Prentice Hall © 2003Chapter 5 Chapter 6 Thermochemistry CHEMISTRY

Prentice Hall © 2003Chapter 5 Thermodynamics - is the study of energy and the transformations it undergoes in chemical reactions. Units: Joules (J) calorie = J calorie – the amount of E needed to raise the temperature of 1 gram of water 1 o C

Prentice Hall © 2003Chapter 5 Energy Examples of energy: Potential E Kinetic E Work – E needed to move an object against a force Heat – E transferred from hot to cold objects - capacity to do work or transfer heat

Prentice Hall © 2003Chapter 5 Today’s Topics 4 Thermodynamic Functions Definition of State Function Internal Energy Point of Views – System, Surroundings Universe Sign Conventions

Prentice Hall © 2003Chapter 5 4 Thermodynamic Functions E-Internal Energy H-Enthalpy S-Entropy G-Gibb’s Free Energy

Prentice Hall © 2003Chapter 5 What is a state function? A state function is a property that depends on the present condition and not on how the change occurs. Derived from calculating the change:  = [Final value – initial value]

Prentice Hall © 2003Chapter 5 4 State Functions Therefore:  E = E f -E i  H = H f - H i  S = S f - S i  G = G f - G i

Prentice Hall © 2003Chapter 5 Internal Energy (E) Internal Energy - is the sum of the kinetic and potential energy of a system Is the sum of heat (q) and work (w)  E = E final – E initial  E = q + w

Prentice Hall © 2003Chapter 5 Point of Views System - the reaction we are studying Surroundings – anything else besides the reaction For example: the container, you, etc…. Universe – system + surroundings

Prentice Hall © 2003Chapter 5 q = heat and w = work If q is (+), system gaining heat from surroundings (endo) If q is (-), system giving up heat to the surroundings (exo) If w is (+), system is the recipient of work from the surroundings. (in short, surroundings is doing work on the system.

Prentice Hall © 2003Chapter 5 Thermodynamic Functions Have value, unit and magnitude. The sign of  E depends on the magnitude of q and w. Knowing the value of  E does not tell us which variable is larger, q or w.

Prentice Hall © 2003Chapter 5 Problem 1 Calculate the change in internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.

Prentice Hall © 2003Chapter 5 Problem 2 Consider the reaction of hydrogen and oxygen gases to produce water. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The expanding gas does 480 J of work on the surroundings as it pushes against the atmosphere. Calculate the change in the internal energy of the system?

Prentice Hall © 2003Chapter 5 Problem 3 Calculate  E and determine whether the process is endothermic or exothermic. 1.)q = 1.62 kJ and w = -874 J 2.)The system releases 113 kJ of heat to the surroundings and does 39 kJ of work. 3.) The system absorbs 77.5 kJ of heat while doing 63.5 kJ of work on the surroundings.

Prentice Hall © 2003Chapter 5 Thermodynamic Functions Have value, unit and magnitude. The sign of  E depends on the magnitude of q and w. Knowing the value of  E does not tell us which variable is larger, q or w.

Prentice Hall © 2003Chapter 5 Problem 4 The  E for a reaction is -30 kJ. Is the process endothermic or exothermic?

Prentice Hall © 2003Chapter 5 Problem 5 Calculate  E and determine whether the process is endothermic or exothermic. 1.) A 50 g sample of water is cooled from 30 o C to 15 o C, thereby losing 3150 J of heat. 2.) A chemical reaction releases 8.65 kJ of heat and does no work on the surroundings. 3.) A balloon is heated by adding 900 J of heat. It expands doing 422 J of work on the atmosphere.

Prentice Hall © 2003Chapter 5 Enthalpy Enthalpy – accounts for heat flow in chemical reactions that occur at constant P when nothing other than P-V work are performed  H = H final - H initial If:H = E + P  V Then:  H =  E + P  V

Prentice Hall © 2003Chapter 5 Ways of Measuring  H Calorimetry Hess’s Law Heats of Formation (  H o f )

Prentice Hall © 2003Chapter 5 Calorimetry - Is the measurement of heat flow Specific Heat capacity (C) - the amount of heat needed to raise the temperature of 1 g of substance 1 o C. The greater the heat capacity, the greater the heat required to produce a rise in temp.

Prentice Hall © 2003Chapter 5 Calorimetry Equation q = mC  T - q system = q surroundings - q system = q water + q calorimeter - q substance = (mC  T solution + mC  T calorimeter ) Simplifies to: - q substance = (mC  T solution + C  T calorimeter )

Prentice Hall © 2003Chapter 5 Calorimetry Problem A 30.0 gram sample of water at 280 K is mixed with 50.0 grams of water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The specific heat capacity of the solution is 4.18 J/g- o C.

Prentice Hall © 2003Chapter 5 Calorimetry Problem A 46.2 gram sample of copper is heated to 95.4 o C and then placed in a calorimeter containing 75.0 gram of water at 19.6 o C. The final temperature of the metal and water is 21.8 o C. Calculate the specific heat capacity of copper, assuming that all the heat lost by the copper is gained by the water.

Prentice Hall © 2003Chapter 5 Calorimetry Problem A 15.0 gram sample of nickel metal is heated to o C and dropped into 55.0 grams of water, initially at 23 o C. Assuming that no heat is lost to the calorimeter, calculate the final temperature of the nickel and water. The specific heat of nickel is J/g- o C. The specific heat of water is 4.18 J/g- o C.

Prentice Hall © 2003Chapter 5 Problem 4 The specific heat of water is 4.18 J/g-K. How much heat is needed to warm 250 g of water from 22 o C to 98 o C? What is the molar heat capacity of water? Molar heat Capacity = C x molar mass

Prentice Hall © 2003Chapter 5 Problem 5 Large beds of rocks are used in solar heated homes to store heat. Assume that the specific heat of rocks is 0.82 J/g-K. A. Calculate the amount of heat absorbed by 50 kg. of rocks if their temperature rose by 12.0 o C. B. What temperature change would these rocks undergo if they emitted 450 kJ of heat?

Prentice Hall © 2003Chapter 5 Work – E needed to move an object against a force When P is constant, P-V work is given by w = - P  V

Prentice Hall © 2003Chapter 5 Enthalpy Enthalpy – accounts for heat flow in chemical reactions that occur at constant P when nothing other than P-V work are performed  H = H final - H initial If:H = E + P  V Then:  H =  E + P  V

Prentice Hall © 2003Chapter 5 When P is constant, P-V work is given by w = - P  V Thus at constant P,  H = q p  H = (+) endothermic  H = (-) exothermic

Prentice Hall © 2003Chapter 5 Enthalpy Enthalpy of a reaction or heat of Reaction:  H = H products - H reactants 1. sign of  H depends on the amount of reactant consumed 2.  H sign is opposite for backwards reaction 3.  H rxn depends on the physical state of the reactants and products.

Prentice Hall © 2003Chapter 5 Problem 1 Given the reaction: 2H 2 (g) + O 2 (g)  2 H 2 O (g)  H = -483 kJ Calculate the  H value for: 2 H 2 O (g)  2H 2 (g) + O 2 (g)

Prentice Hall © 2003Chapter 5 Problem 2 Given the reaction: 2H 2 (g) + O 2 (g)  2 H 2 O (g)  H = -483 kJ How much heat is released when 10.5 grams of H 2 is burned in a constant-pressure system?

Prentice Hall © 2003Chapter 5 Problem 3 Hydrogen peroxide can decompose to water and hydrogen by the following reaction: 2H 2 O 2 (l)  2H 2 O (l) + O 2 (g)  H = -196 kJ Calculate the value of q when 5.00 g of H 2 O 2 decomposes at constant pressure.

Prentice Hall © 2003Chapter 5 Ways of Measuring  H Calorimetry Hess’s Law Heats of Formation (  H o f )

Prentice Hall © 2003Chapter 5 Hess’ Law If a reaction is carried out in steps,  H for the reaction will equal the sum of the enthalpy changes for the individual steps.

Prentice Hall © 2003Chapter 5 Problem 1 Given: C (s) + O 2 (g)  CO 2 (g)  H = kJ CO(g) + ½ O 2 (g)  CO 2 (g)  H = kJ Calculate the enthalpy of combustion for: C(s) + ½ O 2 (g)  CO (g)

Prentice Hall © 2003Chapter 5 Problem 2 Given: C(graphite) + O 2 (g)  CO 2 (g)  H = kJ C(diamond) + O 2 (g)  CO 2 (g)  H = kJ Calculate the enthalpy of combustion for: C(graphite)  C (diamond)

Prentice Hall © 2003Chapter 5 Problem 3 Given: C 2 H 2 (g) + 5/2 O 2 (g)  2CO 2 (g) + H 2 O (l)  H = kJ C (s) + O 2 (g)  CO 2 (g)  H = kJ H 2 (g) + 1/2 O 2 (g)  H 2 O (l)  H = kJ Calculate the enthalpy of combustion for: 2C (s) + H 2 (g)  C 2 H 2 (g)

Prentice Hall © 2003Chapter 5 Enthalpies of Formation - known as heat of formation - gives the energy needed for a compound to form Standard enthalpy - is the enthalpy change (  H) when the reactants and products are in their standard state, usually 1 atm and 25 o C- denoted by  H o ( ex.  H o f )

Prentice Hall © 2003Chapter 5 Standard Enthalpy of Formation, (  H o f ) By definition: The standard enthalpy of formation ( ex.  H o f ) of the most stable form of any element is ZERO because there is no formation reaction needed when the element is in its standard state. - important for diatomic molecules - need knowledge of standard states of compounds

Prentice Hall © 2003Chapter 5 If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H o f. Standard conditions (standard state): 1 atm and 25 o C (298 K). Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states. Enthalpies of Formation

Prentice Hall © 2003Chapter 5 If there is more than one state for a substance under standard conditions, the more stable one is used. Standard enthalpy of formation of the most stable form of an element is zero. Using Enthalpies of Formation of Calculate Enthalpies of Reaction We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation. Enthalpies of Formation

Prentice Hall © 2003Chapter 5 Enthalpies of Formation

Prentice Hall © 2003Chapter 5 Using Enthalpies of Formation of Calculate Enthalpies of Reaction For a reaction Enthalpies of Formation

Prentice Hall © 2003Chapter 5 Next Topic  S = Entropy = disorder

Prentice Hall © 2003Chapter 5 Thermodynamic Question Can a process occur? Spontaneous or Non-spontaneous? –Forward vs. Reverse reactions –Example: Gas Expansion Reversible or irreversible?

Prentice Hall © 2003Chapter 5 Entropy Entropy, S, is a measure of the disorder of a system. Spontaneous reactions proceed to lower energy or higher entropy. In ice, the molecules are very well ordered because of the H-bonds. Therefore, ice has a low entropy.

Prentice Hall © 2003Chapter 5 As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases). Water is more random than ice, so ice spontaneously melts at room temperature. Conclusion: The higher the entropy the more spontaneous the reaction.

Prentice Hall © 2003Chapter 5 Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates. Entropy is a state function. For a system,  S = Sfinal - Sinitial. If  S > 0 the randomness increases, if  S < 0 the order increases.

Prentice Hall © 2003Chapter 5 Entropy Suppose a system changes reversibly between state 1 and state 2. Then, the change in entropy is given by –at constant T where q rev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.) Entropy and the Second Law of Thermodynamics

Prentice Hall © 2003Chapter 5 The Second Law of Thermodynamics Spontaneous processes have a direction. In any spontaneous process, the entropy of the universe increases.  S univ =  S sys +  S surr : the change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings. Entropy is not conserved:  S univ is increasing. Entropy and the Second Law of Thermodynamics

Prentice Hall © 2003Chapter 5 The Second Law of Thermodynamics Reversible process:  S univ = 0. Spontaneous process (i.e. irreversible):  S univ > 0.  S sys for a spontaneous process can be less than 0 as long as  S surr > 0. This would make  S univ still (+). For an isolated system,  S sys = 0 for a reversible process and  S sys > 0 for a spontaneous process. Entropy and the Second Law of Thermodynamics

Prentice Hall © 2003Chapter 5 Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. Entropy changes dramatically at a phase change. As we heat a substance from absolute zero, the entropy must increase.

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5 Absolute entropy can be determined from complicated measurements. Standard molar entropy, S  : entropy of a substance in its standard state. Similar in concept to  H . Units: J/mol-K. Note units of  H: kJ/mol. Standard molar entropies of elements ( S  ) are not 0. For a chemical reaction which produces n moles of products from m moles of reactants: Entropy Changes in Chemical Reactions

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5 For a spontaneous reaction the entropy of the universe must increase. Reactions with large negative  H values are spontaneous. How do we correlate  S and  H to predict whether a reaction is spontaneous? Gibbs free energy, G, of a state is For a process occurring at constant temperature Gibbs Free Energy

Prentice Hall © 2003Chapter 5 Gibb’s Free Energy Is the capacity to do maximum useful work. Heat decreases the amount of useful work done.

Prentice Hall © 2003Chapter 5 WORK At constant P, w sys = - P  V At constant temperature, w sys = - T  S If  G =  H – T  S, for maximum useful work,  H must be = 0.

Prentice Hall © 2003Chapter 5 Three important conditions: –If  G < 0 then the forward reaction is spontaneous. –If  G = 0 then reaction is at equilibrium and no net reaction will occur. –If  G > 0 then the forward reaction is not spontaneous. If  G > 0, work must be supplied from the surroundings to drive the reaction. For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products. Gibbs Free Energy

Prentice Hall © 2003Chapter 5 Problem Correlate  S and  H to predict whether a reaction is non-spontaneous or spontaneous. If spontaneous, determine whether the reaction will be spontaneous at all temperature, at high temperature, or at low temperature. A. If  S is (-) and  H is (+). B. If  S is (-) and  H is (-). C. If  S is (+) and  H is (+). D. If  S is (+) and  H is (-).

Prentice Hall © 2003Chapter 5 Consider the formation of ammonia from N 2 and H 2. Initially ammonia will be produced spontaneously (Q < K eq ). After some time, the ammonia will spontaneously react to form N 2 and H 2 (Q > K eq ). At equilibrium, ∆G = 0 and Q = K eq. Gibbs Free Energy

Prentice Hall © 2003Chapter 5 Standard Free-Energy Changes  G  f Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and  G  = 0 for elements.  G  for a process is given by The quantity  G  for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (  G  > 0) or products (  G  < 0). Gibbs Free Energy

Prentice Hall © 2003Chapter 5 Focus on  G =  H - T  S: –If  H 0, then  G is always negative. –If  H > 0 and  S < 0, then  G is always positive. (That is, the reverse of 1.) –If  H < 0 and  S < 0, then  G is negative at low temperatures. –If  H > 0 and  S > 0, then  G is negative at high temperatures. Even though a reaction has a negative  G it may occur too slowly to be observed. Free Energy and Temperature

Prentice Hall © 2003Chapter 5 Free Energy and Temperature

Prentice Hall © 2003Chapter 5 Recall that  G  and K (equilibrium constant) apply to standard conditions. Recall that  G and Q (equilibrium quotient) apply to any conditions. It is useful to determine whether substances under any conditions will react: Free Energy and The Equilibrium Constant

Prentice Hall © 2003Chapter 5 At equilibrium, Q = K and  G = 0, so From the above we can conclude: –If  G  1. –If  G  = 0, then K = 1. –If  G  > 0, then K < 1. Free Energy and The Equilibrium Constant

Prentice Hall © 2003Chapter 5 Calculating  H,  S,  G Use Hess’ Law Standard Heats of Formation (  H o,  S o,  G o ) Equations

Prentice Hall © 2003Chapter 5 Next Chapter: ELECTROCHEMISTRY

Prentice Hall © 2003Chapter 5 Galvanic Cell – spontaneous Electrochemical Cell – non-spontaneous

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5 Hess’ Law If a reaction is carried out in a series of steps,  H for the reaction will equal the sum of the enthalpy changes for the individual steps

Prentice Hall © 2003Chapter 5

Prentice Hall © 2003Chapter 5 Kinetic Energy and Potential Energy Kinetic energy is the energy of motion: Potential energy is the energy an object possesses by virtue of its position. Potential energy can be converted into kinetic energy. Example: a bicyclist at the top of a hill. The Nature of Energy

Prentice Hall © 2003Chapter 5 Systems and Surroundings System: part of the universe we are interested in. Surroundings: the rest of the universe. We sometimes use the calorie instead of the joule: 1 cal = J (exactly) A nutritional Calorie: 1 Cal = 1000 cal = 1 kcal The Nature of Energy

Prentice Hall © 2003Chapter 5 Transferring Energy: Work and Heat Force is a push or pull on an object. Work is the product of force applied to an object over a distance: Energy is the work done to move an object against a force. Heat is the transfer of energy between two objects. Energy is the capacity to do work or transfer heat. The Nature of Energy

Prentice Hall © 2003Chapter 5 Internal Energy Internal Energy: total energy of a system. Cannot measure absolute internal energy. Change in internal energy, The First Law of Thermodynamics

Prentice Hall © 2003Chapter 5 Relating  E to Heat and Work Energy cannot be created or destroyed. Energy of (system + surroundings) is constant. Any energy transferred from a system must be transferred to the surroundings (and vice versa). From the first law of thermodynamics: when a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system: The First Law of Thermodynamics

Prentice Hall © 2003Chapter 5 Exothermic and Endothermic Processes Endothermic: absorbs heat from the surroundings. Exothermic: transfers heat to the surroundings. An endothermic reaction feels cold. An exothermic reaction feels hot. State function: depends only on the initial and final states of system, not on how the internal energy is used. The First Law of Thermodynamics

Prentice Hall © 2003Chapter 5 Chemical reactions can absorb or release heat. However, they also have the ability to do work. For example, when a gas is produced, then the gas produced can be used to push a piston, thus doing work. Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) The work performed by the above reaction is called pressure-volume work. When the pressure is constant, Enthalpy

Prentice Hall © 2003Chapter 5 Enthalpy, H: Heat transferred between the system and surroundings carried out under constant pressure. Enthalpy is a state function. If the process occurs at constant pressure, Enthalpy

Prentice Hall © 2003Chapter 5 Since we know that We can write When  H, is positive, the system gains heat from the surroundings. When  H, is negative, the surroundings gain heat from the system. Enthalpy

Prentice Hall © 2003Chapter 5 Enthalpy

Prentice Hall © 2003Chapter 5 For a reaction: Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2CH 4 (g) + 4O 2 (g)  2CO 2 (g) + 4H 2 O(g)  H =  1604 kJ Enthalpies of Reaction

Prentice Hall © 2003Chapter 5 When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ Change in enthalpy depends on state: H 2 O(g)  H 2 O(l)  H = -88 kJ Enthalpies of Reaction

Prentice Hall © 2003Chapter 5 Heat Capacity and Specific Heat Calorimetry = measurement of heat flow. Calorimeter = apparatus that measures heat flow. Heat capacity = the amount of energy required to raise the temperature of an object (by one degree). Molar heat capacity = heat capacity of 1 mol of a substance. Specific heat = specific heat capacity = heat capacity of 1 g of a substance. Calorimetry

Prentice Hall © 2003Chapter 5 Constant Pressure Calorimetry Atmospheric pressure is constant! Calorimetry

Prentice Hall © 2003Chapter 5 Hess’s law: if a reaction is carried out in a number of steps,  H for the overall reaction is the sum of  H for each individual step. For example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ Hess’s Law

Prentice Hall © 2003Chapter 5 If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H o f. Standard conditions (standard state): 1 atm and 25 o C (298 K). Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states. Enthalpies of Formation

Prentice Hall © 2003Chapter 5 If there is more than one state for a substance under standard conditions, the more stable one is used. Standard enthalpy of formation of the most stable form of an element is zero. Using Enthalpies of Formation of Calculate Enthalpies of Reaction We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation. Enthalpies of Formation

Prentice Hall © 2003Chapter 5 Enthalpies of Formation

Prentice Hall © 2003Chapter 5 Using Enthalpies of Formation of Calculate Enthalpies of Reaction For a reaction Enthalpies of Formation