1. Topic 15.1 - Standard enthalpy change of a reaction

2. Standard Enthalpy Changes Hana Amir and Madeley

3. Definition
Any reaction that depends on temperature, pressure and state
The enthalpy change happens when all reactants and products are in their standard state.

4. Some enthalpies Enthalpy of reaction Enthalpy of formation
Enthalpy of neutralization
Enthalpy of hydration
Enthalpy of combustion
Enthalpy of solution
Enthalpy of atomization

5. 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

6. 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)

7. 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)

8. Energy Cycle ΔHѳreaction ΔHѳf(reactants) ΔHѳf(products) Products
Elements

9. 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)

10. Example Calculate the enthalpy change for the reaction:
C3H8(g) + 5O2(g) CO2(g) + 4H2O(g)
Standard change of formation: ∆Hѳf /kJ mol-1
C3H8(g) : -105
CO2(g): -394
H2O(l): -286

11. Steps:
Write down the equation with the corresponding enthalpies of formation underneath:
C3H8(g) + 5O2(g) CO2(g) + 4H2O(g)
(-105) (-394) (-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 .

12. ∆Hѳreaction = 3(-394) + 4(-286) -(-105) = -2221 KJ mol-1
∆Hѳreaction = Σ∆Hѳf (Products) - Σ∆Hѳf (Reactants)
∆Hѳreaction = 3(-394) + 4(-286) -(-105)
= KJ mol-1

13. 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 > 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 13

14. 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

15. Answers!
1. – kJ mol–1
2. 3C (Graphite) + 3H2(g) + 1/2O2(g) CH3COCH3

16. Standard enthalpy change of Combustion

17. 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) CO2(g) + 7H2O(l)
The standard enthalpy of combustion is always negative

18. 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) CO2(g) + 4H2O(l)

19. Calculating standard enthalpy change
ΔHѳreaction
Reactants
Products
ΔHѳc(reactants)
ΔHѳc(products)
Combustion Products
ΣΔHѳc(reactants) = ΣΔHѳ(products) + ΔH
ΔH = ΣΔHѳc(reactants) - ΣΔHѳc(products)

20. Question!
May 2010 Paper 1 TZ 2B

21. 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 H2:ΔHѳ = -286 Kjmol-1
C6H12O6: ΔHѳ = Kjmol-1
ΔH = ΣΔHѳc(reactants) - ΔHѳc(products)
ΔH = ( 6(-394) + 6(-286) + 3(0) ) - (-2803)
= Kjmol-1

22. C(s, graphite) C(s, diamond)
Question!
Calculate the enthalpy change for the following reaction!
C(s, graphite) C(s, diamond)
C(s, graphite) + O2(g) CO2(g) ΔHѳ = -393 Kjmol-1
C(s, diamond) + O2(g) CO2(g) ΔHѳ = -395 Kjmol-1
Solution:
ΔHѳ
C(s, graphite) + O2(g)
C(s, diamond) + O2(g)
-393 Kjmol-1
-395 Kjmol-1
CO2(g)
-393 Kjmol-1 = -395 Kjmol-1 + ΔHѳ
ΔH = +2 Kjmol-1

23. Question
May 2008 Paper 1 TZ 1A

24. Question
Nov 2007 Paper 1 D

25. ΔH = ΣΔHѳf(products) - ΔHѳf(reactants)
Comparison
Standard enthalpy of combustion
Standard enthalpy of formation
ΔHѳreaction
ΔHѳreaction
ΔHѳf(products)
ΔHѳf(reactants)
Products
Reactants
Elements
ΔH = ΣΔHѳf(products) - ΔHѳf(reactants)
Reactants
Products
ΔHѳc(reactants)
ΔHѳc(products)
Combustion Products
ΔH = ΣΔHѳc(reactants) - ΔHѳc(products)
Its not CPR in chem! Its CRP!
Think First Price

26. Question!
May 2010 Paper 1 TZ 2A

27. Questions!
May 2010 Paper 1 TZ1A
May 2010 Paper 1 TZ1C

28. 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! :)