Enthalpy (H)

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Enthalpy (H) The heat transferred sys ↔ surr during a chemical rxn @ constant P Can’t measure H, only ΔH At constant P, ΔH = q = mCΔT, etc. Literally, ΔH = H products - H reactants ΔH = + (endothermic) Heat goes from surr into sys ΔH = - (exothermic) Heat leaves sys and goes into surr

description

Enthalpy (H). The heat transferred sys ↔ surr during a chemical rxn @ constant P Can’t measure H, only Δ H At constant P, Δ H = q = mC Δ T, etc. Literally, Δ H = H products - H reactants Δ H = + (endothermic) Heat goes from surr into sys Δ H = - (exothermic) - PowerPoint PPT Presentation

Transcript of Enthalpy (H)

Page 1: Enthalpy (H)

Enthalpy (H)

• The heat transferred sys ↔ surr during a chemical rxn @ constant P

• Can’t measure H, only ΔH• At constant P, ΔH = q = mCΔT, etc.• Literally, ΔH = Hproducts - Hreactants

• ΔH = + (endothermic)• Heat goes from surr into sys

• ΔH = - (exothermic)• Heat leaves sys and goes into surr

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Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓Notice that the total energy does not change

Myers, Oldham, Tocci, Chemistry, 2004, page 41

Reactant + Energy Product Endothermic Reaction

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Surroundings

System

Surroundings

System

En

erg

y

Beforereaction

AfterreactionMyers, Oldham, Tocci, Chemistry, 2004, page 41

In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑Notice again that the total energy does not change

Reactant Product + Energy Exothermic Reaction

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PE

Reaction Coordinate Diagrams: Endothermic Reaction

Progress of the Reaction

Reactants

ProductsD(PE)

Ea

ΔHrxn= +

Activation Energy

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PE

Reaction Coordinate Diagrams: Exothermic Reaction

Progress of the Reaction

Reactants

Products

D(PE)

Ea

Activation Energy

ΔHrxn= -

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Reaction Coordinate DiagramsDraw the reaction coordinate diagram for the following rxn:

C(s) + O2(g) CO2 + 458.1kJ

EXOTHERMIC

PE

Progress of the Reaction

C + O2

CO2

D(PE)

Ea

Activation Energy

ΔHrxn= -458.1 kJ

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• All reactions have some ΔH associated with itH2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

• How can we interpret this ΔH?• Amount of energy released or absorbed per specific

reaction species• Use balanced equation to find several definitions

- 483.6 kJ

½ mol O2

Enthalpies of Reaction

- 483.6 kJ

1 mol H2

- 483.6 kJ

1 mol H2O

Able to use like conversion factors in stoichiometry

or or

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Enthalpies of Reaction

• Formation of waterH2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

• ΔH is proportional to amount used and will change as amount changes

2H2(g) + O2(g) → 2H2O(l)

• For reverse reactions, sign of ΔH changes2H2O(l) → 2H2(g) + O2(g)

• Treat ΔH like reactant or product H2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

ΔH = - 967.2 kJ

ΔH = + 967.2 kJ

H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)

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2.00 mol C

50.0 g O2

Consider the following rxn:

C(s) + 1/2O2(g) CO + 458.1kJIs the ΔH for this reaction positive or negative?

NEGATIVE (E released as a product)What is the ΔH for 2.00 moles of carbon, if all the carbon is used?

= - 916 kJ

What is the ΔH if 50.0g of oxygen is used?

= -1430 kJ

What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction?

= 818 kJ

- 458.1 kJ1 mol C

1 mol O2

0.5 mol O2

- 458.1 kJ32.0 g O2

50.0 g CO 1 mol CO

1 mol CO458.1 kJ

28.0 g CO

Enthalpies of Reaction Practice

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Reactants Products

The change in enthalpy is the same whether the reaction takes place in one step or a series of steps

Why? Because enthalpy is a state function

Hess’s Law

Victor Hess

2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer

2 CH3OH 2 CH4 + O2 ΔHrxn = +328 kJ

2 CH4 + O2 2 CH3OH ΔHrxn = -328 kJ

To review:

1. If a reaction is reversed, ΔH is also reversed

2(CH4 + 2 O2 CO2 + 2 H2O) ΔHrxn = -1605 kJ

CH4 + 2 O2 CO2 + 2 H2O ΔHrxn = -802.5 kJ

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Example: Methanol-Powered Cars

2 CH4(g) + 4 O2(g) 2 CO2(g) + 4 H2O(g) ΔHrxn = -1605 kJ

2 CH3OH(l) 2 CH4(g) + O2(g) ΔHrxn = +328 kJ

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ΔHrxn = -802.5 kJ

2 CH4(g) + O2(g) 2 CH3OH(l) ΔHrxn = -328 kJ

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) ΔHrxn = ?

2 CH3OH + 3 O2 2 CO2 + 4 H2O ΔHrxn = -1277 kJ

3

- ( ) - ( )

2 ( ) 2 ( )

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Tips for applying Hess’s Law…

Look at the final equation that you are trying to create first…

• Find a molecule from that eq. that is only in one of the given equations (i.e. CH3OH, CO2)• Make whatever alterations are necessary to those• Once you alter a given equation, you will not alter it again

• Continue to do this until there are no other options

• Next, alter remaining equations to get things to cancel that do not appear in the final equation

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1. Given the following data:S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ

2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ

.Calculate ΔH for the following reaction: S(s) + O2(g) → SO2(g)

S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ 2SO3(g) → O2(g) + 2SO2(g) ΔH = +198.2 kJ SO3(g) → ½ O2(g) + SO2(g) ΔH = +99.1 kJ

S(s) + O2(g) → SO2(g) ΔH = -296.1 kJ

*

- ½ ( ) - ½ ( )

*

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2. Given the following data:C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300. kJ C(s) + O2(g) → CO2(g) ΔH = -394 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ Calculate ΔH for the following reaction: 2C(s) + H2(g) → C2H2(g)

2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = +1300 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ

2C(s) + H2(g) → C2H2(g) ΔH = +226 kJ

2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ

2( ) 2( )-( )

***

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NO(g) + O(g) → NO2(g) ΔH = - 233 kJ

3. Given the following data:  2O3(g) → 3O2(g) ΔH = - 427 kJ O2(g) → 2O(g) ΔH = + 495 kJNO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ Calculate ΔH for the following reaction:

NO(g) + O(g) → NO2(g)

NO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJO(g) → ½ O2(g) ΔH = - 247.5 kJ

3/2 O2(g) → O3(g) ΔH = + 213.5 kJ

*

**

-½( ) -½( )-½( ) -½( )

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4. Given the following data: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -23 kJ

3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) ΔH = -39 kJ

Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ΔH = 18 kJ

Calculate ΔH for the following reaction: FeO(s) + CO(g) → Fe(s) + CO2(g)

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1. A B ∆H = + 30 kJC B ∆H = - 60 kJ

Calculate ΔH for the following reaction: A C

Hess’s Law HW Questions

2. Suppose you are given the following reactions:4X 2Y DH = - 40 kJX ½ Z DH = - 95 kJ

Calculate ΔH for the following reaction: Y Z

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3. From the following heats of reaction:2 H2 (g) + O2 (g) 2 H2O (g) DH = -483.6 kJ

3 O2 (g) 2 O3 (g) DH = +284.6

kJ

Calculate the heat of the reaction (∆H): 3 H2 (g) + O3 (g) 3 H2O (g)

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4. From the following enthalpies of reaction: H2 (g) + F2 (g) 2 HF (g) DH = -

537kJC (s) + 2 F2 (g) CF4 (g) DH = - 680

kJ2 C (s) + 2 H2 (g) C2H4 (g) DH = + 52.3 kJ

Calculate the DH for the reaction of ethylene with F2.

C2H4 (g) + 6F2 (g) 2 CF4 (g) + 4 HF(g)

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5. Given the following data:N2 (g) + O2 (g) 2 NO (g) DH = + 180.7

kJ2 NO (g) + O2 (g) 2 NO2 (g)DH = - 113.1 kJ

2 N2O (g) 2 N2 (g) + O2 (g) DH = - 162.3

kJ Calculate DH for the reaction below: N2O (g) + NO2 (g) 3 NO (g) DH = ?

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Heats of Formation, ΔH°f

The enthalpy change when one mole of a compound is formed from the elements in their standard states

° = standard conditions• Gases at 1 atm pressure• All solutes at 1 M concentration (remember M = mol/L)• Pure solids and pure liquids

f = a formation reaction

• 1 mole of product formed

• From the elements in their standard states (1 atm, 25°C)

For all elements in their standard states, ΔH°f = 0

C9H12NO3(s)O2(g)N2(g) +H2(g) +Cgr +1/29 6 3/2

What’s the formation reaction for adrenaline, C9H12NO3(s)?

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Thermite Reaction

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

ΔHrxn = ?Welding railroad tracks

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Thermite Reaction

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

2 Al(s)

Fe2O3(s) 2 Fe(s) 2 Fe(l)

2 Al(s)3/2 O2(g) Al2O3(s)

Reactants Elements(standard states)

Products

ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))

ΔHrxn = 2(15 kJ) + (-1676 kJ) - (-822 kJ) – 2(0)

ΔHrxn = - 824 kJ

ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

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∆Hrxn = Σ n∆Hof Products - Σn∆Ho

f Reactants

(- 74.8) (0) (- 106.7) (0)

1. CH4(g) + 2 Cl2(g) CCl4(g) + 2 H2(g)

∆H = [(-106.7) + 0] – [(-74.8)+0]

= -106.7 + 74.8

= - 31.9 kJ

2. 2 KCl(s) + 3 O2(g) 2KClO3(s)

(- 435.9) (0) (- 391.2)

∆H = [(2)(- 391.2)] – [(2)(- 435.9) + (3)(0)]

= - 782.4 + 871.8

= 89.4 kJ

2 3 2

2 2

ΔH°f Example Problems

ΔHrxn = ?

ΔHrxn = ?

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(-124.4) (-407.1) (-127.0) (-446.2)

3. AgNO3(s) + NaCl (aq) AgCl(s) + NaNO3(aq)

∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)]

= -573.2 + 531.5

= - 41.7 kJ

4. C2H5OH(l) + 7/2 O2(g) 2CO2(g) + 3H2O(g)

(-277.7) (0) (-393.5)

∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)]

= -1512.4 + 277.7

= -1234.7 kJ

(3)(7/2) (2) (-241.8)

∆Hrxn = Σ n∆Hof Products - Σn∆Ho

f Reactants

ΔH°f Example Problems

ΔHrxn = ?

ΔHrxn = ?

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Enthalpy Review#2. Calculate H for the following reaction: 

N2H4 (g) + O2 (g) → N2 (g) + 2H2O (g) Given: H (kJ/mol)2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (g) -1010

N2O (g) + 3 H2 (g) → N2H4 (g) + H2O (g) -317

2 NH3 (g) + ½ O2 (g) → N2H4 (g) + H2O (g) -143

H2 (g) + ½ O2 (g) → H2O (g) -286

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Bond Energies

• Chemical reaction ⇔ Bond breakage & bond formation

• Bond energy = energy required to break a bond

• Bond breaking is endothermic (raises potential energy)

• Bond formation is exothermic (lowers PE)

• Average energy for one type of bond in different molecules

• Common Bond Energies

C-H : 413 kJ/mol C=O : 799 kJ/mol

O=O : 495 kJ/mol O-H : 467 kJ/mol

• ΔHrxn = (bonds broken) – (bonds formed)

Energy required Energy released

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Bond Energies• ΔHrxn = (bonds broken) – (bonds formed)

ex. CH4 + 2 O2 CO2 + 2 H2O

H

C

H

HHO

HHO

HH

C

O

OO O

O O+ +

H

C

H

HHO

HHO

HH

C

O

OO O

O O+ +

C-H : 413 kJ/mol C=O : 799 kJ/molO=O : 495 kJ/mol O-H : 467 kJ/mol

ΔHrxn = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)]

ΔHrxn = [4(413 kJ) + 2(495 kJ)] – [2(799 kJ) + 4(467

kJ)]

ΔHrxn = -824 kJCompare to ΔHrxn = -802.5 kJ

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Enthalpy Summary• Enthalpy (ΔH) rxn = heat = q

• All rxns have some ΔH• ΔH = +…endo• ΔH = - …exo

• If given ΔH, can use stoich to quantify

• Three ways to estimate ΔH (if not given)• Hess’s Law

• known eq’s manipulated into desired eq. ΔHrxn

• Heats of Formation (ΔHf): values from appendix

• ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

• Bond energies• ΔHrxn = bonds broken – bonds formed