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Test 2 Study Guide

Comprehensive study guide covering limiting reactants, ionic compounds, molarity, energy, specific heat, enthalpy, and Hess’ Law with detailed examples and explanations. Prepare effectively for your upcoming test!

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Test 2 Study Guide

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  1. Test 2 Study Guide Thursday February 28 8:00 p.m. Dobo 134

  2. Top 7 Topics • Limiting reactants • Ionic compounds in water • Molarity • Energy • Specific heat • Enthalpy • Hess’ Law

  3. Limiting Reactants - 3 • Theoretical Yield • How much is used, formed, left over

  4. Ionic Compounds 5 • In water • Strong / weak electrolytes • Ionic equations • Metathesis reactions • Ion Solubility

  5. Molarity 4 • Interconverting molarity, moles, volume • Dilution

  6. Energy 2 • E = q + w

  7. Specific heat 1 • Calculate energy given specific heat

  8. Enthalpy 3 • Endo , exo, sign of H • Given heat for a certain mass, calculate reaction H • Given H and mass, calculate heat

  9. Hess’ Law 2 • Addition of reactions to obtain H

  10. Top 7 TopicsLet’s do some examples! • Limiting reactants • Ionic compounds in water • Molarity • Energy • Specific heat • Enthalpy • Hess’ Law

  11. Limiting Reactants - 3 • Theoretical Yield – how much you expect based upon the amount of reactants. • If one reactant is is present in excess, then the other limits how much can be made and is used to determine the theoretical yield.

  12. 2H2 + O2 2 H2O • If 10 moles of oxygen and 10 moles of hydrogen…. • The hydrogen limits the amount of water that can be formed • Based upon 10 moles H2, we can make 10 moles water • Based upon 10 moles O2, we can make 20 moles water. • H2 limits. Theoretical Yield = 10 moles ( 180 g) H2O • At end of reaction, will have 5 moles O2 left over

  13. How much is used, formed, left over • Q 1 on Test

  14. Ionic Compounds 5 • In water • Strong / weak electrolytes • Ionic equations • Metathesis reactions • Ion Solubility

  15. Group 1A ( alkali metals) cations soluble • NH4+ soluble • NO3- soluble • CH3CO2- = C2H3O2- = acetate soluble • Will be given a chart for other ions

  16. Soluble ionic compounds in water exist as free ions surrounded by water molecules. • Soluble ionic compounds are strong electrolytes • Strong acids or bases (HCl, H2SO4), NaOH) are strong electrolytes • Weak acids or bases (CH3CO2H, acetic acid) are weak electrolytes • Molecules that do not disassociate (CO2, sugar) are non electrolytes.

  17. potassium sulfate with barium nitrate • 2K+(aq) + SO4-2 (aq) + Ba+2 (aq) + 2 NO3- (aq) 2K+(aq) + 2 NO3- (aq) +BaSO4(s) • SO4-2 (aq) + Ba+2 (aq)  BaSO4(s)

  18. Molarity 4 • Interconverting molarity, moles, volume • M means moles per liter = moles/L • M = moles/L • If you know two things, can determine the third • If you have moles and volume, can determine molarity • 3 moles dissolved in 0.5 L = 3moles/0.5L = 6M

  19. Molarity - dilution • (Vconc )(Mconc ) = (Vdil )(Mdil ) • If you know 3, can solve for the fourth • How many mL of 3M HCl is needed to make 100mL of 1.5 M HCl? • (Vconc)(3M) = (100mL)(1.5M) • Vconc = 50 mL

  20. E = q + w • q > 0 heat transferred from the surroundings to the system (endothermic) • q < 0 heat transferred from the system to the surroundings ( exothermic) • w > 0 work is done by the surroundings on the system • w < 0 work is done by the system on the surroundings • q > 0, w > 0 E > 0 • q < 0, w < 0 E< 0

  21. Specific heat • Calculate heat energy given specific heat • q = (specific heat) ( mass in grams)(T)

  22. 18.      When 72 g of a metal at 97.0 C is added to 100.0 g of water at 25.0 C, the final temperature is 29.1 C. What is the heat capacity of the metal if cwater = 4.184 J/g.K? • Heat lost by metal = heat gained by water • qmetal = - qwater

  23. Enthalpy • Endothermic H > 0 • Exothermic H < 0

  24. Given heat for a certain mass, calculate reaction H • If it takes 60 kJ to melt 180 grams of ice, what is H for the following reaction? • H2O(s) H2O(l) • (60kJ / 180g) ( 18 g/mole) = 6 kJ/mole H = 6kJ

  25. Given H and mass, calculate heat H2O(s) H2O(l) H = 6 kJ How much heat is needed to melt 900 grams of ice? (900g)(6 kJ/mole)(1mole/18 g) = 300 kJ

  26. Hess’ Law • If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

  27. 19.      What is the value of H for IF5(g)  IF3(g) + F2(g) H = ? given the following thermochemical equations? IF(g) + F2(g)  IF3(g) H = -390 kJ IF(g) + 2 F2(g)  IF5(g) H = -745 kJ

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