Friday, Oct. 25th: “A” DayMonday, Oct. 28th: “B” Day

AgendaHomework questions/Quick reviewSection 10.2 Quiz: “Using Enthalpy”Section 10.3: “Changes in Enthalpy During

Chemical Reactions” Calorimetry, calorimeter, adiabatic calorimetry, Hess’s

Law, standard enthalpy of formationHomework

Pg. 15 practice worksheet (MUST show work)Sec. 10.3 review, pg. 357: #1-5Concept Review: “Changes in Enthalpy During

Chemical Reactions”

Homework Questions/Problems

Pg. 349: #1-8

Section 10.2 Quiz: “Using Enthalpy”

You can use both your book and your notes.You’ll need both 10.1 AND 10.2 notes.

May the FORCE bewith you!

#4: M = molar mass#8: Use the “25 J rule”

Changes in Enthalpy Accompany ReactionsChanges in enthalpy occur during chemical

reactions.A change in enthalpy during a reaction

depends on many variables, but temperature is one of the most important variables.

To standardize enthalpies of reaction, data are presented for reactions in which both reactants and products have the standard thermodynamic temperature of 25˚C or 298.15 K.

Chemical CalorimetryCalorimetry: the measurement of heat-

related constants, such as specific heat or latent heat.

Calorimeter: a device used to measure the heat absorbed or released in a chemical or physical change.

Nutritionists Use Bomb Calorimetry

A bomb calorimeter is used to measure enthalpy changes caused by combustion reactions.

Adiabatic Calorimetry is Another Strategy

Instead of using a water bath to absorb the energy generated in a combustion reaction, adiabatic calorimetry uses an insulating vessel that doesn’t allow energy to pass through.

As a result, the temperature of the reaction mixture will change and can be recorded.

Adiabatic calorimetry is used for reactions that are not ignited, such as for reactions in aqueous solution.

Hess’s LawHess’s Law: the law that states that the

amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction.

The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process.

Standard Enthalpies of FormationStandard enthalpy of formation: the enthalpy

change in forming 1 mol of a substance from elements in their standard state.

By definition, the values of the standard enthalpies of formation for elements are zero.

Symbol: ΔH˚f

Unit: kJ/mol

Calculating Enthalpy Change for a Chemical Reaction

Using a list of standard enthalpies of formation, the enthalpy change of any reaction for which there is data available can be calculated:

ΔHreaction = ΔHf°products - ΔHf

°reactants

ΔHreaction is in kJ or Joules (moles cancel out)

Table 2: Standard Enthalpies of Formation

ExampleCalculate the enthalpy change for the following reaction.

Use the standard enthalpies of formation listed in Table A-11 on pg 833-834.

HCl(g) + NH3(g) NH4Cl(s)

ΔHreaction = ΔHf0

products - ΔHf0

reactants

ΔHf0

product = (1 mol)(-314.4 kJ/mol) = -314/4 kJ

ΔHf0

reactants=[(1 mol)(-92.3 kJ/mol)+(1 mol)(-45.9 kJ/mol)]

= -138.2 kJ

ΔHreaction = (-314.4 kJ) – (-138.2 kJ)

-176.2 kJ (exothermic reaction)

Additional PracticeCalculate the enthalpy change for the following reaction.

Use the standard enthalpies of formation listed in Table A-11 on pg 833-834.

N2(g) + 3 H2(g) 2 NH3(g)

State whether the reaction is exothermic or endothermic.

ΔHreaction = ΔH f0

products - ΔH f0

reactants

ΔHf0

prod = [(2 mol)(-45.9 kJ/mol) = -91.8 kJ

ΔHf0

reactants = [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)]

= 0 kJΔHreaction = (-91.8 kJ) – (0 kJ) = -91.8 kJ

*Reaction is exothermic because ΔH is negative.*

Calculating a Reaction’s Change in Enthalpy Sample Prob. E, pg.356

Calculate the change in enthalpy for the reaction below using data from Table 2 on pg 355.

2 H2(g) + 2 CO2(g) 2 H2O(g) + 2 CO(g)

State whether the reaction is exothermic or endothermic.

ΔHreaction = ΔH f0

products - ΔH f0

reactants

ΔHf0

prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)]

= -704.6 kJΔHf

0reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)]

= -787 kJΔHreaction= (-704.6 kJ) – (-787 kJ) = 82.4 kJ

*Reaction is endothermic because ΔH is positive.*

Homework

Pg. 15 practice workshetMUST show work!

Section 10.3 review, pg. 357: #1-5Concept Review: “Changes in Enthalpy During

Chemical Reactions”