Chapter 6. Thermodynamics

6.1  Endothermic and Exothermic Processes

6.1.1

6.1.2

6.1.3

6.1.4

PART A. The process is endothermic because energy is absorbed from the can (surroundings) by the expanding gas.

Detailed explanation here.

PART B. The process is exothermic because the water releases energy to the surroundings while freezing.

Detailed explanation here.

PART C. The process is exothermic because the calcium chloride system releases energy to melt the ice (surroundings).

Detailed explanation here.

PART A. The process is endothermic because energy is absorbed from the surroundings to dissolve ammonium nitrate.

Detailed explanation here.

PART B. The surroundings (beaker and water) will cool down as energy is absorbed during the dissolution process.

Detailed explanation here.

Reactants' Energy: Keep the same in both diagrams (e.g., 150 kJ).

Forward Activation Energy: Maintain 150 kJ in both.

Energy Difference (ΔH):

  • Keep the absolute value the same.

  • Invert the sign: Endothermic (original) to Exothermic (new).

Products' Energy: Set at 50 kJ for the exothermic reaction.

Detailed explanation here.

PART A. LiCl releases more energy during solute-solvent interactions compared to solute-solute interactions. NH4​NO3​ requires more energy to break its solute-solute interactions than the energy released in solute-solvent interactions.

Detailed explanation here.

PART B. If 2 moles of LiCl are dissolved, the temperature increase will approximately double. If 2 moles of NH4NO3​ are dissolved, the temperature decrease will approximately double.

Detailed explanation here.

6.2  Energy Diagrams

6.2.1

PART A. Path 2 requires less energy because it involves a lower activation energy, likely due to the presence of a catalyst.

Detailed explanation here.

PART B. The activation energy of the forward reaction is 220 kJ/mol for Path 1 and 120 kJ/mol for Path 2.

Detailed explanation here.

PART C. The activation energy of the reverse reaction is 160 kJ/mol for Path 1 and 60 kJ/mol for Path 2.

Detailed explanation here.

PART D. ΔH remains unchanged at -60 kJ/mol, regardless of the pathway followed.

Detailed explanation here.

6.3  Heat Transfer and Thermal Equilibrium 

6.3.1

6.3.2

6.3.3

PART A. The temperature of copper decreases, while the temperature of water increases until they reach the same temperature.

Detailed explanation here.

PART B. Copper’s particles have the greater average kinetic energy before thermal equilibrium because it is at a higher temperature.

Detailed explanation here.

PART C. Water molecules have the greater average speed because they are much lighter than copper atoms.

Detailed explanation here.

PART D. Yes, at thermal equilibrium, copper and water have the same average kinetic energy because they are at the same temperature.

Detailed explanation here.

Part A. The energy lost by copper (qCu​) is transferred to water (qH2O​).

Detailed explanation here.

Part B. Copper undergoes a larger temperature change than water because it has a lower specific heat capacity.

Detailed explanation here.

Part C. At thermal equilibrium, both substances have the same final temperature (Tf).

Detailed explanation here.

Part A. Thermal equilibrium means that the energy of particle motion (kinetic energy) has equalized, resulting in the same temperature for both substances.

Detailed explanation here.

Part B. The water molecules will have the greatest average speed.

Detailed explanation here.

6.4  Heat Capacity and Calorimetry

6.4.1

6.4.2

6.4.3

6.4.4

6.4.5

31.7g of water was heated.

Detailed explanation here.

Part A: The energy is absorbed by the dissolving NaCl to break its ionic bonds and form hydrated ions in water.

Detailed explanation here.

Part B: The temperature change is caused by the conversion of thermal energy from the water into potential energy associated with the hydrated ions.

Detailed explanation here.

Part C: In a non-insulated calorimeter, energy could flow into the system from the surroundings, reducing the temperature decrease of the solution. If heat flowed out, the solution’s temperature would drop even further.

Detailed explanation here.

Part D: The energy of the system is conserved because the thermal energy lost by the water is converted into potential energy associated with the interactions between the dissolved ions and water molecules.

Detailed explanation here.

The specific heat of copper is 0.356J/g°C.

Detailed explanation here.

The energy of the water system decreases as it loses thermal energy to the refrigerator's air. This energy is transferred to the surroundings, reducing the kinetic energy of the water molecules and lowering the temperature of the water. The system's total energy is conserved when considering the transfer of energy to the surroundings.

Detailed explanation here.

3,010 J is required to heat the ethanol.

Detailed explanation here.

40.0°C would be the final temperature of a 50.0 g sample of ethanol heated with the same energy (3,010 J) in part (a).

Detailed explanation here.

6.5  Energy of Phase Changes

6.5.1

6.5.2

The energy added during a phase change is used to overcome the intermolecular forces holding the molecules together in the initial phase.

Detailed explanation here.

Part A. 7.51 kJ is needed to melt 22.5 g of ice.

Detailed explanation here.

Part B. Approximately 7.51 kJ of energy is released when 22.5 g of water freezes.

Detailed explanation here.

6.6  Introduction to Enthalpy of Reaction

6.6.1

-136.2 kJ or 136.2 kJ released when 50.0 g of ammonia is produced.

Detailed explanation here.

6.7  Bond Enthalpies

6.7.1

6.7.2

Part A. Total Energy of Reactants: 1648 + 243 = 1891 kJ; Total Energy of Products: 1574 + 431 = 2005 kJ

Detailed explanation here.

Part B. ΔH = −114 kJ

Detailed explanation here.

Part C. The enthalpy change is part (b) is negative (ΔH = −25 kJ) indicating the reaction is exothermic.

Detailed explanation here.

Part A. The bond enthalpy of A–A is 63.5 kJ/mol

Detailed explanation here.

Part B: Sketch the Potential Energy Diagram

The potential energy diagram should:

  1. Start with a higher energy level for the reactants.

  2. Show a peak for the activation energy.

  3. End with a lower energy level for the products since the reaction is exothermic.

  4. The difference between the reactants and products is 25 kJ, representing ΔH.

Detailed explanation here.

6.8  Enthalpy of Formation

6.8.1

The enthalpy change for the synthesis of 2 moles of ammonia is: -91.8 kJ.

Detailed explanation here.

6.9  Hess's Law

6.9.1

The enthalpy of N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(l) is -960.5 kJ.

Detailed explanation here.