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Constant Pressure Calorimetry: Another Example

Constant Pressure Calorimetry: Another Example.

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Constant Pressure Calorimetry: Another Example

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  1. Constant Pressure Calorimetry: Another Example A piece of chromium metal weighing 24.26 g is heated in boiling water to a temperature of 98.3°C and then dropped into a coffee cup calorimeter containing 82.3g of water at 23.3°C. When thermal equilibrium is reached, the final temperature is 25.6°C. Calculate the Cp of chromium.

  2. Constant Volume Calorimetry: It’s The Bomb! • Technique can be used to obtain the heat content of combustion of compounds • Used in the food, fuel and pharmaceutical industries to know how much energy would be released by completely consuming the compound • Uses a BOMB Calorimeter

  3. CALORIMETRY Measuring Heats of Reaction • Place sample of known mass inside the bomb • Place oxygen in the sample chamber and immerse bomb into water • Ignite the bomb and measure temperature of water • Since the volume doesn’t change, no P-V work is done, so the qr is a measurement of the ΔU

  4. Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Calorimetry Total heat evolved = qtotal = qwater + qbomb

  5. Measuring Heats of ReactionCALORIMETRY Calculate energy of combustion (∆U) of octane. 2C8H18 + 25O2-->16CO2 + 18H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200. g water • Heat capacity of bomb = 837 J/K

  6. Measuring Heats of ReactionCALORIMETRY Step 1 Calc. energy transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. energy transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total energy evolved 41,200 J + 6860 J = 48,060 J Energy of combustion (∆U) of 1.00 g of octane = - 48.1 kJ

  7. The First Law of Thermodynamics • Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat • However, these changes rarely occur singly, so we can describe the change in internal energy as: U = q + w (The 1st Law) • The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

  8. The First Law: Put Another Way 1st Law of Thermodynamics A system can store energy. A change in the energy of a system means that there must be a change in the heat or the work done BY or TO the system. OR The Total Energy of the Universe is Constant

  9. The First Law • Internal energy is an example of a state function • A state function is a property that only depends on the current state of the system and is independent of how that state was reached • Pressure, Volume, Temperature and density are all examples of state functions

  10. State Functions and Things that Aren’t • Work and heat ARE NOT state functions • The amount of work done depends on how the change was brought about

  11. Change in Internal Energy (Implications of a State Function) • It doesn’t matter what path we take to get to the final point, the change in internal energy is only dependent on where we started and where we finished • Let’s think about this on a molecular level… • If we expand an ideal gas isothermally, the molecules will have the same kinetic energy and will move at the same speed • Despite the fact that the volume has increased, the potential energy of the system remains the same because there are no forces between molecules (KMT) • Since neither the kinetic nor potential energy has changed, the change in internal energy is…

  12. Zero! U = 0 for the isothermal expansion of an ideal gas

  13. Enthalpy • In a constant volume system in which no work is done (neither expansion nor non-expansion), we can rearrange the first law to: U = q + w (but w=0) U = q • Most systems are constant pressure systems which can expand and contract • When a chemical reaction takes place in such a system, if gas is evolved, it has to push against the atmosphere in order to leave the liquid or solid phase • Just because there’s no piston, it doesn’t mean that no work is done!

  14. Enthalpy • Let’s look at an example: • If we supply 100J of heat to a system at constant pressure and it does 20J of work during expansion, the U of the system is +80J (w=-20J) • We can’t lose energy like this • Enthalpy, H, is a state function that we use to track energy changes at constant pressure H=U + PV • The change in enthalpy of a system (H) is equal to the heat released or absorbed at constant pressure

  15. Enthalpy • Another way to define enthalpy is at constant pressure: H = q

  16. Enthalpy and Chemical Reactions • Enthalpy is a tricky thing to grasp, but we can look at it this way: Enthalpy is the macroscopic energy change (in the form of heat) that accompanies changes at the atomic level (bond formation or breaking) • Enthalpy has the same sign convention as work, q and U • If energy is released as heat during a chemical reaction the enthalpy has a ‘-’ sign • If energy is absorbed as heat from the surrounding during a reaction, the enthalpy has a ‘+’ sign

  17. Heat Transfers at Constant Pressure H = U + PV • The change in the enthalpy of a system is equal to the heat released or absorbed at constant pressure WTF? • In a coffee cup calorimeter (constant pressure calorimeter), the heat, q, that is released or absorbed is equal to the change in enthalpy, H • When we add heat to a constant pressure system, the enthalpy increases by that amount • H<0 for exothermic reactions H>0 for endothermic reactions

  18. USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ Exothermic reaction — energy is a “product” and∆H = – 241.8 kJ

  19. Energy Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) f Liquid water q = (heat of fusion)(mass)

  20. Energy Transfer and Changes of State Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. Liquid --> Vapor + energy

  21. Heating/Cooling Curve for Water Note that T is constant as ice melts

  22. +333 J/g +2260 J/g Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g

  23. Heat & Changes of State What quantity of energy as heat is required to melt 500. g of ice and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total energy = 1.51 x 106 J = 1510 kJ

  24. USING ENTHALPY Making liquid H2O from H2 + O2 involves twoexothermic steps. H2 + O2 gas H2O vapor Liquid H2O

  25. Hess’s Law The ∆rHofor a reaction that is the sum of one or more reactions is the sum of the ∆rHo values for all of the reactions Remember: Since ∆H is a state function, it doesn’t matter how we get to where we end up, all that matters is where we started and where we finished.

  26. USING ENTHALPY Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) f H2O(g) + 242 kJ H2O(g) f H2O(liq) + 44 kJ ----------------------------------------------------------------------- H2(g) + 1/2 O2(g) f H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

  27. Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

  28. USING ENTHALPY Making CO from C (graphite) and O2 is a single step, but the CO reacts with oxygen to form CO2. 1) C(graphite) + ½O2(g) --> CO (g) ? kJ/mole-rxn 2) CO(g) + O2(g) --> CO2(g) + -283 kJ/mole-rxn ----------------------------------------------------------------------- 3) C (graphite) + ½O2(g) --> CO2(g) -393.5 kJ/mole-rxn Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. ∆Ho3= ∆Ho1 + ∆Ho2

  29. Using Enthalpy ∆Ho3= ∆Ho1 + ∆Ho2 -393.5 kJ/mole-rxn = ∆Ho1 + (-283.0 kJ/mole-rxn) ∆Ho1 = -110.5 kJ/mole-rxn

  30. Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Active Figure 5.16

  31. Hess’s Law: Another Example Use Hess’s Law to calculate the enthalpy change for the formation of CS2(l) from C(s) and S(s). The overall equation is: C(s) + 2S(s) --> CS2(l) C(s) + O2(g) --> CO2(g) ∆Ho = -393.5 kJ/mole-rxn S(s) + O2(g) --> SO2(g) ∆Ho = -295.8 kJ/mole-rxn CS2(l) + 3O2(g) --> CO2(g) +2SO2(g) ∆Ho = -1103.9 kJ/mole-rxn

  32. Hess’s Law: Another Example How do we start? -Let’s set our overall reaction up: C(s) + O2(g) --> CO2(g) ∆Ho = -393.5 kJ/mole-rxn S(s) + O2(g) --> SO2(g) ∆Ho = -295.8 kJ/mole-rxn CO2(g) +2SO2(g) --> CS2(l) + 3O2(g) ∆Ho = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS2(l) ∆Ho = ? Note: We reversed reaction 3 and changed the sign Now: We need to make sure that all of the molecules balance and take care to multiply their ∆rHo values if necessary

  33. Hess’s Law: Another Example C(s) + O2(g) --> CO2(g) ∆Ho = -393.5 kJ/mole-rxn S(s) + O2(g) --> SO2(g) ∆Ho = -295.8 kJ/mole-rxn CO2(g) +2SO2(g) --> CS2(l) + 3O2(g) ∆Ho = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS2(l) ∆Ho = ? We need 2 S(s) atoms, so let’s multiply reaction2 by 2: 2S(s) + 2O2(g) --> 2SO2(g) ∆Ho = -591.6 kJ/mole-rxn

  34. Hess’s Law: Another Example Now let’s look at what we’ve got: C(s) + O2(g) --> CO2(g) ∆Ho1 = -393.5 kJ/mole-rxn 2S(s) + 2O2(g) --> 2SO2(g) ∆Ho2 = -591.6 kJ/mole-rxn CO2(g) +2SO2(g) --> CS2(l) + 3O2(g) ∆Ho3 = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS2(l) ∆Ho = ∆Ho1 + ∆Ho2 + ∆Ho3 Everything looks good, so let’s add it up: ∆Ho= (-393.5 kJ/mole-rxn)+(-591.6 kJ/mole-rxn)+(1103.95 kJ/mole-rxn) ∆Ho= 118.9 kJ/mole-rxn

  35. Standard Enthalpy Values Most ∆H values are labeled ∆Ho Measured under standard conditions P = 1 bar Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas

  36. Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆Hfo = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Appendix

  37. ∆Hfo, standard molar enthalpy of formation Equations for Standard Molar Enthalpy of Formation: KNO3 K(s) + ½ N2(g) + 3/2O2(g) --> KNO3 FeCl3 Fe(s) + 3/2 Cl2(g) --> FeCl3(s) C12H22O11 12C(s) + 11H2(g) + 11/2 O2(g) --> C12H22O11(s)

  38. Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known, ∆Ho =  ∆fHo(products) -  ∆fHo(reactants) Remember that ∆ always = final – initial

  39. Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Ho for CH3OH(g) + 3/2 O2(g) -->CO2(g) + 2 H2O(g) ∆Ho =  ∆Hfo(prod) -  ∆Hfo(react)

  40. Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) -->CO2(g) + 2 H2O(g) ∆rHo =  ∆Hfo(prod) -  ∆Hfo(react) ∆Hro = ∆Hfo(CO2) + 2 ∆Hfo(H2O) - {3/2 ∆Hfo(O2) + ∆Hfo(CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} ∆Hro = -675.6 kJ per mol of methanol

  41. Using Standard Enthalpy Values Calculate the heat of reaction of nitric oxide, i.e., ∆Hro for 4NH3 (g) + 5 O2(g) -->4NO(g) + 6 H2O(g) ∆Ho =  ∆Hfo(prod) -  ∆Hfo(react) ∆Ho(prod) = 4(∆Hfo NO) + 6(∆Hfo H2O) = 4(90.29) + 6(-241.830) = -1089 kJ/mole-rxn ∆Hfo(react) = 4(∆Hfo NH3) + 5(∆Hfo O2) = 4(-45.90) + 5(0) = -183.6 kJ/mole-rxn ∆Hro =  ∆Hfo(prod) -  ∆Hfo(react) = -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = -906.2 kJ/mole-rxn

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