Enthalpy (H)
-
Upload
jessica-ashley -
Category
Documents
-
view
233 -
download
11
description
Transcript of 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
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
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
PE
Reaction Coordinate Diagrams: Endothermic Reaction
Progress of the Reaction
Reactants
ProductsD(PE)
Ea
ΔHrxn= +
Activation Energy
PE
Reaction Coordinate Diagrams: Exothermic Reaction
Progress of the Reaction
Reactants
Products
D(PE)
Ea
Activation Energy
ΔHrxn= -
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
• 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
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)
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
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
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 ( )
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
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 ( )
*
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( )-( )
***
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( )
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)
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
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)
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)
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 = ?
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)?
Thermite Reaction
Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)
ΔHrxn = ?Welding railroad tracks
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)
∆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 = ?
(-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 = ?
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
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
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
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