Chapter 10 Section 1 Energy Transfer Energy as Heat

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).

Energy as Heat, continued
Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Visual Concepts Heat

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer 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.

Comparing Extensive and Intensive Properties
Chapter 10 Visual Concepts Comparing Extensive and Intensive Properties

Temperature and the Temperature Scale
Chapter 10 Visual Concepts Temperature and the Temperature Scale

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Molar Heat Capacity
Section 1 Energy Transfer 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.

Molar Heat Capacity, continued
Chapter 10 Section 1 Energy Transfer 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)

Calculating Molar Heat Capacity
Chapter 10 Section 1 Energy Transfer 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.

Calculating Molar Heat Capacity, continued
Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Section 1 Energy Transfer Molar Heat Capacity, continued Molar Heat Capacity Depends on the Number of Atoms

M (g/mol)  cp (J/Kg) = C (J/Kmol)
Chapter 10 Section 1 Energy Transfer 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)

Chapter 10 Visual Concepts Specific Heat

Chapter 10 Section 1 Energy Transfer 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.

Chapter 10 Molar Enthalpy Change
Section 2 Using Enthalpy 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.

Molar Enthalpy Change, continued
Chapter 10 Section 2 Using Enthalpy 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

molar enthalpy change = CT
Chapter 10 Section 2 Using Enthalpy 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.

Chapter 10 Visual Concepts Enthalpy Change

Chapter 10 Calculating Molar Enthalpy Change for Heating
Section 2 Using Enthalpy 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?

T = Tfinal  Tinitial = 5.2 K
Chapter 10 Section 2 Using Enthalpy Calculating Molar Enthalpy Change for Heating Sample Problem B Solution Tinitial = -5.4C = K and Tfinal = -0.2C = K For H2O(s), C = 37.4 J/Kmol T = Tfinal  Tinitial = 5.2 K H = CT

Chapter 10 Calculating Molar Enthalpy Change for Cooling
Section 2 Using Enthalpy 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.

Chapter 10 Calculating Molar Enthalpy Change for Cooling
Section 2 Using Enthalpy 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

Molar Enthalpy Change, continued
Chapter 10 Section 2 Using Enthalpy 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.

Enthalpy of a System of Several Substances
Chapter 10 Section 2 Using Enthalpy 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.

Enthalpy of a System of Several Substances, continued
Chapter 10 Section 2 Using Enthalpy 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 = 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.

Changes in Enthalpy Accompany Reactions
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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 K.

Changes in Enthalpy Accompany Reactions, continued
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Changes in Enthalpy Accompany Reactions, continued Chemists usually present a thermodynamic value for a chemical reaction by using the chemical equation. 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 K, the enthalpy decreases by 36.4 kJ.

Chapter 10 Chemical Calorimetry
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Chemical Calorimetry, continued
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Chapter 10 Bomb Calorimeter
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Bomb Calorimeter

Chapter 10 Visual Concepts Calorimeter

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Section 3 Changes In Enthalpy During Chemical Reactions
Chapter 10 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.

P4(s) + 10Cl2(g)  4PCl5(g) H = -1596 kJ
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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 = kJ Phosphorus pentachloride may also be prepared in a two-step process. Step 1: P4(s) + 6Cl2(g)  4PCl3(g) H = kJ Step 2: PCl3(g) + Cl2(g)  PCl5(g) H = -93 kJ

Chapter 10 Hess’s Law, continued
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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 = kJ 4PCl3(g) + 4Cl2(g)  4PCl5(g) H = 4(-93 kJ) = -372 kJ

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Hess’s Law, continued So, the total change in enthalpy by the two-step process is as follows: (-1224 kJ) + (-372 kJ) = 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.

Chapter 10 Visual Concepts Hess’s Law

Chapter 10 Hess’s Law, continued Using Hess’s Law and Algebra
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Chapter 10 Hess’s Law, continued Using Hess’s Law and Algebra
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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

Hreaction = Hproducts - Hreactants
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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 standard enthalpy of formation of the substance,  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

Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Hess’s Law, continued Standard Enthalpies of Formation, continued Standard Enthalpies of Formation

Calculating a Standard Enthalpies of Formation
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Calculating a Standard Enthalpies of Formation Sample Problem D Calculate the standard enthalpy of formation of pentane, C5H12, using the given information. (1) C(s) + O2(g)  CO2(g)  = kJ/mol (2) H2(g) + ½O2(g)  H2O(l)  = kJ/mol (3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l) H = kJ/mol

Calculating a Standard Enthalpies of Formation
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Calculating a Standard Enthalpies of Formation Sample Problem D Solution

2H2(g) + 2CO2(g)  2H2O(g) + 2CO(g)
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 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.

(2 mol)(0 kJ/mol) - (2 mol)(-393.5 kJ/mol) = 82.4 kJ
Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 Calculating a Reaction’s Change in Enthalpy, continued Sample Problem E Solution Standard enthalpies of formation as follows: H2O(g),  = kJ/mol CO(g),  = kJ/mol H2(g),  = 0 kJ/mol CO2(g),  = kJ/mol. H = (2 mol)( kJ/mol) + (2 mol)( kJ/mol) - (2 mol)(0 kJ/mol) - (2 mol)( kJ/mol) = 82.4 kJ

Chapter 10 Entropy Some reactions happen easily, but others do not.
Section 4 Order and Spontaneity 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.

Chapter 10 Entropy, continued
Section 4 Order and Spontaneity 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.

Chapter 10 Entropy, continued Factors That Affect Entropy
Section 4 Order and Spontaneity 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.

Chapter 10 Entropy, continued Factors That Affect Entropy, continued
Section 4 Order and Spontaneity 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.

2Na(s) + Cl2(g)  2NaCl(s) S = -181 J/K
Chapter 10 Section 4 Order and Spontaneity 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.

NaCl(s)  Na+(aq) + Cl–(aq) S = 43 J/K
Chapter 10 Section 4 Order and Spontaneity 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.

Chapter 10 Entropy, continued Factors That Affect Entropy, continued
Section 4 Order and Spontaneity Entropy, continued Factors That Affect Entropy, continued Standard Entropy Changes for Some Reactions

Relating Enthalpy, Entropy, and Free Energy Changes
Chapter 10 Visual Concepts Relating Enthalpy, Entropy, and Free Energy Changes

Chapter 10 Entropy, continued Hess’s Law Also Applies to Entropy
Section 4 Order and Spontaneity 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.

Section 4 Order and Spontaneity 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.

Section 4 Order and Spontaneity 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.

Sreaction = Sproducts - Sreactants
Chapter 10 Section 4 Order and Spontaneity 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, 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

Section 4 Order and Spontaneity Entropy, continued Hess’s Law Also Applies to Entropy, continued Standard Entropies of Some Substances

Chapter 10 Hess’s Law and Entropy Sample Problem F
Section 4 Order and Spontaneity Hess’s Law and Entropy Sample Problem F Calculate the entropy change that accompanies the following reaction.

Hess’s Law and Entropy, continued
Chapter 10 Section 4 Order and Spontaneity Hess’s Law and Entropy, continued Sample Problem F Solution The standard entropies are as follows. For H2O, S0 = J/K•mol. For CO, S0 = J/K•mol. For H2, S0 = J/K•mol. For CO2, S0 = J/K•mol.

Hess’s Law and Entropy, continued
Chapter 10 Section 4 Order and Spontaneity Hess’s Law and Entropy, continued Sample Problem F Solution, continued

Chapter 10 Section 4 Order and Spontaneity 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?

Gibbs Energy, continued
Chapter 10 Section 4 Order and Spontaneity 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.

Chapter 10 Section 4 Order and Spontaneity 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.

Chapter 10 Section 4 Order and Spontaneity 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 = kJ/K The change in Gibbs energy for the reaction is G = H – T S = -392 kJ - ( K)(0.047 kJ/K) = -406 kJ

Chapter 10 Section 4 Order and Spontaneity 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. The standard Gibbs energy of formation, G0f, of a substance is the change in energy that accompanies the formation of 1 mol of the substance from its elements at K.

Greaction = Gproducts – Greactants
Chapter 10 Section 4 Order and Spontaneity 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

Chapter 10 Section 4 Order and Spontaneity Gibbs Energy, continued Standard Gibbs Energies of Formation

Equation for Free Energy Change
Chapter 10 Visual Concepts Equation for Free Energy Change

C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)
Chapter 10 Section 4 Order and Spontaneity 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)

Calculating a Change in Gibbs Energy from H and S, continued
Chapter 10 Section 4 Order and Spontaneity 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 = ( ) K = 298 K G = H - T S G = (-139 kJ) - (298 K)(277 J/K) G = (-139 kJ) - (83 kJ) G = -222 kJ

C(s) + H2O(g)  CO(g) + H2(g)
Chapter 10 Section 4 Order and Spontaneity Calculating a Gibbs Energy Change using  Values Sample Problem H Calculate G for the following water-gas reaction. C(s) + H2O(g)  CO(g) + H2(g) Is this reaction spontaneous?

Chapter 10 Section 4 Order and Spontaneity Gibbs Energy, continued Sample Problem H Solution For H2O(g),  = kJ/mol. For CO(g),  = kJ/mol. For H2(g),  = 0 kJ/mol. For C(s) (graphite),  = 0 kJ/mol. G = G(products) - G(reactants)

Calculating a Gibbs Energy Change using  Values, continued
Chapter 10 Section 4 Order and Spontaneity Calculating a Gibbs Energy Change using  Values, continued Sample Problem H Solution, continued G = G(products) - G(reactants) = [(mol CO(g))( for CO(g)) + (mol H2(g))( for H2(g))] – [(mol C(s))( for C(s)) + (mol H2O(g))( for H2O(g))] = [(1 mol)( kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)(0 kJ/mol) – (1 mol)( kJ/mol)] = ( ) kJ = kJ The reaction is nonspontaneous under standard conditions.

Chapter 10 Section 4 Order and Spontaneity 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.

Chapter 10 Section 4 Order and Spontaneity 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.

Chapter 10 Section 4 Order and Spontaneity 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.

Understanding Concepts
Chapter 10 Standardized Test Preparation Understanding Concepts 1. Which of these thermodynamic values can be determined using an adiabatic calorimeter? A. ∆G B. ∆H C. ∆S D. ∆T

Chapter 10 Standardized Test Preparation Understanding Concepts 2. Which of these statements about the temperature of a substance is true? F. Temperature is a measure of the entropy of the substance. G. Temperature is a measure of the total kinetic energy of its atoms. H. Temperature is a measure of the average kinetic energy of its atoms. I. Temperature is a measure of the molar heat capacity of the substance.

Chapter 10 Standardized Test Preparation Understanding Concepts 3. Which of the following pairs of conditions will favor a spontaneous reaction? A. a decrease in entropy and a decrease in enthalpy B. a decrease in entropy and an increase in enthalpy C. an increase in entropy and a decrease in enthalpy D. an increase in entropy and an increase in enthalpy

Chapter 10 Standardized Test Preparation Understanding Concepts 4. What is the standard enthalpy of formation of N2?

Chapter 10 Standardized Test Preparation Understanding Concepts 4. What is the standard enthalpy of formation of N2? Answer: Because the nitrogen molecule is the normal form of the element nitrogen, its standard enthalpy of formation is defined as zero.

Chapter 10 Standardized Test Preparation Understanding Concepts 5. What are the circumstances that cause a nonspontaneous reaction to occur?

Chapter 10 Standardized Test Preparation Understanding Concepts 5. What are the circumstances that cause a nonspontaneous reaction to occur? Answer: Nonspontaneous reactions require an input of energy, such as heat or light.

Chapter 10 Standardized Test Preparation Understanding Concepts 6. Based on changes in entropy and enthalpy, predict whether this reaction is spontaneous or nonspontaneous: 2AB(s)  A2(g) + B2(g) kJ

Chapter 10 Standardized Test Preparation Understanding Concepts 6. Based on changes in entropy and enthalpy, predict whether this reaction is spontaneous or nonspontaneous: 2AB(s)  A2(g) + B2(g) kJ Answer: ∆H is negative and ∆S is positive. Therefore ∆G is negative and the reaction is spontaneous.

Chapter 10 Reading Skills
Standardized Test Preparation Reading Skills Read the passage below. Then answer the questions. Almost all of the organisms on Earth rely on the process of photosynthesis to provide the energy needed for the functions of living. During photosynthesis carbon dioxide and water combine to form glucose and oxygen. The photosynthesis reaction is represented by the chemical equation: 6CO2(g) + 6H2O(l) + energy  C6H12O6(s) + 6O2(g). The source of energy for the reaction is light from the sun.

Standardized Test Preparation Reading Skills 7. If the absolute value of T∆S is smaller than the absolute value of ∆H, is photosynthesis a spontaneous reaction? Explain your answer.

Standardized Test Preparation Reading Skills 7. If the absolute value of T∆S is smaller than the absolute value of ∆H, is photosynthesis a spontaneous reaction? Explain your answer. Answer: The reaction is endothermic, requiring an input of energy. If the absolute value of T∆S is smaller than that of ∆H, then the Gibbs free energy is positive, whether ∆S is positive or negative, so the reaction is nonspontaneous.

Chapter 10 Standardized Test Preparation Understanding Concepts 8. Based on the nature of the reactants and products, what can be deduced about the entropy change during photosynthesis? F. Entropy increases because one of the products is an element. G. Entropy decreases because there are substantially fewer molecules of product than of reactants. H. Entropy increases because one of the products is a solid while one of the reactants is a liquid. I. Entropy does not change during the reaction because both sides of the reaction include the same number of atoms.

Standardized Test Preparation Reading Skills 9. If a manufacturing process was developed to make glucose from carbon dioxide and water, using heat as the energy source, how would the amount of energy required compare to that of the process that plants use?

Standardized Test Preparation Reading Skills 9. If a manufacturing process was developed to make glucose from carbon dioxide and water, using heat as the energy source, how would the amount of energy required compare to that of the process that plants use? Answer: The same amount of energy would be needed.

Interpreting Graphics
Chapter 10 Standardized Test Preparation Interpreting Graphics The table below shows molar heat capacities (joules per kelvins  mole) of elements and compounds. Use it to answer questions 10 through 13.

Chapter 10 Standardized Test Preparation Interpreting Graphics 10. What is the specific heat of aluminum chloride, which has a molar mass of 133.4? A J/K•g B J/K•g C J/K•g D J/K•g

Chapter 10 Standardized Test Preparation Interpreting Graphics 11. What is the relationship between the number of atoms per unit of an ionic compound and its molar heat capacity?

Chapter 10 Standardized Test Preparation Interpreting Graphics 11. What is the relationship between the number of atoms per unit of an ionic compound and its molar heat capacity? Answer: The molar heat capacity is about 25 J/K•mol for each atom in a mole of the ionic compound.

Chapter 10 Standardized Test Preparation Interpreting Graphics 12. Which of these statements best describes the general relationship of molar heat capacity of two different metals? F. The molar heat capacity of the two metals is about the same. G. The difference in molar heat capacity of two metals depends on the temperature. H. The molar heat capacity of the metal with the lower atomic mass is generally smaller. I. The molar heat capacity of the metal with the higher atomic mass is generally smaller.

Chapter 10 Standardized Test Preparation Interpreting Graphics 13. How many joules of heat are required to raise the temperature of one mole of liquid water by 2.00°C?

Chapter 10 Standardized Test Preparation Interpreting Graphics 13. How many joules of heat are required to raise the temperature of one mole of liquid water by 2.00°C? Answer: J

Chapter 10 Section 1 Energy Transfer Energy as Heat

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Chapter 10 Section 1 Energy Transfer Energy as Heat

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