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Test 2 Study Guide

Thursday February 28

8:00 p.m. Dobo 134

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Top 7 Topics

• Limiting reactants

• Ionic compounds in water

• Molarity

• Energy

• Specific heat

• Enthalpy

• Hess’ Law

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Limiting Reactants - 3

• Theoretical Yield

• How much is used, formed, left over

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Ionic Compounds 5

• In water

• Strong / weak electrolytes

• Ionic equations

• Metathesis reactions

• Ion Solubility

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Molarity 4

• Interconverting molarity, moles, volume

• Dilution

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Energy 2

E = q + w

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Specific heat 1

• Calculate energy given specific heat

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Enthalpy 3

• Endo , exo, sign of H

• Given heat for a certain mass, calculate reaction H

• Given H and mass, calculate heat

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Hess’ Law 2

• Addition of reactions to obtain H

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Top 7 TopicsLet’s do some examples!

• Limiting reactants

• Ionic compounds in water

• Molarity

• Energy

• Specific heat

• Enthalpy

• Hess’ Law

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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.

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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

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• How much is used, formed, left over

• Q 1 on Test

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Ionic Compounds 5

• In water

• Strong / weak electrolytes

• Ionic equations

• Metathesis reactions

• Ion Solubility

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• Group 1A ( alkali metals) cations soluble

• NH4+ soluble

• NO3- soluble

• CH3CO2- = C2H3O2

- = acetate soluble

• Will be given a chart for other ions

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• 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.

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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)

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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

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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

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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

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Specific heat

• Calculate heat energy given specific heat

• q = (specific heat) ( mass in grams)(T)

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• 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

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Enthalpy

• Endothermic H > 0

• Exothermic H < 0

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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/moleH = 6kJ

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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

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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.

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• 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