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5.1 Exothermic reactions

Topic 5- energetics SL 5.1 Exothermic and endothermic reactions 5.2 Calculation of Energy/Enthalpy Changes 5.3 Hess law 5.4 Bond enthalpy. 5.1 Exothermic reactions. Heat is produced and transferred to surroundings NaOH (s ) + H 2 O  NaOH ( aq ) + heat

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5.1 Exothermic reactions

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  1. Topic 5-energetics SL5.1 Exothermic and endothermic reactions5.2Calculation of Energy/Enthalpy Changes5.3 Hess law5.4 Bond enthalpy

  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 Hproducts- Enthalpyofproducts DH= Hproducts- Hreactants= - 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 ENDOTHERMIC

  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. DHfo: standard enthalpy of formation (cf. page 8 in Data Booklet) The enthalpy difference for the reaction when the substance is formed from it’s elements under standard conditions.

  12. 5.2 Calculation of Energy/Enthalpy Changes • Measurements: Open calorimeter Bomb calorimeter

  13. Calculation of heat of solution • q = c . M . DT (page 1 in Data Booklet) • q= energy (J) • m = mass (g) • DT = temperature change (K) • c = specific heat capacity, different for all substances • E.g. 4.18 J/g*K for Water

  14. Example • The heat energy required to heat 50 g of water from 20oC to 60oC is: q = 50*4.18*(60-40) = 8364 J = 8.364 kJ Energy and heat are always positive

  15. Enthalpy changes Mg + ½ O2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 q=1202*0.025 = 30 kJ (energy is always positive)

  16. 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.

  17. http://www.ausetute.com.au/hesslaw.html C + ½ O2  CO DH1=-283,0 kJ CO +½ O2  CO2 DH2=-110,5 kJ C + O2  CO2 DH3 = DH1+DH2= -393, 5 kJ

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

  19. 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.

  20. 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 % • cf. page 7 in Data Booklet

  21. 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

  22. 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

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

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

  25. N2(g) + 3 H2(g)2 NH3 (g) (Enthalpies involved (see data booklet page 7)) N≡N 945kJ/mol H-H 436 kJ/mol N-H 391 kJ/mol DH = (bonds broken) – (bonds formed) = (945 + 3*436) – (2*3*391) = -93 kJ/mol Exothermic (If using other data DHf = -92kJ/mol)

  26. Enthalpy 6 H + 2 N + 2253 kJ N2+ 3 H2 -2346 kJ 2 NH3 + 93 kJ N2 + 3 H2 2 NH3 The enthalpyof the productsare2346-2253 = 93 kJ lowerthan the reactants The excess enthalpy96 kJ is releasedto the surroundings. Exothermicreaction, DH= -93 kJ/mol

  27. 1/2 N2(g) + O2 (g) NO2(g) Enthalpy 2 O + N ENDOTHERMIC -812 kJ + 970,5 kJ The sumof the bondenthalpies NO2 ½ N2+ O2 - 93 kJ The freenitrogen and oxygen atoms form bondstocreate the products. The bond enthalpyis released. 1 mol N-O = 222kJ 1mol N=O = 590 kJ 812 kJ is formed The bondsof the reactantsare broken, enthalpy is needed 1/2 mol N≡N 1/2* 945= 472,5 kJ 1 mol O=O = 498 kJ Sum970,5 kJ is spent The resultingenthalpy, 93 kJ is taken from the surroundings. Endothermicreaction, DH= 93 kJ/mol

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