Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved
Energy Capacity to do work or to produce heat. Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. The total energy content of the universe is constant. Copyright © Cengage Learning. All rights reserved

Energy is the capacity to do work
Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position

Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy 400C 900C greater thermal energy

Energy Potential energy – energy due to position or composition. Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Copyright © Cengage Learning. All rights reserved

Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Copyright © Cengage Learning. All rights reserved

Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work – force acting over a distance. Energy is a state function; work and heat are not State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Copyright © Cengage Learning. All rights reserved

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

Thermochemistry is the study of heat change in chemical reactions.
The system is the specific part of the universe that is of interest in the study. open closed isolated Exchange: mass & energy energy nothing

Chemical Energy Endothermic Reaction: Heat flow is into a system. Absorb energy from the surroundings. Exothermic Reaction: Energy flows out of the system. Energy gained by the surroundings must be equal to the energy lost by the system. Copyright © Cengage Learning. All rights reserved

2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) Hg (l) + O2 (g)

CONCEPT CHECK! Is the freezing of water an endothermic or exothermic process? Explain. Exothermic process because you must remove energy in order to slow the molecules down to form a solid. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Exo Endo Exothermic (heat energy leaves your hand and moves to the ice) Endothermic (heat energy flows into the ice) Endothermic (heat energy flows into the water to boil it) Exothermic (heat energy leaves to condense the water from a gas to a liquid) Endothermic (heat energy flows into the ice cream to melt it) Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! For each of the following, define a system and its surroundings and give the direction of energy transfer. Methane is burning in a Bunsen burner in a laboratory. Water drops, sitting on your skin after swimming, evaporate. System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic) System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic) Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Hydrogen gas and oxygen gas react violently to form water. Explain. Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted. Copyright © Cengage Learning. All rights reserved

Thermodynamics The study of energy and its interconversions is called thermodynamics. Law of conservation of energy is often called the first law of thermodynamics. Copyright © Cengage Learning. All rights reserved

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings C3H8 + 5O CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

Thermochemistry Thermodynamics is the science of the relationship between heat and other forms of energy. Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions. 2

First Law of Thermodynamics
To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it. Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings. Work, on the other hand, is the energy exchange that results when a force F moves an object through a distance d; work (w) = Fd 2

First Law of Thermodynamics
To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it. Remembering our sign convention. Work done by the system is negative. Work done on the system is positive. Heat evolved by the system is negative. Heat absorbed by the system is positive. 2

Internal Energy Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system. To change the internal energy of a system: ΔE = q + w q represents heat w represents work Copyright © Cengage Learning. All rights reserved

Work vs Energy Flow To play movie you must be in Slide Show Mode
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Internal Energy Sign reflects the system’s point of view. System does work on surroundings: w is negative Surroundings do work on the system: w is positive Copyright © Cengage Learning. All rights reserved

Work Work = P × A × Δh = PΔV P is pressure. A is area.
Δh is the piston moving a distance. ΔV is the change in volume.

Work For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV To convert between L·atm and Joules, use 1 L·atm = J. Copyright © Cengage Learning. All rights reserved

W = -0 atm x 3.8 L = 0 L•atm = 0 joules
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P DV (a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules (b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = L•atm w = L•atm x 101.3 J 1L•atm = J

EXERCISE! Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. They both perform the same amount of work. w = -PΔV a) w = -(2 atm)( ) = -6 L·atm b) w = -(3 atm)( ) = -6 L·atm Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Determine the sign of ΔE for each of the following with the listed conditions: a) An endothermic process that performs work. |work| > |heat| |work| < |heat| b) Work is done on a gas and the process is exothermic. Δ E = negative Δ E = positive a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative Copyright © Cengage Learning. All rights reserved

Heat of Reaction Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings. Heat flows from a region of higher temperature to one of lower temperature; once the temperatures become equal, heat flow stops. 3

Enthalpy and Enthalpy Change
Enthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. An extensive property is one that depends on the quantity of substance. Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system. 3

DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0

EXERCISE! Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. –252 kJ (5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8) ΔH = -252 kJ Copyright © Cengage Learning. All rights reserved

Thermochemical Equations
A thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. 3

Thermochemical Equations
In a thermochemical equation it is important to note phase labels because the enthalpy change, DH, depends on the phase of the substances. 3

Applying Stoichiometry and Heats of Reactions
Consider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH4? 3

Calorimetry Science of measuring heat Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial

Calorimetry If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution. Copyright © Cengage Learning. All rights reserved

How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x J/g • 0C x –890C = -34,000 J

A Problem to Consider Suppose a piece of iron requires 6.70 J of heat to raise its temperature by one degree Celsius. The quantity of heat required to raise the temperature of the piece of iron from 25.0 oC to 35.0 oC is: 3

A Problem to Consider Calculate the heat absorbed when the temperature of grams of water is raised from 20.0 oC to 50.0 oC. (The specific heat of water is J/g.oC.) 3

Heats of Reaction: Calorimetry
A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. The heat absorbed by the calorimeter and its contents is the negative of the heat of reaction. 3

A Coffee–Cup Calorimeter Made of Two Styrofoam Cups
Constant-Pressure Calorimetry A Coffee–Cup Calorimeter Made of Two Styrofoam Cups No heat enters or leaves! Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A g sample of water at 90°C is added to a g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C The correct answer is b). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A g sample of water at 90.°C is added to a g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C The correct answer is c). The final temperature of the water is 23°C. - (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 23°C Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C The correct answer is c). The final temperature of the water is 18°C. - (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.) Tf = 18°C Copyright © Cengage Learning. All rights reserved

A Problem to Consider When 23.6 grams of calcium chloride, CaCl2, was dissolved in water in a calorimeter, the temperature rose from 25.0 oC to 38.7 oC. If the heat capacity of the solution and the calorimeter is 1258 J/oC, what is the enthalpy change per mole of calcium chloride? 3

First, let us calculate the heat absorbed by the calorimeter. Now we must calculate the heat per mole of calcium chloride. 3

Calcium chloride has a molecular mass of g, so Now we can calculate the heat per mole of calcium chloride. 3

Constant-Volume Calorimetry
No heat enters or leaves!

Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol 1 cal = J 1 Cal = 1000 cal = 4184 J Substance DHcombustion (kJ/g) Apple -2 Beef -8 Beer -1.5 Gasoline -34

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Copyright © Cengage Learning. All rights reserved

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ
This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3. N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ ΔH1 = ΔH2 + ΔH3 = 68 kJ Copyright © Cengage Learning. All rights reserved

The Principle of Hess’s Law
Copyright © Cengage Learning. All rights reserved

To play movie you must be in Slide Show Mode
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Characteristics of Enthalpy Changes
If a reaction is reversed, the sign of ΔH is also reversed. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer. Copyright © Cengage Learning. All rights reserved

Problem-Solving Strategy
Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products. Copyright © Cengage Learning. All rights reserved

Example Multiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( ) Desired reaction:

REVIEW: Thermochemical Equations
The following are two important rules for manipulating thermochemical equations: When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by multiplying the DH in the original equation by that same factor. When a chemical equation is reversed, the value of DH is reversed in sign. 3

The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) H2O (l) DH = 2 x 6.01 = 12.0 kJ

The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ

Standard Enthalpy of Formation (ΔHf°)
Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Copyright © Cengage Learning. All rights reserved

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f

Conventional Definitions of Standard States
For a Compound For a gas, pressure is exactly 1 atm. For a solution, concentration is exactly 1 M. Pure substance (liquid or solid) For an Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C. Copyright © Cengage Learning. All rights reserved

A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
ΔH°reaction = –(–75 kJ) (–394 kJ) + (–572 kJ) = –891 kJ Copyright © Cengage Learning. All rights reserved

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Hess’s Law Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. 3

C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g) C (graphite) + O2 (g) CO2 (g)

Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = kJ 2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = (2x-296.1) = 86.3 kJ rxn

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol. 2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x– x–187.6 ] – [ 2x49.04 ] = kJ -5946 kJ 2 mol = kJ/mol C6H6

Hess’s Law For example, suppose you are given the following data:
Could you use these data to obtain the enthalpy change for the following reaction? 3

Hess’s Law If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

Enthalpy - represents heat energy.
Summary: Enthalpy - represents heat energy. Change in Enthalpy (DHo) - energy difference between the products and reactants. Energy released (exothermic), enthalpy change is negative In the combustion of CH4, DHo = -211 kcal Energy absorbed (endothoermic), enthalpy change is positive. In the decomposition of NH3, DHo=+22 kcal

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?

The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

So enthalpy explains heat transfer in a reaction…..
What about predicting if a reaction will occur spontaneously at a given temperature?

Spontaneous and Nonspontaneous Reactions
Spontaneous reaction - occurs without any external energy input. Often, but not always, exothermic reactions are spontaneous. Thermodynamics is used to help predict if a reaction will occur. Another factor is needed.

Entropy (So) - a measure of the randomness of a chemical system.
The second law of thermodynamics - the universe spontaneously tends toward increasing disorder or randomness. Entropy (So) - a measure of the randomness of a chemical system. High entropy - highly disordered system Low entropy - well organized system No such thing as negative entropy. The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.

DSo of a reaction = So(products) - So(reactants)
A positive DSo means an increase in disorder for the reaction. A negative DSo means a decrease in disorder for the reaction.

DSo is positive DSo is negative

If exothermic and positive DSo…
SPONTANEOUS If endothermic and negative DSo… NONSPONTANEOUS For any other situations, it depends on the relative size of DHo and DSo.

Free Energy Concept The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS. This quantity gives a direct criterion for spontaneity of reaction. 2

Free Energy Free energy (DGo) - represents the combined contribution of the enthalpy and entropy values for a chemical reaction. DGo = DHo - TDSo T in Kelvins Predicts spontaneity Negative DGo…Spontaneous Positive DGo…Nonspontaneous

Further summary: Thermodynamics determines if a reaction will occur but tells us nothing about the time it will take Enthalpy - represents heat energy. Entropy (So) - a measure of the randomness of a chemical system. Free energy (DGo) - represents the combined contribution of the enthalpy and entropy values for a chemical reaction. ∆G predicts spontaneity of a chemical reaction at a given temperature. More in next semester.

Spontaneity and Temperature Change
All of the four possible choices of signs for DHo and DSo give different temperature behaviors for DGo. DHo DSo DGo Description – Spontaneous at all T + Nonspontaneous at all T + or – Spontaneous at low T; Nonspontaneous at high T Nonspontaneous at low T; Spontaneous at high T 2

Calculation of DGo at Various Temperatures
In this method you assume that DHo and DSo are essentially constant with respect to temperature. You get the value of DGTo at any temperature T by substituting values of DHo and DSo at 25 oC into the following equation. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Place below each formula the values of DHfo and So multiplied by stoichiometric coefficients. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ You can calculate DHo and DSo using their respective summation laws. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Now you substitute DHo, DSo (= kJ/K), and T (=298K) into the equation for DGfo. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Now you substitute DHo, DSo (= kJ/K), and T (=298K) into the equation for DGfo. So the reaction is nonspontaneous at 25 oC. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Now we’ll use 1000 oC (1273 K) along with our previous values for DHo and DSo. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Now we’ll use 1000 oC (1273 K) along with our previous values for DHo and DSo. So the reaction is spontaneous at 1000 oC. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ To determine the minimum temperature for spontaneity, we can set DGfo =0 and solve for T. 2

A Problem To Consider Find the DGo for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. So: 38.2 92.9 213.7 J/K DHfo: -635.1 kJ Thus, CaCO3 should be thermally stable until its heated to approximately 848 oC. 2

Energy Sources Used in the United States

Fuels Fossil fuels account for nearly 85% of the energy usage in the United States. Anthracite, or hard coal, the oldest variety of coal, contains about 80% carbon. Bituminous coal, a younger variety of coal, contains 45% to 65% carbon. Fuel values of coal are measured in BTUs (British Thermal Units). A typical value for coal is 13,200 BTU/lb. 1 BTU = 1054 kJ 4

Fuels Natural gas and petroleum account for nearly three-quarters of the fossil fuels consumed per year. Purified natural gas is primarily methane, CH4, but also contains small quantities of ethane, C2H6, propane, C3H8, and butane, C4H10. We would expect the fuel value of natural gas to be close to that for the combustion of methane. 4

Fuels Petroleum is a very complicated mixture of compounds.
Gasoline, obtained from petroleum, contains many different hydrocarbons, one of which is octane, C8H18. This value of DHo is equivalent to 44.4 kJ/gram. 4

Fuels With supplies of petroleum estimated to be 80% depleted by the year 2030, the gasification of coal has become a possible alternative. First, coal is converted to carbon monoxide using steam. The carbon monoxide can then be used to produce a variety of other fuels, such as methane. 4

The Earth’s Atmosphere
Transparent to visible light from the sun. Visible light strikes the Earth, and part of it is changed to infrared radiation. Infrared radiation from Earth’s surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere. Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be. Copyright © Cengage Learning. All rights reserved

The Earth’s Atmosphere

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