BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

• UltimaOnline

  12 Apr 12, 02:26

  ^_^
• UltimaOnline

  12 Apr 12, 18:55

  Originally posted by Audi:

  Convert...

  CH3CH2CH2Br to CH3CH(CH3)COOH.

  There should be 3 steps between the reactant and the product.

  DIdn't occur to me that I should have started with elimination of halogenoalkane. And I totally didn't want to give up on such a supposedly simple question.

  
  Thought process :

  CH3CH2CH2Br to CH3CH(CH3)COOH.

  Since the final product is branched, a simple nucleophilic substitution will not suffice.

  Elimination of HX following by addition of HX, is useful for converting terminal alkyl halides to internal alkyl halides.

  Notice that the product has one more C atom than the reactant (which means we need a C nucleophile, such as the cyanide nucleophile, or a Grignard reagent, or a Wittig reagent).

  Hence, substitute away the halogen (good leaving group) with CN by heating under reflux with NaCN(alc).

  Carry out acidic hydrolysis of CN to convert it into (an amide intermediate which is further hydrolyzed to) COOH.

   

  Originally posted by SBS n SMRT:

  btw, TS mentions 3 steps, utima your method is technically 4 steps

  Step 1: Alc KOH, refulx

  Step 2: HX, rtp

  Step 3: Alc KCN, reflux

  Step 4: H+/H2O heat 

  
  There is a way to cut out step 2, but H2 Chem JC students aren't familiar with this process.

  Step 1 (Elimination of HX) : KOH (alc), heat under reflux

  Step 2 (Hydrocyanation) : HCN (alc), Ni/Cu/Pd catalyst, heat under reflux

  Step 3 (Acid-catalyzed Hydrolysis) : H+/H2O, heat under reflux 

• UltimaOnline

  13 Apr 12, 00:26

  ^_^
• UltimaOnline

  15 Apr 12, 12:31

  Kahynickel

   posted :

  I request to review a question i set for AS. It relates to the difference in bonding of CO2 and SiO2 and their relation to bond energy, bond polarity.
  
  In part (a) I have given the relavane melting point of CO2 and SiO2 and asked students to draw the structures of CO2 and SiO2.
  
  Then in part (b) using these diagram I asked from the perspective of bond energy and bond polarity the difference in the bonding of these two oxides using the data from the data booklet. I made two headings bond energy and bond polarity. This is what I expect students to state and the relevant problems I am facing with this question.
  
  "Si=O bonds are not stable" is given in the question.
  
  BOND ENERGY: C-O, Si-O, C=O values quoted from the data booklet.
  
  C=O bond is stronger than C-O in CO2 therefore a simple molcular structure is favoured over a giant covalent. while in SiO2, Si-O bonds are stronger because as given Si=O does not exist therefore a giant -covalent structure is likely to occur than a simple molecular structure. [Now should I give the information regarding Si=O or is there any other way of rephrasing the question?]  
  
  BOND POLARITY: Si-O bonds are more polar in SiO2 than C=O in CO2 since the difference in the electronegativities is larger in case of Si than C because Si has is less electronegative than C. Hence a three-dimensional network of Si-O bonds in SiO2 is favoured. [ I also like to involve the Si=O bond here is it workable?]
  
  I would also like to involve the parameter of bond length here. Since multiple bonds are stable than single bonds but double bonds are shorter than single bond. Here is what i wish to stae w.r.t this case.

  shorter C=O bond is favoured in CO2  as compared to long C-O bonds if CO2 were to be giant covalent because both C and O are small atoms they can easily overlap to form stronger C=O bond. In SiO2 silicon has a lrager atomic radius than C and thus will form weaker Si=O bonds or a lot of  longer Si-O yet strong covalent bonds thus a giant covalent stritcure is more favoured over a simple molecular form.

  I am looking forward to your suggestions.

   

• UltimaOnline

  15 Apr 12, 12:32

  Kahynickel, I agree with all your marking points.

  As for how to get the student to write more about the hypothetical Si=O bond, the question could be structured to give hints to guide the student. For instance, along the lines of :

  a) By comparing the lengths and strengths of single bonds versus double bonds. state the correlation between the two parameters.

  b) By comparing the relative atomic radii of C vs Si, suggest the difference in bond lengths that the C and Si atoms would form with O atoms.

  c) Hence, deduce whether single bonds or double bonds, are favoured when C atoms, and Si atoms, bond with O atoms.

   

  Personally, I would prefer the student explain this in terms of atomic orbitals (for which the question can be similarly structured, as in the example above, to guide the student towards this explanation).

  Alternatively, if you leave your question as more open-ended essay style rather than structured, your mark scheme could accept a variety of points. For such essay style questions, the Cambridge mark scheme often has "accept any 5 of the following 7 points" that will suffice for the student to gain full marks for the question.

   

  Carbon and silicon are both Group IV elements. Why then, under standard conditions, does carbon dioxide exist as a simple covalent molecule, while silicon dioxide exist as a giant covalent lattice?

  To form the pi bond between C and O atoms, involves the more effective sideways or side-on overlap between the (less diffused) unhybridized 2p orbitals of both C and O atoms. To form the pi bond between Si and O atoms, involves the less effective sideways or side-on overlap between the (more diffused) unhybridized 3p orbital of Si and the (less diffused) unhybridized 2p orbital of O.

  Consequently, the C=O pi bond is significantly stronger than the (hypothetical) Si=O pi bond, and hence it is more energetically favourable for C atoms to form (relatively strong) double bonds with O, and in contrast for Si atoms to form (a greater number of stronger) single bonds with O.

  Conclusion : it is energetically favoured that (under standard conditions), carbon dioxide has a simple covalent molecular structure, while silicon dioxide has a giant covalent lattice structure.

• UltimaOnline

  16 Apr 12, 00:33

  Originally posted by Kahynickel:

  Thank you for your invaluable suggestions.

  I would like this question to be structure-one rather than essay-type since CIE exam papers prefers structured question for AS. However essay-type questions cannot be completely ruled out. They are often asked in A2.

  Secondly your "hybridization answer approach" will prove quite challenging for students I have included the comparison of atomic radii of C and Si in the question.

  Finally, after this part (a) and part (b) I have also introduced part (c) which relates to the use and property of three compounds.

  part (c)

  The industrial processes use different compounds carrying out different purposes. Choose one use and property on which that particular use depends of each the following material listed below.

  In each case choose a diffrerent property on which the use depends.

  Al2O3

  Use            Abrasive / catalyst in cracking of hydrocarbons / 

  Property                                  Inert

  SiO2

  use           Ceramics (porcelain, chaina dish) / stationary phase in TLC, HPLC, GLC / 

  Property      high melting point                         inert/ can be derivatized on Silanol groups

  MgO

  use             Refractory material in furnace lining 

  Property     absorbs acidic impurites and electrical insulator

   

  Kahynickel, an excellent exam question that you've written for your students, as usual.

    

  Just a couple of suggestions to add :

   

  Al2O3 - extremely high mineral hardness and hence can be used as an abrasive.

   
  MgO - extremely high melting point or physically and chemically stable at high temperatures, hence can be used as refractory material.

   

• UltimaOnline

  16 Apr 12, 00:38

  Kahynickel wrote this question for his 'A' level students and shared the question here in January 2012 :

  -

  A supercritical fluid (SCF) is a gas that is compressed and heated so that it shows properties of a liquid and a gas at the same time. The fluid is neither a gas nor a liquid and is best described as intermediate to these two extremes.   A supercritical fluid retains solvent power approximating liquids as well as the transport properties common to gases. They have replaced organic solvents in various applications. The table shows some of SFCs and their critical conditions.

  -

  Fluid

  Critical temperature (C)

  Critical pressure (bar)

   -

  Carbon dioxide

  31.1

  73.8

  -

  Water

  374

  221.2

   -

  Ammonia

  405

  113.5

  -

  Supercritical CO₂ (ScCO₂) is the most widely-used fluid on large-scale. It has found applications in the extraction of foodstuff, removal of caffeine and as cleaning agent.

  (i) By referring to the table suggest one advantage for the large-scale of ScCO₂ as compared to ScH₂O and ScNH₃.

  ScCO2 requires low temperature and low pressure which requires less electricity and less time than Sc NH3 and Sc H2O which requires high temperaures needs larger amount of electricity  and longer time to produce this temperature.

  -

  (ii) At room temperature CO₂ is not an ideal gas. Suggest in terms of Kinetics Molecular Theory how this behaviour is changed in ScCO₂. 

  At low temperature CO2 has van der Waal's forces so does not behave as ideal gas.  At high temperature the vander Waal's forces would be overcome CO2 will become a super critical fluid. [I am not taking into account the high pressure because gases become ideal at low pressure]

  -              

  (iii) Give the name of one gas which shows ideal behaviour.

  Helium.

  -

  (iv) Besides CO₂, water is the other commonly applied supercritical solvent. Suggest one limitation of using supercritical water as solvent as compared to ScCO₂ in the separation of organic compounds.

  Organic compound may react with ScH2O (ie. hydrolysis) as it is reactive while ScCO2 is non-polar overall and less reactive than ScH2O.

• UltimaOnline

  16 Apr 12, 02:31

  Originally posted by Kahynickel:

  As you will be aware of the situation CIE exams are appoaching I am preparing two final mock exams. This question was a part of my second mock. I would also request you look into the parts of two questions.

  Question 1 Aluminium is the third most abundant element in the Earth’s crust, occurring combined in many minerals. 

  Give the electronic configuration of Al³⁺ ions in the space below.

   

  (a)(i) Al³⁺……………………………………………………………………………………………………

   

  (ii) Using your answer from (a) sketch the shapes of the orbitals having lowest energy, highest energy and the next available orbital in Al³⁺. Label in each case the name of the orbitals. [Below i have provided the axes forthe answers] the answers are:

           Is                         2p                                 3s

  lowest energy            highest energy          the next available orbital

                                                                      [is this right]

  [How about the approach of this question? Will students get it!!......

  Question 2  [It is also a part of the same mock exam]

  Nitrogen, which makes up about 80% of the Earth’s atmosphere, is very unreactive. 

  (a)(i) Explain the lack of reactivity of nitrogen.

   

  It has strong N-N triple bonds which needs high energy to break. bond energy may be quoted.  

  
  

  (ii) State one use of nitrogen which depends upon its lack of reactivity.

   

  It is used to provide inert atmosphere during synthesis of organic compounds. 
  

    

  (b)(i) On the other hand, Fluorine is most reactive member of group VII. Draw the dot and cross diagram for fluorine.

   

                                                       All lone pairs and bond pair shown                                                          

                                                                         F.xF

   

  (ii) Fluorine reacts explosively with hydrogen. This reaction is performed in dark i.e. in the absence of sunlight.

   

                                                    

                                                    H₂(g)  +  F₂(g)          2HF(g)

   

  The corresponding reaction of chlorine requires sunlight. By using your answer from (c)(i) and choosing suitable data from the Data Booklet to support your answer explain the difference in the reactivity of fluorine and chlorine.

  
  

  Data from data booklet:              F-F 158 kJmol-1    Cl-Cl   244 kj mol-1 

   

  Flourine has samller atomic/ covalent radius than Chlorine but lone pairs in F-F causes repulsion since they are very near to each other (second shell) as comapred Cl-Cl in which lone pairs repulsion are insignificant thus F-F has lower bond energy than C-Cl and reacts violently with hydrogen.   

   

  
  Another well-written question.

  Yes that's correct, 3s is the next available orbital. Yes, the phrasing of the question is fine, students should have no problems understand what's required.

  (Regarding relative energy levels of orbitals, it is the 4s versus 3d orbitals that students have difficulty with. Despite being slightly further away from the nucleus, the 4s orbitals have slightly lower energy compared to the 3d orbitals. The reasons for this are beyond an 'A' level discussion).

  Only suggestion to add to the very last part of your question :

  You might further split your last question "By using your answer from (c)(i) and choosing suitable data from the Data Booklet to support your answer explain the difference in the reactivity of fluorine and chlorine" into two sub-parts, eg. along the lines of :

  a) Using information in the Data Booklet, compare the atomic radii of fluorine and chlorine, and compare the molecular bond dissociation enthalpies of F2 and Cl2. Identify the apparent contradiction and suggest an explanation for it.

  b) Using information in the Data Booklet, calculate the enthalpies of the reactions between (i) hydrogen and fluorine, and (ii) hydrogen and chlorine. Hence explain the difference in the reactivity of fluorine and chlorine, with hydrogen gas.

   

• UltimaOnline

  18 Apr 12, 11:57

  Kahynickel wrote the following question for his 'A' level students. Answers are provided below.

   

  Question

  Elements of group I and II are prepared by the electrolysis of their molten salts. Sodium is manufactured from molten sodium chloride and magnesium from magnesium chloride.

  (a) Give the balanced equations, including state symbols, for the reaction occurring at each electrode during the electrolysis of molten sodium chloride.

  (i) Release of 1 mole of product in its standard state at anode.

  (ii) Gain of all electrons from anode to form product at cathode.

  (b) Give an overall reaction for the process.

  (c)  In a certain electrolysis cell for the manufacturing of sodium, 6500 dm³ of the product was collected at anode. Using your answer from (b) calculate the amount, in tonnes, of the product formed at cathode. 

   

  (d) A white powder contains the three oxides MgO, Al₂O₃ and SiO₂. Three samples were tested each containing only one of these oxides. Read the following description of tests then answer the questions that follow.

  Adding aqueous HCl to a sample of white power produces a colorless solution. To this resultant solution dilute sodium hydroxide was added dropwise, a white precipitate was seen which dissolves in excess of the reagent.  A second sample of white powder produces a white precipitate on adding little water and then in excess. Adding aqueous HCl to this second sample of white powder also gave no reaction. Adding aqueous HCl to the third sample of white power produces a colorless solution which gave a white precipitate on further adding NaOH_(aq) dropwise until in excess

   

   

  (i) Which two oxides might be present in the first sample of white powder?  

  (ii) Which of these samples first, second or third was MgO? 

  (iii) Using your answer from (d)(i) give the exact formulae of the two species present in colorless solution in the first sample stating the pH associated with each species and the total pH of the colorless solution as a result of both these species. Mg(aq)2+ and Al(aq)3+ 6.5, 3 , 2.95

  Formula……………………………     pH……………                   Total pH……………

  Formula…………………………...     pH……………                

  (iv) Identify the oxide which was actually present in the first sample. Also give two reactions resulting in the formation of colorless solution and the addition of dilute NaOH.

  ………………………………………………………………………………………………………………………………….…….

  • Balanced equations for the two reactions occurring when the oxide you have chosen in (iii) produced:    

  a)       Colorless solution………………………………………………………………………………………………………..

   

  b)       Reaction with NaOH(aq)………………………………………………………………………………………….……..

  (v) By considering your reactions a and b what “word” is used for this type of oxide.

  ………………………………………………………………………………………………………………………………….……

  (vi) Identify the oxide present in the second sample and explain why no reaction took place here.

  ……………………………………………………………………………………………………………………………………..…

  (vii) Give the formula of the white precipitate formed in the third sample. Also state the pH of this solution.

  ………………………………………………………………………………………………………………………………………..

   

  Originally posted by Kahynickel:

  Question

  Elements of group I and II are prepared by the electrolysis of their molten salts. Sodium is manufactured from molten sodium chloride and magnesium from magnesium chloride.

  (a) Give the balanced equations, including state symbols, for the reaction occurring at each electrode during the electrolysis of molten sodium chloride.

  (i) Release of 1 mole of product in its standard state at anode.

                                           2Cl-(g)          Cl2(g) + 2e-

  (ii) Gain of all electrons from anode to form product at cathode.

                                        2Na+(l)   + 2e-         2Na(s) 

  (b) Give an overall reaction for the process.

                              2Cl-(g) + 2Na+(l)                   2Na(s) + Cl2(g) 

                                     

  (c)  In a certain electrolysis cell for the manufacturing of sodium, 6500 dm³ of the product was collected at anode. Using your answer from (b) calculate the amount, in tonnes, of the product formed at cathode. 

  moles of Cl2 =6500/24= 270.8 or 271

   Cl2: Na from (b)

      1: 2

  moles of Na= 271.2 x 2 = 542

  mass of Na= 542x23 = 12466g = 12.5 tonnes

   

  d) A white powder contains the three oxides MgO, Al₂O₃ and SiO₂. Three samples were tested each containing only one of these oxides. Read the following description of tests then answer the questions that follow.

  Adding aqueous HCl to a sample of white power produces a colorless solution. To this resultant solution dilute sodium hydroxide was added dropwise, a white precipitate was seen which dissolves in excess of the reagent.  A second sample of white powder produces a white precipitate on adding little water and then in excess. Adding aqueous HCl to this second sample of white powder also gave no reaction. Adding aqueous HCl to the third sample of white power produces a colorless solution which gave a white precipitate on further adding NaOH_(aq) dropwise until in excess.

  (i) Which two oxides might be present in the first sample of white powder?

     MgO and Al2O3

  (ii) Which of these samples first, second or third was MgO?   third 

  (iii) Using your answer from (d)(i) give the exact formulae of the two species present in colorless solution in the first sample stating the pH associated with each species and the total pH of the colorless solution as a result of both these species. Mg(aq)2+ and Al(aq)3+ 6.5, 3 , 2.95

  Formula: [Mg(OH2)6]2+     pH 6.5                   Total pH 2.95-2.98 (must be less than 3.0)

  Formula: [Al (OH2)6]3+     pH 3.0                 

  (iv) Identify the oxide which was actually present in the first sample. Also give two reactions resulting in the formation of colorless solution and the addition of dilute NaOH. 

  Al2O3

  • Balanced equations for the two reactions occurring when the oxide you have chosen in (iii) produced:    

  a)       Colorless solution           Al2O3 + 6HCl           2AlCl3 + 3H2O 

  b)       Reaction with NaOH(aq)  Al(OH)3(s)  + OH-(aq)           AL(OH)4-(aq)

  (v) By considering your reactions a and b what “word” is used for this type of oxide.

  amphoteric 

  (vi) Identify the oxide present in the second sample and explain why no reaction took place here.

  SiO2. Since it is an acidic oxide and does not react with HCl and is insoluble in water forming a precipiatate.

  (vii) Give the formula of the white precipitate formed in the third sample. Also state the pH of this solution.

  Mg(OH)2 (s). pH would be 9 

   

• UltimaOnline

  18 Apr 12, 12:14

  ------------------------------------

  Originally posted by Kahynickel:

  In the oxides test quetions above if I include P4O10 it will make students to think a bit deeper than in case of 3 oxides given (MgO, Al2O3 and SiO2). However,only 3 oxides will be tested positive in the three samples.

  Indeed, that will make for a more challenging, interesting question.

  Not to mention that P4O10, even though an acidic oxide, because it is the anhydride of phosphoric(V) acid, will therefore still 'dissolve' in aqueous acidic solutions (eg. HCl(aq) ), in the sense of hydrolysis of the P4O10 to generate phosphoric(V) acid. Which means this 'dissolving' would be a hydrolysis* reaction, not a Bronsted-Lowry acid-base reaction, and the resulting solution would have two different acids.

  (Technically, hydrolysis may be considered a Lewis acid-base reaction, with P4O10 as the Lewis acid and water as the Lewis base).

  Good one, Kahynickel!

  ---------------------------------------

  UltimaOnline's Bonus Qn :

  Would a reaction occur if NaOH(aq) was added to SiO2 at room temperature? If yes, write an equation for the reaction. If no, explain why.

  Answer :

  SiO2 is chemically inert to both aqueous acids and aqueous alkalis, hence no reaction occurs with NaOH(aq) at room temperature.

  Being a giant covalent acidic oxide, it will react only with fused (ie. molten) NaOH(l) at high temperatures, to generate the sodium silicate salt Na2SiO3 and water.

  Due to SiO2 having a giant covalent lattice structure, this particular acid-base reaction has a very high activation energy, in order to break the multitude of strong covalent bonds in the giant covalent lattice, required for the reaction to proceed.  

• UltimaOnline

  18 Apr 12, 23:01

  Originally posted by SBS n SMRT:

  Do methyl-enols give a positive iodoform test?

  If the methyl group is directly bonded to the sp2 hybridized C atom of the alkene that is not directly bonded to the hydroxyl group, then no, because the molecule tautomerizes to an ethyl ketone, not methyl ketone.

  If the methyl group is directly bonded to the sp2 hybridized C atom also directly bonded to the hydroxyl group, then yes, because the molecule tautomerizes to a methyl ketone.

• UltimaOnline

  22 Apr 12, 12:25

  Qn : Show how you would convert propanone to propane, describing the reagents and conditions as well as giving the structures of the intermediate products.
   
  Ans :

  More-efficient-but-beyond-'A'-levels-answers :
  Clemmensen reduction (ie. Zn(Hg), HCl, heat under reflux), or
  Wolff-Kishner reduction (ie. N2H4, followed by KOH, heat under reflux)
  
  Less-efficient-but-accepted-at-'A'-levels-answer :
  Reduction (LiAlH4 in dry ether, followed by protonation via hydrolysis),
  followed by Dehydration (concentrated H2SO4 catalyst, 170 deg),
  followed by Hydrogenation (H2 gas with either Ni catalyst with heating, or Pt catalyst at room temperature).
  Ketone ---> Alcohol ---> Alkene ---> Alkane.
   
  --------------------------------------------------

  Bonus Qn :
  For hydrogenation, if Ni catalyst requires heating while Pt catalyst doesn't, then why bother using Ni instead of Pt?

  Ans : Coz Ni is a lot cheaper than Pt.
  

• UltimaOnline

  23 Apr 12, 07:40

  The following reaction was carried out in a calorimeter with a heat capacity of 2000 J K-1. The specific heat capacity of the solutions should be taken as 4.2 J K-1 cm-3. You may assume that the mass of excess metal in the experiments are negligible. In the first experiment, excess magnesium powder was added to 100 cm3 of a 1.00 mol dm-3 solution of copper (II) sulfate. The temperature rose from 19.5 °C to 41.2 °C. In the second experiment, excess copper powder was added to 100cm3 of a 0.50 mol dm-3 solution of silver nitrate. The temperature rose from 19.5 °C to 20.9 °C.

  Calculate the enthalpy change of reactions in both experiments, as well as the expected enthalpy change, per mole of magnesium, for the reaction (not carried out experimentally) between magnesium and silver nitrate.

  
  Solution :
   

  The heat capacity of the entire system is the sum of the heat capacities for the individual components, ie. the calorimeter and the aqueous solution.

  Total Heat Capacity = Heat Capacity of Water + Heat Capacity of Calorimeter
  = (100cm3 x 4.2 J K-1 cm3) + (2000 J K-1)
  = 2420 J K-1

  Heat transferred in Experiment 1 = Total Heat Capacity x Change in Temperature
  Q = (2420 J K-1) x (41.2 - 19.5)
  Q = 5.2514 x 10^4 J

  Heat transferred in Experiment 2 = Total Heat Capacity x Change in Temperature
  Q = (2420 J K-1) x (20.9 - 19.5)
  Q = 3.3880 x 10^3 J

  Note that because the temperature rose, the reactions are exothermic, and therefore you must write the negative sign for the enthalpy change value.

  Mg(s) + Cu2+(aq) ---> Mg2+(aq) + Cu(s) 

  Enthalpy change of 1st experiment in regard to either Mg or Cu2+
  = Heat transferred / (moles of reactant of either Mg or Cu2+)
  = - (5.2514 x 10^4 J) / (100/1000 x 1.0)
  = - 5.2514 x 10^2 kJ/mol of Mg or Cu2+

  Cu(s) + 2Ag+(aq) ---> Cu2+(aq) + 2Ag(s) 

  Enthalpy change of 2nd experiment in regard to Ag+
  = Heat transferred / (moles of reactant of Ag+)
  = - (3.3880 x 10^3 J) / (100/1000 x 0.5)
  = - 6.7760 x 10^1 kJ/mol of Ag+

  Enthalpy change of 2nd experiment in regard to Cu
  = Heat transferred / (moles of reactant of Cu)
  = - (3.3880 x 10^3 J) / (100/1000 x 0.5) / 2
  = - 1.3552 x 10^2 kJ/mol of Cu

  By Hess Law, enthalpy change of reaction between magnesium and silver nitrate
  = (enthalpy change of reaction between magnesium and copper)
  + (enthalpy change of reaction between copper and silver nitrate)
  = (-5.2514 x 10^2)kJ + (-1.3552 x 10^2)kJ
  = -6.6066 x 10^2 kJ/mol of Mg

   

• UltimaOnline

  24 Apr 12, 01:49

  Originally posted by cubsarecute:

  Qn.  When we say "solubility of Mg(OH)₂ is 9 mg/dm3 in water", does the "9 mg/dm3" refer to the cations ([Mg²⁺]) only, or does it actually refer to the whole compound K_sp = [Mg²⁺][OH^-] = 9 mg/dm3?

   

   
  Mass solubility refers to the sample mass of the ionic species (ie. Mg(OH)2, not just Mg2+) that can dissolve in one dm3 of solution. Mass solubility is not Ksp (which is the solubility product). To convert mass solubility to solubility product, you need to first obtain the molar solubility.

  Eg. If the mass solubility of Mg(OH)2 is 9mg/dm3, the molar solubility (ie. mass solubility / molar mass) is hence 1.544 x 10^-4 mol/dm3, and therefore the solubility product is (1.544 x 10^-4) ( 2 (1.544 x 10^-4) )^2 = 1.472 x 10^-11 (mol/dm3)^3

• UltimaOnline

  27 Apr 12, 03:33

  ^_^
• UltimaOnline

  29 Apr 12, 01:28

  Originally posted by hoay:

  What factors can affect the value of the activation energy of a reaction?

  1 the presence of a catalyst
  2 changes in temperature

  Catalyst lowers the activation energy of the reaction.

  Temperature raises the average K.E of the molecules hence they achieve Ea. So both these factors are correct. Am I right??

  
  No, activation energy for any particular reaction is only altered by the use of a catalyst, and remains unchanged by temperature.

  Only when activation energy (Ea) remains unchanged, then the % of reactant molecules with energy equals or exceeding Ea can increase as temperature increases, thus increasing the rate of reaction.

  An analogy would be : if your income goes up, but your cost of living also goes up, then the rate at which you can reach your retirement goals won't be able to increase.

  Only when the cost of living (ie. activation energy) remains unchanged, and your income (ie. % of molecules with energy equals or exceeding Ea) goes up, only then will the rate at which you can reach your retirement goals happily increase.

   

  Also see :
  http://en.wikipedia.org/wiki/Activation_energy

  Temperature independence and the relation to the Arrhenius equation
  
  The Arrhenius equation gives the quantitative basis of the relationship between the activation energy and the rate at which a reaction proceeds. From the Arrhenius equation, the activation energy can be expressed as

  

  where A is the frequency factor for the reaction, R is the universal gas constant, T is the temperature (in kelvins), and k is the reaction rate coefficient. While this equation seems to suggest that the activation energy is dependent on temperature, in regimes in which the Arrhenius equation is valid this is cancelled by the temperature dependence of k. Thus, E_a can be evaluated from the reaction rate coefficient at any temperature (within the validity of the Arrhenius equation).

• UltimaOnline

  1 May 12, 23:30

  The Haber process for the manufacture of ammonia is represented by the following equation.

                                   N2(g) + 3H2(g)   <--->   2NH3(g)              ΔH = –92 kJ mol–1

  Which statement is correct about this reaction when the temperature is increased?

  A) Both forward and backward rates increase.
  B) The backward rate only increases.
  C) The forward rate only increases.
  D) There is no effect on the backward or forward rate.

  
  Answer :   A

  When the temperature is increased, the percentage of reactant molecules with energy equal or exceeding the activation energies for both the backward and forward reactions, are increased (along with temperature). Hence, both the forward and backward reaction rates increase.

  However, the percentage increase in rate is greater for the reaction in the direction with the larger activation energy. If the forward reaction is endothermic (ie. the forward reaction will accordingly have a larger activation energy, compared to the backward reaction's smaller activation energy), then increasing the temperature will increase both the forward and backward reaction rates, but will increase the forward reaction rate by a larger magnitude compared to the backward reaction rate. This results in the position of equilibrium shifting towards the right (ie. towards the products), even as the rate of both the forward and backward reactions simultaneously increase.

  In an endothermic forward reaction, heat may be regarded as a reactant (rather than product). Increasing the temperature is thus akin to increasing the concentration of the reactants, which will therefore cause the position of equilibrium to shift towards the right (ie. towards the products), as predicted by Le Chatelier's principle (ie. the system attempting to remove the added heat energy by using some of it to convert the reactants to the products). Mathematically, the Kc (ie. equilibrium constant) value will increase, while the Qc (ie. reaction quotient) value remains initially unchanged.

  Hence (while increasing the temperature of the system increases both the forward and backward rate constants), the rate of the forward reaction (ie. indicated by the new relatively much larger k-forward value) will now exceed the rate of the backward reaction (ie. indicated by the new relatively only slightly larger k-backward value) until equilibrium is eventually re-established for the system, ie. Qc = Kc.

  Once Qc = Kc and the system has re-established equilibrium, the rates of the forward and backward reactions will be exactly the same. This is despite the k-forward rate constant still having a larger value compared to the k-backward rate constant. To understand this, recall that the rate of reaction is given as the rate constant multiplied by the molarities of the reactants, each raised to the power of their orders. Thus at position of equilibrium, having smaller reactant molarities and a larger k-forward rate constant, will be mathematically equal to having larger product molarities and a smaller k-backward rate constant.

• UltimaOnline

  2 May 12, 00:21

  If the Kc value changes only with temperature, then why does the position of equilibrium shift when we change the pressure?

    

  N2(g) + 3H2(g)   <--->   2NH3(g)              ΔH = –92 kJ mol–1

   
   

  Answer :

   
  Increasing the pressure (unless otherwise specified) involves decreasing the volume of the container. Because the temperature remains unchanged, the Kc value does not change. But because the volume of the container decreases, the molarities of all species increase.

   

  For the Qc expression, the numerator which consists of [NH3]^2 increases by a smaller magnitude, compared to the denominator which consists of [N2]^1 x [H2]^3. Mathematically, this results in the Qc value now becoming smaller than the Kc value (which did not change since temperature did not change).

   

  If we change any variable (eg. pressure, volume, moles or molarities) other than temperature, and the Qc value becomes smaller than the Kc value, we say the position of equilibrium shifts to the right.

   

  If we change any variable (eg. pressure, volume, moles or molarities) other than temperature, and the Qc value becomes larger than the Kc value, we say the position of equilibrium shifts to the left.

   

  If we change the temperature (eg. increasing the temperature for an exothermic forward reaction), and the Kc value becomes smaller than the Qc value, we say the position of equilibrium shifts to the left.

   

  If we change the temperature (eg. increasing the temperature for an endothermic forward reaction), and the Kc value becomes larger than the Qc value, we say the position of equilibrium shifts to the right.
• UltimaOnline

  3 May 12, 09:11

  Originally posted by hoay:

  In the reaction:

                              SO2 + 2H2S       2H2O + 3S

  Hhow do we know that sulfur is acting as an oxidizing agent, reducing agent, an acid or a base, when the oxidation number of S has changed to zero (from S+4 and S -2)?.

   

  To be precise, sulfur dioxide is being reduced (and is hence the oxidizing agent) and hydrogen sulfide is being oxidized (and is hence the reducing agent). Since elemental sulfur is not present as a reactant, it is less proper to say "sulfur is a reducing agent" or "sulfur is an oxidizing agent".

  This reaction may be regarded as the opposite of a proper disproportionation reaction, because of the two sulfur-containing reactant species, one is being oxidized and the other being reduced, both generating elemental sulfur as a product. Thus it would be more proper to say, "sulfur in SO2 is being reduced, while sulfur in H2S is being oxidized, both generating elemental sulfur as a product". 

   

  Bronsted-Lowry acid-base reaction

  H2S functions as the Bronsted-Lowry acid and donates protons. The O atoms of SO2 function as the Bronsted-Lowry base and accepts protons.

   

  Lewis acid-base reaction

  The S atoms of H2S function as the Lewis base (ie. nucleophile). The S atoms of SO2 function as the Lewis acid (ie. electrophile).

• UltimaOnline

  6 May 12, 12:21

  Why do Group II metals have increased reactivities going down the group, while Group VII halogens have decreased reactivities going down the group?

   

  When Group II metals react, they lose electrons and are oxidized. Down the group, due to increasing atomic radius and increasing shielding effect, electrostatic attraction between the positively charged nucleus and valence shell electrons decreases, hence going down the group, Group II metals have smaller (ie. less endothermic and thus more favourable) ionization energies and thus have increased reactivities.

   

  When Group VII halogens react, they gain electrons and are reduced. Down the group, due to increasing atomic radius and increasing shielding effect, electrostatic attraction between the positively charged nucleus and valence shell electrons decreases, hence going down the group, Group VII halogens have smaller (ie. less exothermic and thus less favourable) electron affinities and thus have decreased reactivities.

   
• UltimaOnline

  8 May 12, 11:19

  Originally posted by atomos:

  Just need a quick confirmation regarding ideal gas plots.

  pV = nRT.

  For a vertical axis to be density(rho) /pressure,

  I arrived at rho/p= M/RT

  So at constant temperature, plotting rho/p vs p would give us a flat line?

  Wondering if I've overlooked anything.

  
  Density = mass / volume

  Since for a constant n, mass is also constant, hence regard density as constant/V

  Accordingly, density / pressure = constant/(PV)

  Since pressure and volume have an inversely proportional relationship (ie. when you increase pressure volume decreases; when you increase volume pressure decreases), hence PV = constant.

  So if y-axis is density/pressure = constant/(constant) = constant, and since x-axis is P (ie. a variable), hence a straight horizontal line is obtained for the graph.

• UltimaOnline

  14 May 12, 12:48

  Originally posted by Kahynickel:

  I recently got Singapore A-level papers from Nov 2004 to Nov 2010 in exchange with the 2010 Prelims papers from a Singapore student. I still need Nov 2011 papers and June papers.

  The questions are excellent.  One of the things that I have noticed is that papers-setters sometime extend a multiple choice question into a full-length question for theory paper. The first question of my first mock that I e-mailed you was an example of this mind-set.

  The question was:

  When two liquids are mixed, heat may be evolved if intermolecular bonds formed are stronger

  than those broken, even if there is no chemical reaction'

   

  Which pair of liquids, when mixed, will give out heat?

  A    CH₂Cl₂ and (CH₃)₂CO

  B    CHCI₃ and C₆H₁₄

  C    CCl₄ and (CH₃)₂CO

  D    CCl₄ and CH₃CH₂OH

  The answer was A.

  In D CCl4 has VDW forces while ethanol has H-bonding so they don’t mix therefore no heat is evolved. H-bonds are stronger than VDW forces in CCl4 consequently heat must be added to make them mix. You can say that in the context of this question intermolecular forces are not broken. [Is this right]   

  In C propanone has permanent –dipoles as compared to VDW forces in CCl4 again the same explanation as in D.

  B is also same as that of C.    

  In A the intermolecular forces are same in both cases. Propanone has, however, stronger permanent-dipoles forces than CH₂Cl₂ because of the high electronegative O atom against the low-electronegative C-atom. The intermolecular bonds /forces formed here are strong permanent dipoles (I don’t think that VDW forces have a role to play here albeit CH2Cl2 has more electron than propanone.) so heat is given out.

  In addition for the rest of the pairs to be soluble heat will be added but only A is the one which gives out heat.

   I am waiting for your comment as I have to finalize the question in a long format.

  
  Btw and Fyi, the Singapore-Cambridge exams are only offered in November. We do not have June papers.

   

  I've little to add to your excellent explanations on this particular MCQ question, except on two minor points :

   

  1) For options B, C and D, although the two species listed are generally not miscible (due to reasons you've given), but due to favourable entropy, they will be a slight measure of mixing. Moreover, the question specified, "when the two species are mixed", which you may take it to mean "forcibly mixed" together, eg. by mechanical means. In which case, your structured or open-eneded question, may require the student to take this into consideration.

  For options B, C and D, when mixing does occur, no heat is evolved, because the newly formed inter-species (solute-solvent) interactions are permanent dipole - induced dipole van der Waals (Debye) forces, which are weaker than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces (option B & C) or the hydrogen bonding (option D), that must be overcome as part of the mixing process.

   

  2) For option A, as you've pointed out, since the new interactions are of the same type as the old interactions, ie. permanent dipole - permanent dipole van der Waals (Keesom) forces, we cannot be sure, without carrying out the relevant experiments, that A will definitely give out heat. But because this is an MCQ question, we can eliminate B, C, D as the answer, and therefore A is the only remaining viable answer. But if you were to phrase your structured or open-ended question to guide the process, eg. "In the table above, which experiment would be the most likely to have an exothermic reading?" and "For the experiment you identified, suggest with explanation at the molecular level, why this experiment might have an exothermic reading."

  Then perhaps the molecular geometries and sterics might play a part. Dichloromethane is tetrahedral, while propanone is trigonal planar. Perhaps the student can score credit if he/she writes along the lines of, "due to sterics consideration of the molecular geometries involved, the newly formed permanent dipole - permanent dipole van der Waals (Keesom) forces between dichloromethane and propanone, might be stronger than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces between dichloromethane and dichloromethane, (but not necessarily stronger than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces between propanone and propanone)."

• UltimaOnline

  7 Jun 12, 11:02

  What happens to the pH of neutral (ie. non-acidic, non-alkaline) pure water at temperatures colder or warmer than 25 deg C? Does it remain at pH 7? If not, how and why does it change?

  If a solution has a pH of 7, at temperatures colder or warmer than 25 deg C, is it neutral, acidic or alkaline?

  Answers :

  Proton dissociation is endothermic.
  As temperature increases, [H+] and [OH-] simultaneously increase.
  Since [H+] increases, pH decreases.
  Since [OH-] increases, pOH decreases.
  
  Hence :
  Neutral water has pH < 7 at temperatures higher than 25 deg C.
  Neutral water has pH > 7 at temperatures lower than 25 deg C.
  A solution with pH 7 at temperatures higher than 25 deg C is alkaline.
  A solution with pH 7 at temperatures lower than 25 deg C is acidic.

• UltimaOnline

  10 Jun 12, 03:49

  KickMe pleaded :

  How do I find the oxidation state of Fe in [Fe(SCN)(H2O)6]2+ and Mn in HMn(CO)5?

  Memo shrugged :

  Idk about HMn(CO)5.

  The pentacarbonylmonohydridomanganese [Mn(CO)(H)] neutral coordination complex is a bit of a complex matter (pun intended), slightly controversial even at University levels.

  The OS of Mn in [Mn(CO)(H)] may be argued to be either +1, or -1.

  ------------------------------------------------------

  My BedokFunland JC formulae :

  Ionic Charge = Sum of Oxidation States
  Oxidation State = Formal Charge + Electronegativity Consideration

  If we were to consider formal charges after the coordinate dative bonds are formed, and because all 6 ligands donor atoms (hydrogen and carbons) are all more electronegative than manganese (according to standard electronegativity ratings), then accordingly :

  Let OS of the Mn atom be X.
  OS of each O atom = Formal Charge + Electronegativity Consideration = (+1) + (-3) = (-2)
  OS of each C atom = Formal Charge + Electronegativity Consideration = (0) + (+3) + (-1)  = (+2)
  OS of the H atom = Formal Charge + Electronegativity Consideration = (0) + (-1) = (-1)

  Ionic Charge = Sum of Oxidation States = 0 (for a neutral coordination complex)
  this implies 5(-2) + 5(+2) + 1(-1) + X = 0
  hence X = +1

  As illustrated above, if we use the standard electronegativity ratings of hydrogen as being more electronegative than metals, then according to this arguement, the Oxidation State of Manganese in the pentacarbonylmonohydridomanganese [Mn(CO)(H)] neutral coordination complex may be regarded as +1.

  ------------------------------------------------------

  From another perspective, because the pentacarbonylmonohydridomanganese [Mn(CO)(H)] neutral coordination complex may be generated from the protonation of the uninegative ionic charged pentacarbonyl manganate [Mn(CO)5]- ion, in which the Oxidation State of Mn is -1, and because the behaviour of the H ligand varies amongst transition metal hydrides (some behaving as protonic H+, others behaving as hydridic H-), it may be argued that should the H atom herein behave as protonic (for which it has the capacity to do so under certain conditions), meaning that in this particular coordination complex, hydrogen is in effect less electronegative than manganese (due to the strong inductively electron-withdrawing effects of the carbonyl or carbon monoxide ligands), then accordingly the Oxidation State of Manganese in the pentacarbonylmonohydridomanganese [Mn(CO)(H)] neutral coordination complex could be regarded as -1.

• UltimaOnline

  28 Jun 12, 08:20

  Originally posted by hoay:

  A 0.10 mol dm–3 solution of NH2CH2CH2NH2 has a pH of 11.3. When 30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3 solution of NH2CH2CH2NH2, the final pH is 1.6.

  Using the following axes, sketch the pH changes that occur during this addition of
  HCl(aq).

  ANS: Since it is a  diprotic base there will be two curves. But how to calculate the end-point? 

  
  Data given only allows the student to calculate Kb1, but not Kb2. If Kb2 is not given, the pH for the 1st and 2nd equivalence points cannot be determined.

  The final pH of 1.6 is due primarily to the excess HCl, and not due to the acidity of N2H6 2+, since the former acid is much stronger than the latter.

  Since the question says "sketch" and since Kb2 is not given, the student will not be required to determine the exact pH of the 1st and 2nd equivalence points.

  If Kb2 was given, then we can calculate both Ka values (ie. of NH2CH2CH2NH3+ and +H3NCH2CH2NH3+, aka protonated ethylenediamine and diprotonated ethylenediamine).

  Only if we have both Kb (or therefore, both Ka) values, then we can calculate the pH of the 1st and 2nd equivalence points as follows :

  1st equivalence point : only 1 amphiprotic species present, use pH = 1/2 (pKa1 + pKa2)

  2nd equivalence point : only 1 acidic species present, use Ka of diprotonated ethylenediamine.