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Let’s Play Jeopardy !

Let’s Play Jeopardy !. Terms . Fix that Mistake. Enthalpy. Potpourri. Laboratory. Making Connections. Terms. Mistake. Enthalpy. Potpourri. Laboratory. Connections. $100. $100. $100. $100. $100. $100. $200. $200. $200. $200. $200. $200. $300. $300. $300. $300.

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Let’s Play Jeopardy !

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  1. Let’s Play Jeopardy!

  2. Terms

  3. Fix that Mistake

  4. Enthalpy

  5. Potpourri

  6. Laboratory

  7. Making Connections

  8. Terms Mistake Enthalpy Potpourri Laboratory Connections $100 $100 $100 $100 $100 $100 $200 $200 $200 $200 $200 $200 $300 $300 $300 $300 $300 $300 $400 $400 $400 $400 $400 $400 $500 $500 $500 $500 $500 $500

  9. 1-100 1 - 100 Compare and Contrast Exothermic and Endothermic Reactions $100

  10. 1 - 100 $100

  11. 1-200 Define Heat $200

  12. 1-200A 1 - 100 Transfer of kinetic energy from a hotter object to a colder object $200

  13. What is Enthalpy? $300

  14. When pressure is constant, the heat absorbed or released during a chemical reaction $300

  15. 1-400 What is calorimetry? $400

  16. 1-400A Determining the heat of a reaction by measuring temperature changes produced in a calorimeter 1 - 100 $400

  17. 1-500 Define Hess’s Law Hint: Include the two associated rules $500

  18. 1-500A Enthalpy for a target reaction is the sum of the enthalpy changes for the component steps. • If reaction is reversed, sign for enthalpy of that reaction is reversed. • If coefficients of a reaction multiplied by a factor, enthalpy of that reaction multiplied by same factor 1 - 100 $500

  19. 2-100 How is ∆T calculated? $100

  20. 2-100A 1 - 100 ΔT = T2 – T1 $100

  21. 2-200 1 - 100 What is the formula you should use to find the specific heat of a substance? $200

  22. q = m • Cp •ΔT Therefore, Cp = q m •ΔT $200

  23. 2-300 What is the equation for calculating percent error? $300

  24. 2-300A 1 - 100 % Error = exp – actual actual x 100 $300

  25. How are heat, temperature and enthalpy related? $400

  26. 1 - 100 Temperature (°C) is proportional to heat of a reaction (J). q = m • Cp •ΔT Enthalpy (J/mol) is heat of a rxn based on the amt of reactants and/or products you have $400

  27. Given the problem: State what is given, what you’re solving for, and how you would solve. What is the specific heat of gold if the temperature of a 8.21 g sample of gold is goes from 4.2°C to 8°C when 6.51 J of heat is added? $500

  28. 1 - 100 q = 6.51 J M = 8.21 g Cp = ? T1 = 4.2°C T2 = 8°C Use equation: q = m Cp (T2 – T1) $500

  29. 3-100 Does this reaction have a positive or negative enthalpy? Video $100

  30. 3-100A 1 - 100 Reaction is exothermic, therefore the sign for enthalpy is negative $100

  31. 3-200 Draw an enthalpy diagram for the following reaction: C3H8 + 5O2 3CO2 + 4H20 ∆H = -2043 kJ $200

  32. EXOTHERMIC C3H8 + 5O2 ΔH = -2043 kJ H  3CO2 + 4 H20 Reaction Progress  $200

  33. 3-300 Solve: How much heat will be released when 11.8 g of iron reacts with excess oxygen? 3Fe + 2O2 Fe3O4ΔH =-1120.48 kJ $300

  34. 11.8 g Fe 1 mol Fe -1120.48 J 1 55.85 g Fe 3 mol Fe • • $300

  35. Solve: How much energy is transferred when 147 g of NO2is dissolved in 100 g of water? 3 NO2 + H20  2 HNO3 + NO ΔH = -138 kJ $400

  36. Find Limiting Reactant Solve 147 g NO2 1 mol NO21 mol H20 18 g H20 1 46 g NO2 3 mol NO2 1 mol H20 = 19.2 g H20 needed; Given 100 g. NO2 is limiting reactant 147 g NO2 1 mol NO2 -138 kJ 1 46 g NO2 3 mol NO2 = -147 kJ • • • • • EXOTHERMIC! $400

  37. Using the following, CaO + CO2 CaCO3 ΔH = -178 kJ CaO + H2O  Ca(OH)2 ΔH = -65 kJ Calculate the ΔH for the reaction Ca(OH)2 + CO2 CaCO3 + H20 $500

  38. KEEP: CaO + CO2 CaCO3 ΔH = -178 kJ FLIP: CaO + H2O  Ca(OH)2 ΔH = -65 kJ So, CaO + CO2 CaCO3 ΔH = -178 kJ Ca(OH)2 CaO + H2O ΔH = +65 kJ Therefore, ΔH for target reaction = -178 + 65 = -113 kJ EXOTHERMIC! $500

  39. 1 - 100 Draw a diagram that depicts energy flow between the system and the surroundings of an endothermic rxn $100

  40. 1 - 100 SURR SYS $100

  41. 4-200 What is meant by Standard State? $200

  42. 1 - 100 The most stable form of an element under standard conditions (25°C, 1 atm) $200

  43. 4-200 What is the specific heat of water? (HINT: Remember to include units) $300

  44. 4-400A 1 - 100 4.184 Jg°C $300

  45. 4-400 What two major factors affect a substance’s heat capacity? $400

  46. 4-400A 1 - 100 Heat capacity is amt of heat needed to raise the temperature of an object by 1°C. Mass and composition impact a substance’s heat capacity. $400

  47. Solve: When 13.7 g sample of Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4°C to 19.7°C. Calculate ΔH for the reaction. $500

  48. Figure out how many moles of Pb(NO3)2 were involved Find qsurr and qrxn Find Enthalpy 1 - 100 13.7 g Pb(NO3)2 1 mol Pb(NO3)2 1 331.2 g Pb(NO3)2 = 0.041 mol • qsurr = (85.0 g H2O) (4.184 J/g°C) (19.7 – 23.4) = -1315.9 J qrxn = +1315.9 J ΔH = +1315.9 J = 32095 J/mol 0.041 mol ENDOTHERMIC! $500

  49. 5-100 1 - 100 Why is a calorimeter made of styrofoam? $100

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