Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
SlideShare a Scribd company logo

Seminar revision on chapter electrchemistry, carbon compound and thermo chemistry

•Download as PPT, PDF•
•
MRSMPC
MRSMPC

1) The document summarizes topics on electrochemistry, carbon compounds, and thermochemistry covered in a seminar for chemistry form 5 students. It includes diagrams of electrolytic and voltaic cells and questions related to these topics. 2) Organic reactions involving ethene, ethanol and ethanoic acid are summarized including reagents, conditions, and equations. 3) Data from an experiment to determine the heat of precipitation of PbSO4 is presented and used to calculate the heat of precipitation.

1 of 49
Seminar chemistry form 5 ( Revision for topics) i) Electrochemistry ii) Carbon compound iii) Thermo chemistry Date : 5 April 2011 ( Tuesday)
1)
Refer to the diagram above: Electrolytic cell Anode (positive electrode) Cathode (negative electrode) oxygen gas Copper metal Cell P Electrode M Electrode N Name of cell Label anode and cathode State products
Bubble of colourless gas that ignites the glowing wooden splinter 4OH - O 2 + 2H 2 O + 4e Brown solid is formed Cu 2+ +2e Cu State observations Write ionic equations
Electrical energy Chemical energy Cell P State change of energy
Refer to the diagram above: Voltaic cell ( Daniel cell) Anode (NEGATIVE ELECTRODE) Cathode ( POSITIVE ELECTRODE) Zinc ion Copper metal Cell Q Electrode R Electrode S Name of cell Label anode and cathode State products
Zinc Electrode becomes thinner Copper electrode becomes thicker Zn Zn 2+ + 2e Cu 2+ + 2e Cu State observations Write ionic equations
Chemical energy Electrical energy Cell Q State change of energy
2)
Answer the following questions K + , H + , Cl - , OH - Cathode Anode Electrode P Electrode Q a)State all ions present b)Label anode and cathode
H + Cl - Hydrogen gas Chlorine gas K + ,H + Cl - ,OH - Cathode Anode b)Label anode and cathode c)State ions attracted to d)State ions discharged e)State products and what is the factor involved?
Hydrogen gas Chlorine gas Position of ions in ECS Concentrations of ions in solution e)State products and what is the factor involved? Factor involved
Position of hydrogen ion is lower than potassium ion in Electrochemical Series Chloride ion is more concentrated than hydroxide ion f)Give a reason why you have the product as mentioned above?
2H + + 2e H 2 2Cl - Cl 2 + 2e g)Write ionic equations
-Bubble of colourless gas is released. - Bring a lighted wooden splinter to the mouth of test tube, a ‘pop’ sound is produced -Bubble of yellowish gas is released. -Bring a piece of moist blue litmus paper to the mouth of the test tube. -Blue litmus paper turns red then bleached h)State observations and how do you test the products?
Hydrogen gas. Position of ion in Electro chemical series Oxygen gas. Position of ion in Electro chemical series i)If the experiment is repeated with concentration of solution 0.001 mol dm -3 . State the products . And what is the factor involved?
3) Organic reaction Ethene ethanol Ethyl ethanoate CO 2 + H 2 O Ethanoic acid Reaction 2 Reaction 1 Reaction 4 Reaction 3 Reaction 5 Reaction 6
  A) Refer to the above reaction: Combustion Excess oxygen Reaction 1 Name of reactions b)Reagents used
Heat in excess oxygen C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O Reaction 1 c)Conditions (if any) d)Equations
  A) Refer to the above reaction: Dehydration Porcelain chip (Al 2 O 3 ) Reactions 2 Name of reactions b)Reagents used
Pass ethanol over heated porcelain chips C 2 H 5 OH C 2 H 4 + H 2 O Reaction 2 c)Conditions (if any) d)Equations
  A) Refer to the above reaction: Hydration Steam Reactions 3 Name of reactions b)Reagents used
-Catalyst,H 3 PO 4 -Temperature,300 o C -Pressure, 60atm C 2 H 4 + H 2 O C 2 H 5 OH Reaction 3 c)Conditions (if any) d)Equations
  A) Refer to the above reaction: Esterification Glacial Ethanoic acid and concentrated sulphuric acid as the catalyst Reactions 4 Name of reactions b)Reagents used
warm the mixture C 2 H 5 OH+ CH 3 COOH CH 3 COOC 2 H 5 + H 2 O Reaction 4 c)Conditions (if any) d)Equations
  A) Refer to the above reaction: Esterification Absolute Ethanol and concentrated sulphuric acid as the catalyst Reactions 5 Name of reactions b)Reagents used
warm the mixture C 2 H 5 OH+ CH 3 COOH CH 3 COOC 2 H 5 + H 2 O Reaction 5 c)Conditions (if any) d)Equations
  A) Refer to the above reaction: oxidation Acidified K 2 Cr 2 O 7 or acidified KMnO 4 Reactions 6 Name of reactions b)Reagents used
warm the mixture C 2 H 5 OH+ 2[O] CH 3 COOH + H 2 O Reaction 6 c)Conditions (if any) d)Equations
B)Draw diagram for experiment to carry out for reaction 2
Describe the coagulation of latex when exposed to air
 
 
 
5) A student obtained the data to determine heat of precipitation of PbSO4 Highest temperature : 30.5 0 C Solution Vol (cm3) Conc (moldm-3) Initial temp ( 0 C) Pb(NO 3 ) 2 50 0.5 27.4 Na 2 SO 4 50 0.5 27.6
What is meant by heat of precipitation? Heat changed when 1 mole of precipitate is formed from its ions in an aqueous solution
b) Calculate heat of precipitation of PbSO4 Mole Pb 2+ : (0.5)(50)/1000 = 0.025 mol Mole SO 4 2- : (0.5)(50)/1000 = 0.025 mol Pb 2+ + SO 2- 4 PbSO 4 mole PbSO 4 = Mole Pb 2+ or mole SO 2- 4 = 0.025 mol
ϴ = 30.5 - ( 27.4 + 27.6 ) 2 = 3.0 ° C Heat of precipitation = mC ϴ mol = ( 100)(4.2)(3) 0.025 = 50400 J/mol ∆ H = - 50.4 KJ/mol
d)Write thermo chemical equation Pb 2+ + SO 4 2- PbSO 4 ∆ H = - 50.4 KJ/mol
e) Write the ionic equation Pb 2+ + SO 4 2- PbSO 4
E) Construct energy level diagram Pb 2+ + SO 4 2- Energy ∆ H = - 50.4 KJ/mol PbSO 4
f) The experiment is repeated using K 2 SO 4 to replaced Na 2 SO 4 . Heat of precipitation of PbSO 4 remain the same. Explain. Because the same precipitate is formed, which is PbSO 4. Only Pb 2+ ions and SO 4 2- ions react Na + ions and K + ions do not react
6) A student carried out an experiment to determine heat of displacement of copper from CuSO4 solution. He added excess zinc powder to 50 cm 3 of 0.2 moldm -3 CuSO 4 . The thermo chemical equation is shown below : Zn + Cu 2+ Zn 2+ + Cu ∆ H = -80.64 KJ/mol
a) Calculate the change in temperature Mol Copper= mol copper(II) sulphate = (0.2)(50)/1000 = 0.01 mol ∆ H = mC ϴ mol 80 640 J = ( 50)(4.2)( ϴ ) 0.01 ϴ = 3.8 ° C
b) Write the ionic equation Zn + CuSO 4  Cu + ZnSO 4 Zn + Cu 2+  Zn 2+ + Cu
The experiment is repeated with the following changes. What is the effect in the change of temperature when : Concentration of CuSO 4 is doubled, without changing the volume : So, change of temp or ϴ is doubled. Because as the concentration doubled, the number of particle per unit volume also doubled.
The experiment is repeated with the following changes. What is the effect in the change of temperature when : volume of CuSO 4 is halved, without changing the concentration : So, change of temp or ϴ is remain the same. Because, the changes in volume do not affect the number of particles per unit volume
7) State the diff betw heat change and the heat of reaction Heat absorb , heat released -Heat of Precipitation -Heat of displacement -Heat of neutralisation -Heat of combustion HEAT CHANGE HEAT OF REACTION A) OTHER NAMES Depends on name of reactions :
7) State the differences between heat change and heat of reaction HEAT CHANGE HEAT OF REACTION B) FORMULA USED H= mC ϴ ∆ H = mC ϴ mol C)UNIT Joule Kilo Joule/ mol D) SYMBOL none ∆ H E) SIGN No sign Either + for endothermic rex or – for exothermic rex

More Related Content

Seminar revision on chapter electrchemistry, carbon compound and thermo chemistry

  • 1. Seminar chemistry form 5 ( Revision for topics) i) Electrochemistry ii) Carbon compound iii) Thermo chemistry Date : 5 April 2011 ( Tuesday)
  • 2. 1)
  • 3. Refer to the diagram above: Electrolytic cell Anode (positive electrode) Cathode (negative electrode) oxygen gas Copper metal Cell P Electrode M Electrode N Name of cell Label anode and cathode State products
  • 4. Bubble of colourless gas that ignites the glowing wooden splinter 4OH - O 2 + 2H 2 O + 4e Brown solid is formed Cu 2+ +2e Cu State observations Write ionic equations
  • 5. Electrical energy Chemical energy Cell P State change of energy
  • 6. Refer to the diagram above: Voltaic cell ( Daniel cell) Anode (NEGATIVE ELECTRODE) Cathode ( POSITIVE ELECTRODE) Zinc ion Copper metal Cell Q Electrode R Electrode S Name of cell Label anode and cathode State products
  • 7. Zinc Electrode becomes thinner Copper electrode becomes thicker Zn Zn 2+ + 2e Cu 2+ + 2e Cu State observations Write ionic equations
  • 8. Chemical energy Electrical energy Cell Q State change of energy
  • 9. 2)
  • 10. Answer the following questions K + , H + , Cl - , OH - Cathode Anode Electrode P Electrode Q a)State all ions present b)Label anode and cathode
  • 11. H + Cl - Hydrogen gas Chlorine gas K + ,H + Cl - ,OH - Cathode Anode b)Label anode and cathode c)State ions attracted to d)State ions discharged e)State products and what is the factor involved?
  • 12. Hydrogen gas Chlorine gas Position of ions in ECS Concentrations of ions in solution e)State products and what is the factor involved? Factor involved
  • 13. Position of hydrogen ion is lower than potassium ion in Electrochemical Series Chloride ion is more concentrated than hydroxide ion f)Give a reason why you have the product as mentioned above?
  • 14. 2H + + 2e H 2 2Cl - Cl 2 + 2e g)Write ionic equations
  • 15. -Bubble of colourless gas is released. - Bring a lighted wooden splinter to the mouth of test tube, a ‘pop’ sound is produced -Bubble of yellowish gas is released. -Bring a piece of moist blue litmus paper to the mouth of the test tube. -Blue litmus paper turns red then bleached h)State observations and how do you test the products?
  • 16. Hydrogen gas. Position of ion in Electro chemical series Oxygen gas. Position of ion in Electro chemical series i)If the experiment is repeated with concentration of solution 0.001 mol dm -3 . State the products . And what is the factor involved?
  • 17. 3) Organic reaction Ethene ethanol Ethyl ethanoate CO 2 + H 2 O Ethanoic acid Reaction 2 Reaction 1 Reaction 4 Reaction 3 Reaction 5 Reaction 6
  • 18.   A) Refer to the above reaction: Combustion Excess oxygen Reaction 1 Name of reactions b)Reagents used
  • 19. Heat in excess oxygen C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O Reaction 1 c)Conditions (if any) d)Equations
  • 20.   A) Refer to the above reaction: Dehydration Porcelain chip (Al 2 O 3 ) Reactions 2 Name of reactions b)Reagents used
  • 21. Pass ethanol over heated porcelain chips C 2 H 5 OH C 2 H 4 + H 2 O Reaction 2 c)Conditions (if any) d)Equations
  • 22.   A) Refer to the above reaction: Hydration Steam Reactions 3 Name of reactions b)Reagents used
  • 23. -Catalyst,H 3 PO 4 -Temperature,300 o C -Pressure, 60atm C 2 H 4 + H 2 O C 2 H 5 OH Reaction 3 c)Conditions (if any) d)Equations
  • 24.   A) Refer to the above reaction: Esterification Glacial Ethanoic acid and concentrated sulphuric acid as the catalyst Reactions 4 Name of reactions b)Reagents used
  • 25. warm the mixture C 2 H 5 OH+ CH 3 COOH CH 3 COOC 2 H 5 + H 2 O Reaction 4 c)Conditions (if any) d)Equations
  • 26.   A) Refer to the above reaction: Esterification Absolute Ethanol and concentrated sulphuric acid as the catalyst Reactions 5 Name of reactions b)Reagents used
  • 27. warm the mixture C 2 H 5 OH+ CH 3 COOH CH 3 COOC 2 H 5 + H 2 O Reaction 5 c)Conditions (if any) d)Equations
  • 28.   A) Refer to the above reaction: oxidation Acidified K 2 Cr 2 O 7 or acidified KMnO 4 Reactions 6 Name of reactions b)Reagents used
  • 29. warm the mixture C 2 H 5 OH+ 2[O] CH 3 COOH + H 2 O Reaction 6 c)Conditions (if any) d)Equations
  • 30. B)Draw diagram for experiment to carry out for reaction 2
  • 31. Describe the coagulation of latex when exposed to air
  • 32.  
  • 33.  
  • 34.  
  • 35. 5) A student obtained the data to determine heat of precipitation of PbSO4 Highest temperature : 30.5 0 C Solution Vol (cm3) Conc (moldm-3) Initial temp ( 0 C) Pb(NO 3 ) 2 50 0.5 27.4 Na 2 SO 4 50 0.5 27.6
  • 36. What is meant by heat of precipitation? Heat changed when 1 mole of precipitate is formed from its ions in an aqueous solution
  • 37. b) Calculate heat of precipitation of PbSO4 Mole Pb 2+ : (0.5)(50)/1000 = 0.025 mol Mole SO 4 2- : (0.5)(50)/1000 = 0.025 mol Pb 2+ + SO 2- 4 PbSO 4 mole PbSO 4 = Mole Pb 2+ or mole SO 2- 4 = 0.025 mol
  • 38. ϴ = 30.5 - ( 27.4 + 27.6 ) 2 = 3.0 ° C Heat of precipitation = mC ϴ mol = ( 100)(4.2)(3) 0.025 = 50400 J/mol ∆ H = - 50.4 KJ/mol
  • 39. d)Write thermo chemical equation Pb 2+ + SO 4 2- PbSO 4 ∆ H = - 50.4 KJ/mol
  • 40. e) Write the ionic equation Pb 2+ + SO 4 2- PbSO 4
  • 41. E) Construct energy level diagram Pb 2+ + SO 4 2- Energy ∆ H = - 50.4 KJ/mol PbSO 4
  • 42. f) The experiment is repeated using K 2 SO 4 to replaced Na 2 SO 4 . Heat of precipitation of PbSO 4 remain the same. Explain. Because the same precipitate is formed, which is PbSO 4. Only Pb 2+ ions and SO 4 2- ions react Na + ions and K + ions do not react
  • 43. 6) A student carried out an experiment to determine heat of displacement of copper from CuSO4 solution. He added excess zinc powder to 50 cm 3 of 0.2 moldm -3 CuSO 4 . The thermo chemical equation is shown below : Zn + Cu 2+ Zn 2+ + Cu ∆ H = -80.64 KJ/mol
  • 44. a) Calculate the change in temperature Mol Copper= mol copper(II) sulphate = (0.2)(50)/1000 = 0.01 mol ∆ H = mC ϴ mol 80 640 J = ( 50)(4.2)( ϴ ) 0.01 ϴ = 3.8 ° C
  • 45. b) Write the ionic equation Zn + CuSO 4  Cu + ZnSO 4 Zn + Cu 2+  Zn 2+ + Cu
  • 46. The experiment is repeated with the following changes. What is the effect in the change of temperature when : Concentration of CuSO 4 is doubled, without changing the volume : So, change of temp or ϴ is doubled. Because as the concentration doubled, the number of particle per unit volume also doubled.
  • 47. The experiment is repeated with the following changes. What is the effect in the change of temperature when : volume of CuSO 4 is halved, without changing the concentration : So, change of temp or ϴ is remain the same. Because, the changes in volume do not affect the number of particles per unit volume
  • 48. 7) State the diff betw heat change and the heat of reaction Heat absorb , heat released -Heat of Precipitation -Heat of displacement -Heat of neutralisation -Heat of combustion HEAT CHANGE HEAT OF REACTION A) OTHER NAMES Depends on name of reactions :
  • 49. 7) State the differences between heat change and heat of reaction HEAT CHANGE HEAT OF REACTION B) FORMULA USED H= mC ϴ ∆ H = mC ϴ mol C)UNIT Joule Kilo Joule/ mol D) SYMBOL none ∆ H E) SIGN No sign Either + for endothermic rex or – for exothermic rex