CHM 151 - Chapter 5: Thermochemistry Notes :

Chapter 5: Thermochemistry

Energy and Energy Changes

• Thermochemistry: The study of heat (transfer of thermal energy) in chemical reactions.

• Heat: Transfer of thermal energy (kinetic energy) between two bodies at different temperatures.

• SI unit of energy: Joule (J), the kinetic energy of a 2-kg mass moving at 1 m/s.

System, Surroundings, and the Universe

• System: The specific part of the universe of interest.

• Surroundings: The rest of the universe outside the system.

• Universe: The combination of the system and surroundings.

  • $Universe = System + Surroundings$

Exothermic and Endothermic Processes

• Exothermic Process: A process that releases heat.

  • Heat flows from the system to the surroundings.

• Endothermic Process: A process that absorbs heat.

  • Heat flows from the surroundings to the system.

Example - Exothermic Reaction:

• Combustion reaction: $2H<em>2(g) + O</em>2(g) \rightarrow 2H_2O(l)$

  • System: The reaction itself.

  • Surroundings: Everything else around the reaction.

  • Heat is generated as a product.

Example - Endothermic Reaction:

• Decomposition reaction: $2HgO(s) \rightarrow 2Hg(l) + O_2(g)$

  • System: The reaction itself.

  • Surroundings: Everything else around the reaction.

  • Heat is absorbed for the reaction to proceed.

Practice Problems

1. Kinetic Energy Calculation: Calculate the kinetic energy of a helium atom moving at 125 m/s.

   • Given: Helium mass = 4.003 amu, Conversion factor: $1.661 \times 10^{-24} g = 1 amu$

   • Convert mass to kg: $4.003 \, amu \times \frac{1.661 \times 10^{-24} g}{1 \, amu} \times \frac{1 \, kg}{1 \times 10^3 g} = 6.649 \times 10^{-27} kg$

   • Kinetic energy formula: $E_K = \frac{1}{2} m v^2$

   • $E_K = \frac{1}{2} (6.649 \times 10^{-27} kg) (125 \, m/s)^2$

   • $E_K = 5.19 \times 10^{-23} kg \cdot m^2/s^2 = 5.19 \times 10^{-23} J$

2. Exothermic or Endothermic: Determine if $H<em>2O(s) \rightarrow H</em>2O(l)$ is exothermic or endothermic.

   • Heat is applied to melt ice, so energy is absorbed.

   • Endothermic process.

   • Equation: $energy + H<em>2O(s) \rightarrow H</em>2O(l)$

3. System, Surroundings, and Universe: For a round bottom flask with 50mL of water on a lab bench:

   • System: Round bottom flask with water.

   • Surroundings: The lab.

   • Universe: Lab + flask with water.

Introduction to Thermodynamics

• Thermodynamics: The scientific study of the interconversion of heat and other forms of energy.

Laws of Thermodynamics

1. First Law: Energy can be converted from one form to another, but cannot be created or destroyed.

2. Second Law: (Covered in Chp. 18)

3. Third Law: (Covered in Chp. 18)

Systems of Thermodynamics

• Open System: Exchanges both mass and energy with its surroundings.

• Closed System: Allows the transfer of energy but not mass.

• Isolated System: Does not exchange either mass or energy with its surroundings.

Practice Problems

1. Closed System: Identify closed systems from a list.

   • Examples may include:

     • Pot of cold water on a countertop.

     • Helium-filled Mylar balloon.

2. State Function: Identify state functions from a list.

   • Examples may include:

     • Dieter's weight.

     • Distance between two points on a map.

     • Energy required to heat a given amount of water at 25 °C to boiling point.

3. Internal Energy Calculation: Calculate the overall internal energy, $\Delta U$, for a system that absorbs 188 J of heat and does 141 J of work on its surroundings.

   • System absorbs heat: $q$ is positive.

   • System does work: $w$ is negative.

   • $\Delta U = q + w$

   • $\Delta U = 188 \, J + (-141 \, J) = 47 \, J$

4. Heat Calculation: Calculate $q$ and determine if heat is absorbed or released when a system does work on the surroundings equal to 64 J and $\Delta U = 213 \, J$.

   • Work (w) = -64 J (System does work).

   • $\Delta U = q + w$

   • $213 \, J = q + (-64 \, J)$

   • $q = 277 \, J$ (Heat absorbed by the system).

5. Work Calculation: Calculate $w$ and determine whether work is done by the system or on the system when 415 J of heat is released and $\Delta U = 510 \, J$.

   • Heat (q) = -415 J (Heat is released, exothermic).

   • $\Delta U = q + w$

   • $510 \, J = (-415 \, J) + w$

   • $w = 925 \, J$ (Work done on the system).

Work and Enthalpy

Thermochemical Equations

• Representation of chemical reactions, including enthalpy changes.

Physical Changes

• Example: Ice melting to liquid at sea level above 0°C.

  • Reaction: $H<em>2O(s) \rightarrow H</em>2O(l)$

  • $\Delta H_{rxn} = +6.01 \, kJ/mol$ (Endothermic).

  • Every mole of ice converted to liquid absorbs 6.01 kJ of heat.

  • Enthalpy diagram illustrates heat absorbed by the system.

Chemical Changes

• Example: Combustion of methane.

  • Reaction: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(l)$

  • $\Delta H = -890.4 \, kJ/mol$ (Exothermic).

  • 1 mole of $CH<em>4$ reacts with 2 moles of $O</em>2$ to produce 1 mole of $CO<em>2$ and 2 moles of $H</em>2O$.

  • The $\Delta H$ value of -890.4 kJ/mol represents the heat given off by the system to the surroundings.

Practice Problems

1. Work Done During Expansion: Calculate the work done when nitrogen gas expands from 1.6 to 5.4 L at constant temperature.

   • Formula: $W = -P \Delta V$

   • a. Against a vacuum: $W = 0 \, J$ because $P = 0$.

   • b. Against a constant pressure of 0.80 atm:

     • $W = -(0.80 \, atm)(5.4 \, L - 1.6 \, L) = -3.04 \, L \cdot atm$

     • $W = -3.04 \, L \cdot atm \times (101.3 \, J / 1 \, L \cdot atm) = -307.95 \, J$

   • c. Against a constant pressure of 3.7 atm:

     • $W = -(3.7 \, atm)(5.4 \, L - 1.6 \, L) = -14.06 \, L \cdot atm$

     • $W = -14.06 \, L \cdot atm \times (101.3 \, J / 1 \, L \cdot atm) = -1424.3 \, J$

2. Solar Energy Calculation: Calculate the solar energy required to produce 75.0 g of $C<em>6H</em>{12}O_6$ (photosynthesis).

   • Reaction: $6H<em>2O(l) + 6CO</em>2(g) \rightarrow C<em>6H</em>{12}O<em>6(s) + 6O</em>2(g) \quad \Delta H = +2803 \, kJ/mol$

   • Molar mass of $C<em>6H</em>{12}O_6 = 180.2 \, g/mol$

   • Moles of $C<em>6H</em>{12}O_6 = \frac{75.0 \, g}{180.2 \, g/mol} = 0.416 \, mol$

   • Energy required: $0.416 \, mol \times 2803 \, kJ/mol = +1166 \, kJ = +1.17 \times 10^3 \, kJ$

3. Heat Transfer Calculation: Consider the reaction: $2Mg(s) + O_2(g) \rightarrow 2MgO(s) \quad \Delta H = -1204 \, kJ/mol$

   • a) Exothermic or endothermic? Exothermic.

   • b) Heat transferred when 3.55 g of $Mg(s)$ reacts:

     • $\frac{3.55 \, g \, Mg}{24.31 \, g/mol} \times \frac{-1204 \, kJ}{2 \, mol \, Mg} = -87.9 \, kJ$

   • c) Grams of $MgO$ produced during an enthalpy change of -234 kJ:

     • $-234 \, kJ \times \frac{2 \, mol \, MgO}{-1204 \, kJ} \times \frac{40.31 \, g \, MgO}{1 \, mol \, MgO} = 15.7 \, g \, MgO$

   • d) Kilojoules of heat absorbed when 40.3 g of $MgO(s)$ is decomposed:

     • $\frac{40.3 \, g \, MgO}{40.31 \, g/mol} \times \frac{1204 \, kJ}{2 \, mol \, MgO} = 602 \, kJ$

Calorimetry

• Calorimetry: An experimental method to study heat processes in chemical reactions.

• Can be studied under constant-pressure or constant-volume conditions.

• Specific heat and heat capacity equations determine heat exchange during experiments.

• $q = c \cdot m \cdot \Delta T$

  • $q$: heat

  • $c$: specific heat of the surrounding

  • $m$: mass of the surrounding

  • $\Delta T$: change in temperature of the surrounding.

Specific Heat

• Amount of heat required to raise the temperature of 1 g of a substance by 1°C.
  *Values of specific heat for different substances are provided (e.g., water, ice, aluminum).

Constant - Volume Calorimetry

Bomb Calorimeter is used to measure heat released from volatile reactions.

• No work is done on the reaction, so all energy is heat.

• Heat of calorimeter = - heat of the reaction.

• $q<em>{cal} = -q</em>{rxn}$

• $q<em>{rxn} = -q</em>{cal}$

• $q_{cal} = C \Delta T$

• At constant volume: $q_v = \Delta U$

Practice Problems

1. Heat required to heat 255 g of water from 25.2 °C to 90.5 °C:

   • $q = (4.184 \, J/g \cdot °C)(255 \, g)(90.5 \, °C - 25.2 \, °C) = 69698 \, J = 69.7 \, kJ$

2. Heat capacity of a metal pellet:

   • A metal pellet with a mass of 100.0 g, originally at 88.4 °C, is dropped into 125g of water originally 25.1°C. The final temp. of both the pellet and the water is 31.3 °C. Calculate the heat capacity C in of the pellet.

     • $q<em>{sys}=- q</em>{surr}$

     • water = 4.184$J/g°C</p><ul><li><p>m= 125g</p></li><li><p>AT= 31.3°C - 25.1°C<br>*Q=( 4.184 J/g °C) (125g)(31.3°C - 25.1°C)</p></li></ul></li></ul></li></ul></li></ol><ul><li><p>Q= 3241 J</p></li></ul><ol><li><p>Energy content of a cookie:</p><ul><li><p>$q_{cal} = C \Delta T$</p></li><li><p>A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 39.97 kJ/°C. During the combustion the temperature of the water in the calorimeter increased by 3.90°C. Calculate the energy content (in kJ/g) of the cookie.</p><ul><li><p>qcal=39.97 kJ/°C)</p></li><li><p>AT=3.90°C</p></li></ul></li></ul></li></ol><ul><li><p>Q cal = (39.97 kJ/°C)(3.90°C)</p></li><li><p>Q cal= 155. 7kJ</p><ul><li><p>Total the energy Content =155.9 KJ/ 7.35g= 21.3 KJ/g</p></li></ul></li></ul><h4 collapsed="false" seolevelmigrated="true">Hess's Law</h4><ul><li><p>Hess's Law states that the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps, regardless of the path taken.</p></li></ul><h4 collapsed="false" seolevelmigrated="true">Standard Enthalpies of Formation</h4><h4 collapsed="false" seolevelmigrated="true">Practice Problem</h4><ol><li><p>Calculating$\Delta H_f°$for the reaction when:</p><ul><li><p>$\Delta H_f°$for$Ag^+(aq)$is +105.9 kJ/mol</p></li><li><p>$\Delta H_f°$for$Cl^-(aq)$is -167.2 kJ/mol</p></li><li><p>$\Delta H_f°$for$AgCl(s)$$ is -127.0 kJ/mol