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Title: Lesson 3 Hess’s Law and Enthalpy of Formation and Combustion Learning Objectives: – Define Hess’s Law – Using calculations show that enthalpies.

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Presentation on theme: "Title: Lesson 3 Hess’s Law and Enthalpy of Formation and Combustion Learning Objectives: – Define Hess’s Law – Using calculations show that enthalpies."— Presentation transcript:

1 Title: Lesson 3 Hess’s Law and Enthalpy of Formation and Combustion Learning Objectives: – Define Hess’s Law – Using calculations show that enthalpies for formation of products are the same regardless of the route taken – Calculate change in enthalpy of reactions using enthalpy of formation or combustion data

2 Main Menu Refresh  How much energy, in joules, is required to increase the temperature of 2.0 g of aluminium from 25 to 30 °C? (Specific heat of Al = 0.90 J g –1 K –1 ). A. 0.36 B. 4.5 C. 9.0 D. 54

3 Main Menu Calculations linked to Enthalpy Cycles (Hess’s Law)  Enthalpy of Formation  Enthalpy of Combustion  Bond Enthalpy  You must know the difference between these!

4 Main Menu Hess’ Law  The enthalpy change of a reaction is independent of the pathway of that reaction  i.e. All that matters is the start and finish points  Note: add when going ‘with’ an arrow, subtract when going against an arrow. A + BC + D E + F ∆H 1 ∆H 2 ∆H 3 ∆H 1 = ∆H 2 + ∆H 3 A + BC + D E + F ∆H 1 ∆H 2 ∆H 4 ∆H 1 = ∆H 2 - ∆H 3 + ∆H 4 G + H ∆H 3 The first law of thermodynamics relates to the conservation of energy. It is sometimes expressed in the following form: Energy cannot be created or destroyed, it can only change form.

5 Main Menu Why is this useful  It is not always possible to directly measure the enthalpy change we want.  It may be an endothermic reaction that needs a constant heat supply  It may be that the reaction doesn’t ‘stop’ where you need it to  It may be that the reaction is simply too slow  Hess cycles allow us to measure enthalpy changes indirectly

6 HRHR HESS’S LAW THE TOTAL ENTHALPY CHANGE OF A REACTION H1H1 REACTANTS PRODUCTS INTERMEDIATES H2H2 By Hess’s Law:  H R = Intermediates = IS INDEPENDENT OF THE REACTION ROUTE. oxides if ΔH C known, or gaseous atoms if E[X-Y] (Bond Enthalpy) known elements if ΔH f known, Alternative route  H 1 +  H 2 Direct measurement of ΔH R may not be possible because : 1. Reaction incomplete 2. Other reactions occur 3. Reaction too slow EXAMPLE OF CYCLE 1

7 HRHR HESS’S LAW H1H1 REACTANTS PRODUCTS INTERMEDIATES H2H2 Therefore By Hess’s Law:  H R = Intermediates = oxides if ΔH C known, or gaseous atoms if E[X-Y] (Bond Enthalpy) known elements if ΔH f known, Alternative route -  H 1 +  H 2 We can consider situations where the conversions are the other way round. In this case the intermediate will convert to both reactants and products. (Look at the direction of the arrows!) EXAMPLE OF CYCLE 2 Pathways: I  R =  H 1 I  P =  H 2

8 HRHR HESS’S LAW H1H1 REACTANTS PRODUCTS INTERMEDIATES H2H2 Intermediates = oxides if ΔH C known, elements if ΔH f known, Alternative route We can consider situations where the conversions are the other way round. In this case the reactants and products will both convert to intermediates. (Look at the direction of the arrows!) Therefore By Hess’s Law:  H R =  H 1 + -  H 2 EXAMPLE OF CYCLE 3 or gaseous atoms if E[X-Y] (Bond Enthalpy) known Pathways: R  I =  H 1 P  I =  H 2

9 LEARN!!! STANDARD ENTHALPY CHANGE OF FORMATION,  H  f The heat change when ONE MOLE of a substance is FORMED from its ELEMENTS in their standard states at 298K and 100kPa NB  H o f [ELEMENT in its standard state] = ZERO Q P C STANDARD ENTHALPY CHANGE OF COMBUSTION,  H  C The heat produced when ONE MOLE of a substance is burned in excess oxygen measured at 298K and 100kPa Q P C BOND ENTHALPY or BOND ENERGY E[X-Y] The average heat needed when ONE MOLE of covalent bonds are broken, measured in the gaseous state at 298K and 100kPa Q P C ELEMENTS OXIDES GAS ATOMS

10 Hess’s Law states that the total enthalpy change is independent of the route taken 2NO 2(g) 2NO (g) + O 2(g) N 2(g) +2O 2(g) ΔH r = -66.4kJmol -1 Route 1 Route 2 -180.8kJ+114.4kJ Route 2 ΔH r = +114.4 + (-180.8) = -66.4kJmol -1 The total enthalpy change for route 1 is the same as for route 2 7.4 HESS’S LAW http://www.youtube.com/watch?v=y 01ePN0Hr-Y This is a thermo- chemical cycle

11 Enthalpy Cycles and Hess’s Law Think of a hike to your nice wood cabin in the French Alps. The wood cabin is your destination – but you can either walk through the forest or over the mountains. Either way you will still end up at the cabin!

12 Calculate the potential energy of each climber taking route 1 and route 2 -137KJ +125KJ +87KJ -193KJ +102KJ -163KJ +52KJ -147KJ -269KJ +7KJ Regardless of the route the climber and the miner took they ended up having the same amount of potential energy!!

13 Main Menu Using Hess’s Law  Although carbon and hydrogen do not combine directly to form propane, C 3 H 8, the enthalpy change for the reaction is:  This can be calculated from the enthalpy of combustion data of the elements and the compound  The steps in the cycle may be hypothetical and may refer to reactions that do not take place. The only requirement is that the individual chemical reactions in the sequence must balance.

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18 Main Menu Complete the Hess’s Law Worksheet


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