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Calorimetry

Calorimetry. The energy released or absorbed during a chemical reaction. Some definitions. Exothermic: energy is given off (temperature increases) Endothermic: energy is required (temperature decreases) Enthalpy, D H: Internal energy in kJ/mol - D H: exothermic + D H: endothermic.

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Calorimetry

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  1. Calorimetry The energy released or absorbed during a chemical reaction.

  2. Some definitions Exothermic: energy is given off (temperature increases) Endothermic: energy is required (temperature decreases) Enthalpy, DH: Internal energy in kJ/mol - DH: exothermic + DH: endothermic

  3. Types of Calorimeters Coffee cup Bomb

  4. Coffee-cup calorimetry q = -mcDT q: quantity of heat in joules, J m: mass of liquid, g c: specific heat capacity, 4.18J/goC DT: change in temperature in oC, Tfinal - Tinitial

  5. There are two kinds of coffee-cup calorimetry problems. 1st kind: Grams of a chemical are added to a given mass or volume of water. What to do: Grams of water will be plugged into m. DT, Last temperature minus first temperature. Solve for q and change joules to kJ. Grams of the other chemical convert to moles. Solve for DH by putting kJ over moles and divide.

  6. Example #1 In a coffee-cup calorimeter, 5g of sodium are dropped in 300g of water at 20oC the temperature rises to 35oC. Calculate the enthalpy change for this reaction.

  7. Example #2 A 1.5g sample of calcium is dropped into 275mL of water at 20oC in a coffee-cup calorimeter. The temperature rises to 23oC. The density of water is 1g/mL. What is the energy change for this reaction in kJ/mol?

  8. There are two kinds of coffee-cup calorimetry problems. 2nd kind: Two volumes of a solution are given. What to do: Add them and plugged into m (the density of most solutions = 1g/mL, so the mL = the grams) DT, Last temperature minus first temperature. Solve for q and change joules to kJ. Pick one of the chemicals and multiply (L)(M) to find moles. Solve for DH by putting kJ over moles and divide.

  9. Example #3 450mL of 2M HCl are mixed with 450mL of 2M NaOH in a coffee-cup calorimeter. The temperature rises from 23oC to 25oC. Calculate the enthalpy change for this reaction.

  10. Example #4 In a coffee-cup calorimeter, 275mL of 0.5M Ca(OH)2 react with 275mL of 0.5M H2SO4. The temperature increases from 25oC to 30oC. Calculate the energy in kJ/mol.

  11. Bomb Calorimetry q = -CDT q: quantity of heat, kJ C: heat capacity, kJ/oC DT: change in temperature (increases positive, decreases negative)

  12. Two kinds of Bomb also: 1st kind: Enthalpy (kJ/mol) is given What to do: Convert grams to moles Multiply the moles by the enthalpy. This answer plugs in for q Find C or DT (whichever it asks for)

  13. Example #5 A 5g sample of hydrogen is burned in a bomb calorimeter. The enthalpy of combustion for hydrogen is -242kJ/mol. What is the heat capacity of the calorimeter if the temperature rises 5oC?

  14. Example #6 30g of magnesium are burned in a bomb calorimeter that has a heat capacity of 18kJ/oC. If the enthalpy of combustion for magnesium is -602kJ/mol, how much will the temperature rise?

  15. Still Bomb 2nd kind: Enthalpy is asked for (or energy change in kJ/mol) What to do: Convert grams to moles. Plug in C and DT and solve for q. Solve for DH by putting kJ over moles and divide.

  16. Example #7 If 10g of Ca are burned in a bomb calorimeter that has a heat capacity of 12.8kJ/oC, the temperature increases by 12.38oC. What is the enthalpy of combustion of calcium?

  17. Example #8 An 18g sample of sulfur is burned in a bomb calorimeter. The heat capacity of the calorimeter is 8.9kJ/oC and the temperature increases 18.74oC. What is the enthalpy of combustion for sulfur?

  18. DHrxn using DHf values DHrxn = SDHfproducts – SDHfreactants In Appendix Four (pages A19-A24) you will find the DHf values. The values are listed alphabetically by element. If it is an acid, i.e. HCl, look under the second element to find your chemical. Pay attention to the phase your chemical is in.

  19. Example #9 Calculate DH for the following reactions: 1. 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s) 2. 2CH3OH(l) + 3O2(g) 2CO2(g)+ 4H2O(l) 3. 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

  20. Hess’s Law You will use the reaction given, rearrange them, multiply them by a number (whole or fraction) to make the reaction asked for. The energy given for each reaction will be multiplied by the same number you used for the reaction. If the reaction is reversed the sign for the energy number will change to the opposite sign.

  21. Example #10 Using the reactions: C(s) + O2(g) CO2(g)DH = -393.5kJ/mol CO(g) + ½ O2(g)  CO2(g) DH = -283kJ/mol Calculate the enthalpy for: C(s) + ½ O2(g)  CO(g)

  22. Example #11 Calculate DH for the reaction: 2C(s) + H2(g) C2H2(g) Given the following: 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O (l) DH = -2599.2kJ/mol C(s) + O2(g) CO2(g)DH = -393.5kJ/mol 2H2(g) + O2(g)  2H2O (l)DH = -571.6kJ/mol

  23. Example #12 Given the following data: H2(g) + ½ O2(g) H2O(l) DH = -286kJ N2O5(g) + H2(l)  2HNO3(l)DH = -77kJ ½ N2(g) + 3/2 O2(g) + ½ H2(g)  HNO3(l) DH = -174kJ Calculate the DH for: 2 N2(g) + 5 O2(g)  2 N2O5(g)

  24. Energy required for a phase change Slanted lines: use the formula q=mcDT Slanted lines occur below 0oC, between 0oC and 100oC and above 100oC. cice = 2.1, cliq = 4.18 cvap = 1.8 J/goC Flat lines: Use the formula q = DH(mol) Flat lines occur at 0oC and 100oC. At 0oC use DHfus = 6kJ/mol and at 100oC use DHvap = 40.7kJ/mol

  25. Example #12 How much energy is required to take 25g of water from -35oC to 120oC?

  26. Example #13 How much energy will be needed to take 30g of water from 20oC to 115oC?

  27. Example #14 How much energy will be needed to take 100g of ice from 0oC to 90oC?

  28. Example #15 How much energy will be needed to take 45g of water from 19oC to 85oC?

  29. Phase changes cont. You may be given an amount of energy and have to find the final temperature the water sample will reach. You will make a graph again, check to see how much energy you will need to get to the next place (up a slant or across a flat line). If you have enough, subtract that number from the energy you have. Continue until you do not have enough energy.

  30. More on phase changes. If you do not have enough energy to make it across a flat line, that is the final temperature (either 0oC or 100oC). If you do not have enough energy to make it up a slanted line, plug the energy in (change to Joules) and calculate DT. Add that to the temperature that line came from.

  31. Example #16 What is the highest temperature a 50g sample of water at 5oC will reach after 18kJ of energy are added?

  32. Example #16 What is the highest temperature 25g of ice at 0oC will reach after 16kJ of energy have been added?

  33. Example #17 What temperature will 20g of ice at -10oC reach after 20kJ of energy have been applied?

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