Thermochemistry UNIT 10 Chapter 16.

Thermochemistry UNIT 10 Chapter 16

Thermodynamics The branch of science that studies the transformation of energy from one form to another

Thermochemistry Specifically studies heat changes that accompany chemical reactions and phase changes

Energy and Its Forms Defined  the ability to do work or produce heat
Two Forms of Energy  Potential Energy (PE) Energy due to composition or position Energy stored in a substance because of its chemical composition Examples: plants, gasoline Plays important role in chemical reactions Chemical Potential Energy

Energy and Its Forms Two Forms of Energy  Kinetic Energy (KE)
The energy of motion KE: directly related to motion of the atoms in a substance and its temperature measured as OR temperature heat

Law of Conservation of Energy
Energy can be converted from one form to another, but it is neither created nor destroyed.

Heat Thermal energy flowing from warmer objects to cooler objects
A way of transferring energy between objects and their surroundings

Temperature Measure of the average kinetic energy of particles of matter More heat → more motion → higher temperature Lose energy → temperature decreases Relative scales: °C, °F Absolute scale: Kelvin (K) Temperature is an intensive property; it does not depend on the sample size.

Units for Heat Metric System unit: calorie (cal)
One calorie = the amount of heat required to raise the temperature of one gram of water by 1°C Calorie (Cal): nutritional or food Calorie 1 Cal = 1 kcal SI unit: joule (J) 1 cal = J One kilojoule (kJ) = 1000 joules (J) = 1000 cal

Units for Heat Relationships between units leads to conversion factors
1 cal = J 1 cal 4.184 J 4.184 J 1 cal

Conversion Practice Convert 108.7 calories to Calories.
Conversion factor? 1 Cal = 1000 cal 108.7 cal 1 Cal 1000 cal Cal = = Cal

Units are cal/g°C or J/g°C
Specific Heat The heat needed to raise the temperature of one gram of a substance by 1°C Units are cal/g°C or J/g°C

Specific Heat Specific heat of water is 1.00 cal/g°C or 4.184 J/g°C
Water has a high specific heat

Heat Absorbed/Released
Calorimetry is a procedure used to measure thermal energy in a chemical reaction. A calorimeter is an insulated device used for measuring the amount of heat released or absorbed during a chemical or physical process.

Heat Absorbed/Released
Known mass of water is placed in chamber and initial temperature is recorded. Tested matter sample is heated and placed in calorimeter. Heat flows from heated sample to cooler water and temperature of water rises. Heat flow stops when temperatures of sample and water are equal. Final temperature of water is recorded. Known masses and temperatures are used to calculate the heat released by sample and absorbed by water.

Heat Equation q = cm∆T Variable Representation Unit

The heat absorbed or released
Heat Equation q = cm∆T Variable Representation Unit q The heat absorbed or released J

Heat Equation q = cm∆T Variable Representation Unit q The heat absorbed or released J c The specific heat J/g°C

Heat Equation q = cm∆T Variable Representation Unit q The heat absorbed or released J c The specific heat J/g°C m The mass of substance g

Heat Equation q = cm∆T Variable Representation Unit q
The heat absorbed or released J c The specific heat J/g°C m The mass of substance g ∆T The temperature change °C

Problem-Solving Strategy
Write the heat equation. Draw a data table for variables. Assign values to the variables. Solve for the unknown variable.

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c m ∆T TI TF

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c m 10.0 g ∆T TI TF

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c m 10.0 g ∆T TI 10.0°C TF

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c m 10.0 g ∆T TI 10.0°C TF 15.0°C TI = Initial temperature ∆T = TF – TI TF = Final temperature

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c m 10.0 g ∆T 5.0°C TI 10.0°C TF 15.0°C TI = Initial temperature ∆T = TF – TI TF = Final temperature

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q c 4.184 J/g°C m 10.0 g ∆T 5.0°C TI 10.0°C TF 15.0°C q = (4.184 J/g°C)(10.0 g)(5.0°C) q = J

Example How much heat is needed to raise the temperature of 10.0 g of water from 10.0°C to 15.0°C? q = cm∆T Water q 209 J c 4.184 J/g°C m 10.0 g ∆T 5.0°C TI 10.0°C TF 15.0°C q = (4.184 J/g°C)(10.0 g)(5.0°C) q = J

Heat Equation…Again qwater = –qsample Recall that
In a calorimeter, heat flows from the hotter unknown sample to the cooler water The flow of heat stops when the temperatures of the sample and water are equal Therefore, the energy absorbed by the water is equal to the energy released by the heated sample In other words, qwater = –qsample

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 2. Draw data table for variables. 1. Write the heat equation. Water Metal q c m ∆T TI TF

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m ∆T TI TF

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m 125 g ∆T TI TF

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m 125 g ∆T TI 25.6°C TF

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m 125 g 50.0 g ∆T TI 25.6°C TF

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m 125 g 50.0 g ∆T TI 25.6°C 115.0°C TF

Example q = cm∆T ∆ T = TF – TI ∆ Twater = 29.3°C - 25.6°C
A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T 3. Assign values to variables. Water Metal q c m 125 g 50.0 g ∆T TI 25.6°C 115.0°C TF 29.3°C 4. Calculate ∆T for water and metal and add to table. ∆ T = TF – TI ∆ Twater = 29.3°C °C

A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q c m 125 g 50.0 g ∆T 3.7°C TI 25.6°C 115.0°C TF 29.3°C 4. Calculate ∆T for water and metal and add to table. ∆ T = TF – TI ∆ Twater = 29.3°C °C ∆ Tmetal = 29.3°C °C

A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q c m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C 4. Calculate ∆T for water and metal and add to table. 5. For which substance – water or metal – is there enough information to use the heat equation? ∆ T = TF – TI ∆ Twater = 29.3°C °C ∆ Tmetal = 29.3°C °C

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q c m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C 5. For which substance – water or metal – is there enough information to use the heat equation? Water

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q c 4.184 J/g°C m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C 5. For which substance – water or metal – is there enough information to use the heat equation? Water

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q J c 4.184 J/g°C m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C

Example q = cm∆T q = (4.184 J/g°C)(125 g)(3.7°C) q = 1935.1 J
A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T q = (4.184 J/g°C)(125 g)(3.7°C) q = J

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q J – J c 4.184 J/g°C m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C 6. Once q for water is known, use heat equation to solve for c of the metal.

Example q = cm∆T c = c = 0.452 J/g°C q m ∆T –1935.1 J
A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T c = c = J/g°C q m ∆T – J (50.0 g)(–85.7°C)

Example A 125-gram sample of water is placed in a calorimeter with an initial temperature of 25.6°C. A 50.0-gram sample of an unknown metal is heated to 115.0°C and is placed in the water. The final temperature of the water and the metal is 29.3°C. q = cm∆T Water Metal q J – J c 4.184 J/g°C 0.452 J/g°C m 125 g 50.0 g ∆T 3.7°C –85.7°C TI 25.6°C 115.0°C TF 29.3°C

Nutrients Eating a balanced diet is fundamental to good health. Energy that keeps your brain and body functioning comes from the foods you eat. Your digestive system and cells in your body break down and gradually oxidize food to release energy that your cells can use and store. Taking in more energy than your body expends results in an increase of stored fuel and, consequently, an increase in body mass. How can you calculate your energy expenditure and plan your caloric intake to maintain your body weight within a specific range?

Nutrients and Calorimetry
Energy is released through chemical reactions during metabolism. Different nutrients are used as fuel for burning or as building material for the body. Carbohydrates are nutrients burned as fuel. 1 gram carbohydrate = 4.0 Cal energy Building nutrients include fats, proteins, vitamins and minerals. 1 gram fat = 9.0 Cal energy 1 gram protein = 4.0 Cal energy

Nutrients and Calorimetry
The energy unit for measuring energy in foods is the nutritional or food Calorie (Cal). 1 Cal = 1 kcal = 1000 cal Recall: The specific heat of water is equal to 1.00 cal/g°C

Heat Equation with Foods
q = cm∆T Variable Representation Unit

q = cm∆T Variable Representation Unit q The heat absorbed or released cal

q = cm∆T Variable Representation Unit q The heat absorbed or released cal c The specific heat cal/g°C

q = cm∆T Variable Representation Unit q The heat absorbed or released cal c The specific heat cal/g°C m The mass of substance g

q = cm∆T Variable Representation Unit q The heat absorbed or released cal c The specific heat cal/g°C m The mass of substance g ∆T The temperature change °C

Problem-Solving Strategy – Take 2
Write the equation for calculating heat. Draw the table for equation variables. Assign values to the variables using the problem statement. Solve for the unknown variable.

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c m ∆T TI TF For what substance are we calculating heat?

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c m ∆T TI TF For what substance are we calculating heat? Bread

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c m ∆T TI TF For what substance are we calculating heat? Bread For what substance do we have measurements?

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c m ∆T TI TF Water For what substance are we calculating heat? Bread For what substance do we have measurements? Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c 1.00 cal/g°C m ∆T TI TF Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] 150 mL H2O 1.00 g H2O 1 mL H2O g H2O = = 150 g H2O

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c 1.00 cal/g°C m 150 g ∆T TI TF Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c 1.00 cal/g°C m 150 g ∆T TI 25°C TF Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c 1.00 cal/g°C m 150 g ∆T TI 25°C TF 65°C Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q c 1.00 cal/g°C m 150 g ∆T 40°C TI 25°C TF 65°C Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q 6000 cal c 1.00 cal/g°C m 150 g ∆T 40°C TI 25°C TF 65°C Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T 6000 cal 1 Cal 1000 cal Cal = = 6 Cal

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q 6 Cal c 1.00 cal/g°C m 150 g ∆T 40°C TI 25°C TF 65°C Water

Practice with Foods A small piece of bread heats 150 mL of water from 25°C to 65°C. How many Calories are stored in the bread? [The density of water is 1.00 g/mL.] q = cm∆T q 6.0 Cal c 1.00 cal/g°C m 150 g ∆T 40°C TI 25°C TF 65°C Water

Practice with Foods If the small piece of bread that produced 6.0 Cal of heat energy had a mass of 10.0 grams, what is the energy content, in Cal/g, of the bread? 6.0 Cal 10.0 g Cal/g = = 0.60 Cal/g energy content

Nutrition Labels RECALL 1 gram carbohydrate provides 4 Cal of energy
1 gram fat provides 9 Cal of energy 1 g protein provides 4 Cal of energy Nutrition labels give information about the grams of carbohydrates, fats, and proteins in each serving.

Nutrition Labels According to the nutrition label at the top of p. 11 of your Note Packet, one serving contains 10 grams of fat. How many Calories of energy will be provided by the total fat in one serving? 10 g fat 9 Cal 1 g fat Cal from fat = = 90 Cal from fat

Nutrition Labels One serving contains 23 grams of carbohydrates. How much energy can be provided by the total carbohydrates in one serving? 23 g carb 4 Cal 1 g carb Cal from carb = = 92 Cal from carb

Nutrition Labels Calculate the heat energy that can be provided by the grams of protein in one serving. 4 g protein 4 Cal 1 g protein Cal from protein = = 16 Cal from protein

Nutrition Labels The nutrition label at the bottom of p. 11 of your Note Packet is copied from a box of chocolate chunk granola bars. Calculate the energy, in Calories, provided by one chocolate chunk granola bar. Use grams provided by fat, carbohydrate, and protein only. Then, calculate the percentage of Calories provided by fat, by carbohydrates, and by protein. % = x 100 whole part

Such reactions and phase changes are either endothermic or exothermic
Thermochemistry Concerned with the heat changes accompanying chemical reactions and phase changes Such reactions and phase changes are either endothermic or exothermic

Chemical Reactions Chemical compounds contain chemical potential energy found in the bonds between atoms and molecules Chemical potential energy holds particles together and can be converted to work or heat In chemical reactions, bonds are broken and formed to take apart compounds and create new compounds Exothermic reactions release energy Example: Heat is released when wood burns or water is formed. Endothermic reactions absorb energy Example: Heat must be added when cooking food or decomposing water

Defined as the heat content of a system at constant pressure
Enthalpy (H) Defined as the heat content of a system at constant pressure

∆Hrxn = Hproducts – Hreactants
Enthalpy The change in enthalpy (not the actual enthalpy) is measured and represents the heat absorbed and released in a chemical reaction. The enthalpy (heat) of reaction (∆Hrxn) is the change in enthalpy for a reaction. ∆Hrxn = Hproducts – Hreactants

Enthalpy ∆Hrxn = Hproducts – Hreactants Endothermic: Reactants absorb heat; therefore, Hproducts > Hreactants ∆Hrxn is positive

Enthalpy ∆Hrxn = Hproducts – Hreactants Exothermic: Reactants release heat; therefore, Hproducts < Hreactants ∆Hrxn is negative

∆Hrxn = q (at constant pressure)
Enthalpy ∆Hrxn = q (at constant pressure)

Thermochemical Equations
Express the amount of heat absorbed or released during reactions Balanced thermochemical equations include physical states of reactants and products and the energy change, expressed as the change in enthalpy

Endothermic Energy is absorbed.
Certain chemical reactions and phase changes require energy. Activation energy is the minimum amount of energy required by the reactants in order for reaction to proceed. Energy must be continuously added for reaction to proceed.

Endothermic Reactants have less energy than products.
Energy can be written as a reactant in the equation.

27 kJ + NH4NO3 (s) → NH4+ (aq) + NO3– (aq)
Endothermic EXAMPLE: A cold pack can be used to ice down a knee injury. When the pack is opened and squeezed, an inner membrane releases ammonium nitrate into water. Heat is transferred away from the knee and is absorbed by the cold pack. NH4NO3 (s) → NH4+ (aq) + NO3– (aq) ∆Hrxn = +27 kJ 27 kJ + NH4NO3 (s) → NH4+ (aq) + NO3– (aq)

Endothermic Rxns 2 H2O (l) → 2 H2 (g) + O2 (g) ∆Hrxn = +572 kJ
572 kJ + 2 H2O (l) → 2 H2 (g) + O2 (g)

Exothermic Energy is released.
According to the Law of Conservation of Energy, energy cannot be destroyed; the released energy must go somewhere. Once sufficient activation energy is added, the reaction proceeds to completion.

Exothermic Reactants have more energy than products.
Energy can be written as a product in the equation.

Exothermic EXAMPLE: A heat pack can warm your hands on a cold day. When the pack is opened, oxygen from the air reacts with iron in the pack. Energy is released as heat. 4 Fe (s) O2 → 2 Fe2O3 (s) ∆Hrxn = –1625 kJ 4 Fe (s) O2 → 2 Fe2O3 (s) kJ

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) + 802 kJ
Exothermic Rxns CH4 (g) O2 (g) → CO2 (g) H2O (g) ∆Hrxn = –802 kJ CH4 (g) O2 (g) → CO2 (g) H2O (g) kJ

Phase changes are changes in the state of matter of a substance.
Phase changes always occur with changes in energy and can be endothermic or exothermic. Phase Changes

Phase Changes for Water
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ Steam ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ Vaporization Enthalpy Condensation Sublimation Deposition ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ Water ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ Melting Freezing ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ Ice ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪

Molar Enthalpies Molar enthalpy (heat) of vaporization: heat required to vaporize one mole of a liquid For water, ∆Hvap = kJ/mol Vaporization or boiling (liquid to gas): H2O (l) → H2O (g) Endothermic process; ∆Hvap is positive Molar enthalpy (heat) of fusion: heat required to melt one mole of a solid substance For water, ∆Hfus = kJ/mol Fusion or melting (solid to liquid): H2O (s) → H2O (l) Endothermic process; ∆Hfus is positive

Molar Enthalpies Molar enthalpy (heat) of condensation: heat required to condense one mole of a gaseous substance For water, ∆Hcond = –40.7 kJ/mol Condensation (gas to liquid): H2O (g) → H2O (l) Exothermic process Same value but opposite sign of ∆Hvap. ∆Hcond is negative. Molar enthalpy (heat) of solidification: heat required to freeze one mole of a liquid For water, ∆Hsolid = –6.01 kJ/mol Solidification or freezing (liquid to solid): H2O (l) → H2O (s) Same value but opposite sign of ∆Hfus. ∆Hsolid is negative .

Molar Enthalpies Sublimation (solid to gas) and deposition (gas to solid) encompass two phase changes. Molar enthalpy (heat) of sublimation: heat required to sublimate one mole of a solid substance For water, ∆Hsub = 6.01 kJ.mol kJ/mol = kJ/mol Endothermic process: melting + vaporization Molar enthalpy (heat) of deposition: heat required to deposit one mole of a gaseous substance For water, ∆Hsolid = –40.7 kJ/mol + –6.01 kJ/mol = –46.71 kJ/mol Exothermic process: condensation + freezing

Phase Change Graph or Heating Curve for Water
140 120 100 80 60 40 20 -20 -40 ENDOTHERMIC Gas sublimation vaporization → ← condensation Temperature °C Liquid melting → ← freezing deposition Solid EXOTHERMIC Energy →

KE vs. PE Recall that kinetic energy (KE) of particles in matter is measured by temperature KE changes as energy is added and the temperature changes On the heating curve, changes in KE are shown by the inclined lines Energy is used to change the particle or molecular motion, which directly affects the temperature

PE  plateaus  phase changes
KE vs. PE Recall that potential energy (PE) is stored energy PE changes as energy is added but the temperature remains constant On the heating curve, changes in PE are shown by the plateaus PE changes occur with phase changes The energy is used to form or break apart the intermolecular forces rather than to increase the temperature PE  plateaus  phase changes

Phase Changes Phase Change State Change IMF Energy Endo/Exo
Sign on q or ∆H

Sign on q or ∆H Vaporization Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Released Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Released Exo Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Released Exo – Freezing Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Released Exo – Freezing Liquid to solid Deposition

Sign on q or ∆H Vaporization Liquid to gas Breaking Required Endo + Melting Solid to liquid Sublimation Solid to gas Condensation Gas to liquid Forming Released Exo – Freezing Liquid to solid Deposition Gas to solid

Calculating Heat During Phase Changes
To calculate heat released or absorbed during phase changes, use the equation q = molΔH q is the heat released or absorbed (J or kJ) mol is the number of moles of the substance (mol) ΔH is the change in enthalpy associated with the specific phase change (kJ/mol)

Calculating Heat During Phase Changes
Example: How much energy is required to melt 36.0 grams of water? q = molΔH ΔH = 6.01 kJ/mol q = (2.00 mol H2O)(6.01 kJ/mol) q = 12.0 kJ 36.0 g H2O 1 mol H2O 18.0 g H2O mol H2O = = 2.00 mol H2O

Heat Capacity for H2O Different specific heat for different phases of water cwater = J/g°C cice = 2.03 J/g°C csteam = 2.01 J/g°C Example: How much heat is required to heat 5 grams of ice from –10°C to 70 °C? Water is going through three processes:  Ice is being heated from –10°C to 0°C.  Ice melts to liquid water at 0°C.  Liquid water is being heated from 0°C to 70°C.

Phase Diagram Defined: graph of pressure vs. temperature showing the phases in which a substance exists under different conditions of temperature and pressure

Phase Diagram Diagrams differ for different substances because of different freezing and boiling points Boiling point: temperature at which vapor pressure of a liquid equals atmospheric pressure Diagrams differ for different substances because of different freezing and boiling points Freezing point: temperature at which a liquid is converted into a crystalline solid

Phase Diagram Triple point: represents temperature and pressure at which three phases of substance can coexist Critical point: point indicating temperature and pressure above which substance cannot exist as liquid

Combustion Combustion is the reaction of a fuel with oxygen, often producing water as a byproduct For biological systems, foods containing glucose or carbohydrates that are easily converted to glucose are used as fuel Fossil fuels, such as petroleum, natural gas, and coal, also undergo combustion reactions and provide warmth and fuel for transportation Combustion is exothermic Standard enthalpy (heat) of combustion: the enthalpy change for the complete combustion for one mole of a substance Symbol ∆H0 indicates standard conditions (1 atm pressure and 298 K or 25°C)

AKA a measure of randomness or disorder
Entropy (S) Measure of the number of possible ways that energy of a system can be distributed AKA a measure of randomness or disorder Increases as the number of possible arrangements of particles increases

Entropy and Spontaneous Processes
Entropy of Solid < Entropy of Liquid < Entropy of Gas Processes that result in greater entropy are spontaneous, like ice melting For example, messing up your bedroom is not hard to do! Processes that result in less entropy are not spontaneous; energy must be supplied For example, cleaning up your messy room takes energy; it doesn’t happen automatically!

The 2nd Law of Thermodynamics states that spontaneous processes always proceed in such a way that the entropy of the universe increases The change in entropy (∆S) is similar to change in enthalpy ∆Ssystem = Sproducts – Sreactants If entropy increases, Sproducts > Sreactants ∆Ssystem is positive If entropy decreases, Sproducts < Sreactants ∆Ssystem is negative Spontaneity can sometimes be predicted

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0)

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic)

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic)

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent Dissolving gas in solvent

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent Dissolving gas in solvent More gaseous product particles than gaseous reactant particles

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent Dissolving gas in solvent More gaseous product particles than gaseous reactant particles More gaseous reactant particles than gaseous product particles

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent Dissolving gas in solvent More gaseous product particles than gaseous reactant particles More gaseous reactant particles than gaseous product particles Increase in temperature

Increases entropy (∆Ssystem > 0) Decreases entropy (∆Ssystem < 0) Changing phase from solid to liquid to gas (endothermic) Changing phase from gas to liquid to solid (exothermic) Dissolving solid or liquid in solvent Dissolving gas in solvent More gaseous product particles than gaseous reactant particles More gaseous reactant particles than gaseous product particles Increase in temperature Decrease in temperature

Thermochemistry UNIT 10 Chapter 16.

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Thermochemistry UNIT 10 Chapter 16.

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