Molar Heat Capacity • Molar Heat Capacity Can Be Used to Determine the Enthalpy Change of a Sample

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Chapter 10 Section 1 Energy Transfer

Energy as Heat

• A sample can transfer energy to another sample.

• One of the simplest ways energy is transferred is as heat.

• Heat is the energy transferred between objects that are at different temperatures

• Though energy has many different forms, all energy is measured in units called joules (J).

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Energy as Heat, continued

• Energy is never created or destroyed.

• The amount of energy transferred from one sample must be equal to the amount of energy received by a second sample.

• The total energy of the two samples remains exactly the same.

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Heat

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Energy as Heat, continued Temperature • Temperature is a measure of how hot (or cold) something is; specifically, a measure of the average kinetic energy of the particles in an object.

• When samples of different temperatures are in contact, energy is transferred from the sample that has the higher temperature to the sample that has the lower temperature.

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Energy as Heat, continued Temperature, continued • If no other process occurs, the temperature of a sample increases as the sample absorbs energy.

• The temperature of a sample depends on the average kinetic energy of the sample’s particles.

• The higher the temperature of a sample is, the faster the sample’s particles move. • The temperature increase of a sample also depends on the mass of the sample.

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Energy as Heat, continued Heat and Temperature are Different • Temperature is an intensive property, which means that the temperature of a sample does not depend on the amount of the sample.

• Heat is an extensive property, which means that the amount of energy transferred as heat by a sample depends on the amount of the sample.

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Comparing Extensive and Intensive Properties

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Temperature and the Temperature Scale

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Energy as Heat, continued A Substance’s Energy Can Be Measure by Enthalpy • Measuring the total amount of energy present in a sample of matter is impossible, but changes in energy content can be determined. • These changes are determined by measuring the energy that enters or leaves the sample of matter. • If 73 J of energy enter a piece of silver and no change in pressure occurs, we know that the enthalpy of the silver has increased by 73 J. • Enthalpy, which is represented by the symbol H, is the total energy content of a sample.

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Energy as Heat, continued A Substance’s Energy Can Be Measure by Enthalpy, continued • Enthalpy is the sum of the internal energy of a system plus the product of the system’s volume multiplied by the pressure that the system exerts on its surroundings • If pressure remains constant, the enthalpy increase of a sample of matter equals the energy as heat that is received. • This relationship remains true even when a chemical reaction or a change of state occurs.

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Energy as Heat, continued A Sample’s Enthalpy Includes the Kinetic Energy of Its Particles

• The particles in a sample are in constant motion. • These particles have kinetic energy. • The enthalpy of a sample includes the total kinetic energy of its particles. • Both the total and average kinetic energies of a substance’s particles are important to chemistry, because these quantities account for every particle’s kinetic energy.

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Molar Heat Capacity • Molar heat capacity can be used to determine the enthalpy change of a sample. • The molar heat capacity of a pure substance is the energy as heat needed to increase the temperature of 1 mol of the substance by 1 K. • Molar heat capacity has the symbol C and the unit J/Kmol. • Molar heat capacity is accurately measured only if no other process, such as a chemical reaction, occurs.

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Molar Heat Capacity, continued

• The following equation shows the relationship between heat and molar heat capacity, where q is the heat needed to increase the temperature of n moles of a substance by T.

q = nCT heat = (amount in moles)  (molar heat capacity)  (change in temperature)

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Calculating Molar Heat Capacity

Sample Problem A

Determine the energy as heat needed to increase the temperature of 10.0 mol of mercury by 7.5 K. The value of C for mercury is 27.8 J/K•mol.

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Calculating Molar Heat Capacity, continued

Sample Problem A Solution

q = nCT q = (10.0 mol )(27.8 J/K•mol )(7.5 K) q = 2085 J

The answer should only have two significant figures, so it is reported as 2100 J or 2.1  103 J.

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Molar Heat Capacity, continued Molar Heat Capacity Depends on the Number of Atoms • One mole of tungsten has a mass of 184 g, while one mole of aluminum has a mass of only about 27 g. • You might expect that much more heat is needed to change the temperature of 1 mol W than is needed to change the temperature of 1 mol Al. • The molar heat capacities of all of the metals are nearly the same. • The temperature of 1 mol of any solid metal is raised 1 K when the metal absorbs about 25 J of heat.

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Molar Heat Capacity, continued Molar Heat Capacity Depends on the Number of Atoms

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Molar Heat Capacity, continued Molar Heat Capacity Is Related to Specific Heat

• The specific heat of a substance is represented by cp and is the energy as heat needed to raise the temperature of one gram of substance by one Kelvin. • The molar heat capacity of a substance, C, is related to moles of a substance not to the mass of a substance. • Because the molar mass is the mass of 1 mol of a substance, the following equation is true.

M (g/mol)  cp (J/Kg) = C (J/Kmol) (molar mass)(specific heat) = (molar heat capacity)

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Specific Heat

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Molar Heat Capacity, continued Heat Results in Disorderly Particle Motion • When a substance receives energy in the form of heat, its enthalpy increases and the kinetic energy of the particles that make up the substance increases.

• The direction in which any particle moves is not related to the direction in which its neighboring particles move. The motions of these particles are random.

• Other types of energy can produce orderly motion or orderly positioning of particles.

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Molar Enthalpy Change

• Because enthalpy is the total energy of a system, it is an important quantity.

• The only way to measure energy is through a change. • There’s no way to determine the true value of H. • H can be measured as a change occurs.

• The enthalpy change for one mole of a pure substance is called molar enthalpy change.

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Molar Enthalpy Change, continued Molar Heat Capacity Governs the Changes • When a pure substance is only heated or cooled, the amount of heat involved is the same as the enthalpy change.

• H = q for the heating or cooling of substances

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Molar Enthalpy Change, continued Molar Heat Capacity Governs the Changes, continued • The molar enthalpy change is related to the molar heat capacity by the following equation.

molar enthalpy change = CT molar enthalpy change = (molar heat capacity)(temperature change)

• This equation does not apply to chemical reactions or changes of state.

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Enthalpy Change

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Calculating Molar Enthalpy Change for Heating

Sample Problem B

How much does the molar enthalpy change when ice warms from 5.4C to 0.2C?

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Calculating Molar Enthalpy Change for Heating

Sample Problem B Solution

Tinitial = -5.4C = 267.8 K and Tfinal = -0.2C = 273.0 K For H2O(s), C = 37.4 J/Kmol

T = Tfinal  Tinitial = 5.2 K H = CT J  37.4 5.2 K K mol J 1.9 102 K mol

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Calculating Molar Enthalpy Change for Cooling

Sample Problem C

Calculate the molar enthalpy change when an aluminum can that has a temperature of 19.2C is cooled to a temperature of 4.00C.

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Calculating Molar Enthalpy Change for Cooling

Sample Problem C Solution

Tinitial = -19.2C = 292 K and Tfinal = 4.00C = 277 K For Al(s), C = 24.2 J/Kmol.

T = Tfinal  Tinitial = 277K  292 K =  15 K

H = CT

 H = (24.2 J/Kmol)( 15 K ) =  360 J/Kmol

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Molar Enthalpy Change, continued Enthalpy Changes of Endothermic or Exothermic Processes

• Enthalpy changes can be used to determine if a process is endothermic or exothermic.

• Processes that have positive enthalpy changes are endothermic.

• Processes that have negative enthalpy changes are exothermic.

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Enthalpy of a System of Several Substances

• Thermodynamics is the branch of science concerned with the energy changes that accompany chemical and physical changes.

• Enthalpy is one of three thermodynamic properties.

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Enthalpy of a System of Several Substances, continued Writing Equations for Enthalpy Changes • An equation can be written for the enthalpy change that occurs during a change of state or a chemical reaction. • The following equation is for the hydrogen and bromine reaction.

H2(g, 298K) + Br2(l, 298K)  2HBr(g, 298K) H = -72.8 kJ

• The enthalpy change for this reaction and other chemical reactions are written using the symbol H. • The negative enthalpy change indicates the reaction is exothermic.

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Changes in Enthalpy Accompany Reactions

• A change in enthalpy during a reaction depends on many variables.

• Temperature is one of the most important variables.

• To standardize the enthalpies of reactions, data are often presented for reactions in which both reactants and products have the standard thermodynamic temperature of 25.00C or 298.15 K.

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Changes in Enthalpy Accompany Reactions, continued • Chemists usually present a thermodynamic value for a chemical reaction by using the chemical equation. 11 Hg + Br l  HBr g H = -36.4 kJ 2222

• This equation shows that when 0.5 mol of H2 reacts with 0.5 mol of Br2 to produce 1 mol HBr and all have a temperature of 298.15 K, the enthalpy decreases by 36.4 kJ.

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Chemical Calorimetry

• For the H2 and Br2 reaction, in which H is negative, the total energy of the reaction decreases.

• The energy is released as heat by the system.

• If the reaction was endothermic, energy in the form of heat would be absorbed by the system and the enthalpy would increase.

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Chemical Calorimetry, continued • The experimental measurement of an enthalpy change for a reaction is called calorimetry. • Calorimetry is the measurement of heat-related constants, such as specific heat or latent heat • Combustion reactions are always exothermic. • The enthalpy changes of combustion reactions are determined using a bomb calorimeter. • A calorimeter Is a device used to measure the heat absorbed or released in a chemical or physical change

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Chemical Calorimetry, continued • A calorimeter is a sturdy, steel vessel in which the sample is ignited electrically in the presence of high- pressure oxygen. • The energy from combustion is absorbed by a surrounding water bath and by the calorimeter. • The water and the other parts of the calorimeter have known specific heats. • A measured temperature increase can be used to calculate the energy released in the combustion reaction and then the enthalpy change.

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Bomb Calorimeter

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Calorimeter

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Chemical Calorimetry, continued Nutritionists Use Bomb Calorimetry • Inside the pressurized oxygen atmosphere of a bomb calorimeter, most organic matter, including food, fabrics, and plastics, will ignite easily and burn rapidly.

• Sample sizes are chosen so that there is excess oxygen during the combustion reactions.

• Under these conditions, the reactions go to completion and produce carbon dioxide, water, and possibly other compounds.

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Chemical Calorimetry, continued Adiabatic Calorimetry Is Another Strategy • Instead of using a water bath to absorb the energy generated by a chemical reaction, adiabatic calorimetry uses an insulating vessel. • The word adiabatic means “not allowing energy to pass through.” • No energy can enter or escape this type of vessel. • The reaction mixture increases in temperature if the reaction is exothermic or decreases in temperature if the reaction is endothermic.

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Chemical Calorimetry, continued Adiabatic Calorimetry Is Another Strategy, continued • If the system’s specific heat is known, the reaction enthalpy can be calculated.

• Adiabatic calorimetry is used for reactions that are not ignited, such as for reactions in aqueous solution.

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Hess’s Law

• Any two processes that both start with the same reactants in the same state and finish with the same products in the same state will have the same enthalpy change.

• Hess’s law states that the overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process.

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Hess’s Law, continued

• When phosphorus is burned in excess chlorine 4 mol of

phosphorus pentachloride, PCl5, is synthesized.

P4(s) + 10Cl2(g)  4PCl5(g) H = -1596 kJ

• Phosphorus pentachloride may also be prepared in a two-step process.

Step 1: P4(s) + 6Cl2(g)  4PCl3(g) H = -1224 kJ

Step 2: PCl3(g) + Cl2(g)  PCl5(g) H = -93 kJ

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Hess’s Law, continued

• The second reaction must take place four times for each occurrence of the first reaction in the two-step process.

• This two-step process is more accurately described by the following equations.

P4(s) + 6Cl2(g)  4PCl3(g) H = -1224 kJ

4PCl3(g) + 4Cl2(g)  4PCl5(g) H = 4(-93 kJ) = -372 kJ

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Hess’s Law, continued

• So, the total change in enthalpy by the two-step process is as follows:

(-1224 kJ) + (-372 kJ) = -1596 kJ

• This enthalpy change, H, for the two-step process is the same as the enthalpy change for the direct route of

the formation of PCl5.

• This example is in agreement with Hess’s law.

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Hess’s Law

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Hess’s Law, continued Using Hess’s Law and Algebra • Chemical equations can be manipulated using rules of algebra to get a desired equation.

• When equations are added or subtracted, enthalpy changes must be added or subtracted.

• When equations are multiplied by a constant, the enthalpy changes must also be multiplied by that constant.

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Hess’s Law, continued Using Hess’s Law and Algebra

• The enthalpy of the formation of CO, when CO2 and solid carbon are reactants, is found using the equations below.

2C(s) + O2(g)  2CO(g) H = -221 kJ C(s) + O2(g)  CO2(g) H = -393 kJ

• You cannot simply add these equations because CO2 would not be a reactant.

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Hess’s Law, continued Using Hess’s Law and Algebra, continued • You subtract or reverse the second equation, carbon dioxide will be on the correct side of the equation.

–C(s) – O2(g)  – CO2(g) H = –(– 393 kJ)

CO2(g)  C(s) + O2(g) H = 393 kJ

• Reversing an equation causes the enthalpy of the new reaction to be the negative of the enthalpy of the original reaction.

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Hess’s Law, continued Using Hess’s Law and Algebra, continued • Adding the two equations gives the equation for the formation of CO by using CO2 and C.

2C(s) + O2(g)  2CO(g) H = – 221 kJ CO2(g)  C(s) + O2(g) H = 393 kJ

2C(s) + O2(g) + CO2(g)  2CO(g) + C(s) + O2(g) H = 172 kJ

• Oxygen and carbon that appear on both sides of the equation can be canceled. C(s) + CO2(g)  2CO(g) H = 172 kJ

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Hess’s Law, continued Standard Enthalpies of Formation • The enthalpy change in forming 1 mol of a substance from elements in their standard states is called the 0 standard enthalpy of formation of the substance,  Hf • The values of the standard enthalpies of formation for elements are 0.

• The following equation is used to determine the enthalpy change of a chemical reaction from the standard enthalpies of formation.

Hreaction = Hproducts - Hreactants

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Hess’s Law, continued Standard Enthalpies of Formation, continued Standard Enthalpies of Formation

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Calculating a Standard Enthalpies of Formation Sample Problem D

Calculate the standard enthalpy of formation of pentane,

C5H12, using the given information. 0 (1) C(s) + O2(g)  CO2(g) Hf = -393.5 kJ/mol

(2) H2(g) + ½O2(g)  H2O(l)  = -285.8 kJ/mol

(3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l) H = -3535.6 kJ/mol

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Calculating a Standard Enthalpies of Formation Sample Problem D Solution

(1) 5C(s) + 5O22 (g)  5CO (g) H = 5(-393.5 kJ/mol)

(2) 6H2 (g) + 3O 2 (g)  6H 2O(l) H = 6(-285.8 kJ/mol)

(3) 5CO2 (g) + 6H 2 O(l)  C 5 H 12 (gH ) + 8O 2 (g) = 3536.6 kJ/mol

0 5C(s) + 6H2 (g)  C 5 H 12 (g) Hf = -145.7 kJ/mol

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Calculating a Reaction’s Change in Enthalpy

Sample Problem E

Calculate the change in enthalpy for this reaction.

2H2(g) + 2CO2(g)  2H2O(g) + 2CO(g)

State whether the reaction is exothermic or endothermic.

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Calculating a Reaction’s Change in Enthalpy, continued Sample Problem E Solution Standard enthalpies of formation as follows: 0 H2O(g),  Hf = -241.8 kJ/mol 0 CO(g),  Hf = -110.5 kJ/mol

H2(g),  = 0 kJ/mol

CO2(g),  = -393.5 kJ/mol.

H = (2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol) - (2 mol)(0 kJ/mol) - (2 mol)(-393.5 kJ/mol) = 82.4 kJ

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Entropy • Some reactions happen easily, but others do not. • Sodium and chlorine react when they are brought together. • Nitrogen and oxygen coexist in the air you breathe without forming poisonous nitrogen monoxide, NO. • One factor you can use to predict whether reactions will occur is enthalpy. A reaction is more likely to occur if it is accompanied by a decrease in enthalpy or if H is negative.

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Entropy, continued • Another factor known as entropy can determine if a process will occur. • Entropy, S, is a measure of the disorder in a system and is a thermodynamic property. • Entropy is not a form of energy and has the units joules per kelvin, J/K. • A process is more likely to occur if it is accompanied by an increase in entropy. • S is positive.

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Entropy, continued Factors That Affect Entropy • Ions in a solution disperse throughout the solution. • This process of dispersion is called diffusion and causes the increase in entropy. • Entropy also increases as solutions become more dilute or when the pressure of a gas is reduced. • In both cases, the molecules fill larger spaces and so become more disordered. • Entropies also increase with temperature, but this effect is not great unless a phase change occurs.

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Entropy, continued Factors That Affect Entropy, continued • The entropy can change during a reaction. • The entropy of a system can increase when the total number of moles of product is greater than the total number of moles of reactant. • Entropy can increase in a system when the total number of particles in the system increases. • Entropy also increases when a reaction produces more gas particles, because gases are more disordered than liquids or solids.

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Entropy, continued Factors That Affect Entropy, continued • Entropy decreases as sodium chloride forms: 2 mol of sodium combine with 1 mol of chlorine to form 2 mol of sodium chloride.

2Na(s) + Cl2(g)  2NaCl(s) S = -181 J/K

• This decrease in entropy is because of the order present in crystalline sodium chloride.

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Entropy, continued Factors That Affect Entropy, continued • Entropy increases when 1 mol of sodium chloride dissolves in water to form 1 mol of aqueous sodium ions and 1 mol of aqueous chlorine ions.

NaCl(s)  Na+(aq) + Cl–(aq) S = 43 J/K

• This increase in entropy is because of the order lost when a crystalline solid dissociates to form ions.

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Entropy, continued Factors That Affect Entropy, continued Standard Entropy Changes for Some Reactions

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Relating Enthalpy, Entropy, and Free Energy Changes

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Entropy, continued Hess’s Law Also Applies to Entropy • The decomposition of nitrogen triiodide to form nitrogen and iodine creates 4 mol of gas from 2 mol of a solid.

2NI3(s)  N2(g) + 3I2(g)

• This reaction has such a large entropy increase that the reaction proceeds once the reaction is initiated by a mechanical shock.

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Entropy, continued Hess’s Law Also Applies to Entropy, continued • Molar entropy has the same unit, J/K•mol, as molar heat capacity.

• Molar entropies can be calculated from molar heat capacity data.

• Entropies can also be calculated by using Hess’s law and entropy data for other reactions.

• You can manipulate chemical equations using rules of algebra to get a desired equation.

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Entropy, continued Hess’s Law Also Applies to Entropy, continued • When equations are added or subtracted, entropy changes must be added or subtracted.

• When equations are multiplied by a constant, the entropy changes must also be multiplied by that constant.

• Atoms and molecules that appear on both sides of the equation can be canceled.

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Entropy, continued Hess’s Law Also Applies to Entropy, continued • The standard entropy is represented by the symbol S0. • The standard entropy of the substance is the entropy of 1 mol of a substance at a standard temperature, 298.15 K. • Elements can have standard entropies that have values other than zero. • Most standard entropies are positive; this is not true of standard enthalpies of formation. • The entropy change of a reaction can be calculated by using the following equation.

Sreaction = Sproducts - Sreactants

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Entropy, continued Hess’s Law Also Applies to Entropy, continued Standard Entropies of Some Substances

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Hess’s Law and Entropy

Sample Problem F

Calculate the entropy change that accompanies the following reaction.

1 1 1 1 H (g ) + CO ( g )  H O( g ) + CO( g ) 22 2 2 2 2 2

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Hess’s Law and Entropy, continued

Sample Problem F Solution

The standard entropies are as follows.

0 For H2O, S = 188.7 J/K•mol. For CO, S0 = 197.6 J/K•mol.

0 For H2, S = 130.7 J/K•mol. 0 For CO2, S = 213.8 J/K•mol.

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Hess’s Law and Entropy, continued

Sample Problem F Solution, continued SSS =  (products) +  (reactants) 11    S =   mol  (188.7 J/K mol) +  mol  (197.6 J/K mol) 22    11    - mol  (130.7 J/K mol) -  mol  (213.8 J/K mol) 22    S = 94.35 J/K + 98.8 J/K - 65.35 J/K - 106.9 J/K S = 193.1 J/K - 172.2 J/K S = 20.9 J/K

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Gibbs Energy

• The tendency for a reaction to occur depends on both H and S.

• If H is negative and S is positive for a reaction, the reaction will likely occur.

• If H is positive and S is negative for a reaction, the reaction will not occur.

• How can you predict what will happen if H and S are both positive or both negative?

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Gibbs Energy, continued

• Gibbs energy is the energy in a system that is available for work represented by the symbol G.

G = H – TS

• Another name for Gibbs energy is free energy.

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Gibbs Energy, continued Gibbs Energy Determines Spontaneity • A spontaneous reaction is one that does occur or is likely to occur without continuous outside assistance, such as input of energy. • A reaction is spontaneous if the Gibbs energy change is negative. • A nonspontaneous reaction will never occur without assistance. • If a reaction has a G greater than 0, the reaction is nonspontaneous. • If a reaction has a G of exactly zero, the reaction is at equilibrium.

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Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy • Reactions that have large negative G values often release energy and increase disorder. • The vigorous reaction of potassium metal and water is an example of this type of reaction. + – 2K(s) + 2H2O(l)  2K (aq) + 2OH (aq) + H2(g) H = -392 kJ S = 0.047 kJ/K • The change in Gibbs energy for the reaction is G = H – T S = -392 kJ - (298.15 K)(0.047 kJ/K) = -406 kJ

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Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy, continued • You can calculate G in another way because lists of standard Gibbs energies of formation exist.

0 • The standard Gibbs energy of formation, G f, of a substance is the change in energy that accompanies the formation of 1 mol of the substance from its elements at 298.15 K.

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Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy, continued • The standard Gibbs energies of formation can be used to find the G for any reaction.

• Hess’s law also applies when calculating G.

Greaction = Gproducts – Greactants

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Gibbs Energy, continued

Standard Gibbs Energies of Formation

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Equation for Free Energy Change

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Calculating a Change in Gibbs Energy from H and S Sample Problem G

Given that the changes in enthalpy and entropy are – 139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then, state whether the reaction is spontaneous at 25C.

C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)

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Calculating a Change in Gibbs Energy from H and S, continued Sample Problem G Solution

H = –139 kJ S = 277 J/K T = 25C = (25 + 273.15) K = 298 K G = H - T S G = (-139 kJ) - (298 K)(277 J/K) G = (-139 kJ) - (83 kJ) G = -222 kJ

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Calculating a Gibbs Energy Change using 0 Gf Values Sample Problem H

Calculate G for the following water-gas reaction.

C(s) + H2O(g)  CO(g) + H2(g)

Is this reaction spontaneous?

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Gibbs Energy, continued

Sample Problem H Solution

0 For H2O(g), Gf = -228.6 kJ/mol.

0 For CO(g), Gf = -137.2 kJ/mol.

For H2(g),  = 0 kJ/mol. For C(s) (graphite),  = 0 kJ/mol. G = G(products) - G(reactants)

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Calculating a Gibbs Energy Change using 0 Gf Values, continued Sample Problem H Solution, continued G = G(products) - G(reactants) =

0 [(mol CO(g))(Gf for CO(g)) + (mol H2(g))( for H2(g))] – 0 [(mol C(s))(Gf for C(s)) + (mol H2O(g))( for H2O(g))] = [(1 mol)(-137.2 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)(0 kJ/mol) – (1 mol)(-228.6 kJ/mol)] = (-137.2 + 228.6) kJ = 91.4 kJ

The reaction is nonspontaneous under standard conditions.

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Gibbs Energy, continued Predicting Spontaneity • Does temperature affect spontaneity? • Consider the equation for G. G = H - TS • The terms H and S change very little as temperature changes. • The presence of T in the equation for G indicates that temperature may greatly affect G.

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Gibbs Energy, continued Predicting Spontaneity, continued • Suppose a reaction has both a positive H value and a positive S value.

• If the reaction occurs at a low temperature, the value for TS will be small and will have little impact on the value of G.

• The value of G will be similar to the value of H and will have a positive value.

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Gibbs Energy, continued Predicting Spontaneity • When the same reaction proceeds at a high enough temperature, TS will be larger than H and G will be negative.

• Increasing the temperature of a reaction can make a nonspontaneous reaction spontaneous.

• A nonspontaneous reaction cannot occur unless some form of energy is added to the system.

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