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Chapter 5 Thermochemistry

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1 Chapter 5 Thermochemistry
Lecture Presentation Chapter 5 Thermochemistry LO 1.15, 5.4, 5.5, 5.6, 5.7 Ashley Warren Kings High School AP Chemistry © 2012 Pearson Education, Inc.

2 Thermodynamics v. Thermochemistry
The study of energy and its transformations is known as thermodynamics. Industrial Revolution Relationships between chemical reactions and energy changes that involve heat = thermochemistry Greek – therme = heat, dynamis = power Thermochem is simply a portion of thermodynamics.

3 © 2012 Pearson Education, Inc.
Energy Energy is the ability to do work or transfer heat. Energy used to cause an object that has mass to move is called work. Energy used to cause the temperature of an object to rise is called heat. Why is a pitcher able to throw a baseball faster than he could throw a bowling ball? © 2012 Pearson Education, Inc.

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Definitions: Work Energy used to move an object over some distance is work: w = F  d where w is work, F is the force, and d is the distance over which the force is exerted. In the picture, work is done as energy and is transferred from the pitchers arm to the ball. We describe a force as any push or pull exerted on an object. If we define the object as the system…then we (surroundings) are performing work on that system by transferring energy to it. © 2012 Pearson Education, Inc.

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Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects. Combustion reaction – releases chemical energy stored in the molecules of fuels. If we define the system as the substances in the reaction then the released energy causes the temperature of the system to increase. Energy in the form of heat is then transferred from the hotter system to the cooler surroundings. © 2012 Pearson Education, Inc.

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Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 2 Ek =  mv2 As speed increases, the KE increases! For a given speed, the KE increases with increasing mass. Thus a more massive truck traveling at 55 mph has more KE than a smaller car at the same speed. © 2012 Pearson Education, Inc.

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Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. It arises why????*** The most important form of potential energy in molecules is electrostatic potential energy, Eel: Potential energy is especially found in the chemical bonds. Potential energy is in essence “stored” energy that arises because of attractions and repulsions an object experiences in relation to other objects. Go back to last slide and look at cyclist easy to think about potential converting into kinetic…bike down a hill. It is this interconversion of energy, with one form decreasing and the other increasing….this is the cornerstone of thermodynamics. Gravitational forces play a negligible role in the ways that atoms and molecules interact with one another….forces that arise from electrical charges are more important. One of the most important forms of potential energy in chemistry is electrostatic potential energy which arises from the interaction between charged particles. Let’s talk about what happens to the energy if charges increase? What happens if distance increases? K is a proportionality constant that = 8.99 x 109 J m/C2. As the electrostatic potential energy goes to zero as d becomes infinite; in other words, the zero of electrostatic potential energy is defined as infinite separation of the charged particles. The figure illustrates how the energy behaves for charges of the same and different signs. We can see when the particles have the same charge, they repel each other and a repuslive force pushes them apart….in this case the energy is positive and the potential energy decreases as the particles move farther and farther apart. When the particles have opposite signs….in this case the energy is negative and the potential energy increases (becomes less negative) as the particles move apart. (Many substances release chemical energy when they react, energy due to the potential energy stored in their arrangements of their atoms….the energy a substance possesses because of its temperature (thermal energy) is associated with the kinetic energy of the molecules in the substance) KQ1Q2 d Eel = © 2012 Pearson Education, Inc.

8 © 2012 Pearson Education, Inc.
Units of Energy The SI unit of energy is the joule (J): An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = J 1 J = 1  kg m2 s2 Because a joule is not a large amount of energy we often use kJ. A calorie is defined as the amount of energy required to raise the temperature of 1 gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius. A calorie is now defined in terms of joules. A related energy unit used in nutrition is the Calorie. 1Cal = 1000 cal = 1kcal © 2012 Pearson Education, Inc.

9 Definitions: System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). 3 types The surroundings are everything else (here, the cylinder and piston). If we think about system and surroundings with chemical reactions, the reactants and products would be the system and the container and everything else would constitute the surroundings. Systems are classified as open, closed, or isolated. An open system is one in which matter and energy can be exchanged with the surroundings. A closed system (what we study most) can exchange energy but not matter with their surroundings. We can think about it as mass….the system should not lose or gain mass because matter is conserved. It can exchange energy though in the forms of work and heat. An isolated system is one in which neither energy nor matter can be exchanged with the surroundings. An insulated thermos is an example of this. © 2012 Pearson Education, Inc.

10 Describing and Calculating Energy Changes
A bowler lifts a 5.4 kg (12 lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. a.) What happens to the potential energy of the ball as it is raised? b.) What quantity of work, in J, is used to raise the ball? c.) The ball is dropped. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground? © 2012 Pearson Education, Inc.

11 First Law of Thermodynamics
Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. This important observation….that energy is conserved is known as the First Law of Thermodynamics! © 2012 Pearson Education, Inc.

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Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. To analyze energy changes in a chemical system we need to define internal energy. We generally do not know the numerical value of a system’s internal energy…in thermodynamics we are mainly concerned with the change in E. Imagine that we start with a system with an initial internal energy Einitial. The system undergoes a change, which might involve work being done or heat being transferred. After the change, the final internal energy is Efinal. © 2012 Pearson Education, Inc.

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Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial Again, we generally can’t determine the actual values of Efinal and Einitial for any system….but we can ultimately find delta E. In a chemical reaction, the initial state refers to the reactants and the final state refers to the products. When hydrogen and oxygen react to form water, the system loses energy to the surroundings. Because energy is lost from the system, the internal energy of the products is less than the reactants. The diagram above is called an energy diagram and it shows that the internal energy of the mixture of the reactants is greater than that of the products. © 2012 Pearson Education, Inc.

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Energy Diagram The internal energy for Mg (s) and Cl2 (g) is greater than that of MgCl2 (s). Sketch an energy diagram that represents the following reaction: MgCl2 (s)  Mg (s) + Cl2 (g) © 2012 Pearson Education, Inc.

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Internal Energy Thermodynamic quantities such as ΔE have 3 parts: A # A Unit -- A sign that signifies direction A positive value indicates that delta E results when Efinal is greater than Einitial indicating that the system has gained energy from its surroundings. A negative value of delta E is when Efinal is less than Einitial indicating that the system has lost energy to its surroundings. © 2012 Pearson Education, Inc.

16 Changes in Internal Energy
If E > 0, Efinal > Einitial Therefore, the system absorbed energy from the surroundings. (meaning that E is positive) This energy change is called endergonic. © 2012 Pearson Education, Inc.

17 Changes in Internal Energy
If E < 0, Efinal < Einitial Therefore, the system released energy to the surroundings. This energy change is called exergonic. © 2012 Pearson Education, Inc.

18 Changes in Internal Energy
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, E = q + w. Let’s use an analogy! The internal energy of a system changes in magnitude as heat is added to or removed from the system or as work is done on or by the system. If we think of internal energy as the system’s bank account of energy, we see that deposits or withdrawals can be made in increments of heat or in increments of work. Deposits increase the energy of the system (positive delta E) whereas withdrawals decrease the energy of the system (negative delta E) When heat is added to a system or work is done on a system, its internal energy increases!!!!! Therefore, when heat is transferred to the system from the surroundings, q is positive. Adding heat to the system is like making a deposit to the energy account….the energy of the system increases. Likewise, when work is done on the system by the surroundings, w has a positive value. Conversely, both heat lost by the system and the work done by the system on the surroundings have negative values that is they lower the internal energy. These are energy withdrawals and these lower the amount of energy in the system’s account. This is our algebraic expression for the first law of thermodynamics. © 2012 Pearson Education, Inc.

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E, q, w, and Their Signs We have sign conventions for q, w, and E Notice that any energy entering the system as either heat or work carries a positive sign!!! © 2012 Pearson Education, Inc.

20 Relating Heat & Work to Internal Energy
Gases A and B are confined in a cylinder and piston arrangement and they react to form a solid product C. As the reaction occurs, the system loses 1150J of heat to the surroundings. The piston moves downward as the gases react to form the solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? Hint! If you do not show the formula for the first first law of thermodynamics it’s wrong If you do not show + or – signs it is wrong If your answer does not have a sign it is wrong © 2012 Pearson Education, Inc.

21 Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic. Using the terms endothermic and exothermic describes the direction of heat transfer. During an endothermic process, like the melting of ice, heat flows INTO the system from the surroundings. If we touch the container in which the ice is melting the container feels cold to us because heat has passed from our hand to the container. In an exothermic, heat EXITS or flows out of the system. (combustion is a good example of this) © 2012 Pearson Education, Inc.

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State Functions We do know that the internal energy of a system is independent of the path by which the system achieved that state. In the system depicted in the figure below, the water could have reached room temperature from either direction. Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. The conditions that influence internal energy include the temperature and pressure. Furthermore, the internal energy of a system is proportional to the total quantity of matter in the system because energy is an extensive property. Suppose we have defined out system as 50 grams of water at 25 degrees Celsius. The system could have reached this state by cooling 50 grams of water from 100 degrees Celsius to 25 degrees Celsius or by melting 50 grams of ice and subsequently warming the water to 25 degrees Celsius. The internal energy of the water at 25 degrees Celsius is the same in either case. © 2012 Pearson Education, Inc.

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State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, E depends only on Einitial and Efinal. A state function is a property of a system that is determined by specifying the system’s condition or state (in terms of pressure, temperature, and so forth) Analogy = suppose you drive from Chicago, which is 596 ft above sea level, to Denver, which is 5280 ft above sea level. No matter which route you take, the altitude change is 4684 ft. The distance you travel, however, depends on your route. Altitude is analogous to a state function because the change in altitude is independent of the path taken. Distance traveled is not a state function because it is dependent on the path traveled. © 2012 Pearson Education, Inc.

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State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its E is the same. But q and w are different in the two cases. If a wire shorts out the battery no work is accomplished because nothing is moved against a force. All of the energy lost from the battery is in the form of heat….the wire will get warmer and release heat to surroundings. If the battery is used to make a motor turn, the discharge produces work…some heat is also released, but not as much as when the battery is shorted out. We can see that the magnitudes of the q and w for the two situations is different…however delta E is the same because delta E is a state function. © 2012 Pearson Education, Inc.

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Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). © 2012 Pearson Education, Inc.

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Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PV The work involved in the expansion or compression of gases is called pressure-volume work (or PV work). When pressure is constant in a process, the sign an magnitude of the pressure volume work are given by the formula above. The pressure is always either a positive number or 0. If the volume expands then delta V is positive. Because the expanding system does work on the surroundings, w is negative because energy leaves the system as work. If the gas is compressed, delta V is negative and that indicates that w is positive because work is done on the system by the surroundings. © 2012 Pearson Education, Inc.

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Enthalpy A system that consists of a gas confined to a container can be characterized by several different properties. If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy (denoted with symbol, H) is the internal energy plus the product of pressure and volume: Pressure of the gas, volume of the container…. Like internal energy, E, both Pressure and volume are state functions…meaning that they are only dependent on their current state and not the path taken to that state. We can combine these 3 state functions E, P, and V to define a new state function called enthalpy! (comes from a Greek word “to warm”). H = E + PV © 2012 Pearson Education, Inc.

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Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + PV If a system does not change its volume during the course of a process, does it do pressure-volume work? Change in enthalpy = change in internal energy plus the product of the constant pressure times the change in volume. © 2012 Pearson Education, Inc.

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Enthalpy Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV H = (q + w) − w H = q So, at constant pressure, the change in enthalpy is the heat gained or lost. Because we can measure heat or readily calculate heat….using enthalpy is more of interest to us than change in internal energy. © 2012 Pearson Education, Inc.

30 Endothermicity and Exothermicity
A process is endothermic when H is positive. A process is exothermic when H is negative. When delta H is positive this means it gained heat from the surroundings. When negative it means the system has released heat to the surroundings. To continue our analogy, under constant pressure an endothermic process deposits energy in the system in the form of heat and an exothermic process withdraws energy in the form of heat. © 2012 Pearson Education, Inc.

31 Determining the Sign of ΔH.
Indicate the sign of the enthalpy change, ΔH, in the processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic. a.) An ice cube melts. b.) 1 gram of butane (C4H10) is combusted in sufficient oxygen to give complete combustion of CO2 and H2O. Molten gold poured into a mold solidifies at atmospheric pressure. With the gold defined as the system, is the solidification an exothermic or endothermic process? Explain. © 2012 Pearson Education, Inc.

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Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants The enthalpy change that accompanies a reaction is called either the “enthalpy of reaction” or the “heat of reaction” and it is sometimes written ΔHrxn. © 2012 Pearson Education, Inc.

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Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. When we give a numerical value for the heat of reaction we must specify the reaction involved. When hydrogen and oxygen form water the system releases kJ of heat. Since the system is releasing heat, delta H reaction has a – value. ΔHrxn = kJ because the reaction is exothermic. We always report the ΔHrxn at the end of the balanced equation. A reaction that shows the balanced equation and the ΔHrxn value is called a thermochemical equation. We can also see the nature of the exothermic reaction in the enthalpy diagram above. © 2012 Pearson Education, Inc.

34 The Truth about Enthalpy
The following guidelines are helpful when using thermochemical equations and enthalpy diagrams: Enthalpy is an extensive property. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. H for a reaction depends on the state of the products and the state of the reactants. 1. The magnitude of ΔH is proportional to the amount of reactant consumed in the process. Ex: 890 kJ of heat is produced when 1 mol of CH4 is burned in a constant pressure system. This means the combustion of 2 mol of CH4 and 4 mol of oxygen would release twice as much, 1780 kJ. 2. Can see from diagram 3. If water was produced in the gas state instead of the liquid state from above, the heat of reaction would be -802 kJ instead of -890 kJ. Less heat would be available to transfer to the surroundings because the enthalpy of water in the gaseous state is greater than that of the liquid state. © 2012 Pearson Education, Inc.

35 Relating ΔH to Quantities of Reactants and Products
How much heat is released when 4.50 grams of methane gas is burned in a constant pressure system? CH4 (g) + 2 O2 (g)  CO2 (g) + 2H2O (l) ΔH = -890 kJ Hydrogen Peroxide can decompose to water and oxygen. Calculate the quantity of heat released when 5.00 g of hydrogen peroxide decomopses at constant pressure. 2H2O2 (l)  2H2O (l) + O2 (g) ΔH = -196 kJ © 2012 Pearson Education, Inc.

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Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. A device used to measure heat flow is a calorimeter. © 2012 Pearson Education, Inc.

37 Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. All substances change temperature when they are heated, but the magnitude of the temperature change produced by a given quantity of heat varies from substance to substance. The temperature change experienced by an object when it absorbs a certain amount of heat is determined by its heat capacity denoted, C. The greater the heat capacity, the greater the heat required to produce a given increase in temperature. © 2012 Pearson Education, Inc.

38 Heat Capacity and Specific Heat
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C). Molar heat capacity is the heat capacity one mole of a substance. We denote this with the symbol Cm. Specific heat capacity is denoted with the symbol Cs. © 2012 Pearson Education, Inc.

39 Heat Capacity and Specific Heat
Specific heat, then, is Specific heat = heat transferred mass  temperature change s = q m  T When a sample absorbs heat (positive q), the temperature increases (positive delta T) Let’s rearrange the formula above and we get q = mcΔt…with using this formula we can calculate the quantity of heat a substance gains or loses. © 2012 Pearson Education, Inc.

40 Relating Heat, Temperature Change, and Heat Capacity
How much heat is needed to warm 250 grams of water from 22 degrees Celsius to 98 degrees Celsius? What is the molar heat capacity of water? Assume the specific heat of rocks is 0.82 J/g – K. Calculate the quantity of heat absorbed by the 50.0 kg of rocks if their temperature increases by 12.0 degrees Celsius. What temperature change would these rocks undergo if they emitted 450 kJ of heat? © 2012 Pearson Education, Inc.

41 Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Let’s imagine we add two aqueous solutions each containing a reactant to a coffee cup calorimeter. Once mixed, the reactants can react to form products. In this case there is no physical boundary between the system and surroundings. The reactants and the products are the system, and the water in which they are dissolved is part of the surroundings. If we assume that the calorimeter is perfectly insulated, then any heat released or absorbed by the reaction will raise or lower the temperature of the water in the solution. Thus, we measure the temperature change of the solution and assume that any changes are due to heat transferred from the reaction to the water (for exo) or transferred from the water to the reaction (endo) By monitoring the temperature of the solution, we are seeing the flow of heat between the system and the surroundings. © 2012 Pearson Education, Inc.

42 Constant Pressure Calorimetry
Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m  s  T qsoln = -qrxn For an exo rxn. Heat is lost by the reaction and gained by the water so the temperature of the solution rises. The opposite occurs for endo rxns. The heat gained or lost by the solution, qsoln, is equal in magnitude but opposite in sign to the heat absorbed or released by the reaction, qrxn. We can calculate the qsoln. By using q = mcΔt © 2012 Pearson Education, Inc.

43 © 2012 Pearson Education, Inc.
Measuring ΔH When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.0° Celsius to 27.5° Celsius. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g – K. © 2012 Pearson Education, Inc.

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Measuring ΔH #2 When 50.0 mL of M AgNO3 and 50.0 mL of M HCl are mixed in a constant pressure calorimeter, the temperature of the mixture increases from 22.30° Celsius to 23.11° Celsius. Calculate ΔH for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of grams and a specifi heat of 4.18 J/g-°C. © 2012 Pearson Education, Inc.

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Bomb Calorimetry Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. qrxn = -Ccal x ΔT Combustion reactions are commonly studied using calorimetry….most accturately a bomb calorimeter. The substance is placed in a small cup within an insulated sealed vessel called a bomb. The bomb, which is made to withstand high pressures, has an inlet valve for adding oxygen and electrical leads. After the sample has been placed in the bomb, the bomb is sealed and pressurized with oxygen. It is then placed in the calorimeter and covered with an accurately measured quantity of water. The combustion reaction is initiated by passing an electrical current through a fine wire in contact with the sample. When the wire becomes sufficiently hot, the sample will ignite. The heat released when combustion occurs is absorbed by the water and the various components of the calorimeter causing the water temperature to rise. To calculate the heat of combustion from the measured temperature increase, we must know the total heat capacity of the Calorimeter, Ccal. © 2012 Pearson Education, Inc.

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Bomb Calorimetry The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2 (g), CO2 (g), and H2O (l). When 4.00 grams of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00° C to 39.50°C. In a separate experiment the heat capacity of the calorimeter is measured to be kJ/°C. Calculate the heat of reaction for the combustion of a mole of CH6N2. © 2012 Pearson Education, Inc.

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Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is very small. © 2012 Pearson Education, Inc.

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Hess’s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy. © 2012 Pearson Education, Inc.

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Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” Whether the reaction is carried out in one step or in a series of steps, the sum of the enthalpy changes associated with the individual steps must be the same as the enthalpy change associated with a one step process….this is because enthalpy is a state function! © 2012 Pearson Education, Inc.

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Hess’s Law Examples Pg. 181 and 182… © 2012 Pearson Education, Inc.

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Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. © 2012 Pearson Education, Inc.

52 Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Or otherwise called the heat of formation. The magnitude of any enthalpy change depends on the temperature, pressure, and state of the reactants and products. © 2012 Pearson Education, Inc.

53 Standard Enthalpies of Formation
Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). To compare enthalpies of different reactions, we must define a set of conditions, called a standard form at atmospheric pressure (1 atm) and the temperature of interest which is 298 K. The standard enthalpy change of a reaction is defined as the enthalpy change when all reactants and products are in their standard states. We denote standard enthalpy change as Hf The standard enthalpy of formation of a compound is the change in enthalpy for the reaction that forms one mole of the compound from its elements with all substances in their standard states: If: elements (in standard state)  compund (1 mol in standard state) Then: H = Hf Note: the standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state. Thus, C(graphite), hydrogen, oxygen, are 0. © 2012 Pearson Education, Inc.

54 Enthalpies of Formation
For which of these reactions at 25°C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose H is an enthalpy of formation? 2 Na (s) + ½ O2 (g)  Na2O (s) 2 K (l) + Cl2 (g)  2KCl (s) C6H12O6 (s)  6C (diamond) + 6H2 (g) + 3 O2 (g) © 2012 Pearson Education, Inc.

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Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Consider the reaction above Imagine this as occurring in three steps: C3H8(g)  3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) 3CO2(g) 4H2(g) + 2O2(g)  4H2O(l) We can use Hess’s Law and tabulations of Hf values to calculate the standard enthalpy change for any reaction for which we know the Hf values for all reactants and products. In the first reaction the Hf = -Hf [C3H8(g)] because remember we want the elements making a compound….however this is decomp. Which is equal in magnitude but opposite sign. In the second reaction the Hf = 3[CO2(g)] and the third reaction would be 4 x water (l) © 2012 Pearson Education, Inc.

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Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) The sum of these equations is C3H8(g)  3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) 3CO2(g) 4H2(g) + 2O2(g)  4H2O(l) C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) © 2012 Pearson Education, Inc.

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Calculation of H We can use Hess’s law in this way: H = nHf,products – mHf°,reactants where n and m are the stoichiometric coefficients. © 2012 Pearson Education, Inc.

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Calculation of H C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)] = [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)] = (− kJ) – (− kJ) = − kJ Or See pg. 185 © 2012 Pearson Education, Inc.

59 Guidelines to Remember for this Calculation
Decomposition – the reverse formation reaction; use the negative version of the standard enthalpy of formation. Stoichiometric Coefficients – if there are 3 moles of CO2, then we must multiply the standard enthalpy of formation by 3. © 2012 Pearson Education, Inc.

60 Enthalpy of Reaction from Enthalpies of Formation
Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6 (l) to CO2 (g) and H2O (l). Compare the quantity of heat produced by combustion of 1.00 g propane with that produced by 1.00 g benzene. © 2012 Pearson Education, Inc.

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Example #2 The standard enthalpy change for the reaction CaCO3 (s)  CaO (s) + CO2 (g) is kJ. Use Table 5.3 to calculate the standard enthalpy of formation of CaCO3 (s). © 2012 Pearson Education, Inc.

62 © 2012 Pearson Education, Inc.
Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats. The starches (carbs) are decomposed in the intestines into glucose. © 2012 Pearson Education, Inc.

63 © 2012 Pearson Education, Inc.
Fuels A 28 g serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbs, and 2 g fat. Using the average fuel values of these substances, estimate the fuel value (caloric content) of this serving. A person of average weights uses about 100 Cal/mi when running or jogging. How many servings of this cereal provide the fuel value requirements to run 3 mi? © 2012 Pearson Education, Inc.

64 © 2012 Pearson Education, Inc.
Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels. In 2008, the US consumed 1.05 x 1017 kJ of energy. This value corresponds to an average daily energy consumption per person of 9.4 x 105 kJ. Although the US is only 4.5% of the world’s population, we account for nearly 20% of the world’s total energy consumption. © 2012 Pearson Education, Inc.

65 Chapter 5 Homework Problems
Conceptual Problems: 3, 4, 6, 8, 9 Problems: 16, 24, 26, 28, 32, 34, 37, 39, 41, 44, 47, 50, 53, 58, 62, 64, 73, 75, 78, 95 © 2012 Pearson Education, Inc.


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