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Chapter 5. Thermochemistry.

Chapter 5. Thermochemistry. Chapter 5. Thermochemistry. Thermochemistry deals with changes in energy that occur in chemical reactions. The study of energy and its transformations is known as thermodynamics . 5.1 The nature of energy.

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Chapter 5. Thermochemistry.

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  1. Chapter 5. Thermochemistry.

  2. Chapter 5. Thermochemistry. Thermochemistry deals with changes in energy that occur in chemical reactions. The study of energy and its transformations is known as thermodynamics. 5.1 The nature of energy. Different types of energy are kinetic energy and potential energy.

  3. Different forms of energy are interconvertible: Potential Energy Kinetic Energy Chemical Energy

  4. Kinetic energy (Ek) Ek = ½ mv2 Where m is the mass of the object in kg, and v is its speed in m/s (meters per second, also written as m.s-1). Kinetic energy is important in chemistry because molecules are constantly in motion and so have kinetic energy.

  5. Units of Energy: The unit of energy is the joule (J). This is the energy required to move an object weighing 2 kg at 1 m per second. units of joules = kg x m2 s2 Example. What is the energy in J of an O2 molecule moving at 200 m/sec? Mass of O2 molecule = 2 x 16.0 = 32.0 amu (1 kg = 1000 g) One amu = 1.66 x 10-24 g = 1.66 x 10-24 g x _1 kg_ 1000 g = 1.66 x 10-27 kg.

  6. Mass of O2 molecule = 32.0 x 1.66 x 10-27 kg = 5.31 x 10-26 kg Energy in Joules is calculated as: Ek = ½ mv2 Ek = ½ x 5.31 x 10-26 kg x (200 m/sec)2 = 1.06 x 10-21 J. mass velocity What is the kinetic energy of a person (50kg) moving at a speed of 1m/s?:

  7. Weight of person Walking speed

  8. example What is the kinetic energy in Joules of a 45 g golf-ball moving at 61 m/s? Ek = ½ mv2 Note: m has units of kg, v of m/s 45 g = 45 g x 1 kg/1000 g = 0.045 kg. Ek = ½ x 0.045 kg x 61 m x 61 m = 83.7 J. ss What happens to this energy when the ball lands in a sand-trap? Ans. It is converted to heat. units check out

  9. Units of energy A joule is a very small amount of energy, and so one commonly uses the kJ. Energies for bonds are usually expressed in kJ/mol. The calorie: Amount of energy required to raise temperature of 1 g of water by 1 ºC. Use kcal.mol-1 (=kcal/mol) for bonds. Nutritional Calorie (note upper case) = 1000 cal.

  10. Types of energy Potential energy: Object has this by virtue of its position. Electrostatic energy (not covered in CHM 101). Potential energy due to electrostatic attraction or repulsion. Chemical Energy: Due to arrangement of atoms, e.g. gasoline, glucose Thermal Energy:Due to kinetic energy of molecules.

  11. Calculation of work or potential energy: The potential energy equals the work done to raise the object to the height it is above the ground. e.g. a 5.4 kg bowling ball is raised to a height of 1.6 m above the ground. What is its potential energy? Note: The force is the gravitational constant is g = 9.8 m/s2. Work = m x g x d = 5.4 kg x 9.8 m x 1.6 m s2 = 85 kg.m2/s2 = 85 J (check: J = units of kg.m2/s2so our calculation produces the right units). Units Check out

  12. System and surroundings In thinking about thermodynamics, we cannot think about the whole universe at one time. We have to think about the system of interest to us, which for chemistry is usually the contents of something the size of a beaker. Thermodynamics is the book-keeping of energy, and so we are concerned with how much heat goes in or out of the system from the surroundings. everything else = ‘surroundings’ energy out energy in system an example of a system – a beaker plus a solution A system is like a bank account –see below

  13. The System and its Surroundings: Energy can also be transferred from the surroundings to the system Energy can be transferred from the system to the surroundings OR heat in heat out Hot coffee Cold soda Heat is transferred from the hotter to the colder object … until their temperatures are equal

  14. The sign of the loss or gain of energy: the signs of the energy changes are rather like a bank account: +ve for money in, -ve for money out We are interested in how much energy goes in or out of a system because this is what causes a chemical reaction to take place. If energy is lost from the system into the surroundings the sign of the energy change is negative, and if energy is gained, it is positive. energy out = negative system energy in = positive A system is like your bank account. You only worry about what goes in or out of it, not what happens to the money in the surroundings, i.e. the rest of the world.

  15. Work and Heat: Energy can be transferred from one object to another either as work or as heat. Energy used to make an object move against a force is called work. w = F x d ( work = force x distance) Heat is energy transferred from a hotter to a colder object.

  16. Energy can be transferred as heat (q) Energy can also be transferred as work (w) When an object is moved by a force, F, over a distance, d, energy (work) is transferred the soccer player is doing work on the ball

  17. energy ENERGY IS THE CAPACITY TO DO WORK OR TO TRANSFER HEAT. Heat is the non-ordered transfer of energy due to random collisions between particles, whereas work is the ordered transfer of energy.

  18. 5.2. The first Law of Thermodynamics. Energy is conserved This means that energy cannot be created or destroyed, but only converted from one kind of energy to another.

  19. Internal energy molecules also have vibrational energy The internal energy of a system (E) is the sum of all kinetic and potential energies We don’t know the internal energy of the system, and can generally only calculate ΔE, the change in E that accompanies a change in the system. ΔE = Efinal - Einitial system kinetic energy is energy of molecules rapidly moving about

  20. Relating ΔE to heat and work. ΔE = q + w Where: q is the heat transferred to the system, and w is the work done on the system Heat transferred into the system, and work done on the system, are positive.

  21. Practice exercise: Calculate the change in internal energy where the system absorbs 140 J of heat from the surroundings, and does 85 J of work on the surroundings: q = + 140 J (absorbs heat = +ve) w = - 85 J (does work on ______________ surroundings = -ve) ΔE = + 55 J ______________________

  22. Endothermic and Exothermic processes: When a chemical process absorbs heat as it occurs, it is referred to as endothermic. When heat is given off, it is exothermic. The burning of H2 in O2 is exothermic, because a large amount of heat is given off: 2 H2 + O22 H2O + heat

  23. Exothermic / Endothermic Processes H2O (l) H2O (g) Endothermic: system gains heat Water Water vapor H2O (l) H2O (s) Exothermic: system loses heat Water ice

  24. State functions. The value of a state function depends only on the present conditions, not on how it got there. Examples of state functions are temperature and ΔE, the change in internal energy. Q and w are not state functions, because one can get to a particular value of ΔE by a variety of combinations of q and w. E itself is also a state function. ΔE = q + w The same value of ΔE can be achieved e.g. with large q and small w, or small q and large w

  25. The height difference between Denver and Chicago is a state function. Denver Route A 4684 ft Chicago Route B The height difference between Denver and Chicago is a state function because it is independent of the route taken to travel from one to the other. The travel distance is not a state function because one can travel by different routes.

  26. Work done in a chemical reaction: The work done in a chemical reaction at constant pressure is given by PΔV, where V is the change in volume during the reaction. For a reaction involving a gas, this can be a considerable contribution. piston } increase in volume due to H2 gas given off Zn metal H2 formed plus air air Zn dissolving HCl Bubbles of H2 HCl

  27. Enthalpy (ΔH): If a reaction is carried out at constant P, which is true for all reactions open to the atmosphere, e.g. in a beaker, then the work done by the system is equal to -P ΔV. However, the change in volume of a solution will generally be very small, and so this can be ignored. The heat given off or absorbed during a reaction at constant pressure is known as the enthalpy, and is given the symbol ΔH.

  28. Enthalpy (ΔH): It can be shown that the change at constant pressure is given by ΔH = qp where the subscript ‘p’ denotes constant pressure. When ΔH is positive, the system has gained heat from the surroundings, and is endothermic. When ΔH is negative, the process isexothermic.

  29. 5.4. Enthalpies of reaction. In a thermochemical equation, the heat of reaction for the equation is written as: 2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ This is the heat given off when 2 moles of H2 combine with one mole of O2 to give 2 moles of water, all in the gas phase. Note that a large negative value of ΔH such as we have here is associated with a very exothermic reaction.

  30. 1. Heat is an extensive property With an extensive property such as heat, the amount of heat given off is proportional to the amount of substance reacted. small log log twice as big burning = twice as much heat

  31. Enthalpy is an extensive property. If we burn one mol of H2 with ½ mol of O2, we will get ­483.6/2 = -241.8 kJ, as shown below: 2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ 2 moles 1 mole ΔH = -483.6 kJ 1 mole ½ mole ΔH = -241.8 kJ Factor = _____moles we have______ moles in balanced equation = 1 mole = 0.5 2 moles So multiply everything in the equation by the factor of 1/2 , including the enthalpy.

  32. 2. The enthalpy of a reaction is of opposite sign to its reverse reaction. If we burn H2 a large amount of heat is given off: If we break H2O up into H2 and O2, an equal amount of heat energy has to be put into this reverse reaction: ΔH = - 483.6 kJ Heat given off 2 moles H2 1 mole O2 2 moles H2O ΔH = + 483.6 kJ Heat put back in 2 moles H2O 2 moles H2 1 mole O2

  33. The enthalpy of a reaction is equal in magnitude but opposite in sign for the reverse reaction. 2 H2(g) + O2(g) → 2 H2O(g) ΔH = - 483.6 kJ but for the reverse reaction: 2 H2O(g) → 2 H2(g) + O2(g) ΔH = + 483.6 kJ For the reverse reaction one simply changes the sign of ΔH.

  34. 3.The enthalpy change for a reaction depends on the state of the reactants. 2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ but 2 H2(g) + O2(g) → 2 H2O(l) ΔH = -659.6 kJ since H2O(l) → H2O(g) ΔH = +88 kJ or H2O(g) → H2O(l) ΔH = -88 kJ Note that -659.6 + (2 x 88) = -483.6 kJ (discussed later) water vapor liquid water liquid water water vapor water vapor liquid water

  35. Example: How much heat is given off by burning 3.4 g of H2 in excess O2? 2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ 2 moles 1 mole ‘Excess O2’ means that H2 is the limiting reagent, and so we don’t need to bother with the O2. So we know that 2 moles of H2 burns in O2 to give off -483.6 kJ, so we need to know how many moles of H2 we have in 3.4 g.

  36. Problem (contd.) Molecular mass H2 = 1.0 + 1.0 = 2.0 g/mol Moles H2 = 3.4 g x 1 mole = 1.7 moles 2.0 g 2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ 2 moles 1 mole 1.7 moles ΔH = ? ΔH = - 483.6 kJ x moles we have moles in balanced equation = - 483.6 kJ x 1.7 moles 2 moles = - 411.1 kJ

  37. Specific heat Specific heat is the amount of heat in joules it takes to raise the temperature of a substance by 1 K. The units of specific heat are J/g.K. Some examples are: SubstanceSpecific heat (J/g.K) H2O(l) 4.184 N2(g) 1.04 Al(s) 0.90 Fe(s) 0.45 Hg(l) 0.14

  38. Calculating heat produced from rise in temperature and a knowledge of the specific heat: Example: 5 ml of H2SO4 (at 21.2 ºC) is added to 50 ml of water in a coffee-cup calorimeter. The temperature of the solution in the calorimeter rises from 21.2 to 27.8 ºC. How much heat was liberated by the dissolution of the H2SO4? (assume all 55 ml of solution has specific heat of water = 4.184 J/g.K, and density of water = 1g/ml). thermometer Temperature = 21.2 ºC Temperature = 27.8 ºC add 5 ml H2SO4 55 ml H2O plus H2SO4 50 ml H2O Coffee-cup calorimeter

  39. rises in K or ºC will be the same Problem (contd.) 55 ml x 1g = 55 g of solution 1 ml 4.184 = heat in J (q) weight (g) x temperature rise (K or ºC) = ________q (J)__________ 55 g x (27.8 – 21.2) ºC q = 4.184 J x 55 g x 6.6 ºC 1 g x 1 ºC = -1519 J specific heat of water (q is negative because heat is evolved)

  40. Example on calculating heat evolved per mole: When 9.55 g of NaOH dissolves in 100.0 g of water in a coffee-cup calorimeter, the temp. rises from 23.6 to 47.4 oC. Calculate ΔH for the process: (Assume specific heat is as for pure water = 4.18 J/g.K.) NaOH (s) → Na+ (aq) + OH- (aq) We assume that when presented with the balanced equation we need to calculate ΔH for the numbers of moles indicated by the coefficients, i.e. 1 mole NaOH

  41. Problem (contd.) Wt. of solution = (100.0 g + 9.95 g) = 109.95 g change in K = 47.4 - 23.6 = 23.8 K. q = specifc heat x mass in g x temp. rise in K q = 4.18 J x 109.95 g x 23.8 K = -10938 J g x K = -10.9 kJ F. Wt. NaOH = 23 + 16 + 1 = 40 g/mol Moles NaOH = 9.95 g x 1 mole 40 g = 0.249 mol

  42. Problem (contd.) NaOH (s) → Na+ (aq) + OH- (aq) 1 mole 1 mole 1 mole 0.249 mole 0.249 mole 0.249 mol ΔH = -10.9 kJ ΔH = -10.9 kJ x 1 mol/0.249 mol = - 43.8 kJ/mol Note: If the temperature rises in a process, then ΔH will be negative.

  43. 5.3 Calorimetry (covered in labs): stirrer thermometer The ‘system’ in a coffee cup calorimeter is usually the solution in the calorimeter. A thermometer is used to monitor the temperature rise due to the chemical reaction being studied. One assumes the heat capacity of the solution is that of water. two nested coffee cups to provide better insulation lid the solution = the ‘system’ Coffee-cup calorimeter

  44. 5.6 Hess’s Law. Hess’s Law states: If a reaction is carried out in a series of steps, ΔH for the overall reaction will equal the sum of the enthalpy changes for the individual steps. Hess’s Law provides a method for calculating ΔH values that are impossible to measure directly.

  45. -50kJ Hess’s law Increasing Enthalpy (H) The enthalpy of going from H2O (g) (water vapor) to H2O (s) (ice) in one step (-50 kJ) is the sum of the two steps of going first from H2O (g) to H2O (l) (-44kJ) and then from H2O (l) to H2O (s) (-6 kJ) H2O (g) -44kJ H2O (l) -6 kJ H2O (s)

  46. Adding enthalpies following Hess’ Law: We cannot measure directly the heat of burning graphite to give CO. However, we can calculate this by combining two equations: C(s) + O2(g) → CO2(g) ΔH = -395.5 kJ CO2(g) → CO(g) + ½O2(g) ΔH = 283.0 kJ ____________________________________________________________________ C(s) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g) ΔH = -110.5 kJ C(s) + ½O2(g) → CO(g) ΔH = -110.5 kJ add the ΔH values { } add the equations Cancel things that occur on both sides of equation net equation

  47. The enthalpy of conversion of graphite to diamond from Hess’ Law Graphite is the stable form of carbon: C(graphite) → C(diamond) ΔH = +1.9 kJ This value of ΔH could not be measured directly, but could be obtained from the enthalpy of combustion of graphite and diamond using Hess’ Law: C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ C(diamond) + O2(g) → CO2(g) ΔH = -395.4 kJ ___________________________________________________________________ C(diamond) → C(graphite) ΔH = -1.9 kJ or C(graphite) → C(diamond) ΔH = +1.9 kJ subtract

  48. Using Hess’ Law to calculate the energy of formation of ethylene from C (graphite) and H2 gas: An impossible (so far) reaction to carry out would be: 2 C(s) + H2(g) = C2H2(g) (acetylene). We can calculate the energy of the above by combining the heats of combustion of the components in the reaction: C2H2(g) + 2½O2(g) → 2 CO2(g) + H2O(l) ΔH = -1299.6 kJ C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ H2(g) + ½ O2(g) → H2O(l) ΔH = -285.8 kJ

  49. We first want to get the products on the right hand side, so we reverse the first equation: C2H2(g) + 2½ O2(g) → 2 CO2(g) + H2O(l) ΔH = -1299.6 kJ 2 CO2(g) + H2O(l) → C2H2(g) + 2½O2(g) ΔH = +1299.6 kJ Then we double the second equation because there are two C-atoms in the desired reaction: C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ 2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = -787.0 kJ

  50. We now add them together in two steps (it’s easier that way): 2 CO2(g) + H2O(l) → C2H2(g) + 2½ O2(g) ΔH = +1299.6 kJ 2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = -787.0 kJ _______________________________________________________________________ 2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ 2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ H2(g) + ½ O2(g) → H2O(l) ΔH = -285.8 kJ _______________________________________________________________________ 2 C(s) + H2(g) → C2H2(g) ΔH = +226.8 kJ

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