C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = KJ/mol - SlidePlayer

Presentation on theme: "C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = kJ/mol"— Presentation transcript:

1 C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = -2803 kJ/molHave you ever wondered how your body regulates its own temperature? The human body has the ability to maintain a constant temperature within the range °F ( °C) This is controlled by your own thermostat, the hypothalamus gland. This gland controls the rate at which the body metabolizes glucose and other heat related functions C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = kJ/mol

2 C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = -2803 kJ/molFor every mole of glucose you burn, you are burning _____ calories! This process releases energy, therefore it is exothermic. The hypothalamus gland controls other mechanisms in the body to help retain or release heat. Thermochemistry at work!!!

3 Chapter 5 ThermochemistryChemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Hmwk pg , # 7, 12, 15, 19, 21, 23, 29, 33, 41, 45, 51, 53, 59, 61, 67, 71, 77, 83, 97, 103 Chapter 5 Thermochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

4 Energy The ability to do work or transfer heat.Work: Energy used to cause an object that has mass to move. Heat: Energy used to cause the temperature of an object to rise.

5 Potential Energy Energy an object possesses by virtue of its position or chemical composition. (Chemicals have potential within the bonds)

6 Kinetic Energy Energy an object possesses by virtue of its motion. KE = ½ mv2

7 Units of Energy kg m2 1 J = 1 -------- s2The SI unit of energy is the joule (J). An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = J Calories that we eat are actually kilocalories are abbreviated as Cal 1 J = kg m2 s2

8 System and SurroundingsThe system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

9 Work Energy used to move an object over some distance. w = F  d,where w is work, F is the force, and d is the distance over which the force is exerted.

10 Heat Energy can also be transferred as heat.Heat flows from warmer objects to cooler objects. Recall, cold is simply a lack of heat.

11 Transferal of Energy The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

12 Transferal of Energy The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. As the ball falls, its potential energy is converted to kinetic energy.

13 Transferal of Energy The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. As the ball falls, its potential energy is converted to kinetic energy. When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

14 First Law of ThermodynamicsEnergy 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. Use Fig. 5.5

15 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. Use Fig. 5.5

16 Internal Energy By definition, the change in internal energy, ΔE, is the final energy of the system minus the initial energy of the system: ΔE = Efinal − Einitial Use Fig. 5.5

17 Changes in Internal EnergyIf ΔE > 0, Efinal > Einitial Therefore, the system absorbed energy from the surroundings. This energy change is called endergonic.

18 Changes in Internal EnergyIf ΔE < 0, Efinal < Einitial Therefore, the system released energy to the surroundings. This energy change is called exergonic.

19 Changes in Internal EnergyWhen energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, ΔE = q + w.

20 ΔE, q, w, and Their Signs

21 Exchange of Heat between System and SurroundingsWhen heat is absorbed by the system from the surroundings, the process is endothermic.

22 Exchange of Heat between System and SurroundingsWhen heat is released by the system to the surroundings, the process is exothermic.

23 State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

24 State Functions However, 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 below, the water could have reached room temperature from either direction.

25 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, ΔE depends only on Einitial and Efinal.

26 State Functions However, q and w are not state functions.Whether the battery is shorted out or is discharged by running the fan, its ΔE is the same. But q and w are different in the two cases.

27 State Functions Yes! Your bank account(the balance is $25 whether you simply deposited $25 or deposited $100 and withdrew $75) No! The distance traveled from Penncrest to K o P mall. The distance will depend on the route taken.

28 Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

29 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 = −PΔV

30 Enthalpy 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 is the internal energy plus the product of pressure and volume: H = E + PV

31 Enthalpy When the system changes at constant pressure, the change in enthalpy, ΔH, is ΔH = Δ (E + PV) This can be written ΔH = ΔE + PΔV

32 Enthalpy Since ΔE = q + w and w = −PΔV, we can substitute these into the enthalpy expression: ΔH = ΔE + PΔV ΔH = (q+w) − w ΔH = q So, at constant pressure the change in enthalpy is the heat gained or lost.

33 Endothermicity and ExothermicityA process is endothermic, then, when ΔH is positive.

34 Endothermicity and ExothermicityA process is endothermic when ΔH is positive. A process is exothermic when ΔH is negative.

35 Endo or Exo? State whether ΔH is positive or negativeThe combustion of fuel in a car engine Salting winter roads Making snow on the ski trails Skiing down the trails Snow-covered trails melt in the spring time. Placing an ice pack on an injury

36 The combustion of fuel in a car engineHeat is leaving, ΔH is negative, exothermic Salting winter roads Energy between NaCl molecules will leave, ΔH is negative, exothermic Making snow on the ski trails Skiing down the trails Kinetic energy is dissipated in the snow under the skis in the form of heat, ΔH is negative, exothermic Snow covered trails melt in the spring time Heat is absorbed, ΔH is positive, endothermic Placing an ice pack on an injury Heat released by injury is absorbed, ΔH is positive, endothermic

37 Enthalpies of ReactionThe change in enthalpy, ΔH, is the enthalpy of the products minus the enthalpy of the reactants: ΔH = Hproducts − Hreactants

38 Enthalpies of ReactionThis quantity, ΔH, is called the enthalpy of reaction, or the heat of reaction.

39 The Truth about EnthalpyEnthalpy is an extensive property. ΔH for a reaction in the forward direction is equal in size, but opposite in sign, to ΔH for the reverse reaction. ΔH for a reaction depends on the state of the products and the state of the reactants.

40 Enthalpy Calculation Sucrose is oxdized to carbon dioxide and water. The enthalpy change can be measured in the laboratory as shown C12H22O O2 → 12CO2 + 11H2O ΔH = kJ What is the enthalpy change for the oxidation of 5.00 g of sugar? q = kJ

41 Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure ΔH through calorimetry, the measurement of heat flow.

42 Heat Capacity and Specific HeatThe amount of energy required to raise the temperature of a substance by 1 K (1oC) is its heat capacity. 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.

43 Heat Capacity and Specific HeatSpecific heat, then, is Specific heat = heat transferred mass x temperature change s = q m x ΔT

44 Constant Pressure CalorimetryBy 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.

45 Constant Pressure CalorimetryBecause the specific heat for water is well known (4.184 J/g-K), we can measure ΔH for the reaction with this equation: q = m x s x ΔT

46 Using Specific Heat CapacityA 88.5-g piece of iron has a temperature of 78.8°C is placed in a beaker containing 244 g of water at 18.8°C. When thermal equilibrium is reached, what is the final temperature? (Assume no heat is lost to warm the beaker or the surroundings.) The specific heat of iron is 0.45 J/g-K. Tfinal = 289 K = 16.2°C

47 Another Example A 15.5-g piece of chromium, heated to 100.0°C, is dropped into 55.5 g of water at 16.5°C. The final temperature of the metal and water is 18.9°C. What is the specific heat capacity of chromium? (Assume no heat is lost to warm the container or the surroundings.) smetal = 0.44 J/g·K

48 Bomb Calorimetry Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water. The bomb and its contents are defined as the system: qrxn = -Ccal x ΔT Ccal is the heat capacity of the bomb calorimeter.

49 Bomb Calorimetry Because the volume in the bomb calorimeter is constant, work cannot occur. What is measured is really the change in internal energy, ΔE, not ΔH. For most reactions, the difference is very small.

50 Using a Bomb CalorimeterMethylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of methylhydrazine is as follows: 2CH6N2 + 5O2 → 2N2 + 2CO2 + 6H2O When 4.00 g 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. What is the heat of reaction for the combustion of a mole of CH6N2 in this calorimeter? Was the reaction endo- or exothermic? -1.30 x103 kJ/mol CH6N2

51 Hess’s Law ΔH is well known for many reactions, and it is inconvenient to measure ΔH for every reaction in which we are interested. However, we can estimate ΔH using ΔH values that are published and the properties of enthalpy.

52 Recall…The Truth about EnthalpyEnthalpy is an extensive property. ΔH for a reaction in the forward direction is equal in size, but opposite in sign, to ΔH for the reverse reaction. ΔH for a reaction depends on the state of the products and the state of the reactants.

53 Hess’s Law Hess’s law states that “If a reaction is carried out in a series of steps, ΔH for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

54 Hess’s Law Because ΔH is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

55 Enthalpies of FormationAn enthalpy of formation, ΔHf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

56 Standard Enthalpies of FormationStandard enthalpies of formation, ΔHf, are measured under standard conditions (25°C and 1.00 atm pressure) and represent the change in enthalpy associated with the reaction that forms 1 mole of the compound from its elements, with all substances in their standard states. If the element exists in more than one form, the most stable form of the element is used for the formation reaction.

57 Calculation of ΔH C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)Imagine this as occurring in 3 steps: C3H8 (g)  3 C(graphite) + 4 H2 (g) 3 C(graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

58 Calculation of ΔH C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)Imagine this as occurring in 3 steps: C3H8 (g) 3 C(graphite) + 4 H2 (g) 3 C(graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

59 Calculation of ΔH C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)Imagine this as occurring in 3 steps: C3H8 (g) 3 C(graphite) + 4 H2 (g) 3 C(graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l)

60 Calculation of ΔH C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)The sum of these equations is: C3H8 (g) 3 C(graphite) + 4 H2 (g) 3 C(graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

61 Calculation of ΔH We can use Hess’s law in this way:ΔH = n x ΔHf(products) – m x ΔHf(reactants) where n and m are the stoichiometric coefficients.

62 Calculation of ΔH C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)ΔH = [3( kJ) + 4( kJ)] - [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] - [( kJ) + (0 kJ)] = ( kJ) - ( kJ) = kJ Notice, that by definition, the standard enthalpy of formation of the most stable form of any element is zero, and therefore, the standard enthalpy of O2 is zero.

63 Example Calculate ΔH for this reaction 2C (s) + H2 (g) → C2H2 (g)Given the following reactions and their respective enthalpy changes C2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l) ΔH = kJ C (s) + O2 (g) → CO2 (g) ΔH = kJ H2 (g) + 1/2O2 (g) → H2O (l) ΔH = kJ 226.8 kJ

64 Use these energies to obtain ΔH for the formation of methane. You Try This One Suppose you want to know the enthalpy change for the formation of methane, CH4, from solid carbon (as graphite) and hydrogen gas: C(s) + 2H2 (g) → CH4 (g) ΔH = ? The enthalpy change for this reaction cannot be measured in the laboratory because the reaction is very slow. We can, however, measure the enthalpy changes for the combustion of carbon, hydrogen and methane. C (s) + O2 (g) → CO2 (g) ΔH = kJ H2 (g) + ½ O2 (g) → H2O (l) ΔH = kJ CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = kJ Use these energies to obtain ΔH for the formation of methane. ΔH = kJ

65 Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats. The energy content in foods in reported in Calories and is determined with the use of a bomb calorimeter.

66 Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

67 Thermochemistry is Clearly Important to Our Everyday LivesPersonal Nutrition – nutritional value labels Creation of alternative fuel sources Addressing Global Warming

68 Chapter Summary Thermodynamics is the study of energy and its transformations. All chemical changes involve a transfer of energy, be it into the reaction or out of the reaction. Transformed energy in a chemical reaction comes from or forms chemical bonds and is exchanged with the surroundings as heat and/or work. When a gas is produced or consumed in a chemical reaction at constant pressure, we call the energy change enthalpy. All substances have a characteristic enthalpy. Calorimeters and bomb calorimeters are devices that allow us to measure the enthalpy change of a chemical reaction by measuring the temperature change associated with such a chemical reaction. Because enthalpy is a state function, the enthalpy of a reaction depends only on the final and initial states of the system and therefore, the enthalpy of a process is the same whether a reaction is carried out in one or in multi-steps. Hess’s Law allows us the calculate the unknown enthalpy of a reaction from a series of know enthalpies of reactions.