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Thermodynamics Chapters 5 and 19

Chapters 5 and 19. Energy is commonly defined as the capacity to do work or transfer heat. Work is defined as the energy used to cause and object

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  • Slide 1
  • Chapters 5 and 19
  • Slide 2
  • Energy is commonly defined as the capacity to do work or transfer heat. Work is defined as the energy used to cause and object to move. Heat is defined as the energy used to cause the temperature of an object to increase.
  • Slide 3
  • Kinetic Energy: Energy of motion KE = mv 2 Example: A car moving at 55 mph has a greater kinetic energy than a car moving at 40 mph. All atoms and molecules are in motion and have kinetic energy. Potential Energy: Associated with the position of an object. Chemical potential energy is the energy stored in chemical bonds.
  • Slide 4
  • The SI unit of energy is the joule (J) This is derived from the formula for kinetic energy. Another unit of energy is the calorie. 1 cal = 4.184 J
  • Slide 5
  • Things in the universe we single out to study and observe are referred to as systems. Everything else is the surroundings. Systems can be: Open: Where matter and energy can be exchanged with the surroundings. Closed: Where energy but not matter can be exchanged with the surroundings. Isolated: Where neither matter nor energy can be exchanged with the surroundings.
  • Slide 6
  • The two ways that we experience energy changes are work and heat. Work is defined as the energy used to move an object. = F x d The other way to transfer energy is heat. Heat is the transfer of energy from a warmer object to a cooler one.
  • Slide 7
  • The first law of thermodynamics states that energy cannot be created or destroyed, only transferred. Any energy lost by the system must have been transferred to the surroundings.
  • Slide 8
  • The internal energy of a system is the sum of ALL kinetic and potential energy of all of its components. The internal energy of a system is defined as E. We can usually never calculate the actual numerical value of E. We can calculated the change in the internal energy of a system ( E).
  • Slide 9
  • A closed system may exchange energy with the surroundings as heat or work. This would obviously change the amount of internal energy resulting in a positive or negative E. The first law of thermodynamics can be written algebraically as: E = q +
  • Slide 10
  • When heat is added to a system or work is done on the system, its internal energy would increase ( E = +) If heat is released from the system or the system does work on the surroundings, its internal energy would decrease ( E = -)
  • Slide 11
  • For q+ Means system gains heat - means system loses heat For w+ means work is done on the system - means work is done by the system For E+ means net gain of energy by system - means net loss of energy by system
  • Slide 12
  • Two gases, A(g) and B(g), are confined in a cylinder and piston arrangement. Substances A and B react to form a solid product: A(g) + B(g) C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. As the volume of the gas decreases under constant atmospheric pressure, the surroundings do 480 J of work on the system what is the change in the internal energy of the system?
  • Slide 13
  • When a process occurs in which the system absorbs heat, the process is called endothermic. A process in which the system loses heat is called exothermic.
  • Slide 14
  • Internal energy is an extensive property. Meaning a 25 g sample of water at 25 o C would contain less total internal energy than a 50 g sample of water at 25 o C. Suppose we define our system as 50 g of water at 25 o C. This system could have arrived at this state by cooling 50 g of water from 100 o C, or by melting ice from 0 o C. The internal energy after each process would be the same. A state function is a property of a system that depends only on its current state, not how it got there.
  • Slide 15
  • Suppose you drive from Chicago to Denver. Chicago is 596 ft above sea level; Denver is 5280 ft above sea level. No matter what rout you take your altitude change will be 4684 ft. Altitude is a state function. The rout you take however can have a big difference on the distance your travel.
  • Slide 16
  • Consider a battery
  • Slide 17
  • Consider the reaction between zinc and an acid Zn(s) + 2 H + (aq) Zn 2+ (aq) + H 2 (g) If we carry this reaction out in an open beaker, we can see the hydrogen gas being evolved. It may not be obvious but the hydrogen gas is doing work on the surroundings. This is more evident if we add a cylinder and piston set up to the experiment.
  • Slide 18
  • This is the most common form of work discussed in chemistry. This is called pressure-volume (PV) work. When pressure is constant PV work is described as: w = -P V The thermodynamic function called enthalpy accounts for both the follow of heat into or out of a system as well as PV work. Positive enthalpy indicates a total increase in energy. Negative enthalpy indicates a total decrease in energy.
  • Slide 19
  • Indicate the sign of the enthalpy change, H, in each of the following processes carried out under atmospheric pressure, and indicate whether the process is endothermic or exothermic. (a) An ice cube melts. (b) 1 g of butane (C 4 H 10 ) is completely combusted in excess oxygen.
  • Slide 20
  • Just like internal energy: H = H final H inital In a chemical reaction: H = H products H reactants When the combustion of hydrogen gas is controlled so that 2 mol of H 2 (g) burn to form 2 mol of H 2 O(g) at constant pressure the system releases 483.6 kJ of energy.
  • Slide 21
  • Enthalpy is an extensive property. The magnitude of H, therefore, is directly proportional to the amount of reactant consumed. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) H = -890 kJ 2 CH 4 (g) + 4 O 2 (g) 2 CO 2 (g) + 4 H 2 O(l) H = -1780 kJ
  • Slide 22
  • The enthalpy of a reaction is equal in magnitude but opposite in sign to the enthalpy of its reverse reaction. CO 2 (g) + 2 H 2 O(l) CH 4 (g) + 2 O 2 (g) H = 890 kJ The enthalpy of a reaction depends on the state of the reactants and products. CH 4 (g) + 2 O 2 (g) CO 2 (g) + H 2 O(g) H = - 802 kJ
  • Slide 23
  • How much heat is released when 4.50 g of methane gas is burned in a constant pressure system?
  • Slide 24
  • Calorimetry is the experimental determination of the H for any physical or chemical process. In order to do a calorimetry experiment we need to know some background. Heat Capacity (C): Heat capacity, C, is the amount of heat energy required to raise the temperature of a substance by 1 K (1 o C).
  • Slide 25
  • A substance with a high heat capacity requires more heat energy to raise its temperature than a substance with a low heat capacity. Heat capacity is an extensive property. 10 mL of water heat up much faster than 1 L of water. The heat capacity of one mole of a substance is called its molar heat capacity C m. The heat capacity of one gram of a substance is called its specific heat capacity or just specific heat C s.
  • Slide 26
  • C s = Example: It requires 209 J of heat to increase the temperature of 50.0 g of water by 1.00 K. What is the specific heat of water?
  • Slide 27
  • (a) How much heat is needed to warm 250 g of water from 22 o C to 98 o C? The specific heat of water is 4.18 J/g-K. (b) What is the molar heat capacity of water?
  • Slide 28
  • Many enthalpies of reactions have been tabulated by scientists. Having this information available and being able to use it allows us to calculate the heat of reaction for almost every chemical reaction. Because enthalpy change ( H) is a state function, reactions can be thought of as happening in one step or in a series of steps.
  • Slide 29
  • The combustion of methane (CH 4 ) gas to produce carbon dioxide and liquid water happens in two steps.
  • Slide 30
  • The enthalpy of reaction for the combustion of C to CO 2 is: C(s) + O 2 (g) CO 2 (g) H = -393.5 kJ The enthalpy of reaction for the combustion of CO to CO 2 is: CO(g) + O 2 (g) CO 2 (g) H = -283.0 kJ Use this information to calculate the enthalpy of reaction for the combustion of C(s) to CO(g)
  • Slide 31
  • Calculate H for the reaction: 2 C(s) + H 2 (g) C 2 H 2 (g) Given the following chemical equations and their respective enthalpy changes C 2 H 5 (g) + 5/2 O 2 (g) 2 CO 2 (g) + H 2 O(l) H = -1299.6 kJ C(s) + O 2 (g) CO 2 H = -393.5 kJ H 2 (g) + O 2 (g) H 2 O(l)
  • Slide 32
  • Many different enthalpies of many different processes have been tabulated. We have already used the enthalpy of fusion and the enthalpy of vaporization. One other important type of enthalpy change that has been tabulated for many compounds in the enthalpy of formation. This is the heat change associate with the formation of a compound from its constituent elements.
  • Slide 33
  • Enthalpies of formation change depending on things like, pressure, temperature, and the physical states of the reactants and products. In order to compare different enthalpies of formations we need to study then under the same conditions. Most enthalpies of formation (an of other reactions as well) are tabulated under standard conditions standard conditions are defined as 1 atm and 298 K. Standard enthalpy change is that of a reaction where all of the components are in the physical state they would be in at standard conditions.
  • Slide 34
  • The standard enthalpy of formation of a compound ( H o f ) is the enthalpy change for the reaction that forms one mole of the compound at standard state. Elements (in standard state) 1 mole of Compound (in standard state) The value of H o f is always reported as kJ/mol The standard enthalpy change for any element in its standard state is 0 kJ/mol.
  • Slide 35
  • The enthalpy of any reaction can be calculated from the addition of the tabulated H o f for all the components. Example: Calculate the standard heat of reaction for the combustion of propane (C 3 H 8 ) with oxygen to from CO 2 and water.
  • Slide 36
  • A spontaneous process is one that occurs without any outside assistance. The melting of ice at temperatures higher than 0 o C is a spontaneous process. Any process that is spontaneous is one direction is not spontaneous in the reverse direction. Think about trying to freeze water at 25 o C. Not impossible but not easy. The majority of spontaneous processes are exothermic. All spontaneous processes are considered to be irreversible. A reversible process is one where the system and the surroundings can be converted back to their original states without any net change.
  • Slide 37
  • Now that we know what a spontaneous process is we can start to predict if an unfamiliar process will be spontaneous or not. We will use the thermodynamic quantity of entropy (S) to do this. In general we will define entropy as randomness. The more free the molecules are to move in random order, the higher the entropy of the system.
  • Slide 38
  • Just like E, and H, entropy (S) is a state function. We can also not calculate the specific value of S for any system. We can only calculate the change in S. S = S final S initial In the special case of an isothermal process S is equal to the heat that would be transferred in the reverse process divided by the temperature.
  • Slide 39
  • The melting of a substance at its melting point and the vaporization of a substance at its boiling point are isothermal processes. We achieve these changes by adding a certain amount of heat from the surroundings. ( H fus or H vap ) The S for these processes can be calculated easily because these are isothermal processes.
  • Slide 40
  • The normal freezing point of mercury is - 38.9 o C, and its molar enthalpy of fusion is H fusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?
  • Slide 41
  • We saw that in the first law of thermodynamics energy is always conserved. Entropy is different. The entropy change in any spontaneous process will always be positive. If we calculate the entropy change of one mole of ice melting to liquid water we get:
  • Slide 42
  • In the last example we saw that the total entropy change of the universe was positive. This corresponds to a spontaneous and irreversible process. If the sum of the S of both the system and surroundings equal zero the process is reversible.
  • Slide 43
  • There are three properties of matter that we can use to predict the sign (+ or -) of S: Temperature Volume The number of independently moving particles. Think of these three factors and how they affect the movement and disorder of molecules.
  • Slide 44
  • When water vaporizes the molecules spread out into a larger volume. This results in S = + Consider the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) In this process we are combining two molecules into one. The result of this is less total movement of molecules in the system and a negative entropy.
  • Slide 45
  • In general we expect entropy to increase when: Gases are formed from either solids or liquids. Liquids or solutions are formed from solids. The number of gas molecules increases during a chemical reaction.
  • Slide 46
  • Predict whether S would be negative or positive for each of the following processes. H 2 O(l) H 2 O(g) Ag + (aq) + Cl - (aq) AgCl(s) 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (g) N 2 (g) + O(g) 2 NO(g)
  • Slide 47
  • As we cool a substance down the molecules begin to move slower. If we continue to cool a substance down to absolute zero (0 K) the molecules will have no kinetic energy. The third law of thermodynamics states that the entropy of a pure crystalline solid substance at absolute zero is 0.
  • Slide 48
  • We talked about how calorimetry is used to determine the enthalpy change for a reaction. No such type of experiment exists to determine the entropy change of a reaction. The absolute value of S can however be determined through complex experimentation.
  • Slide 49
  • The entropies of substances in their standard states are usually tabulated as molar quantities: As you can see, unlike enthalpies of formation, the molar entropy of a pure element is NOT zero. The standard molar entropies of gases are, in general, greater than those of liquids and solids. Standard molar entropies generally increase with increasing molar mass. Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance.
  • Slide 50
  • Just like using the standard heats of formation to calculate the enthalpy change in a reaction, we can use the standard molar entropies in the same way. S o =
  • Slide 51
  • Calculate the S o for the synthesis for ammonia from N 2 (g) and H 2 (g) at 298 K.
  • Slide 52
  • The change in entropy of the surroundings during any process will depend on how much heat is given off or absorbed by the system. for a reaction at constant pressure
  • Slide 53
  • So far we have studied two thermochemical concepts, entropy and enthalpy. These two quantities must be used together to tell definitively if a reaction is spontaneous or not. Gibbs free energy tells us if a reaction is truly spontaneous.
  • Slide 54
  • If G is negative, the reaction is spontaneous in the forward direction. If G is zero, the reaction is at equilibrium If G is positive the reaction is nonspontaneous in the forward direction but the reverse reaction is spontaneous.
  • Slide 55
  • Calculate the standard free energy change for the formation of NO(g) from N 2 (g) and O 2 (g) at 298 K. Given that H o = 180.7 kJ and S o = 24.7 J/K. Is the reaction spontaneous under these circumstances?
  • Slide 56
  • Just like the standard enthalpies of formation there is a tabulated list of standard free energies of formation. We use this tabulated data in the same way we use the standard enthalpies of formation.
  • Slide 57
  • Use the tabulated standard free energy of formation to calculate. The standard free- energy change for the following reaction at 298 K: P 4 (g) + 6 Cl 2 (g) 4 PCl 3
  • Slide 58
  • We saw that G o f values are tabulated. These all correspond to the formation reactions at 298 K (25 o C). Many reactions do not occur at 298 K.
  • Slide 59
  • HH SS-T S GGReaction Character
  • Slide 60
  • For a certain reaction, H o = -35.4 kJ and S o = -88.5 kJ/K. Is this reaction exothermic or endothermic? Does the reaction lead to an increase or increase in randomness or disorder of the system? Calculate G o for this reaction at 298K. Is the reaction spontaneous at 298K? Would the spontaneity of this reaction change if we changed the temperature? How?