Thermochemistry Part 2

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Thermochemistry Part 2

HESS’S LAW

BLB 5.6–7; 8.8 BLB 5:63, 67a,b, 69, 73, 75, 83, 85; BLB 8:65a, 67a,c, 92a BLB 18:72ab, 74

The sum of the !H values for each step is the same as !H for the overall process.

Supplemental 5:1–7; Supplemental 8:11–13

This is true because !H is a state function.

Know: • Hess!s Law • heats of formation • enthalpy of reactions • estimating reaction enthalpy from bond energies

C !H2 !Hrxn B

Check out the grade finder on Angel; what do you need to get on your final?? FINAL DEADLINE for credit on skill check tests: (you must get 100% on any test to receive credit for that test): _______ Final Exam: __________________

Dr. L. S. Van Der Sluys

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!H1 A A B

B C

A+B "# B + C


!H1 !H2 !H1 + !H2= !Hrxn

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Example: Given the following information A B

H2(g) + F2(g) # 2HF(g) 2H2(g) + O2(g) # 2 H2O(g)

Given the following information: !HA = –537kJ !HB = –572kJ

2SO2(g)+ O2(g) 2S(s) + 3 O2(g)

Determine !H for the reaction: C

# 2SO3(g)

!H $196kJ

# 2SO3(g)

$790kJ

What is !Hrxn for the following reaction?

2F2(g) + 2H2O(g)# 4HF(g) + O2(g) !HC =?

IDEA: find combinations of reactions such that nA+mB=C then

S(s) + O2(g) A. B. C. D.

# SO2(g)

– 986 kJ – 594 kJ + 594 kJ – 297 kJ

n !HA + m !HB = !HC Here: 2xA –1xB

2H2(g) + 2F2(g) # 4HF(g) !H = 2(–537kJ) 2 H2O(g) # 2H2(g) + O2(g) !H = –(–572kJ)

_

2F2(g) + 2H2O(g)# 4HF(g) + O2(g) !HC = 2(–537kJ) –1(–572kJ) = –502kJ


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Standard States of the Elements

Heat of Formation !Hf

enthalpy of formation heat given off (or absorbed) when elements combine to give one mole or one molecule of a compound

combine

Elements

Compounds

!Hf

!H° f standard enthalpy of formation enthalpy of formation when all substances are in their Standard State

DEFINITION OF STANDARD STATE 1. P = 1 atm 2. T = 25°C (298K) 3. element is in its most stable state (gas/liquid/solid)

For an element in its standard state:

!H° f = ___


(by definition)

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(Most stable phase under standard conditions of 298K and 1 atm)

1. Metals: all are _______ at 298K and 1 atm except one (Which one?) 2. Semi metals (metalloids): all are _________ at 298K and 1 atm 3. Nonmetals at 298K and 1 atm i: Noble gases (Group 8): atomic ___________ ii: Diatomics: H2, N2, O2, Group 7 (F2, Cl2, Br2, I2) H2, N2, O2, F2, Cl2, are ______ Br2 is a __________ I2 is a ___________ iii: all other non-metals are _____ C (graphite), S8, P(s), Se Dr. L. S. Van Der Sluys

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Are these Elements in their standard states? YES

NO (What is?)

O2(g) $

N3 (s) F2(g) N2(g) O3(g) C(diamond) C(graphite) Fe(s) Br2(g) H(g)


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For which of the following reactions (at 25° C and 1 atm) is !Hrxn = !H° f ?

We know that !H° rxn is the Standard Enthalpy of reaction, or Heat of reaction; all elements must be in their ________________.

1) H2(g) + F2(g) "# 2HF(g) 2) NO(g)+ 1/2 O2(g)

We know that !H° f is the Heat of formation; formation of _____________(how much?) from ______________ when all reactants are in their __________________

# NO2(g)

3) 2C(graphite)+3H2(g)+1/2O2(g)"#C2H5OH(l) 4) Pb(s) + Cl2(g) "#PbCl2(s)

How are they related? Obtaining !H° rxn from !H° f: !H° rxn = %n !H° f (prod) $ %m !H° f (react)

5) S(s) + O3(g) "# SO3(g)

where n and m are stoichiometric coefficients of products and reactants.

6) Br2(l) "# Br2(g)

(This is an application of Hess’s Law.)


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Using the following information, what is !H° rxn for the reaction:

C2H5OH(l) + 3O2(g)#2CO2(g) + 3H2O(l) !H° f (in kJ/mole): -277.7

a) b) c) d)

-393.5

-285.8

– 115.8 kJ + 115.8 kJ – 1366.7 kJ + 13366.7 kJ

If a piece of fruit contains 16.0 g of fructose, and !Hrxn = $2803 kJ, how many food calories does it contribute to the body? 1 cal = 4.184J 1kcal = 1 Cal (= 1 food calorie)

Combustion! C6H12O6(s) +6____# 6____+ 6_____

NOTE: Appendix C in the text has a more extensive table of !H° f Dr. L. S. Van Der Sluys

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Changes in Enthalpy with Breaking and Formation of Bonds

Bond Properties Review COVALENT BOND LENGTHS and ENERGIES Bond length: distance between nuclei

bond

Bond energy Bond length pm kJ/mol

348 154 C"C 614 134 C=C 839 121 C&C more electrons shared, shorter bond length bond

!

Bond Enthalpies (bond energy) can be used to estimate !Hrxn.

Bond energy

Bond length pm

kJ/mol 413 110 C"H 328 176 C"Cl 276 196 C"Br Shorter bond length, stronger the bond

!

The overall reaction has two steps; breaking the original bonds, and forming new ones.


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Estimating !Hrxn • Bond energies provide estimates of reaction enthalpies:

!Hrxn ' %nDbroken $ %mDformed n, m = # of bonds Energy is given off ($) when bonds form.

Helps understand origins of !Hrxn


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Example: Predict the heat of reaction for the decomposition of hydrogen peroxide.

2 H2O2 # 2 H2O + O2 !Hrxn = ? Draw Lewis structures of reactants and products:

Peroxide Decomposition: Elephant Toothpaste Demonstration

2 H2O2(l) # 2 H2O(l) + O2(g)

!Hrxn = ?

Add KI to catalyze the reaction: I-(aq) + H2O2(l) # H2O(l) + IO-(aq) IO-(aq) + H2O2(l) # H2O(l) + O2(g)+ I-(aq)

Reactants (broken)

Products

bond #

bond #

D

(formed)

D

2 H2O2(l) # 2 H2O(l) + O2(g) Can also catalyze the reaction with MnO2

Compare to Actual value of !H° rxn = -196 kJ/mole Dr. L. S. Van Der Sluys

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