1 / 43

∆H = mcO/mol = 2940/1000x 0.05 = - 58.8 kJ mol -1

20) In an experiment, 50 cm 3 of 1 mol dm -3 NaCl is added to 50 cm 3 of 1 mol dm -3 AgNO 3 solution . The following data is obtained. Given Density of solution : 1 g cm -3 , specific heat of solution is 4.2 J g -1 o C -1. Precipitation reaction .

don
Download Presentation

∆H = mcO/mol = 2940/1000x 0.05 = - 58.8 kJ mol -1

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. 20) In an experiment, 50 cm3 of 1 mol dm-3 NaCl is added to 50 cm3 of 1 mol dm-3 AgNO3 solution . The following data is obtained Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1o C -1 Precipitation reaction a) What is the name of this reaction? mcO = 100 x 4.2 x 7 = 2940 J b) Calculate heat change c) Calculate heat of reaction ∆H = mcO/mol = 2940/1000x 0.05 = - 58.8 kJ mol-1 d) If NaCl solution is replaced with HCl solution, predict the heat of reaction . Justify your answer Heat of rex is the same ,because the precipitate is still the same, that is AgCl e) If NaCl solution is replaced with Na2CO3 solution, predict the heat of reaction . Justify your answer Heat of rex is the diff , because the precipitate is NOT the same, that is Ag2CO3

  2. 21) In an experiment, 100 cm3 of 1.0 mol dm-3 CaCl2 solution is added to 100 cm3 of 1.0 mol dm-3 Na2CO3 solution . The following data is obtained Ca 2+ + CO32- CaCO3∆H = + 12.6 kJ mol-1 Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1o C -1 Calculate the temperature change of the mixture ∆H = mcO / mol therefore O = ∆H x mol / mc = 12.6 x 1000 x 0.1 / 200 x 4.2 = 1.5 oC

  3. In an experiment, 1 g of zinc powder is added to 50 cm3 of 0.2 mol dm-3 CuSO4 solution . The following data is obtained Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1o C -1 Displacement reaction a) What is the name of this reaction? mcO = 50 x 4.2 x 5 = 1050 J b) Calculate heat change c) Calculate heat of reaction ∆H = mcO / mol = 1050 / 1000x 0.01 = - 105 kJ mol-1

  4. 23) In an experiment, 100 cm3 of 2.0 mol dm-3 HCl is added to 100 cm3 of 2.0 mol dm-3 NaOH solution . The following data is obtained Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1o C -1 a) What is the name of this reaction? Neutralisation reaction b) Calculate heat change mcO = 200 x 4.2 x 12 = 10080 J ∆H = mcO / mol = 10080 / 1000x 0.2 = - 50.4 kJ mol-1 c) Calculate heat of reaction

  5. 21) The thermochemical equation for the reaction between ethanoic acid and sodium hydroxide solution is given below CH3COOH + NaOH  CH3COONa + H2O ∆H = -55 kJ mol-1 Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1o C -1 Calculate the heat given out when 200 cm 3 of ethanoic acid is added to 100 cm3 of sodium hydroxide solution . Both solutions have a concentration of 0.5 mol dm-3 Mol of ethanoic acid = MV/1000 = 0.5 x 200/1000 = 0.1 mol Mol of sodium hydroxide = MV/1000 = 0.5 x 100/1000 = 0.05 mol ∆H = mcO/mol Therefore mcO =∆H x mol = 55x1000 x 0.05 = 2750 J

  6. 25) In an experiment, the following data is obtained [Given Density of water : 1 g cm-3 , specific heat capacity of water is 4.2 J g-1o C -1 RAM : H,1; C,12; O,16 ] Combustion a) What is the name of this reaction? b) Calculate heat change mcO = 100x 4.2 x 22 = 9240 J c) Calculate heat of reaction ∆H = mcO/mol = 9249 /1000x 0.0156 = - 592.3 kJ mol-1

  7. 25) The heat of combustion of ethanol is -1376 kJ mol-1 What is its fuel value? Fuel value = heat of combustion/ Molar mass Fuel value = 1376/ 46 = - 29.9 kJ g -1

  8. C3H7OH + O2 CO2 + H2O 9/2 3 4 Heat released = mcO =200 x 4.2 x 31 =26040 Joule

  9. Mole = mass/molar mass = 0.84/60 = 0.014 0.014 mole of propanol released heat 26040 joule So 1 mol propanol  1x 26040/0.014 =1860 kJ Heat of combustion = -1860 kJ mol-1

  10. = -1860 kJ mol-1

  11. Heat change ……………………………………….when 1 mol of …………………………………. ….. is formed from ……………………………………………..in solution Magnesium carbonate magnesium ions and carbonate ions

  12. Mg 2+ + CO32- MgCO3 mco = 50 x 4.2 x 5.5 = 1155J

  13. Mole = MV/1000 = 25 x2 /1000 = 0.05 mol No of mole of MgCO3 = no of mole of magnesium ion = 0.05 mol In thermochemistry, do not use from equation 0.05 mol of MgCO3 release heat 1155 J So 1 mole of MgCO3  1 x 1155/0.05 Joule = 23.1 kJ Heat of precipitation = + 23.1 kJ mol-1

  14. H= positive

  15. Same precipitate is formed, that is magnesium carbonate / sodium ion or potassium ion does not take part in the reaction

  16. Oxidised: iron (II) sulphate / Fe 2+ Reduced : Acidified potassium manganate(VII)

  17. Oxidation: Reduction:

  18. K Mn O4 = 0

  19. AIM OF EXPERIMENT Able to give the aim of the experiment correctly Sample answers To investigate the effect of Metal Mg and Cu when it in contact with iron in the rusting of iron// To study the effect of Metal Cu can faster the rusting of iron while metal Mg prevent the iron from rusting

  20. a. variable Manipulated variable : Metal X and metal Y //two different metals(one metals is less electropositive and one is more electropositive than iron) //pairs of metal X / iron and Y/iron

  21. Responding variable : Rusting of iron //iron rust //the formation of brown solid //formation of blue spot Constant variables: Iron nail //jelly solution //temperature

  22. c. hypothesis Able to state the relationship between the manipulated variable and the responding variable with direction correctly. Sample answer When a more electropositive metal is in contact with iron, the metal inhibits rusting. // When a less electropositive metal is in contact with iron, the metal speeds up rusting.

  23. d. Material & apparatus Materials/substances Two Iron nails, Magnesium/zinc/aluminium strip, tin/copper/lead/silver strip Potassium hexacyanoferrate(III) solution + phenolphthalein Apparatus Test tube/boiling tube, Sand paper

  24. e. procedure Sample answer: Clean the iron nails and metals strip with sand paper. Coil iron nails with magnesium ribbon and copper strips. Put/place the coiled iron nail into different test tube. Pour/add/fill the hot jelly solution containing potassium hexacyanoferrate(III)solution and phenolphthalein into the test tube.

  25. e. procedure Sample answer: Leave the test tube in a test tube rack for few days. Record the observation. Steps 1 to 6 are repeated using different metal/Y with iron(if steps 2 does not mention two different test tube).

  26. f. Tabulated data

  27. Revision redox

  28. Redox reaction or non redox? • Which one is REDOX? State your reason • AgNO3 + NaCl AgCl + Na NO3 • Cl2 +2 KI  2KCl + I2 Reaction a) is not a redox reaction because oxidation number of Ag does not change, that is from +1 to +1 Reaction b) is a redox reaction because oxidation number of Cl change, from 0 to -1

  29. State oxidation number for the underlined (+1 + Mn + 4O = 0 ) , (+1 + Mn -8 = 0 ), Mn = +7 (+1 + I = 0 ) , I = -1 • KMnO4 • K I • Fe Cl2 • I2 • Cu • CuSO4 • MnO4 – • H3O+ +2 0 0 +2 (Mn + 4O = -1) , Mn = +7 (3H -2 = + 1) , H = +1

  30. 0 oxidation +2 Mg + 2FeCl3 2FeCl2 + MgCl2 +2 reduction +3 Which is oxidized ? reason Which is reduced? Reason Which is oxidizing agent, why Which is reducing agent?why Mg because ON of Mg increase from 0 to +2 FeCl3 because ON of Fe decrease from +3 to +2 FeCl3 because it oxidise Mg to MgCl2 Mg because it reduce FeCl3 to FeCl2

  31. 0 reduction -1 Br2 + 2NaI  2NaBr + I2 oxidation -1 0 Which is oxidized ? reason Which is reduced? Reason Which is oxidizing agent, why Which is reducing agent? why Sodium iodide because ON of iodine increase from -1 to 0 Bromine because ON of Br decrease from 0 to -1 Bromine because it oxidise NaI to I2 NaI because it reduce Br2 to NaBr

  32. 0 +2 oxidation Zn + CuSO4 ZnSO4 + Cu +2 reduction 0 Which is oxidized ? reason Which is reduced? Reason Which is oxidizing agent, why Which is reducing agent? why Zinc because ON of zinc increase from 0 to +2 CuSO4 because ON of copper decrease from +2 to 0 CuSO4 because it oxidise Zn to ZnSO4 Zinc because it reduce CuSO4 to Cu

  33. Reaction A is NOT a redox reaction because Oxidation number of Na does not change, that is from +1 to +1 Reaction B is a REDOX reaction because Oxidation number of Mg change from 0 to +2

  34. In P, +2 In Q, +1 Q is copper(I) oxide because ON of copper is +1 P is copper(II) oxide Because ON of copper is +2

  35. +2 reduction 0 Oxidising agent Reducing agent 0 oxidation +1

  36. reduction +2 0 0 +1 oxidation Because ON of hydrogen increased from 0 to +1 H 2 Because ON of copper decreased from +2 to 0 CuO Because it oxidise hydrogen to hydrogen oxide CuO H 2 Because it reduce CuO to Cu

  37. State the fuction of • Glass wool • Potassium manganate (VII) powder Prevent metal from mixing with KMnO4 ( can cause explosion) release oxygen when heated to react with metal powder

More Related