Thermochemistry The Nature of Energy Kinetic Energy and

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Thermochemistry

Thermochemistry

The Nature of Energy Kinetic Energy and Potential Energy • Kinetic energy is the

The Nature of Energy Kinetic Energy and Potential Energy • Kinetic energy is the energy of motion: • Potential energy is the energy an object possesses by virtue of its position. • Potential energy can be converted into kinetic energy. Example: a bicyclist at the top of a hill.

The Nature of Energy Units of Energy • SI Unit for energy is the

The Nature of Energy Units of Energy • SI Unit for energy is the joule, J: sometimes the calorie is used instead of the joule: 1 cal = 4. 184 J (exactly) A nutritional Calorie: 1 Cal = 1000 cal = 1 kcal

Thermochemistry Terminology • • • System: part of the universe we are interested in.

Thermochemistry Terminology • • • System: part of the universe we are interested in. Surrounding: the rest of the universe. Boundary: between system & surrounding. Exothermic: energy released by system to surrounding. Endothermic: energy absorbed by system from surr. • Work ( w ): product of force applied to an object over a distance. • Heat ( q ): transfer of energy between two objects

The First Law of Thermodynamics • • Internal Energy: total energy of a system.

The First Law of Thermodynamics • • Internal Energy: total energy of a system. Involves translational, rotational, vibrational motions. Cannot measure absolute internal energy. Change in internal energy, DQ Hyper. Chem

The First Law of Thermodynamics Relating DE to Heat(q) and Work(w) • Energy cannot

The First Law of Thermodynamics Relating DE to Heat(q) and Work(w) • Energy cannot be created or destroyed. • Energy of (system + surroundings) is constant. • Any energy transferred from a system must be transferred to the surroundings (and vice versa). • From the first law of thermodynamics:

The First Law of Thermodynamics

The First Law of Thermodynamics

First Law of Thermodynamics • Calculate the energy change for a system undergoing an

First Law of Thermodynamics • Calculate the energy change for a system undergoing an exothermic process in which 15. 4 k. J of heat flows and where 6. 3 k. J of work is done on the system. E = q + w

The First Law of Thermodynamics Exothermic and Endothermic Processes • Endothermic: absorbs heat from

The First Law of Thermodynamics Exothermic and Endothermic Processes • Endothermic: absorbs heat from the surroundings. • An endothermic reaction feels cold. • Exothermic: transfers heat to the surroundings. • An exothermic reaction feels hot.

Endothermic Reaction Ba(OH)2 • 8 H 2 O(s) + 2 NH 4 SCN(s) Ba(SCN)2(s)

Endothermic Reaction Ba(OH)2 • 8 H 2 O(s) + 2 NH 4 SCN(s) Ba(SCN)2(s) + 2 NH 3(g) + 10 H 2 O(l)

The First Law of Thermodynamics State Functions • State function: depends only on the

The First Law of Thermodynamics State Functions • State function: depends only on the initial and final states of system, not on how the internal energy is used.

The First Law of Thermodynamics

The First Law of Thermodynamics

Enthalpy • Chemical reactions can absorb or release heat. • However, they also have

Enthalpy • Chemical reactions can absorb or release heat. • However, they also have the ability to do work. • For example, when a gas is produced, then the gas produced can be used to push a piston, thus doing work. Zn(s) + 2 H+(aq) Zn 2+(aq) + H 2(g) • The work performed by the above reaction is called pressure-volume work. • When the pressure is constant,

Enthalpy

Enthalpy

Enthalpy • Enthalpy, H: Heat transferred between the system and surroundings carried out under

Enthalpy • Enthalpy, H: Heat transferred between the system and surroundings carried out under constant pressure. • Enthalpy is a state function. • If the process occurs at constant pressure,

Enthalpy • Since we know that • We can write • When H is

Enthalpy • Since we know that • We can write • When H is positive, the system gains heat from the surroundings. • When H is negative, the surroundings gain heat from the system.

Enthalpy => Heat of Reaction

Enthalpy => Heat of Reaction

Enthalpies of Reaction • For a reaction: • Enthalpy is an extensive property (magnitude

Enthalpies of Reaction • For a reaction: • Enthalpy is an extensive property (magnitude H is directly proportional to amount): CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) H = -802 k. J 2 CH 4(g) + 4 O 2(g) 2 CO 2(g) + 4 H 2 O(g) H = -1604 k. J

Enthalpies of Reaction • When we reverse a reaction, we change the sign of

Enthalpies of Reaction • When we reverse a reaction, we change the sign of H: CO 2(g) + 2 H 2 O(g) CH 4(g) + 2 O 2(g) H = +802 k. J • Change in enthalpy depends on state: H 2 O(g) H 2 O(l) H = -44 k. J

Calorimetry • • • Heat Capacity and Specific Heat Calorimetry = measurement of heat

Calorimetry • • • Heat Capacity and Specific Heat Calorimetry = measurement of heat flow. Calorimeter = apparatus that measures heat flow. Heat capacity = the amount of energy required to raise the temperature of an object (by one degree). Molar heat capacity = heat capacity of 1 mol of a substance. Specific heat = specific heat capacity = heat capacity of 1 g of a substance.

Table 5. 2: Specific Heats (S) of Some Substances at 298 K Substance N

Table 5. 2: Specific Heats (S) of Some Substances at 298 K Substance N 2(g) Al(s) Fe(s) Hg(l) H 2 O(s) CH 4(g) CO 2(g) Wood , Glass S ( J g-1 K-1 ) 1. 04 0. 902 0. 45 0. 14 4. 184 2. 06 2. 20 0. 84 1. 76 , 0. 84

If 24. 2 k. J is used to warm a piece of aluminum with

If 24. 2 k. J is used to warm a piece of aluminum with a mass of 250. g, what is the final temperature of the aluminum if its initial temperature is 5. 0 o. C? Cyber. Chem - Pizza

Calorimetry Constant Pressure Calorimetry • Atmospheric pressure is constant!

Calorimetry Constant Pressure Calorimetry • Atmospheric pressure is constant!

Calorimetry Constant Pressure Calorimetry

Calorimetry Constant Pressure Calorimetry

Aqueous Solution Assumptions: Calorimetry

Aqueous Solution Assumptions: Calorimetry

Calorimetry Examples Cyber. Chem - Pizza 1. In an experiment similar to the procedure

Calorimetry Examples Cyber. Chem - Pizza 1. In an experiment similar to the procedure set out for Part (A) of the Calorimetry experiment, 1. 500 g of Mg(s) was combined with 125. 0 m. L of 1. 0 M HCl. The initial temperature was 25. 0 o. C and the final temperature was 72. 3 o. C. Calculate: (a) the heat involved in the reaction and (b) the enthalpy of reaction in terms of the number of moles of Mg(s) used. Ans: (a) – 25. 0 k. J (b) – 406 k. J/mol 2. 50. 0 m. L of 1. 0 M HCl at 25. 0 o. C were mixed with 50. 0 m. L of 1. 0 M Na. OH also at 25. 0 o. C in a styrofoam cup calorimeter. After the mixing process, thermometer reading was at 31. 9 o. C. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2. 9 k. J , -58 k. J/mol [heat of neutralization for strong acid/base reactions]

In an experiment similar to the procedure set out for Part (A) of the

In an experiment similar to the procedure set out for Part (A) of the Calorimetry experiment, 1. 500 g of Mg(s) was combined with 125. 0 m. L of 1. 0 M HCl. The initial temperature was 25. 0 o. C and the final temperature was 72. 3 o. C. Calculate: (a) the heat involved in the reaction and (b) the enthalpy of reaction in terms of the number of moles of Mg(s) used.

qrxn= - S x msoln x ∆T ∆H = -406 k. J/mol

qrxn= - S x msoln x ∆T ∆H = -406 k. J/mol

2. 50. 0 m. L of 1. 0 M HCl at 25. 0 o.

2. 50. 0 m. L of 1. 0 M HCl at 25. 0 o. C were mixed with 50. 0 m. L of 1. 0 M Na. OH also at 25. 0 o. C in a styrofoam cup calorimeter. After the mixing process, thermometer reading was at 31. 9 o. C. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2. 9 k. J , -58 k. J/mol [heat of neutralization for strong acid/base reactions]

Hess’s Law • Hess’s law: if a reaction is carried out in a number

Hess’s Law • Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step. • For example: CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) H = -802 k. J 2 H 2 O(g) 2 H 2 O(l) H= - 88 k. J CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(l) H = -890. k. J

Another Example of Hess’s Law Given: C(s) + ½ O 2(g) CO(g) H =

Another Example of Hess’s Law Given: C(s) + ½ O 2(g) CO(g) H = -110. 5 k. J CO 2(g) CO(g) + ½ O 2(g) H = 283. 0 k. J Calculate H for: C(s) + O 2(g) CO 2(g)

Enthalpies of Formation • If 1 mol of compound is formed from its constituent

Enthalpies of Formation • If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof. • Standard conditions (standard state): Most stable form of the substance at 1 atm and 25 o. C (298 K). • Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state. • Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

Enthalpies of Formation • If there is more than one state for a substance

Enthalpies of Formation • If there is more than one state for a substance under standard conditions, the more stable one is used. • Standard enthalpy of formation of the most stable form of an element is zero.

Enthalpies of Formation Substance DHof (k. J/mol) C(s, graphite) 0 O(g) 247. 5 O

Enthalpies of Formation Substance DHof (k. J/mol) C(s, graphite) 0 O(g) 247. 5 O 2(g) 0 N 2(g) 0

Bucky Ball drawn by Hyper. Chem

Bucky Ball drawn by Hyper. Chem

Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction • For

Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction • For a reaction • Note: n & m are stoichiometric coefficients. • Calculate heat of reaction for the combustion of propane gas giving carbon dioxide and water. C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O( )

C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H

C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O( ) ∆Hof (k. J/mol): -103. 85 -393. 5 ∆Horxn = -2, 219. 9 k. J/mol -285. 8

Foods and Fuels Foods • 1 nutritional Calorie, 1 Cal = 1000 cal =

Foods and Fuels Foods • 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. • Energy in our bodies comes from carbohydrates and fats (mostly). • Intestines: carbohydrates converted into glucose: C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O, H = -2816 k. J • Fats break down as follows: 2 C 57 H 110 O 6 + 163 O 2 114 CO 2 + 110 H 2 O, H = -75, 520 k. J Fats contain more energy; are not water soluble, so are good for energy storage.

Foods and Fuels • • • Fuels Fuel value = energy released when 1

Foods and Fuels • • • Fuels Fuel value = energy released when 1 g of substance is burned. Most from petroleum and natural gas. Remainder from coal, nuclear, and hydroelectric. Fossil fuels are not renewable. In 2000 the United States consumed 1. 03 1017 k. J of fuel. • In 2005 the United States consumed 1. 05 1017 k. J of energy. • In 2008 the United States consumed 1. 05 x 1017 k. J of energy. • Hydrogen has great potential as a fuel with a fuel value of 142 k. J/g. [ gasoline ≈ 35 k. J/g ]

[8. 5%] [8. 1%] [6. 0%] Foods and Fuels [22. 9%] [22. 6%] [7.

[8. 5%] [8. 1%] [6. 0%] Foods and Fuels [22. 9%] [22. 6%] [7. 4%] [40. 4%] [37. 5%] (% for 2000) [% for 2005] [% for 2008] [24. 0%] [22. 6%]

Energy in Fuels The vast majority of the energy consumed in this country comes

Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels. © 2012 Pearson Education, Inc.

Cyber. Chem - Pizza Thermochemistry

Cyber. Chem - Pizza Thermochemistry