Transcript 15.1 Standard enthalpy change of a reaction

Chapter 15 - Standard enthalpy
change of a reaction
Standard Enthalpy Changes
Hana Amir and Madeley
Definition
•Any reaction that depends on temperature,
pressure and state
•The enthalpy change happens when all
reactants and products are in their standard
state.
Some enthalpies
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Enthalpy of reaction
Enthalpy of formation
Enthalpy of neutralization
Enthalpy of hydration
Enthalpy of combustion
Enthalpy of solution
Enthalpy of atomization
Standard Enthalpy Changes of
Reaction
•Definition:
The heat change when molar quantities of reactants as
specified by the chemical equation react to form products
at standard conditions
It depends in the physical state of reactants and the
products and the conditions under which the reaction
occurs.
Standard conditions are: 298K (25o C) and 1.00*105 Pa
Standard Enthalpy change of
Formation
Enthalpy change that occurs when one mole of substance
is formed 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)
• All elements in their standard states (oxygen
gas, 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)
Energy Cycle
Reactants
ΔHѳreaction
ΔHѳf(reactants)
Products
ΔHѳf(products)
Elements
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.
•Σ∆Hѳf (Products)= Σ∆Hѳf (Reactants) + ∆Hѳreaction
This gives the general expression of:
∆Hѳ reaction = Σ∆Hѳf (Products) - Σ∆Hѳf (Reactants)
Example
Calculate the enthalpy change for the reaction:
C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)
Standard change of formation: ∆Hѳf /kJ mol-1
C3H8(g) : -105
CO2(g): -394
H2O(l): -286
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 .
∆Hѳreaction = Σ∆Hѳf (Products) - Σ∆Hѳf (Reactants)
∆Hѳreaction = 3(-394) + 4(-286) -(-105)
= -2221 KJ mol-1
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
Questions!
1 .Use the table of standard enthalpies of formation at 25°C 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
Answers!
1. –1031.76 kJ mol–1
2. 3C (Graphite) + 3H2(g) + 1/2O2(g)-----------CH3COCH3
Standard enthalpy change of
Combustion
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
Exercise!
Write down the enthalpy of combustion equations for
the 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)
Calculating standard enthalpy change
ΔHѳreaction
Reactants
ΔHѳc(reactants)
Products
ΔHѳc(products)
Combustion Products
ΣΔHѳc(reactants) = ΣΔHѳ(products) + ΔH
ΔH = ΣΔHѳc(reactants) - ΣΔHѳc(products)
Question!
May 2010 Paper 1 TZ 2B
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: ΔHѳ = -394 Kjmol-1 H2:ΔHѳ = -286 Kjmol-1
C6H12O6: ΔHѳ = -2803 Kjmol-1
ΔH = ΣΔHѳc(reactants) - ΔHѳc(products)
ΔH = ( 6(-394) + 6(-286) + 3(0) ) - (-2803)
= -1277 Kjmol-1
Question!
Calculate the enthalpy change for the following reaction!
C(s, graphite)
C(s, diamond)
CO2(g) ΔHѳ = -393 Kjmol-1
CO2(g) ΔHѳ = -395 Kjmol-1
C(s, graphite) + O2(g)
C(s, diamond) + O2(g)
Solution:
C(s, graphite) + O2(g)
ΔHѳ
C(s, diamond) + O2(g)
-393 Kjmol-1
-395 Kjmol-1
CO2(g)
-393 Kjmol-1 = -395 Kjmol-1 + ΔHѳ
ΔH = +2 Kjmol-1
Question
May 2008 Paper 1 TZ 1A
Question
Nov 2007 Paper 1 D
Comparison
Standard enthalpy of combustion
Reactants
ΔHѳreaction
ΔHѳc(reactants)
Products
Standard enthalpy of formation
Reactants
ΔHѳreaction
Products
ΔHѳf(reactants)
ΔHѳc(products)
Combustion Products
ΔH = ΣΔHѳc(reactants) - ΔHѳc(products)
Its not CPR in chem! Its CRP!
ΔHѳf(products)
Elements
ΔH = ΣΔHѳf(products) - ΔHѳf(reactants)
Think First Price
Question!
May 2010 Paper 1 TZ 2A
Questions!
May 2010 Paper 1 TZ1A
May 2010 Paper 1 TZ1C
Question!
Nov 2009 Paper 1 TZ1C
Answers to all MCQ questions is the last letter
in the identification of the paper from which the
question was taken! :)