Thermodynamics: Heat, Energy, and Temperature

Thermodynamics I

• Thermodynamics studies the relationship between chemical reactions and heat
• It examines what causes chemical reactions to occur
• Enthalpy, entropy, and free energy are key concepts in thermodynamics

Heat vs. Temperature

• Heat is a form of energy measured in Joules (J)
• Temperature measures the average kinetic energy of atoms and molecules in a system
• The Kelvin (K) temperature scale is proportional to the average kinetic energy
• If kinetic energy doubles, the Kelvin temperature also doubles

Boltzmann Distribution and Temperature

• The distribution of kinetic energies increases as temperature increases
• The average kinetic energy of particles in a system increases as temperature increases

Energy Transfer Between Systems

• Two systems at different temperatures in thermal contact exchange energy (heat)
• Energy transferred to one system equals the energy transferred from the other system

Conservation of Energy

• Energy is neither created nor destroyed, but transformed from one form to another
• A system’s energy changes during chemical reactions, phase changes, or temperature changes
• These changes result in energy transfer into or out of the system as heat or work

Systems and Surroundings

• The system references the actual chemical reaction taking place
• The surroundings encompass the entire universe outside the chemical reaction

Change in a System's Internal Energy (ΔE)

• ΔE = q + w, where ΔE is the change in potential and kinetic energy of particles
• q = Heat transferred into (+ value) or out of (- value) the system
• w = Work done on the system by the surroundings (+ value) or work done on the surroundings by the system (- value)
• Work stems from changes in gas volume

Heat and Endothermic Processes

• In an isolated system, energy is conserved
• In endothermic reactions, products contain more potential energy (PE) but less kinetic energy (KE)
• Example reaction: CH3OH(g) → CO(g) + 2H2(g), ΔH = +90.7 kJ
• Products of endothermic reactions are at a lower temperature
• Heat flows from the surroundings into the system until thermal equilibrium is reached because the products are at a lower temperature than the surroundings

Heat and Exothermic Processes

• In an isolated system, energy is conserved
• During exothermic reactions, products contain less potential energy (PE) but more kinetic energy (KE)
• Example reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
• Products of exothermic reactions are at a higher temperature
• Heat flows from the system into the surroundings until thermal equilibrium is established because the products are at a higher temperature than the surroundings

Work and Endothermic Processes

• If the pressure caused by gas particles colliding with the piston on the outside of the cylinder is greater than that inside the cylinder, the gas outside does work on the piston
• Energy is transferred from the gas to the piston, and the gas inside the cylinder contracts

Work and Exothermic Processes

• If the pressure caused by gas particles colliding with the piston on the inside of the cylinder is greater than that outside the cylinder, the gas inside does work on the piston
• Energy is transferred from the gas to the piston, and the gas inside the cylinder expands

Transfer of Energy Through Work

• Associated with changes in the volume of a gas
• Work is calculated as w = -PΔV, where P is pressure and ΔV is the change in volume
• The negative sign ensures the correct sign for work

Ex 1) Transfer of Energy Through Work

• A gas expands from 0.87 L to 2.46 L in a cylinder under a constant external pressure of 1.05 atm.
• The work associated with the expansion is calculated as follows:
  • w = -PΔV = -P(Vf - Vi)
  • w = -1.05 atm (2.46 L - 0.87 L)
  • w = -1.67 L·atm
  • w = -1.67 L·atm × (101.3 J / 1 L·atm)
  • w = -169 J
• The system expands into the surroundings in an exothermic process

Ex 2) Transfer of Energy Through Work

• 1142 J of heat are added to a cylinder, resulting in the contained gas expanding from 1.35 L to 4.18 L under a constant external pressure of 1.03 atm
• Internal energy change (ΔE) is calculated below
  • w = -PΔV = -P(Vf - Vi)
  • w = -1.03 atm (4.18 L - 1.35 L)
  • w = -2.91 L·atm
  • w = -2.91 L·atm × (101.3 J / 1 L·atm)
  • w = -295 J
  • ΔE = q + w
  • ΔE = +1142 J + (-295 J)
  • ΔE = 847 J
• This process is endothermic

Enthalpy

• Enthalpy change (ΔH) is heat released or absorbed in a chemical reaction
• Enthalpy change calculation: ΔH = H(products) - H(reactants)

Enthalpy change

• For burning hydrogen fuel: 2 H2(g) + O2(g) → 2 H2O (l) ; ΔH = -572 kJ
• For making hydrogen gas: 2 H2O (l) → 2 H2(g) + O2(g) ; ΔH = 572 kJ

Methods for Finding ΔH

• Calculate using average bond enthalpies
• Measure using a calorimeter
• Calculate using Hess's Law
• Calculate it using enthalpies of formation

Bond Energy

• The potential energy of valence electrons decreases as they approach the nucleus of another atom.

Bond Enthalpy

• Energy is always released during bond formation
  • As atoms move closer together, their potential energy decreases
• The same amount of energy must be added to break a specific bond
  • As atoms move away from each other, their potential energy increases

ΔH from Bond Enthalpies

• ΔH = Σ BE(bonds broken) – Σ BE(bonds formed), where BE = Bond Enthalpy representing the amount of energy needed to break a bond and equal to the energy released when said bond is formed

Exothermic Processes

• More energy is released during the formation of bonds in the products than is required to break the bonds in the reactants
• Products are at a lower potential energy than the reactants

Endothermic Processes

• Less energy is released during the formation of bonds in the products than it requires to break the bonds in the reactants
• Products are at a higher potential energy than the reactants

Ex 1) ΔH from Bond Enthalpies

• Calculate the standard enthalpy change for the following reaction using average bond enthalpies: CH4 + 2 O2 → CO2 + 2 H2O

• ΔH = Σ BE(bonds broken) – Σ BE(bonds formed)
• ΔH = [4(BEC-H) + 2(BEO=O)] - [2(BEC=O) + 4(BEO-H)]
• ΔH = [4(413) + 2(495)] – [2(799) + 4(467)]
• ΔH = 2642 kJ/mol – 3466 kJ/mol
• ΔH = - 824 kJ/mol

Ex 2) ΔH from Bond Enthalpies

• Calculate the standard enthalpy change for the following reaction using average bond enthalpies: 2 H2 + O2 → 2 H2O (ΔH = – 572 kJ) average bond energy in an H – H bond

• ΔH = Σ BE(bonds broken) – Σ BE(bonds formed)
• ΔH = [2(BEH-H)+(BEO=O)]-4(BEO-H)
• 2(BEH-H) = - [(BEO = O)-4(BEO-H) – ΔH]
• 2(BEH-H)= -(BEO=0)+4(BEO-H) + ΔH
• 2(BEH-H) = - 495 kJ/mol + 4(467 kJ/mol) – 572 kJ/mol
• BEH-H = 801 kJ/mol / 2 = 401 kJ/mol