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 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 ΔHѳreaction ΔHѳf(reactants) ΔHѳf(products) 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 = 3(-394) + 4(-286) -(-105) = -2221 KJ mol-1 ∆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 13

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 Products ΔHѳc(reactants) 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

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

Question May 2008 Paper 1 TZ 1A

Question Nov 2007 Paper 1 D

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

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! :)