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Energy and Fuels Chemical reactions releasing energy Exothermic!

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Presentation on theme: "Energy and Fuels Chemical reactions releasing energy Exothermic!"— Presentation transcript:

1 Energy and Fuels Chemical reactions releasing energy Exothermic!

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3 Exothermic Energy released is measured in joules (J) Or Calories (1 cal = 4.2J) – hence ‘Calorimitry’ Q= mc Δ T Q is energy transferred(J/g), m is mass of water (g), c is ‘specific heat capacity’ of water (J/g o C) and Δ T is the temperature change Multiply Q by RAM and you know the energy released in J/mol or kJ/mol

4 Calorimitry Issues Energy will be lost to the surroundings, so most simple Calorimitry is wildly inaccurate!

5 Energy Changes in Reactions Just use the same equation! Q= mc Δ T Any reactions using liquids will be assumed to behave like water. (SHC of 4.2 J/g) You will be given the information to plug into an equation…. You may have to do some working Q Δ TΔ Tmc

6 Where does the Energy come from? – Bond Energy Bond Energy is the energy needed to break/make a bond The difference in bond energy between reactants and products shows us whether the reaction is Exo or Endo thermic

7 Q1. Hydrogen could be the fuel used in all cars. One advantage is that when hydrogen reacts with oxygen only water is produced. The chemical equation for this reaction is: 2 H 2 + O 2 2H 2 O This equation can be written showing the structural formulae. 2 H ― H + O ═ O 2 H ― O ― H (a) Use the bond energies in the table to calculate the energy change for this reaction......................................................................................................................... Energy change =.............................................. kJ BondBond Energy KJ H ― H436 O ═ O498 O ― H464

8 Reactant Bond Energy Product Bond Energy Energy Released Energy EXOTHERMIC

9 Reactant Bond Energy Product Bond Energy Energy Required Energy ENDOTHERMIC

10 Strike a match?

11 Reactant Bond Energy Product Bond Energy Energy Released Energy EXOTHERMIC Activation Energy

12 Catalytic Convertor

13 Reactant Bond Energy Product Bond Energy Energy Released Energy EXOTHERMIC Activation Energy lowered by a catalyst

14 Polluted Water Supply

15 What….. Positive Ions Flame tests Hydroxide Tests Negative Ions Limewater test (carbonates) Nitric Acid / Silver Nitrate test (Halides) Ionic equations… Ie Cu 2+ (aq) +2OH - (aq) Cu(OH) 2(s)

16 How much… Idea… If a known volume of a know concentration of acid is neutralised by a known volume of an unknown concentration of alkali, we can work out the unknown concentration This means we need a very precise indicator

17 Concentration mol/dm 3 mol dm -3 dm 3 =10x10x10=1000 cm 3 =1000ml=1litre! So… mol Vol.Conc.

18 How Much…. (b) The acid in 25.0 cm 3 of the champagne reacted completely with 13.5 cm 3 of sodium hydroxide of concentration 0.10 moles per cubic decimetre. Calculate the concentration in moles per cubic decimetre of acid in the champagne. Assume that 1 mole of sodium hydroxide reacts completely with 1 mole of acid..................................................................................................................................... Concentration =......................... moles per cubic decimetre

19 Key info in question… (b) The acid in 25.0 cm 3 of the champagne reacted completely with 13.5 cm 3 of sodium hydroxide of concentration 0.10 moles per cubic decimetre. Calculate the concentration in moles per cubic decimetre of acid in the champagne. Assume that 1 mole of sodium hydroxide reacts completely with 1 mole of acid..................................................................................................................................... Concentration =......................... moles per cubic decimetre

20 Working 1.1 mol NaOH to 1 mol acid 2.If we know number of mols of NaOH, we know the number of mols of acid 3.We have 13.5cm 3 of NaOH at a conc. of 0.10mol per dm 3 (1000cm 3 ) 4.0.10/1000 = 0.0001 mol in 1cm 3 5.0.0001 x 13.5 = 0.00135 mol NaOH in total 1.If we have 0.00135 mol NaOH, we have 0.00135 mol acid also. 2.Volume is 25cm 3 3.So, 0.00135/25= 0.000054 mol in 1cm 3 4.0.0054 x 1000 = 0.054 mol dm -3

21 Order Conc of substance 1 in mol dm -3 Number of mol in 1cm 3 (conc/volume) multiply number of mol by volume to give mol in known volume Find ratio of one substance to the other Use ratio to find number of mol in substance 2 Divide number of mol by volume to get mol in 1cm 3 Multiply number of mol by 1000 to get mol dm -3 of substance 2!

22 What next… Internet research to complete the analysis sheet you have been given Use this presentation to complete the questions in your booster booklet Remember focus on the skills in rearranging equations to work out titrations Don’t worry too much about why This is one small section of C3!


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