1 / 25

Topic 5 Energetics Thermochemistry

Topic 5 Energetics Thermochemistry. SL. 5.1 Exothermic reactions. Heat is produced and transferred to surroundings NaOH (s ) + H 2 O  NaOH ( aq ) + heat HCl + NaOH  NaCl + H 2 O + heat Neutralisation Wood + O 2  CO 2 + H 2 O + heat Combustion.

mechelle-hinton
Download Presentation

Topic 5 Energetics Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 5 EnergeticsThermochemistry SL

  2. 5.1 Exothermic reactions • Heat is produced and transferred to surroundings • NaOH(s) + H2O NaOH(aq) + heat • HCl + NaOHNaCl + H2O + heat Neutralisation • Wood + O2 CO2 + H2O + heat Combustion

  3. Exothermic- compare explosion

  4. Endothermic reactions • Heat is consumed from surroundings- it gets colder or you need to heat • Ba(OH)2(s) + 2 NH4SCN(s) + heat  Ba2+(aq) + 2 SCN-(aq) + 2 H2O(l) + NH3(aq)

  5. Enthalpy, H • H = internal energy. The total chemical energy of a system. Some of the energy is stored in chemical bonds.

  6. DH = enthalpy change • There is no “absolute zero” for enthalpy => enthalpy for a particular state cannot be measured but changes in enthalpy during reactions can be measured. • DH = Hproducts – Hreactants

  7. In the reaction 2 H2 (g) + O2 (g) 2 H2O (g) Hreactants- Enthalpyofreactants- 1856 kJ Hproducts- Enthalpyofproducts- 1370 kJ DH= Hproducts- Hreactants=1370-1856 = - 486 kJ Enthalpy, H 2 H2 + O2 - 486 kJ EXOTHERMIC 2 H2O

  8. In the reaction 1/2 N2(g) + O2 (g) NO2(g) DH= Hproducts- Hreactants= 33,9 kJ/mol Enthalpy, H NO2 33,9 kJ 1/2 N2 + O2 ENDTHERMIC

  9. Exothermic reaction CH4 + 2 O2  CO2 + 2H2O + heat Energy rich Energy poor • DH = (Energy poor) – (Energy rich) => negative value => In Exothermic reactions: DH < 0 => Gives more stable products => In Endothermic reactions DH > 0 => Gives more reactive products.

  10. DHo: standard enthalpy change of reaction Standard conditions: p =101.3 kPa, T =298 K Factors affecting DHo • The nature of the reactants and products • The amount • Changing state involves the enthalpy change • The temperature and pressure of the reaction surroundings

  11. 5.2 Calculation of Energy/Enthalpy Changes • Measurements: Open calorimeter Bomb calorimeter

  12. E = c. m. DT • E= energy • m = mass • DT = temperature change • c = specific heat capacity, different for all substances • E.g. 4.18 J/g*K for Water

  13. Example • The heat energy required to heat 50 g of water from 20oC to 60oC is: • E = 50*4.18*40 = 8364 J = 8.364 kJ

  14. Enthalpy changes • 2 Mg + O2 2 MgODH = -1202 kJ/mol Exothermic • The amount of energy released when 0.6 g of Mg is burnt? Mg m 0.6 g M 24.3 g/mol n 0.025 mol • 1202*0.025 = 30 kJ

  15. 5.3 Hess’s law • The principle of conservation of energy states that energy cannot be created or destroyed. • The total change in chemical potential (enthalpy change) must be equal to the energy gained or lost. • The total enthalpy change on converting a given set of reactants to a particular set of products is constant, irrespective of the way in which the change is carried out.

  16. C + ½ O2 CO DH1 CO +½ O2 CO2DH2 C + O2 CO2 DH3 = DH1+DH2 http://www.ausetute.com.au/hesslaw.html

  17. http://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HessLaw/Energy04.htmhttp://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HessLaw/Energy04.htm

  18. 5.4 Bond Enthalpies • Break chemical bonds requires energy => Endothermic process • Form chemical bonds => Exothermic process • Approximate enthalpy change, DH, can be calculated by looking at bonds being broken and formed in the reaction.

  19. Average bond enthalpies • gaseous molecule into gaseous atoms (not necessary the normal state) • approx in different molecules => not so precise data, but normally within 10 %

  20. CalculateDH for the reaction: 2 H2 (g) + O2 (g) 2 H2O (g) H-H 436 kJ/mol H-O 464 kJ/mol O=O 498 kJ/mol

  21. Enthalpy 4 H + 2 O + 1370 kJ 2 H2 + O2 The bondsof the reactantsare broken, enthalpy is needed 2 mol H-H = 2* 436= 872 kJ 1 mol O=O = 498 kJ Sum 1370 kJ is spent

  22. Enthalpy 4 H + 2 O + 1370 kJ 2 H2 + O2 - 1856 kJ 2 H2O 2.The free hydrogen and oxyden atoms form bondstocreate the products. The bond enthalpy is released 2*2 mol H-O = 4* 464= 1856 kJ is formed

  23. Enthalpy 4 H + 2 O + 1370 kJ 2 H2 + O2 - 1856 kJ 2 H2O + 486 kJ 3. The enthalpyof the productsare1856-1370 = 486 kJ lowerthan the reactants The excess enthalpy 486 kJ is releasedto the surroundings. Exothermicreaction The ”extra” enthalpyneeded(1370 kJ) is called ACTIVATION ENERGy

  24. N2 + 3 H2 2 NH3 (Energies involved (see data booklet)) • N≡N 944kJ/mol • H-H 436 kJ/mol • N-H 388 kJ/mol DH = (bonds broken) – (bonds formed) = (944 + 3*436) – (2*3*388) = -76 kJ/mol Exothermic • (If using other data DHf = -92kJ/mol)

More Related