Chapter 15 - Standard enthalpy change of a reaction

1. Chapter 15 - Standard enthalpy change of a reaction

2. Standard Enthalpy Changes Hana Amir and Madeley

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

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

5. 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 Changes of Reaction

6. 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) Standard Enthalpy change of Formation

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 Reactants Products ΔHѳf(reactants) ΔHѳf(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: ∆Hѳreaction =Σ∆Hѳf (Products)- Σ∆Hѳf (Reactants)

10. 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 Example

11. 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 .

12. ∆Hѳreaction =Σ∆Hѳf (Products)- Σ∆Hѳf (Reactants) ∆Hѳreaction = 3(-394) + 4(-286) -(-105) = -2221 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 > -227 kjmol -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

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. –1031.76 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) 6CO2(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) 3CO2(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-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

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

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! :)