coffee-cup calorimetry:</strong> Occurs at constant pressure and measures$q_P = ext{ extDelta}H$for a reaction.</p></li></ul><h5 id="c53ccc2d-3729-4597-96cc-ae582e4048a9" data-toc-id="c53ccc2d-3729-4597-96cc-ae582e4048a9" collapsed="false" seolevelmigrated="true">Example 6.5: Coffee-Cup Calorimetry Calculation</h5><ul><li><p><strong>Problem:</strong>$0.158 ext{ g}$of Mg metal reacts completely with enough HCl to make$100.0 ext{ mL}$of solution in a coffee-cup calorimeter. The solution's temperature rises from$25.6^ ext{oC}$to$32.8^ ext{oC}$. Assume the dilute aqueous solution has density and specific heat capacity equal to water ($d = 1.00 ext{ g/mL}$,$Cs = 4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1}$). Find$ ext{ extDelta}rH$for the reaction:$Mg(s) + 2HCl(aq)
  ightarrow MgCl2(aq) + H2(g)$, in$ ext{J/mol Mg}$.</p></li><li><p><strong>Given:</strong>$m{ ext{Mg}} = 0.158 ext{ g}$,$V{ ext{soln}} = 100.0 ext{ mL}$,$Ti = 25.6^ ext{oC}$,$Tf = 32.8^ ext{oC}$,$d{ ext{soln}} = 1.00 ext{ g/mL}$,$C{s, ext{soln}} = 4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1}$.</p></li><li><p><strong>Calculations:</strong></p><ul><li><p>Mass of solution:$m{ ext{soln}} = V{ ext{soln}} imes d_{ ext{soln}} = (100.0 ext{ mL})(1.00 ext{ g/mL}) = 100.0 ext{ g}$.</p></li><li><p>$ ext{ extDelta}T = Tf - Ti = 32.8^ ext{oC} - 25.6^ ext{oC} = 7.2^ ext{oC}$.</p></li><li><p>Heat absorbed by solution:$q{ ext{soln}} = m{ ext{soln}} C_{s, ext{soln}} ext{ extDelta}T = (100.0 ext{ g})(4.184 ext{ J g}^{-1} {}^ ext{oC}^{-1})(7.2^ ext{oC}) = 3012.48 ext{ J}$.</p></li><li><p>Heat of reaction ($qr$) for$0.158 ext{ g Mg}$:$qr = -q_{ ext{soln}} = -3012.48 ext{ J}$.</p></li><li><p>Molar mass of Mg =$24.31 ext{ g/mol}$.</p></li><li><p>$ ext{ extDelta}_rH = rac{-3012.48 ext{ J}}{0.158 ext{ g Mg}} imes rac{24.31 ext{ g Mg}}{1 ext{ mol Mg}} = -463590.2 ext{ J/mol}$.</p></li><li><p>Rounding:$ ext{ extDelta}_rH = -4.64 imes 10^5 ext{ J/mol Mg}$.</p></li></ul></li></ul><h4 id="2e5244ce-d31f-4764-8159-ee1e42075ba5" data-toc-id="2e5244ce-d31f-4764-8159-ee1e42075ba5" collapsed="false" seolevelmigrated="true">6.8 - Relationships Involving$ ext{ extDelta}_rH$</h4><p>There are several useful relationships concerning$ ext{ extDelta}_rH$values and chemical reactions:</p><ul><li><p><strong>Multiple of reaction:</strong> If a chemical equation is multiplied by a factor ($n$), its$ ext{ extDelta}_rH$value is also multiplied by that same factor.</p><ul><li><p>Example: If$A + 2B
  ightarrow C$has$ ext{ extDelta}H1$, then$2A + 4B ightarrow 2C$has$ ext{ extDelta}H2 = 2 imes ext{ extDelta}H_1$.</p></li></ul></li><li><p><strong>Reverse of reaction:</strong> If a chemical equation is reversed, the sign of its$ ext{ extDelta}_rH$value is also reversed.</p><ul><li><p>Example: If$A + 2B
  ightarrow C$has$ ext{ extDelta}H1$, then$C ightarrow A + 2B$has$ ext{ extDelta}H2 = - ext{ extDelta}H_1$.</p></li></ul></li><li><p><strong>Sum of reactions (Hess's Law):</strong> If a chemical equation can be expressed as the sum of a series of individual steps, then the$ ext{ extDelta}_rH$for the overall reaction is the sum of the enthalpies of reactions for each individual step.</p></li></ul><h5 id="63c9305c-930a-4cc0-baeb-b743537546b9" data-toc-id="63c9305c-930a-4cc0-baeb-b743537546b9" collapsed="false" seolevelmigrated="true">Hess's Law</h5><ul><li><p><strong>Hess's Law:</strong> States that if a chemical equation can be expressed as the sum of a series of steps, then$ ext{ extDelta}_rH$for the overall equation is the sum of the enthalpies of reactions for each step.</p></li></ul><h5 id="65b7e3ab-143b-42c4-a810-b3b67c229296" data-toc-id="65b7e3ab-143b-42c4-a810-b3b67c229296" collapsed="false" seolevelmigrated="true">Example 6.6: Hess's Law Calculation</h5><ul><li><p><strong>Problem:</strong> Find$ ext{ extDelta}rH$for the reaction:$C(s) + H2O(g)
  ightarrow CO(g) + H_2(g)$.</p></li><li><p><strong>Given reactions with known$ ext{ extDelta}_rH$'s:</strong></p><ol><li><p>$C(s) + O2(g) ightarrow CO2(g) ext{ } ext{ extDelta}H_1 = -393.5 ext{ kJ mol}^{-1}$</p></li><li><p>$2CO(g) + O2(g) ightarrow 2CO2(g) ext{ } ext{ extDelta}H_2 = -566.0 ext{ kJ mol}^{-1}$</p></li><li><p>$2H2(g) + O2(g)
  ightarrow 2H2O(g) ext{ } ext{ extDelta}H3 = -483.6 ext{ kJ mol}^{-1}$</p></li></ol></li><li><p><strong>Manipulation of reactions:</strong></p><ol><li><p>Keep Reaction 1 as is:$C(s) + O2(g) ightarrow CO2(g) ext{ } ext{ extDelta}H_1 = -393.5 ext{ kJ mol}^{-1}$</p></li><li><p>Reverse Reaction 2 and multiply by$ rac{1}{2}$:$ rac{1}{2} (2CO2(g) ightarrow 2CO(g) + O2(g))$<br>$CO2(g) ightarrow CO(g) + rac{1}{2}O2(g) ext{ } ext{ extDelta}H_{2'} = -( rac{1}{2} imes -566.0 ext{ kJ mol}^{-1}) = 283.0 ext{ kJ mol}^{-1}$</p></li><li><p>Reverse Reaction 3 and multiply by$ rac{1}{2}$:$ rac{1}{2} (2H2O(g) ightarrow 2H2(g) + O2(g))$$H2O(g)
  ightarrow H2(g) + rac{1}{2}O2(g) ext{ } ext{ extDelta}H_{3'} = -( rac{1}{2} imes -483.6 ext{ kJ mol}^{-1}) = 241.8 ext{ kJ mol}^{-1}$</p></li></ol></li><li><p><strong>Summing the manipulated reactions:</strong><br>$C(s) + O2(g) ightarrow CO2(g)$<br>$CO2(g) ightarrow CO(g) + rac{1}{2}O2(g)$<br>$H2O(g) ightarrow H2(g) + rac{1}{2}O2(g)$-------------------------------------$C(s) + H2O(g)
  ightarrow CO(g) + H_2(g)$</p></li><li><p><strong>Summing the$ ext{ extDelta}H$values:</strong><br>$ ext{ extDelta}H{ ext{overall}} = ext{ extDelta}H1 + ext{ extDelta}H{2'} + ext{ extDelta}H{3'}$<br>$ ext{ extDelta}H{ ext{overall}} = (-393.5 ext{ kJ mol}^{-1}) + (283.0 ext{ kJ mol}^{-1}) + (241.8 ext{ kJ mol}^{-1})$$ ext{ extDelta}H{ ext{overall}} = 131.3 ext{ kJ mol}^{-1}$</p></li></ul><h5 id="983beefe-b081-418f-b87d-c49324b78251" data-toc-id="983beefe-b081-418f-b87d-c49324b78251" collapsed="false" seolevelmigrated="true">Example Problem: Hess's Law Application</h5><ul><li><p>A multiple-choice question on the slides implies manipulation of chemical equations (reverse, multiply by 3, multiply by 2) to find an overall$ ext{ extDelta}rH$. The solution given is$ ext{ extDelta}rH = 105 ext{ kJ mol}^{-1}$, implying steps similar to the detailed example above would be followed.</p></li></ul><h4 id="0503e271-e086-4e12-b067-6867a8582184" data-toc-id="0503e271-e086-4e12-b067-6867a8582184" collapsed="false" seolevelmigrated="true">6.9 - Determining Enthalpies of Reaction from Standard Enthalpies of Formation</h4><ul><li><p>The enthalpy of a substance is considered zero ($ ext{ extDelta}H = 0$) when it is in its <strong>standard state ($ ext{^ ext{o}}$)</strong>.</p></li></ul><h5 id="dca522e1-9eab-4ba3-9590-ffee0e5a2da3" data-toc-id="dca522e1-9eab-4ba3-9590-ffee0e5a2da3" collapsed="false" seolevelmigrated="true">Standard States Definitions</h5><ul><li><p><strong>Gas:</strong> Pure gas at$1 ext{ bar}$pressure and a specified temperature (often$25^ ext{oC}$).</p></li><li><p><strong>Liquid or Solid:</strong> Pure substance at$1 ext{ bar}$pressure and a specified temperature (often$25^ ext{oC}$).</p></li><li><p><strong>Solution:</strong> Concentration of exactly$1 ext{ mol L}^{-1}$.</p></li></ul><h5 id="c3579974-f7c1-44de-bfe0-da82f0bcb2e9" data-toc-id="c3579974-f7c1-44de-bfe0-da82f0bcb2e9" collapsed="false" seolevelmigrated="true">Standard Enthalpy Change ($ ext{ extDelta}_rH^ ext{o}$)</h5><ul><li><p><strong>Standard Enthalpy Change ($ ext{ extDelta}_rH^ ext{o}$):</strong> The change in enthalpy for a reaction when all reactants and products are in their standard states.</p></li></ul><h5 id="476e29c5-8b81-4e3e-9619-a63b44fc93c9" data-toc-id="476e29c5-8b81-4e3e-9619-a63b44fc93c9" collapsed="false" seolevelmigrated="true">Standard Enthalpy of Formation ($ ext{ extDelta}_fH^ ext{o}$)</h5><ul><li><p><strong>Standard Enthalpy of Formation ($ ext{ extDelta}_fH^ ext{o}$):</strong> The change in enthalpy when$1 ext{ mole}$of a pure compound forms from its constituent elements in their standard states.</p></li><li><p>For a pure element in its standard state,$ ext{ extDelta}_fH^ ext{o} = 0$.</p><ul><li><p>Example: Formation of methane:$C(s, ext{graphite}) + 2H2(g) ightarrow CH4(g) ext{ } ext{ extDelta}_fH^ ext{o} = -74.6 ext{ kJ mol}^{-1}$.</p></li></ul></li></ul><h5 id="4c39b47a-6eee-442b-b54f-2e483b69f47d" data-toc-id="4c39b47a-6eee-442b-b54f-2e483b69f47d" collapsed="false" seolevelmigrated="true">Using$ ext{ extDelta}fH^ ext{o}$to Calculate$ ext{ extDelta}rH^ ext{o}$</h5><ul><li><p>The standard enthalpy of reaction can be calculated using published standard enthalpies of formation values and the following equation, which is derived from Hess's Law:</p><ul><li><p>$ ext{ extDelta}rH^ ext{o} = ext{ extSigma} np ext{ extDelta}fH^ ext{o}( ext{products}) - ext{ extSigma} nr ext{ extDelta}_fH^ ext{o}( ext{reactants})$</p></li><li><p>Where$np$and$nr$are the stoichiometric coefficients for the products and reactants, respectively.</p></li></ul></li></ul><h5 id="5625e33c-6fe4-4368-a30d-8c9fb4c61a27" data-toc-id="5625e33c-6fe4-4368-a30d-8c9fb4c61a27" collapsed="false" seolevelmigrated="true">Example 6.7: Calculating$ ext{ extDelta}_rH^ ext{o}$from Standard Enthalpies of Formation</h5><ul><li><p><strong>Problem:</strong> What is$ ext{ extDelta}_rH^ ext{o}$for the fermentation of glucose?</p></li><li><p><strong>Reaction:</strong>$C6H{12}O6(s) ightarrow 2C2H5OH(l) + 2CO2(g)$</p></li><li><p><strong>Given standard enthalpies of formation ($ ext{ extDelta}_fH^ ext{o}$) from Table 6.5:</strong></p><ul><li><p>$CO_2(g): -393.5 ext{ kJ mol}^{-1}$</p></li><li><p>$C2H5OH(l): -277.6 ext{ kJ mol}^{-1}$</p></li><li><p>$C6H{12}O_6(s, ext{glucose}): -1273.3 ext{ kJ mol}^{-1}$</p></li></ul></li><li><p><strong>Calculation:</strong></p><ul><li><p>$ ext{ extDelta}rH^ ext{o} = [2 ext{ extDelta}fH^ ext{o}(C2H5OH(l)) + 2 ext{ extDelta}fH^ ext{o}(CO2(g))] - [1 ext{ extDelta}fH^ ext{o}(C6H{12}O6(s))]$</p></li><li><p>$ ext{ extDelta}_rH^ ext{o} = [2(-277.6 ext{ kJ mol}^{-1}) + 2(-393.5 ext{ kJ mol}^{-1})] - (-1273.3 ext{ kJ mol}^{-1})$</p></li><li><p>$ ext{ extDelta}_rH^ ext{o} = [-555.2 ext{ kJ mol}^{-1} - 787.0 ext{ kJ mol}^{-1}] - (-1273.3 ext{ kJ mol}^{-1})$</p></li><li><p>$ ext{ extDelta}_rH^ ext{o} = -1342.2 ext{ kJ mol}^{-1} + 1273.3 ext{ kJ mol}^{-1}$</p></li><li><p>$ ext{ extDelta}_rH^ ext{o} = -68.9 ext{ kJ mol}^{-1}$</p></li></ul></li></ul><h4 id="51d492de-2315-4873-b919-021b40e42ca1" data-toc-id="51d492de-2315-4873-b919-021b40e42ca1" collapsed="false" seolevelmigrated="true">Energy Use and the Environment</h4><ul><li><p><strong>Fossil fuels (coal, natural gas, petroleum):</strong> Are highly useful energy sources because their combustion reactions have large negative enthalpies of reaction ($ ext{ extDelta}_rH < 0$), releasing significant heat. This heat is converted to mechanical energy in turbines to generate electricity.</p><ul><li><p><strong>Coal (combustion of carbon):</strong>$C(s) + O2(g) ightarrow CO2(g) ext{ } ext{ extDelta}_rH^ ext{o} = -393.5 ext{ kJ mol}^{-1}$</p></li><li><p><strong>Natural Gas (combustion of methane):</strong>$CH4(g) + 2O2(g)
  ightarrow CO2(g) + 2H2O(g) ext{ } ext{ extDelta}_rH^ ext{o} = -802.3 ext{ kJ mol}^{-1}$</p></li><li><p><strong>Petroleum (combustion of octane):</strong>$C8H{18}(l) + rac{25}{2}O2(g) ightarrow 8CO2(g) + 9H2O(g) ext{ } ext{ extDelta}rH^ ext{o} = -5074.1 ext{ kJ mol}^{-1}$</p></li></ul></li><li><p><strong>Environmental Concerns:</strong></p><ul><li><p>The release of excess carbon dioxide ($CO_2$) from fossil fuel combustion is linked to climate change.</p></li><li><p>Fossil fuels are non-renewable resources, and global energy demands are continuously growing.</p></li></ul></li><li><p><strong>Future Energy Solutions:</strong></p><ul><li><p>Science has the potential to address the world's energy needs by developing more efficient chemical processes for renewable resources.</p></li><li><p>Example: The sun irradiates Earth with approximately$4.2 imes 10^6 ext{ exajoules}$($4.2 ext{ million } imes 10^{18} ext{ J}$) of energy annually. Harvesting just$0.01 ext{%}$$ of this solar radiation could potentially meet global energy demands and eliminate the reliance on fossil fuels and nuclear power. This highlights the importance of research into sustainable energy technologies.