Thermodynamics, Enthalpy, Internal Energy, and Calorimetry

Relationship Between Enthalpy and Internal Energy

• The fundamental thermodynamic relationship connecting enthalpy change ($\Delta H$) and internal energy change ($\Delta E$) is defined by the equation:   $\Delta H = \Delta E + P\Delta V$   where $P$ represents constant external pressure and $\Delta V$ represents the change in volume of the system.

• Sign Conventions and Energetics:

  • Endothermic Reactions: Heat enters the system from the surroundings ($q > 0$). On an energy level diagram, the reaction begins at a lower enthalpy state and proceeds upward to a higher energy state; thus, $\Delta H$ is positive ($\Delta H > 0$).
  • Exothermic Reactions: Heat exits the system into the surroundings ($q < 0$). On an energy level diagram, the reaction starts at a higher energy level and decreases to a lower energy state; thus, $\Delta H$ is negative ($\Delta H < 0$).

• Conditions Where Enthalpy Approximates Internal Energy ($\Delta H \approx \Delta E$):

  • Reactions Involving No Gases ($\Delta V = 0$):
  • In aqueous-phase reactions or liquid/solid reactions where no gases are consumed or produced (e.g., $\text{aqueous solution} + \text{aqueous solution} \rightarrow \text{solid} + \text{aqueous solution}$), there is no volume change ($\Delta V = 0$).
  • Because $\Delta V = 0$, the pressure-volume work term ($P\Delta V$) equals zero ($W = 0$), making $\Delta H = \Delta E$ exactly.
  • Reactions With No Net Change in Moles of Gas ($\Delta n_{\text{gas}} = 0$):
  • If the number of moles of gaseous reactants equals the number of moles of gaseous products, no expansion or compression occurs ($\Delta V = 0$).
  • Without a change in volume, work is zero ($W = 0$), and $\Delta H = \Delta E$.

• Reactions With Significant Volume Changes:

  • Consider the decomposition of water into hydrogen and oxygen gas:     $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$
  • Converting $2\text{ moles}$ of liquid water into $3\text{ moles}$ of gas ($2\text{ moles } H_2$ and $1\text{ mole } O_2$) at $P = 1.00\,atm$ and $T = 298\,K$ produces a net volume expansion of approximately $\Delta V = 70\,L$.
  • Calculating the work term yields:     $P\Delta V = 7.4\,kJ$
  • For this reaction, the internal energy change is:     $\Delta E = 564\,kJ$
  • Calculating the total enthalpy change yields:     $\Delta H = \Delta E + P\Delta V = 564\,kJ + 7.4\,kJ = 572\,kJ$
  • Even with a volume change as large as $70\,L$, the $P\Delta V$ work term ($7.4\,kJ$) constitutes only a tiny fraction (a few percent) of the overall energy change ($564\,kJ$). Therefore, $\Delta H$ remains a reliable approximation for $\Delta E$ across standard chemical reactions.

Calorimetry Principles and Heat Capacity

• Fundamental Thermal Balance:

  • Heat lost by a system equals heat gained by its surroundings:     $q_{\text{system}} = -q_{\text{surroundings}}$
  • To measure energy changes accurately, calorimetry places an insulating barrier around a localized portion of the surroundings to isolate it from the greater environment.

• Factors Determining Temperature Change ($\Delta T$):

  • The magnitude of temperature change in a substance depends upon three variables:
  1. The quantity of heat transferred ($q$).
  2. The amount of material absorbing heat (measured in moles $n$ or mass $m$).
  3. The intrinsic heat capacity of the material.

• Heat Capacity Metrics and Mathematical Equations:

  • Molar Heat Capacity ($C_m$):
  • Definition: The quantity of heat required to raise the temperature of $1\text{ mole}$ of a substance by $1\,^\circ C$ (or $1\,K$).
  • Units: $J\,mol^{-1}\,^\circ C^{-1}$ or $J\,mol^{-1}\,K^{-1}$.
  • Heat Equation:       $q = n \times C_m \times \Delta T$
  • Specific Heat Capacity ($C_s$):
  • Definition: The quantity of heat required to raise the temperature of $1\text{ gram}$ of a substance by $1\,^\circ C$ (or $1\,K$).
  • Units: $J\,g^{-1}\,^\circ C^{-1}$ or $J\,g^{-1}\,K^{-1}$.
  • Heat Equation:       $q = m \times C_s \times \Delta T$

• Temperature Change Equivalence Between Celsius and Kelvin:

  • One unit degree Celsius is equal in size to one Kelvin.
  • Example Comparison:
  • Initial temperature: $T_{\text{initial}} = 10\,^\circ C = 283\,K$
  • Final temperature: $T_{\text{final}} = 20\,^\circ C = 293\,K$
  • Temperature difference in Celsius:       $\Delta T = 20\,^\circ C - 10\,^\circ C = 10\,^\circ C$
  • Temperature difference in Kelvin:       $\Delta T = 293\,K - 283\,K = 10\,K$
  • Consequently, temperature changes ($\Delta T$) expressed in Celsius and Kelvin are identical and numerical values can be substituted directly.

• Practical and Environmental Applications of Heat Capacity:

  • Liquid water exhibits an exceptionally high specific heat capacity, requiring significant thermal energy to increase its temperature (e.g., heating water for cooking or tea).
  • On an environmental scale, large bodies of water such as Lake Ontario moderate regional temperatures (such as in Toronto) by absorbing vast quantities of heat during warm periods and releasing it slowly during cold periods without undergoing extreme temperature shifts.

• Comparison of Heat Transfer in Metals (Aluminum vs. Iron):

  • Scenario: A $3\text{ gram}$ sample of aluminum and a $3\text{ gram}$ sample of iron are both heated to an initial temperature of $80\,^\circ C$ and submerged into identical water baths at lower temperatures until thermal equilibrium is achieved.
  • Specific heat capacity comparison: Aluminum has a specific heat capacity roughly twice as large as that of iron.
  • Outcome: For the exact same mass ($3\,g$) and starting temperature ($80\,^\circ C$), aluminum transfers approximately twice as much heat ($q$) to the water as iron does, heating its surrounding water to a higher final equilibrium temperature. Thermal energy transfer depends on heat capacity, not the speed of heat release.

Constant-Pressure Calorimetry (Coffee-Cup Calorimetry)

• Experimental Instrumentation and Setup:

  • Constructed using two nested polystyrene/styrofoam coffee cups equipped with a lid, a stirrer, and a thermometer.
  • Operates under constant atmospheric pressure ($P = \text{constant}$).
  • Because pressure is constant, measured heat transfer equals the enthalpy change ($q_p = \Delta H$).
  • System Boundary: The chemical species taking part in bond breaking and formation.
  • Surroundings Boundary: The water/solvent in which the species are dissolved, together with the calorimeter walls, stirrer, and thermometer.

• Core Calculations:

  • Heat absorbed or released by the solution:     $q_{\text{solution}} = m_{\text{solution}} \times C_s \times \Delta T_{\text{solution}}$
  • Heat of the system:     $q_{\text{system}} = -q_{\text{solution}}$

• Reporting Molar Enthalpies:

  • Heat ($q$) and enthalpy ($\Delta H$) are extensive properties that scale directly with the quantity of material reacting.
  • Standard enthalpy changes ($\Delta H_{\text{rxn}}$) are reported as intensive molar quantities ($kJ\,mol^{-1}$ or $J\,mol^{-1}$) by dividing system heat by the moles of limiting reactant involved in the reaction:     $\Delta H_{\text{rxn}} = \frac{q_{\text{system}}}{n_{\text{limiting}}}$

Constant-Volume Calorimetry (Bomb Calorimetry)

• Experimental Instrumentation and Setup:

  • Constructed with a heavy-walled, rigid steel reaction container ("bomb") immersed inside a known volume of water contained in an insulated outer jacket.
  • Equipped with electrical ignition wires to initiate reactions and an internal sample cup.
  • Primarily utilized for combustion reactions supplied with pure oxygen gas ($O_2$).
  • Because the rigid steel container cannot expand ($\Delta V = 0$), expansion work is zero ($W = -P\Delta V = 0$).
  • Heat measured at constant volume equals internal energy change directly ($q_v = \Delta E$).

• Core Calculations:

  • The entire bomb calorimeter apparatus (rigid container + surrounding water bath) is evaluated as a single unit using an overall calorimeter heat capacity ($C_{\text{cal}}$):     $q_{\text{calorimeter}} = C_{\text{cal}} \times \Delta T$
  • Heat of reaction:     $q_{\text{system}} = -q_{\text{calorimeter}}$
  • Molar internal energy change:     $\Delta E_{\text{rxn}} = \frac{q_{\text{system}}}{n_{\text{sample}}}$

Detailed Calorimetry Calculations and Problem Walkthroughs

• Problem 1: Neutralization Reaction in a Coffee-Cup Calorimeter:

  • Reaction: Aqueous hydrochloric acid reacts with aqueous sodium hydroxide:     $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$
  • Experimental Data:
  • Volume of $HCl$ solution: $25\,mL$
  • Volume of $NaOH$ solution: $50\,mL$
  • Total volume of solution: $25\,mL + 50\,mL = 75\,mL$
  • Solution density: $1.00\,g/mL$
  • Total solution mass ($m_{\text{solution}}$): $75\,mL \times 1.00\,g/mL = 75\,g$
  • Specific heat capacity of solution ($C_s$): $4.18\,J\,g^{-1}\,^\circ C^{-1}$
  • Concentration of $HCl$: $0.50\,M$
  • Concentration of $NaOH$: $0.50\,M$
  • Step 1: Calculate Heat Absorbed by Solution ($q_{\text{solution}}$):     $q_{\text{solution}} = m_{\text{solution}} \times C_s \times \Delta T$$q_{\text{solution}} = 75\,g \times 4.18\,J\,g^{-1}\,^\circ C^{-1} \times \Delta T = 693\,J = 0.693\,kJ$
  • Step 2: Determine System Heat ($q_{\text{rxn}}$):     $q_{\text{rxn}} = -q_{\text{solution}} = -0.693\,kJ$
  • Step 3: Determine Moles and Limiting Reactant:
  • Moles of $HCl$:       $\text{Moles } HCl = 0.50\,mol/L \times 0.0250\,L = 0.0125\,mol$
  • Moles of $NaOH$:       $\text{Moles } NaOH = 0.50\,mol/L \times 0.0500\,L = 0.0250\,mol$
  • Stoichiometric ratio is $1:1$. Thus, $HCl$ is the limiting reactant ($0.0125\,mol$).
  • Step 4: Calculate Molar Enthalpy of Reaction ($\Delta H_{\text{rxn}}$):     $\Delta H_{\text{rxn}} = \frac{q_{\text{rxn}}}{n_{\text{limiting}}} = \frac{-0.693\,kJ}{0.0125\,mol} = -55.4\,kJ\,mol^{-1}$

• Problem 2: Combustion of Benzene in a Bomb Calorimeter:

  • Experimental Data:
  • Sample mass of benzene ($C_6H_6$): $0.187\,g$
  • Molar mass of benzene ($C_6H_6$): $78.11\,g\,mol^{-1}$
  • Observed temperature rise of water bath/calorimeter ($\Delta T$): $7.48\,^\circ C$ ($7.48\,K$)
  • Step 1: Calculate Heat Absorbed by Calorimeter ($q_{\text{calorimeter}}$):     $q_{\text{calorimeter}} = C_{\text{calorimeter}} \times \Delta T = 7807\,J = 7.807\,kJ$
  • Step 2: Determine System Heat ($q_{\text{rxn}}$):     $q_{\text{rxn}} = -q_{\text{calorimeter}} = -7.807\,kJ$
  • Step 3: Calculate Moles of Benzene Reacted:     $\text{Moles } C_6H_6 = \frac{0.187\,g}{78.11\,g\,mol^{-1}} = 0.002394\,mol$
  • Step 4: Calculate Molar Internal Energy Change ($\Delta E_{\text{rxn}}$):     $\Delta E_{\text{rxn}} = \frac{q_{\text{rxn}}}{n_{C_6H_6}} = \frac{-7.807\,kJ}{0.002394\,mol} = -3261\,kJ\,mol^{-1} \approx -3200\,kJ\,mol^{-1}$

Questions & Discussion

• System vs. Surroundings Distinction in Aqueous Solutions:
  • Question: In an aqueous solution reaction between $HCl$ and $NaOH$, what constitutes the system and what constitutes the surroundings? Is the water included in the system or surroundings?
  • Response: The system consists exclusively of the chemical reactants and products—specifically the solute molecules and ions ($HCl$ and $NaOH$) undergoing bond breaking and formation to form $NaCl$ and $H_2O$. The surroundings consist of everything else in physical contact with the system. This includes the bulk solvent water molecules in which the ions are dissolved, as well as the coffee cup calorimeter container, the thermometer, and the stirrer. The heat given off by the system raises the temperature of the water solvent, which is part of the surroundings.