1 / 52

Chapter 5 Thermochemistry

Chemistry, The Central Science , 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 5 Thermochemistry. John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Keys to Success in Ch 5.

bettybarnes
Download Presentation

Chapter 5 Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 5Thermochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

  2. Keys to Success in Ch 5 • Learn the vocab (make flashcards!) – there’s a lot. • AND learn how the words are related (a word web can really help) • Units and abbreviations are essential! • Watch your + and - signs

  3. Do Now – Jan 18 Use section 5.1 in your book to answer the questions. Then put answers on poster paper to share with the class. • Define “energy”, “work”, and “heat” and state the units of each one. • There are two types of energy discussed. What are they and how are they related? • Which of the two types of energy in #2 is contained in chemical bonds?

  4. Energy, E • Chemical reactions and phase changes either release or absorb energy. • Energy is 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. • What are the units? Where do they come from?

  5. kg m2 1 J = 1  s2 Units of Energy • The US unit of energy are calories • The metric units of energy are Joules • 1 Joule = 4.184 calories

  6. 2 kinds of energy An object has potential energy by virtue of its position or chemical composition (energy stored in chemical bonds). An object has Kinetic energy by virtue of its motion (molecules or atoms moving in chemistry). Potential and kinetic energy can be converted into each other through chemical and physical processes. How does this energy conversion happen in a chemical reaction?

  7. 2 kinds of energy An object has Kinetic energy by virtue of its position or chemical composition (energy stored in chemical bonds). An object has Potential energy by virtue of its motion (molecules or atoms moving in chemistry). Potential and kinetic energy can be converted into each other through chemical and physical processes • Potential energy in a reactant molecules’ bonds is released and converted to kinetic energy in a chemical reaction • Kinetic energy is converted to potential energy into make the products’ bonds

  8. System and Surroundings • The system includes the molecules we want to study, the reactants and products (here, the hydrogen and oxygen molecules). • The surroundings are everything else (here, the cylinder and piston). • A closed system can exchange energy but not matter w/ surroundings • Can the gas particles escape? • Can the piston move? Or get hotter?

  9. Work, w • 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. In chemistry work is …

  10. Heat, q • Energy can also be transferred as heat. • Heat flows from warmer objects to cooler objects. • In chemistry heat is…

  11. 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

  12. 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 (the chemical reaction in our case) loses energy, it must be gained by the surroundings, and vice versa. • By definition, the change in internal energy of the system, E, is the final energy of the system minus the initial energy of the system: E = Efinal−Einitial • And E for the surroundings will be equal in amount but opposite in sign!

  13. E, q, w, and Their Signs • E = q+ w is the equation for the 1st law. • This means that the change in energy E is due to work w (either done by the system or done on the system) and/or heat q (either released or absorbed by the system) • What is E for a systemthat does 300J of work on the surroundings and absorbs 150 J of heat? Is it endergonic or exergonic? What is E for the surroundings in this case?

  14. Enthalpy • If a process takes place at constant pressure (in an open container, as most of the reactions we study do) and the only work done is pressure-volume work (expanding or contracting gases) • THEN we can measure heat flow during the process by measuring the enthalpy H of the system. H = q • So, at constant pressure the change in enthalpy is the heat gained or lost. • Basically this means we can measure the energy change of a chemical reaction by measuring the heat change, which is easy, and ignore work.

  15. Exchange of Heat between System and Surroundings • When heat is absorbed by the system from the surroundings, the process is endothermic and H is positive. • When heat is released by the system to the surroundings, the process is exothermic and H is negative.

  16. Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = H(products)−H(reactants) An enthalpy diagram shows this graphically Write the equation for the combustion of methane. Is this reaction endo or exo? Which has more internal energy, the reactants or the products

  17. Enthalpies of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. For the reaction pictured below: 2H2 (g) + O2 (g)  2 H2O (g),H =-483.6 kJ A thermochemical equation shows the heat as a reactant (if endo) or product (if exo) 2H2 (g) + O2 (g) 2 H2O (g) + 483.6 kJ ,

  18. The Truth about Enthalpy • Enthalpy is an extensive property (it depends on amount!!!). • 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 and temperature of the products and the state of the reactants. • The solid phase of a substance has less enthalpy then the liquid which has less than the gas • Higher temp = higher enthalpy • More particles = more enthalpy

  19. Thermochem Example The thermochemical equation for the formation of water is: 2H2 (g) + O2 (g) 2 H2O (g) + 483.6 kJ How much energy is absorbed or released in the formation of 4.5g of water?

  20. Thermochem Example The thermochemical equation for the formation of water is: 2H2 (g) + O2 (g) 2 H2O (g) + 483.6 kJ How much energy is absorbed or released in the formation of 4.5g of water? Use stoich with H as a part of the mole ratio! 4.5 g H2O /(18.02 g/mol) = 0.25 mol H2O 0.25 mol (483.6 kJ/2 mol H2O) = 60kJ released The 60kJ is released (I know the energy is released because it is a product in the equation)

  21. 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

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

  23. State Functions • However, we do know that the internal energy of a system is independent of how the system got to where it is now. • In the system below, the water could have reached room temperature from either direction.

  24. State Functions • Therefore, internal energy E 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, change in energy E depends only on Einitial and Efinal.

  25. 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.

  26. 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).

  27. 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

  28. 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

  29. Endothermicity and Exothermicity • A process is endothermic, then, when H is positive.

  30. Endothermicity and Exothermicity • A process is endothermic when H is positive. • A process is exothermic when H is negative.

  31. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. Q = mcT

  32. Heat Capacity and Specific Heat • The amount of energy required to raise the temperature of a sample of substance by 1 K (1C) is its heat capacity. • Heat capacity depends on both the mass of the sample and it’s identity. • Specificheat capacity (specific heat) is the amount of energy required to raise the temperature of 1 g of a substance by 1 K. • Specific heat capacity ( s or c) is characteristic of a substance. The higher it is the more energy a substance requires to heat it. Compare water and metal – which would have a higher c?

  33. heat transferred Specific heat = mass  temperature change q c = m T Heat Capacity and Specific Heat Specific heat, s (or c), is This equation can be used to solve for any of the four variables! • EXAMPLE A piece of iron metal requires 220 J to raise its temperature from 22C to 40C. What is the mass of the iron?

  34. q c = mT Specific heat example: A piece of iron metal requires 220 J to raise its temperature from 22C to 40C. What is the mass of the iron? Rearrange To solve for m, m =q/(cT) = 220 J/[(0.45J/g*C)*(40C-22C)] = 27g Fe

  35. Constant Pressure Calorimetry We can measure H for a reaction or dissolving occurring in aqueous solution using a calorimeter and calculating with the specific heat equation: H =qH2O = m  c  T You must measure the mass and temp. change in the lab and use the specific heat of water (4.184J/g*C) The heat lost (or gained) by the water is the heat gained (or lost) by the chemical system being studied. qH2O = -qrxn

  36. 10mL of 1M NaOH solution and 10mL of 1MHCl solution at 22°C are added to 100mL of water in a coffee-cup calorimeter at 22°C. • The final temperature of the system is • 37°C. • What is qrxn? What is qH2O? • Is this reaction endo or exo? EXPLAIN! How many moles of HCl were used? • What is the enthalpy change (H ) per mole of HCl?

  37. 10mL of 1M NaOH solution and 10mL of 1MHCl solution at 22°C are added to 100mL of water in a coffee-cup calorimeter at 22°C. • The final temperature of the system is • 37°C. • What is qrxn? What is qH2O? • qH2O = m  c T = 120g X 4.184J/(g*C) * (37-22)C • = 7524J • qrxn = -qH2O = -7524J • Is this reaction endo or exo? EXPLAIN! qrxnIs negative • How many moles of HCl were used? • 1M*0.01L = 0.01moles • What is the enthalpy change (H ) per mole of HCl? --7524J/0.01moles=-752000J/mol = -752kJ

  38. Bomb Calorimetry • Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water. • 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 between E and H is very small so we ignore it. • This is how the calories in food are determined

  39. Hess’s Law • H is well known for many reactions, and it is inconvenient to directly measure H for every reaction in which we are interested. • However, we can estimate H for any reaction using H values that are already known and the additive properties of enthalpy.

  40. Hess’s Law “If 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.” H1 = H2 + H3 What happens when you add equations 2 and 3 together? TRY IT! So H depends only on the intial reactants and the final products. Not how you get there!

  41. Using Hess’s Law IF you are given an overall equation and a set of equations to arrange so they add up to the overall equation, you need to reverse them or multiply them to get the overall equation. EXAMPLE: Given this information: 2 CO(g) 2 C(s) + O2(g) H1= 221 kJ CO(g) + 2 H2(g)  CH3OH(g)H2= 91 kJ Find Hfor 2 C(s) + O2(g) + 4 H2(g) 2 CH3OH(g)

  42. To get the final equation, reverse the first equation and change sign of H1 : 2 C(s) + O2(g) 2 CO(g) -DH1 = 221 kJ Double the second equation and double H2 2[CO(g) + 2 H2(g) CH3OH(g)] 2 DH2 = 2(91 kJ) 2 C(s) + 4 H2(g) + O2(g) 2 CH3OH(g) • DHrxn = -DH1 + 2 DH2 • DHrxn = 221 kJ + 2(91 kJ) = 403 kJ

  43. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) Imagine this reaction as occurring in 3 steps which add up to the overall equation C3H8 (g) 3 C(graphite) + 4 H2 (g) H1 = + 103.85 kJ 3 C(graphite) + 3 O2 (g)  3 CO2 (g)H2 = -1181kJ 4 H2 (g) + 2 O2 (g)  4 H2O (l) H3 = -1143 kJ C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) Then H = H1+ H2+ H3 = 103.85 – 1181 -1143 = -2220

  44. Enthalpies of Formation • The enthalpy (or heat) of formation of a substance, Hf, is the heat released or absorbed when making a 1 mole of a compound or element “from scratch” • “From scratch” means from its elements in their naturally occurring elemental forms and phases at 25C. • By definition the heat of formation of a pure element in its normal form and phase at 25C is zero (e.g. O2 (g) and H2 (g) but NOT O(g) or H2 (l)) EXAMPLES OF FORMATION REACTIONS: For CO -- C(s) + 1/2O2(g)  CO(g), DH= 110.5 kJ For water -- H2 (g) + 1/2O2 (g)  H2O (l) H = -285 kJ What is the formation reaction for C2H6(g)?

  45. Standard Enthalpies of Formation  Standard enthalpies of formation, Hf, are measured under standard conditions (25°C and 1.00 atm pressure). An element in its standard form at these conditions has an enthalpy of formation of zero. Watch states carefully when looking up these values! H2O (g) has a different from H2O(l).

  46. Calculation of H using Hf We can use heat of formation to calculate the enthalpy change H for any reaction using: H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients. Basically you add up all of the heats of formation of each product times its coefficient in the equation. Do the same for the reactants. Subtract the reactant sum from the product sum and that is H!  

  47. Calculation of H What is H for this reaction? C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) What is H for this reaction? CH4 (g) + 2O2(g)  2 H2O (g)+ CO2 (g)

  48. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ What is H for this reaction? CH4 (g) + 2O2(g)  2 H2O (g)+ CO2 (g)

  49. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ What is H for this reaction? CH4 (g) + 2O2(g)  CO2 (g) +2 H2O (g) H= [(-393.5 kJ) + 2(-285.8 kJ)] - [1(-79.8 kJ) + 2(0 kJ)] = (-965.1 kJ) - (-79.8kJ) = -885.3kJ

More Related