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Chapter 15 - Standard enthalpy
change of a reaction
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Standard Enthalpy ChangesHana Amir and Madeley
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Definition
Any reaction that depends on temperature,
pressureandstate
The enthalpy change happens when all
reactants and products are in their standard
state.
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Some enthalpies
Enthalpy of reaction
Enthalpy of formation
Enthalpy of neutralization
Enthalpy of hydration
Enthalpy of combustion
Enthalpy of solution Enthalpy of atomization
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Standard Enthalpy Changes of
ReactionDefinition:
The heat change when molar quantities of reactants asspecified by the chemical equation react to form productsat standard conditions
It depends in the physical state of reactants and theproducts and the conditions under which the reactionoccurs.
Standard conditions are: 298K (25oC) and 1.00*105Pa
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Standard Enthalpy change of
Formation
Enthalpy change that occurs when one mole of substanceisformed from its elements in the standard state under
standard conditions.
Standard conditions
Temperature: 298K (25o C)
Pressure: 1.00*105 Pa
2C(graphite) + 3H2(g) + 1/2O2(g)-------> C2H5OH (I)
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All elements in their standard states (oxygengas, solid carbon in the form of graphite, etc.)
have a standard enthalpy of formation of zero,
as there is no change involved.
Eg:
O(g) + O(g) ---->O2(g)
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Energy Cycle
Hreaction
Hf(products)Hf(reactants)
ProductsReactants
Elements
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From the diagram we get:
The chemical change elements to products
can either occur directly or indirectly .
The Total enthalpy change must be the same for both routes.
Hf (Products)=Hf (Reactants) +Hreaction
This gives the general expression of:
Hreaction =Hf (Products)- Hf (Reactants)
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Example
Calculate the enthalpy change for the reaction:
C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)
Standard change of formation: Hf /kJ mol-1
C3H8(g) : -105
CO2(g): -394H2O(l): -286
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Steps:
Write down the equation with the corresponding
enthalpies of formation underneath:
C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)
(-105) 0 3(-394) 4(-286)
Note: The standard enthalpies of formation are given in
per mole , hence, they should be multiplied by the
numbers of moles in the balance equation .
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Hreaction =Hf (Products)- Hf (Reactants)
Hreaction =3(-394) + 4(-286) -(-105)
= -2221 KJ mol-1
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Thermochemical Equations
Balanced chemical equation for a reaction including the enthalpy of the reaction
shown immediately after the equation.In a thermochemical equation, the coefficients represent moles and can
therefore be fractional. The following is an example
IB Data booklet > -227 kj mol -1
Ethanol (C2H5OH) is made from the elements (C (Graphite)) and hydrogen
(H2(g)) and oxygen (O2(g))
__C (graphite) + __H2(g) +___O2(g)-----------> C2H5OH(I) H=227kj mol -1
Balance the C, H and 0
2C (graphite) + 3H2(g) +1/2 O2(g)-------------- C2H5OH(I) H=227 kj mol -1
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Questions!
1 .Use the table of standard enthalpies of formation at 25C to calculate
enthalpy change for the reaction
4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g)
2 .Write the thermodynamically equation for the standard enthalpy of
formation of propanone enthalpy change
CH3COCH3
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Answers!
1.1031.76 kJ mol1
2. 3C (Graphite) + 3H2(g) + 1/2O2(g)-----------CH3COCH3
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Standard enthalpy change of
Combustion
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What is standard enthalpy change of
combustion?
The standard enthalpy of combustion is the
enthalpy change that occurs when one mole of
substance burns completely under the standard
conditions of 25 and 1 atm.
Eg: C6H14(l) + 9O2(g) 6CO2(g) + 7H2O(l)
The standard enthalpy of combustion is always negative
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Exercise!
Write down the enthalpy of combustion equations forthe following reactions!
Methane
CH4(g)+ 2O2(g) CO2(g) + 2H2O(l)
Ethanol
C2H5OH(l) + 3O2(g) CO2(g) + 2H2O(l)
Propane
C3H8(g)+ 5O2(g) 3CO2(g) + 4H2O(l)
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Calculating standard enthalpy change
Hreaction
Hc(products)Hc(reactants)
ProductsReactants
Combustion Products
H
c(reactants) = H
(products) + H
H = Hc(reactants) - H
c(products)
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Question!
May 2010 Paper 1 TZ 2B
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Question!
Give an equation for the formation of glucose.6C(graphite)+ 6H2(g) + 3O2(g) C6H12O6(s)
Calculate the enthalpy of formation of glucose
C: H
= -394 Kjmol-1 H
2:H
= -286 Kjmol-1
C6H12O6: H
= -2803 Kjmol-1
H = Hc(reactants) - H
c(products)
H = ( 6(-394) + 6(-286) + 3(0) ) - (-2803)= -1277 Kjmol-1
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Question!
Calculate the enthalpy change for the following reaction!
C(s, graphite) C(s, diamond)
C(s, graphite) + O2(g) CO2(g) H
= -393 Kjmol-1
C(s, diamond) + O2(g) CO2(g) H
= -395 Kjmol-1
Solution:
C(s, graphite) + O2(g) C(s, diamond) + O2(g)
CO2(g)
H
-395 Kjmol-1-393 Kjmol-1
-393 Kjmol-1= -395 Kjmol-1+ H
H = +2 Kjmol-1
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Question
May 2008 Paper 1 TZ 1A
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Question
Nov 2007 Paper 1 D
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Comparison
Hreaction
Hc(products)Hc(reactants)
ProductsReactants
Combustion Products
Standard enthalpy of combustion Standard enthalpy of formation
H = Hc(reactants) - H
c(products)
Hreaction
Hf(products)
Hf(reactants)
ProductsReactants
Elements
H = Hf(products) - H
f(reactants)
Its not CPR in chem! Its CRP! Think First Price
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Question!
May 2010 Paper 1 TZ 2A
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Questions!
May 2010 Paper 1 TZ1C
May 2010 Paper 1 TZ1A
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Question!
Nov 2009 Paper 1 TZ1C
Answers to all MCQ questions is the last letterin the identification of the paper from which the
question was taken! :)