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

• UltimaOnline

  28 Jun 15, 06:38

  Originally posted by gohby:

  Regarding Q4, wouldn't the emission of 4 alpha particles result to the emission of 12 protons and 12 electrons, how does that gel with the answer of 10 protons and 10 electrons?

  Regarding Q1: Thanks, I overlooked the fulfillment of the octet structure for beryllium. If I may be inquisitive about it - how do I know that there wouldn't be any steric hindrance that would prohibit the formation of the addition product, given that there are 3 methyl groups in trimethylamine?

  ACJC Prelim 12:

  The empirical formula of a fluorocarbon is CF2. At the same temperature and pressure, 1dm³ of the fluorocarbon weighs 8.93g while 1dm³ of fluorine gas weighs 1.80g.

  What is the molecular formula of the fluorocarbon?

  A. CF2,

  B. C2F4

  C. C4F8

  D. C5H10

   

  Remarks: I worked out the Mr of the fluorocarbon to be ~188 --> [(1.8/38) moles of fluorocarbon weighing 8.93g]. Whilst C is definitely the best answer, I have some disquiet about the choice as 188 vs 200 is a bit of a stretch.

  I have found a website dated July 2000 of a similar question (http://www2.hmc.edu/www_common/chemistry/TUTORIALS/reviewSolution07Frame.html), except that the 1dm³ of fluorine gas weighs 1.70g. This will result to the Mr of the fluorocarbon being 200. Hence I am wondering if there is something wrong with this question.

  
  Regarding ur radioactive decay query, if it were a straightforward case, then yes your answer would be right (and if it were a free response question, the student should certainly write your answer). However, radioactive decay is a complex matter, and is really beyond the scope of the H2 Chemistry syllabus (thus this isn't a really fair question). The existence of positrons (not covered in the H2 Chemistry syllabus) makes it possible for protons and neutrons to interconvert during radioactive decay. As always, choose the best answer (in this context, one that fits with conservation of mass and charge).
  
  Regarding ur BeF2 qn, methyl groups are small and pose little steric hindrance. Cambridge will set questions that are sufficiently clear cut (ie. as long as the question successfully tests students on concepts within the syllabus, the variables within the question won't be overly sadistic). If Cambridge wanted the able student to say, "I know, it's coz of steric hindrance!", then Cambridge would use obviously bulky groups such as phenyl groups, versus small groups such as methyl groups. They won't use something whose size is directly in-between, that even Cambridge examiners themselves wouldn't know for certain (if the steric hindrance present is sufficient to alter the outcome) without actually carrying out the experiment, let alone Singapore JC teachers, let alone Singapore (or International for that matter) A level Chemistry students. Also, if it were a free response question, then advise your students to cover both possibilities : eg. during electrophilic aromatic substitution, "if the major product turned out to be the ortho product, it's due to there being 2 ortho positions versus 1 para position; if the major product turned out to be the para product, it's due to steric hindrance and van der Waals repulsion between the existing bulky substituent and the bulky incoming electrophile".
  
  Regarding ur ACJC qn, you're right. Either the ACJC teacher made a typo, or he/she thinks it's a good chance to test the student on whether they know to use the principle "always choose the *closest* option available, regardless of discrepancies".
• UltimaOnline

  29 Jun 15, 05:51

  Originally posted by hoay:

  Q .  For each of the following reactions predict the sign of G. If a prediction is not possible because the sign of G will be temperature dependent, describe how G will be affected by raising the temperature.

  (a) An endothermic reaction for which the system exhibits an increase in entropy

  (b) An exothermic reaction for which the system exhibits an increase in entropy

   

  Ans (a) For the reaction to be spontaneous and feasible the enthalpy change must be exothermic and the entropy change must be positive i.e G negative. The first factor does not support here so for G to be –ve the reaction should be carried out at high temperature.

  Ans (b) The sign of G would be negative since both the conditions are right for the reaction to be feasible and spontaneous

  One point if we forget here the need for the temperature control is to look at the sign of enthalpy change if it is +ve then high temperature is applied or -ve low temperature should be applied. Are my answers correct??

  
  

  
  Yes, you're correct. To be more precise :
  
  if enthalpy change is endothermic and entropy change is positive, then high temperature needs to be applied ; or if enthalpy change is exothermic and entropy change is negative then low temperature needs to be applied.
• UltimaOnline

  29 Jun 15, 06:22

  Originally posted by hoay:

  are feasibility and spontaneity of a reaction same terms? spontanous means the reaction must occur without the help of external indicators. Feasibility is about what?

  
  The correct usage of these 2 closely related terms is as follows :
  
  A chemical reaction is spontaneous if it is thermodynamically feasible, ie. Gibbs free energy change is negative.
• UltimaOnline

  29 Jun 15, 20:12

  Originally posted by Metanoia:

  Would a system be expected to experience an increase in entropy when it undergoes an endothermic reaction?

  
  For a chemical reaction that is thermodynamically feasible and hence spontaneous, if the reaction is endothermic, the reaction *must* have a positive entropy change, eg. dissolving solid ammonium salts. As far as the A level Chemistry syllabus is concerned, that's about it.
  
  However, at higher levels and in practice, it's still possible for a thermodynamically non-feasible (ie. non-spontaneous) reaction to occur, eg. in the worst case scenario (thermodynamically speaking), an endothermic reaction with negative entropy change, resulting in a positive Gibbs free energy change regardless of temperature, can still occur *provided* that such a thermodynamically non-feasible reaction is *coupled* with a separate, external chemical reaction that *is* highly thermodynamically feasible and favorable, and hence highly spontaneous, ie. a large magnitude of negative Gibbs free energy change.
  
  Such a coupling would ensure that the *overall* Gibbs free energy change of both coupled chemical reactions would, in totality, still be negative, thermodynamically feasible and hence spontaneous.
  
  Examples of thermodynamically non-feasible or non-spontaneous reactions from H2 Biology that certainly continue to occur (otherwise you wouldn't be alive to read this), would be photosynthesis and protein synthesis.
• UltimaOnline

  29 Jun 15, 21:11

  Originally posted by gohby:

  Hi UltimaOnline, thank you for your response. :) I have 2 further questions on Chem Bonding.

   

  Q1: Which of the following is most likely to be true for sodium ethanoate?

   

  A: It is more soluble in water than sodium chloride

  B: A solution of sodium ethanoate has a higher electrical conductivity than a solution of sodium chloride.

   

  Remarks: Both choices are wrong. For A, I reckon the ion-dipole interaction between Cl- and water is more exothermic than that of CH3CO2- and water because Cl- has a higher charge density, so the ion-dipole complex is more stable. Is this correct? As for B, how do I evaluate the relative electrical conductivity of 2 ionic compounds (esp in this case where both ions in both compounds are singly charged?)

   

  Q2: This question is regarding hydrogen bonding in ice/water.

   

  (i) Why is the bond angle about oxygen in ice is 109.5?

   

  Is it not possible to form the lattice arrangement if the bond angle is 105?

   

  Wikipedia states: “This tetrahedral bonding angle of the water molecule essentially accounts for the unusually low density of the crystal lattice – it is beneficial for the lattice to be arranged with tetrahedral angles even though there is an energy penalty in the increased volume of the crystal lattice.� My question: Okay, but why is it beneficial for the lattice to be arranged with tetrahedral angles?

  
  (ii) In my notes, it states: “Hydrogen bonding is maximised (in ice) because the water molecules are arranged in an ordered way to form a regular lattice�. My question: what is hindering hydrogen bonding in water from being arranged in an ordered way to form a regular lattice?

  
  No problem, keep ur qns coming :)
  
  Q1. Yes you're right about the relative strengths of the ion-dipole interactions. Any difference in entropy effect can be taken to be negligible, unless otherwise specified by the question for A level purposes.
  
  For A level purposes, the electrolytic strength and electrical conductivity of an ionic compound is directly proportional to the *number* of aqueous ions (regardless of magnitude of charges, eg. Na+ vs Mg2+, or charge densities, eg. F- vs Cl-) present per unit volume of solution, ie. the total *molarity* of aqueous ions. Since both sodium chloride and sodium enthanoate are completely soluble, and both dissolve to generate the same *number* of ions (per mole of ionic compound), hence electrolytic strength and electrical conductivity of both sodium ethanoate and sodium chloride are regarded as the same.
  
  Beyond A levels, the electrical conductivity of an ionic species is also further affected by : activity coefficients, ionic atmosphere, asymmetry/relaxation effect, electrophoretic effect, electrostriction, etc.
  
  Q2. You seem to be confusing the 105 deg bond angle *within* a water molecule, and the 109.5 deg hydrogen bond angle *between* water molecules in solid ice. Because of the molecular geometry of water (the 2 lone pairs on the O atom, and the 2 partial +ve H atoms, are tetrahedral relative to each other), and therefore ice *naturally* (hence it's "beneficial", ie. without incurring significant angle deviation or strain) takes on a tetrahedral lattice structure (we say ice is a "solid hydrogen bonded lattice structure"). The so-called slight "energy penalty" (in the Wikipedia article) refers to the work done against the surroundings when water increases its volume during the freezing process.
  
  At temperatures and pressures of liquid water, the H2O molecules have sufficient kinetic energy to overcome the rigid hydrogen bonds that hold the H2O molecules in fixed positions in the solid ice state, and thus the H2O molecules will gladly move about as much as the temperature and pressure allows them, ie. a (thermodynamically favorable) positive entropy change (relative to the solid ice state). In the liquid state (unlike the solid and gaseous states), the hydrogen bonds between H2O molecules are continuously broken and formed, allowing the H2O molecules to move about.
  
  So to address your question directly : it's (positive) entropy that prevents water at liquid state temperatures and pressures from remaining as a solid. Depending on the different temperatures and pressures that permits and favors water to be in the different states, it's advantageous entropy-wise (hydrogen bond breaking) for water to transit from solid to liquid to gaseous states, and it's advantageous enthalpy-wise (hydrogen bond forming) for water to transit from gaseous to liquid to solid states. Assuming constant pressure, the main determining factor (for whether water choose to be solid, liquid or gas) is hence temperature (as seen in Gibbs free energy formula).
• UltimaOnline

  30 Jun 15, 09:52

  Originally posted by hoay:

  the Enthalpy change of formation of elements in their standard states have a zero value. Why the same is not true for standard entropy change? 

  
  By definition, *enthalpy change* of formation of elements in their standard states have a zero value, while the *entropy* of a perfect crystal at absolute zero is exactly equal to zero.
  
  Here's why standard entropy change for most reactions isn't zero : consider (as an example) the formation of ammonia : notice that in the balanced equation for N2 and H2 elements to combine chemical to form NH3, there are 4 moles of gases on the LHS, and only 2 moles of gas on the RHS, which therefore results in a negative entropy change (whether at standard or non-standard conditions).
• UltimaOnline

  12 Jul 15, 09:35

  ^_^
• UltimaOnline

  19 Jul 15, 01:18

  Ghoby posted :
  
  

  Hi UltimaOnline, I have some questions on Reation Kinetics :)

   

  Q1: http://img.photobucket.com/albums/v700/gohby/Chemistry/Q34%20ACJC_zps63m4r5pe.jpg

   

  Response: The order of reaction wrt X is 1 and wrt Y is 1. But how to proceed from there?

   

  Q2: 2SO2(g) + O2(g) ⇔ 2SO3 (g) â–²H = -98kJ/mol

   

  Which of the following conditions increase both the rate of reaction and the yield of products?

   

  1. Add a suitable catalyst and add a suitable solvent to dissolve SO3

  2. Decrease the volume of the vessel

  3. Increase the temperature and decrease pressure

  
  Remarks: The answer is 1 and 2. I don’t understand why the answer is 1 because if we use a suitable solvent to dissolve SO3, wouldn’t the yield of sulphur trioxide drop instantly?

   

  Q3: If the rate equation (and its concomitant rate order(s)) is experimentally derived - how come we are able to deduce it from mechanisms - isn’t that being theoretically derived then?

   

  Q4: 2O3(g) → 3O2(g)

  This reaction is thought to occur via a two–step mechanism:

   

  Step 1 O3(g) O2(g) + O(g) fast, reversible

  Step 2 O3(g) + O(g) → 2O2(g) slow

   

  The rate law that is consistent with this mechanism is k[O3]²/[O2].

  
  Remarks: I know how to get the answer, but I am wondering why the rate equation contains the product as well (shouldn’t it only be expressed in terms of the reactants)? How can the concentration of the product affect the rate of reaction when it is an not a reversible reaction?

  Thank you very much, UltimaOnline :)

• UltimaOnline

  19 Jul 15, 01:18

  Hi Gohby,
  
  For Q1, if you interpret the question to imply molecule Z to be the only product (ie. no other products formed on RHS in the overall equation), then options 2 and 3 are both possible. Option 1 is ruled out because the energy profile diagram indicates an intermediate.
  
  For Q2, it is implied that you can subsequently retrieve your SO3 from your solvent.
  
  For Q3, you have to experimentally derive the orders of reaction first, then use them to deduce the mechanism. Orders cannot be deduced by inspection of the overall equation, but can be deduced by inspection of the rate determining elementary step of the mechanism.
  
  For Q4, the rate equation must be in terms of only reactants and products, but not intermediates. The product O2 can affect the rate of reaction because in the reversible elementary step, the molarity of O2 affects the molarity of O, the intermediate which is involved as a reactant in the rate determining elementary step.
  
  You're welcome, Gohby :)
• UltimaOnline

  21 Jul 15, 01:27

  Hi UltimaOnline,

  Thank you for your response. :)

  For Q1, why does the presence of an intermediate rule out option 1 being the possible overall equation? Am I right to say any option that follows: a X + b Y → Z, where a and b are not 1 would be a plausible option?

  
  For Q3: Am I right to say that mechanisms are merely guesses at how a reaction takes place step by step at a molecular level? Which is why when even we derive the orders from the slow step in a mechanism we are still deriving it experimentally (rather than it being theoretical since we already know the mechanism already)? At the H3 level (or even at times at H2 Chem!) one would have to deduce a mechanism from a given reaction, and deducing mechanisms requires certain theoretical knowledge (e.g. start from an electron rich region, hydride shifts etc.), so isn’t mechanisms deduced theoretically then (and thus rate equations could be said to be deduced theoretically)?

  I have two other further questions:

   

  Q5: The diagrams  P, Q, R and S show how a change in conditions affects the Maxwell-­â€� Boltzmann distribution of molecular energies for gas G. In each case, the original distribution is shown by a solid line and the distribution after a change has been made is shown by a dashed line. Which one of the following statements is correct?

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/Q26%20TPJC_zpsa9xv8jje.jpg

   

  Answer: The change shown in diagram S occurs when the pressure of G is decreased at constant temperature

  
  Remarks: How will a decrease of pressure lead to a decrease in the total number of molecules?

  Q6: Substances R, S and T react according to the following equation:

   

  R(aq) + 2S(aq) + T(aq) → 2U(aq) + V(aq)

   

  To find the rate equation for the above reaction, two sets of separate experiments were  performed and the results are shown  below.

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/Q31%20RI_zpsurt489vr.jpg

   

  What can be deduced from the information  shown?

   

  3 The overall order of the reaction is two.

  
  Remarks: How do I tell whether 3 is correct?

• UltimaOnline

  21 Jul 15, 01:28

  Yo Gohby,
  
  Q1. If there wasn't an intermediate, then option 1 would be true, ie. the overall reaction is simply the single (and thus rate determining) elementary step. But since there's an intermediate, option 1 could be the rate determining step, but can't be the overall equation.
  
  Q3. Yes, you're right. For complex reactions, even after carrying out experiments to determine the order of reaction for the reactants, there may be 2 or more plausible mechanisms that fit the experimentally determined orders of reaction. For such reactions, kinetics experiments alone cannot narrow down or confirm exactly which mechanism is the correct mechanism, and further experiments (beyond the scope of A levels) must be carried out.
  
  Q5. You're right, a decrease of pressure does *NOT* lead to a decrease in the total number of molecules (unless this was an equilibrium question with different no. of moles of gas on the LHS vs RHS). Even Cambridge A level papers have errors (not just typo, sometimes conceptual errors), let alone Singapore JC Prelim papers, which you certainly cannot trust.
  
  Kinetic energy is directly proportional to temperature, and since temperature remains unchanged, hence average kinetic energy of the gas particles remain unchanged, and hence there would be no change to the shape of the Maxwell–Boltzmann distribution curve, and certainly no change to the total number of molecules (ie. area under the curve).
  
  Put in another way, PV=nRT, since pressure is decreased and temperature is constant, and n is also constant (since you can't change the number of molecules unless you're talking about equilibrium reactions which is not what this question is about), it means volume must increase, which doesn't affect the Maxwell–Boltzmann distribution curve (only temperature can affect the Maxwell–Boltzmann distribution curve) for a single gas species (with a fixed mass).
  
  Q6. Option 3 is wrong because the overall order of the reaction is 3. From inspection of the 1st graph, we deduce the reaction is 0th order w.r.t. R *and* 2nd order w.r.t. S (can you figure out why?). From inspection of the 2nd graph, we deduce the reaction is 1st order w.r.t. T (can you figure out why?).
• UltimaOnline

  21 Jul 15, 19:03

  Gohby wrote :
  
  

  Q1: Can I say that any option for a X + b Y → Z, where a and b are not 1, can be accepted (since we don’t know how does the mechanism works?)

   

  Q5: I redacted the options but in the interest proving your conjecture here’s the question in full:

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/Q26%20TPJC_zpszfdhvewd.jpg

   

  Yes I was very perplexed by the choices; A-C are evidently wrong. I thought that D was an option only when I inject a sample of gas into the original sample at the same temperature (or an equilibrium system as you’ve suggested). If this has been transcribed accurately from the actual Prelim Paper, I would be worried as it would be a difficult task to read the minds of the question setters. I am surprised that even in A Levels there are examples of conceptual errors, are you able to point out a couple?

   

  Response to Q6:

   

  The order of reaction wrt R is 0 because with reference to the any of the graphs in the first diagram, [R] decreases linearly, i.e. a decrease in [R] does not lead to a decrease in the rate of reaction → Order wrt R = 0.

   

  The order of reaction wrt T is 1 because the with reference to the curve in the second diagram, the half life is constant.

   

  I wasn’t too sure about the order of reaction wrt S though, I know it can’t be 0 because from the graphs it affects the rate of depletion of R and T when [S] is increased, but here’s my take (don’t know if it’s the most efficient way of thinking): the rate of depletion of [R] is increased by a factor of 4 (this is determined by comparing the gradients of the two graphs in diagram 1) when [S] is doubled. Given that the concentrations of other reactants remain unchanged, the order of reaction wrt S has to be 2. (2²=4).

   

  Q7: To identify an oxide of nitrogen, 0.10 mol of the oxide was mixed with 10 dm³ of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4 dm³ of hydrogen gas remained. The ammonia produced required 125 cm3 of 1.6 mol/dm³ HCl for neutralisation.

   

  All gaseous volumes were measured at room temperature and pressure. What is the formula of the oxide of nitrogen?

   

  A:NO   B: N2O C: NO2 D:N2O4

   

  Remarks: Essentially from the calculations, the ratio of nitrogen oxide: H : NH3 = 1:4:2. Hence, B is the answer because it is the only choice which fulfills this ratio (N2O + 4H2 → 2NH3 + 2 H2O). However, my qn is - how do I know that a nitrogen oxide, when mixed with a hydrogen gas will produce water (in addition to ammonia as stated in the question)? If I am unable to ascertain the other product in addition to ammonia, how do I then write an equation for all the choices to obtain the answer?

  
  Thank you UltimaOnline :)

• UltimaOnline

  21 Jul 15, 19:03

  Q1. Yes, you can say that.
  
  Q5. Yup, all 4 options are wrong.
  
  As for Cambridge A level papers, at the very least, every year there are at least a couple of A level qns that are *ambiguous* (and *sometimes* arguably conceptually erroneous if you've higher level understanding of chemistry, but it's usually not at a level that will affect students, so not really an issue), in which Singapore JC teachers disagree amongst themselves on what the best answer should be. For examples, just buy the 3 or 4 TYS publications available on the market, and note the ambiguous questions for which different authors (many of them ex-MOE teachers themselves) provide different answers for.
  
  Of course, Singapore JC Prelim papers are far worse in this regard. It is common... nay, invariable, that every year, a teacher from a JC will tell their students to "ignore this other JC's terrible mark scheme for this Prelim paper qn, and use our own JC's notes instead". In other words, most Singapore JC teachers won't be able to get an A grade, if they sat for another JC's prelim papers and had their papers marked by the rival JC's teachers.
  
  Heck, it's common even *within* a JC, for teacher A to tell his class to "don't use teacher B's horrible lectures/teachings/notes/mark schemes". Hence, don't trust Prelim papers and their mark schemes.
  
  Rest assured, that Cambridge is far more *reasonable* in this regard (compared to Singapore JC teachers and their Prelim papers and mark schemes), and as long as your students write reasonable answers in a reasonable way, they'll get reasonable marks from Cambridge.
  
  Q6. Yes you're correct, well done. Hence it's overall 3rd order.
  
  Q7. You're right in your calculations and deduction. Since this is an MCQ qn, students should use all available data (including options) to arrive at the best answer, as you've done. You could say this qn would be somewhat unfair to the student, if asked as a non-MCQ qn. As for your query "how do I know that a nitrogen oxide, when mixed with a hydrogen gas will produce water", this is a redox reaction (which more capable students should be able to recognize based on the description given), and by default (unless otherwise specified by the question), water is often a reactant or product of redox reactions (recall the method for balancing redox reactions).
  
  No problem, Gohby! :)
• UltimaOnline

  31 Jul 15, 03:04

  Gohby asked :
  

  Hi UltimaOnline, I have some further questions (on ionic equilibria this time):

   

  Re Q7 @ 21 Jul 15: this is a redox reaction (which more capable students should be able to recognize based on the description given)

   

  Remarks: How can I recognise that it’s a redox reaction? Just because there’s reduction when a nitrogen oxide is reduced to ammonia? But how does that ensure oxidation will occur?

   

  Q1:

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/CJC2012_zpsnohp6cyi.jpg

   

  Answer: 1 and 2 are correct.

   

  Remarks:

   

  1. For 2, what is meant by the point when the slope of the curve is at maximum at its centre?? Is it the point as indicated by the arrow (which I have inserted) in the diagram? If so then 2 is correct because the first neutralisation reaction would have been complete and the salt produced has no net charge.

  2. For 3, is it wrong because at pH 9.7, there are equal concentrations of H3N+CH(CH3)COO- and H2NCH(CH3)COO- (because pH= pKa → max buffer capacity → where the concentrations of the weak acid and its conjugate base is the same)

   

  Q2: Does this statement “H+ (aq) + OH¯ (aq) ions react exothermically” give a correct explanation for “The ionic product of water, Kw increases with increasing temperature”?

   

  Remarks: I am of the opinion that the first statement gives a correct explanation for the second statement. Given the water dissociation equation: H2O ↔ H+ + OH¯, if H+ (aq) + OH¯ (aq) ions react exothermically, the forward reaction must be endothermic. Hence, when the temperature increases, by LCP, the equilibrium will be shifted to the right to minimise the disturbance, which would increase the concentration of H+ + OH¯ at equilibrium, which will increase the ionic product of water. However, according to the answer, the statement statement “H+ (aq) + OH¯ (aq) ions react exothermically” does not give a correct explanation for “The ionic product of water, Kw increases with increasing temperature” and I don’t understand why.

   

  Q3:

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/CJC20122_zps4ftdajiy.jpg

   

  Remarks: Why is B wrong? I thought the degree of dissociation, which is related to the Ka, will only be affected by changes in temperature? How is D the correct answer? (Kb of ammonia is larger than water → it is a stronger base than water → what’s the relation of the strength of HA in liquid ammonia?

   

  Thank you for your help, much appreciated. :)

• UltimaOnline

  31 Jul 15, 03:16

  Q7 - Oxides of N have OS of N in the positive OS (since O is more electronegative than N). NH3 has N in the negative OS (since N is more electronegative than H). Hence N is reduced, and when somebody's reduced, somebody's gotta be oxidized. Here, obviously H is being oxidized (since H is a reactant with OS of 0) to H2O.
  
  Q1. Yes, you're correct. The point referred to is the 1st equivalence point.
  
  Q2. An ambiguous qn. Don't trust Singapore JC Prelim papers. You can argue the relevancy of 1st statement vis-à-vis the 2nd statement either way. Your thought process is fine, just that the 1st statement rather *indirectly* relates to the 2nd statement. The given answer's option (which you didn't reveal) probably relates more *directly* to the 2nd statement.
  
  Q3. The Ka of an acid depends on both temperature and solvent. By default, the solvent is always water, unless otherwise specified. Which is why, based on the solvent as water, temperature is thus the only (other) factor that affects the Ka value. Since NH3 is a stronger base than H2O, NH3 is able to deprotonate HA to a greater degree, and hence the Ka value of any acid HA, would be larger in NH3 than in H2O.
  
  You're welcome to keep your questions coming, Gohby! :Þ
• UltimaOnline

  1 Aug 15, 02:48

  Light5 asked :
  

  Hi, i am studying Cambridge Internation A level chemistry and have found that the questions asked are easier but very similar to the H2 chemistry papers. And as CIE is increasing the difficulty of papers each year, the questions will be more like the ones in the singapore prelim papers. Hence, i would like to ask questions that i have found tricky and difficult from both CIE and H2(i cant even solve some of the H2 prelim questions). 

  Q1:)

  Please look at Q3(d)(iii) of the following paper

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w13_qp_23.pdf

  How will i know what products will be formed when BrCl reacts with KI....i know its supposed to be displacement reaction but how can i be sure that KBr and KCl will be formed instead of Br2.....

  Q2.) Similarly, how would i know the products of the reaction of NaH(sodium hydride) with water if i only that MgH,NaH both react with H2O to give the same colourless guess?

  Is there any general way of knowing the products of chemical reactions such as above in keeping the fact that we know all the reactions of the chapter of periodicity(such as chlorides and oxides with water,etc)?

  Thank You for answering!

• UltimaOnline

  1 Aug 15, 02:49

  Hi Light5, welcome to the forum! :)
  
  Q1. Even if you suspect Br2 may be formed as a possible intermediate (in the reduction of BrCl, which may be conceptualized as Br+ and Cl-, with OSes of +1 and -1 respectively), but knowing full well (as required by the A level syllabus) that the redox reaction between Br2 and I- is (of course) thermodynamically feasible under standard conditions, you should always give the final equation with the final product of the completed reaction, ie. in effect acknowledging that any Br2 formed (eg. as possible intermediate) would be further reduced to Br- by I- (which is in turn oxidized only to I2... any further oxidation would require a stronger oxidizing agent, and clues would be given in the question).
  
  Q2. If you (ie. as an A level candidate) correctly understand the nature of metal hydrides, you'll understand that it consists of the metal cation and the hydride anion H-, which being highly unstable (due to the high anionic charge density due to small ionic radius exacerbated by the low electronegativity of H) is thus a strong Lewis base (ie. strong nucleophile and strong Bronsted-Lowry base), and hence readily abstracts a proton from water (which you should be aware is a protic species) to generate H2 gas, leaving behind the metal hydroxide as the by-product (not to be confused with side-product).
  
  Edited to add : The forward reaction is favored both thermodynamically (due to positive entropy change) as well as by Le Chatelier's principle (due to gaseous product leaving the solvated reaction mixture).
  
  If your understanding and familiarity of reaction mechanisms (more commonly employed in organic chemistry, but chemistry is chemistry, and mechanisms are actually every bit as relevant to inorganic chemistry, despite the lack of exposure in this area for all A level students, whether international or Singapore) is strong, you're advised to draw out the mechanism for this (and other inorganic chemistry) reactions, which will enable you to truly understand (rather than just blindly memorize) why and how such reactions occur, and as such also helps you to correctly determine the identity and structure of the products.
  
  Feel free to continue posting your A level Chem qns here all the way till your A levels at the end of the year, Light5! ;)
• UltimaOnline

  1 Aug 15, 04:41

  Q1. No worries about focusing on Group vs Period, just do your best to *understand* (with use of mechanisms whenever possible; ie. you can attempt drawing out the mechanisms for most, if not) all of the inorganic reactions required by your own A level syllabus, instead of blind memorization. Make sense of it in your own way (eg. by use of analogies), own it for yourself.
  
  Q2. HX does indeed react in a nucleophilic substitution reaction with alcohols (you can attempt to draw the mechanism for this... fact is, online help via forums is by its nature limited, it's not feasible for me to draw out all the mechanisms you may be interested in; I can and will do this for my own tuition students when they're physically present, but it's not feasible and too time-consuming to do this online... if you're really interested in learning how to draw mechanisms for all A level reactions organic and inorganic, you might want to consider engaging a private tutor, in whichever country you're residing in).
  
  Nucleophilic substitutions of HX with RX can and does occur (of course, this is only meaningful with two different halides are involved), with the position of equilibrium favoring the thermodynamically more stable species; specifically the species with the stronger C-X covalent bond (ie. a more effective head-on or end-on overlap by less diffused electron orbitals to form a stronger sigma bond results in a more exothermic bond formation enthalpy).
  
  However, a twist to this was given in the Singapore-Cambridge 2013 A levels P3 Q1, in which (if you have access to the Singapore A level papers, you might like to check this out) the strongly exothermic (thus thermodynamically favorable) lattice formation enthalpy involved in the ionic precipitation of the lower Ksp sodium halide in propanone (polar aprotic) solvent, together with shifting of position of equilibrium as predicted by Le Chatelier's principle, results in R-I generated at the expense of R-Cl, despite R-Cl being a stronger covalent bond.
  
  However for aldehydes and ketones, any nucleophilic addition of the halide ion will be followed by it's elimination (the mechanism for nucleophilic acyl substitution is addition-elimination) to re-generate the aldehyde or ketone, as the halide ions are good leaving groups due to their low anionic charge densities (as opposed to say, hydride ions from LiAlH4 or alkyl nucleophiles from Grignard reagents).
  
  As for any potential Bronsted-Lowry acid-base reactions between HX and alky halides or aldehyde/ketones, alkyl halides and aldehyde/ketones aren't Bronsted-Lowy basic to begin with (aldehydes and ketones are actually slightly Bronsted-Lowry acidic, as the conjugate bases generated by the deprotonation of their alpha protons, are stabilized by having their negative formal charges delocalized by resonance over both the alpha C atom and the electronegative carbonyl O atom of the enolate ion conjugate base... if you're up for the challenge, draw the mechanism between both resonance contributors and elucidate the resonance hybrid).
  
  Q3. This is simultaneously a redox reaction as well as a Bronsted-Lowry acid-base reaction. It is a first and foremost a redox reaction because Mg is oxidized to Mg2+, and H+ from the acid is reduced to H2 gas. It is secondarily a Bronsted-Lowry acid-base reaction because HCl (if in aqueous state, more precisely hydroxonium chloride) is being deprotonated to generate it's conjugate base the chloride ion (or more precisely, water and the chloride ion) and thus generating the salt MgCl2(aq). For a P1 MCQ, choose the best option (psychologically profile the Cambridge setter's intentions if you're able to). For a P2 or P3 qn (or whatever equivalent in your syllabus), give *both* descriptive alternative answers with qualifying explanations to secure full credit.
  
  Q4. Every year, many students routinely ask, "is there a shortcut to drawing all possible isomers?". The short answer is : no there isn't. There can be systematic approaches (it's up to you to formulate your own), but there's no magical shortcut to drawing all possible isomers, especially when stereoisomerism (optical and geometrical) is also to be considered.
  
  But there is indeed a method (in fact, this is the *only* correct method, though it's far from being a shortcut) to reliably identifying any repeated structures. That's to name each of your structures. If they've the same name, then they're the same molecule.
  
  No problem Light5. Go ahead and feel free to continue asking your questions from both your CIE A level syllabus, as well as from Singapore Prelim papers. Though when quoting questions from past CIE papers or Singapore Prelim papers, it'll be preferred if you can provide the (eg. http://papers.xtremepapers.com) links to not only the question paper, but to the mark scheme as well (whenever available), for the sake of efficiency and convenience.
  

  Originally posted by Light5:

  Q1.)Thanks alot, couldnt have even thought of bronsted lowry theory for Q2..do you have any advice as to how to study reaction mechanisms for Inorganic chemistry(any book,webisite,etc) that can be of help? Instead of the Period 3 chlorides,oxides and group2,group 7 chapters, should i learn more compounds formed by period 3 elements and their corresponding reactions and if so, which ones should i focus on?

  Q2.) I know that HBr is an acid and Br, being more electronegative than H, has a partial negative charge..Also, Br has few lone pairs on it...if this is the case why cant HBr participate in a nucleophilic substitution or nucleophilic addition reaction with halogenoalkanes and aldehydes/ketones respectively?

  Q3.) How is Mg + 2HCL = MgCl2 + H2 an acid- reaction? I know that HCl is giving out H+ ions, but it isnt donating these protons to Mg, in fact it is donating Cl- ions to Mg...so how does this reaction stand true with bronsted-lowry acid-base theory?

  Q4.) Is there any general strategy for drawing correct structural isomers ensuring that none of them are repeated...how can i do this on a piece of question paper(since we dont get a rough sheet for the MCQ paper 1). Even if you can demonstrate it with an example, that will be fine?

  Again, thank you for answering my questions. I know my queries arent as advanced as those of singapore students but some of these misconceptions have arised while solving CIE past year questions...i will start asking questions related to singapore prelim papers once i start solving some of the difficult P1 and P3s. 

  
  
• UltimaOnline

  1 Aug 15, 04:56

  Originally posted by Light5:

  Q5.) What could be the products and the mechanism for the following reaction :

  NO2 + H2O = ...?

  All i can predict is that nitric acid will be formed as NO2 is an acidic oxide however, that does not balance the equation so what will be the other product?

  
  Oxides of nitrogen are a complicated and messy affair, and for A level purposes, Cambridge would likely provide in-context hints if such is asked.
  
  The competent and astute A level candidate would notice that the OS of N in NO2 is +4, but yet the OS of N in the common N containing acids HNO2 and HNO3 (Latin names : nitrous and nitric acid; Stock names : nitric(III) and nitric(V) acids) are +3 and +5, which should lead you to a reasonable deduction that this isn't a straightforward case of hydrolysis of an electrophilic covalent oxide to generate an aqueous acid in which the OS of the heteroatom is conserved, as would usually be the case, eg. CO2 + H2O --> H2CO3, or SO3 + H2O --> H2SO4. At the very least, Cambridge would expect the competent and astute A level candidate to suggest the existence of a disproportionation reaction, and Cambridge more than likely would provide in-context hints within the question as to the exact identities of the products and intermediates (see links below).
  
  If you're interested in a more in-depth look into this reaction (it's entirely possible Cambridge might provide some of the following info in a data based question, and query the A level candidate accordingly across several related topics in the A level syllabus, possibly even asking for the reaction mechanism to be drawn out, ala Singapore-Cambridge A level 2012 P3 Q3), you can check out :
  http://pubs.rsc.org/en/content/articlelanding/2003/cp/b208564j#!divAbstract
  http://www.researchgate.net/publication/234036456_The_Heterogeneous_Hydrolysis_of_NO2_in_Laboratory_Systems_and_in_Outdoor_and_Indoor_Atmospheres_An_Integrated_Mechanism
• UltimaOnline

  6 Aug 15, 18:36

  Gohby asked :
  
  

  Hi UltimaOnline,

   

  Continuing from the previous thread, I have further questions on equilibria to seek clarification:

   

  Q1:

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/SRJC_zpswcv8i3zn.jpg

   

  Answer: C

  Remarks: Where pH=pKa, [weak acid] = [conjugate base]. Number of moles of CH3COONa = 0.0001. So shouldn’t the answer be D, because at point D the number of moles of excess ethanoic acid when 20cm³ of it has been added be 0.0001 too?

  
  

  Q2: Bleaching solutions are manufactured by dissolving chlorine gas in sodium hydroxide solution to give the following reaction. Cl2 (g) +  2OH­â€� (aq) ↔ OCl­â€� (aq) + Cl­â€�  (aq) + H2O (l) Users are warned not to mix the bleach with other cleaning solutions to prevent evolution  of hazardous chlorine gas. Which of the following actions will lead to liberation of chlorine  gas?

   

  A Addition of water to bleach

  B Mixing of an alkali with bleach

  C Shaking bleach with table salt, NaCl

  D Subjecting bleach to high pressure

   

  Answer: C

   

  Remarks: Why doesn’t A result to the liberation of chlorine gas too since the equilibrium will shift to the left upon the addition of water? Wrt choice D, if I were to subject the equilibrium system to high pressure, am I right to say that the equilibrium will shift to the right so as to reduce pressure by reducing the number of gaseous particles instead?

   

  Q3: A car burning lead­â€�free fuel has a catalytic converter fitted to its exhaust. On analysis, its exhaust gases are shown to contain small  quantities of nitrogen oxides.  Which modifications would result in lower exhaust concentrations of nitrogen oxides?

  1 an increase in the surface area of the catalyst  in the converter.

  2 an increase in the rate of flow  of the exhaust gases through the converter.

  3 a much higher temperature of combustion in the engine

   

  Answer: 1 only

   

  Remarks: My understanding is this: 2 is wrong because an increase in the rate of flow of gases does not increase the number of active sites on the catalyst, which will not lead to an increase in the rate of reaction. 3 is also wrong because combustion is not a reversible reaction so the higher temperature will not affect the concentration of products. Am I right?

   

  Q4: Pure NOCl gas, was heated at 320°C in a 2 dm³ vessel. At equilibrium, 30% of the NOCl  gas had dissociated according to the equation below and the total pressure was p atm. 2NOCl (g) ↔ 2NO (g) + Cl2 (g)  What is value of Kp?

   

  Answer: 0.0120p

   

  Remarks:

   

  My workings:

   

                       2NOCl (g) ↔ 2NO (g) + Cl2 (g)

  Initial no of moles        a                  0        0       

  Change “        -0.3a        0.3a        0.15a

  Eqm “            0.7a        0.3a        0.15a

  Mole Ratio        0.7/1.15    0.3/1.15    0.15/1.15

  Kp = (0.15/1.15)p  x  (0.3/1.15)²p²

          _______________________

             (0.7/1.15)²p²                

   

  That works out to ~0.024p, and I can’t figure out where the mistake lies.

   

  Thank you for your time, UltimaOnline :)

• UltimaOnline

  6 Aug 15, 18:36

  Hi Gohby,
  
  As always, you've a very good style of asking questions, with your own mathematical working and thought processes indicated clearly.
  
  Q1. You're right, the correct answer is D.
  
  Q2. While you're right that adding water will shift the p.o.e to the LHS, but the effect is less significant than option C, because as indicated by (aq) state symbol, water is already present in very large excess. For D, notice the qn states increasing pressure on bleach, not the entire system. Since bleach is aqueous, increasing pressure on an aqueous species has no effect on p.o.e.
  
  Q3. Options 2 & 3 will *increase* the molarities of the oxides of nitrogen. Option 2 affords less time and opportunity available for the catalysts to do their work. Option 3 provides more energy above the activation energy barrier for oxides of nitrogen to be generated.
  
  Q4. Your working and answer is correct. Volume of container affects the answer only for Kc, not Kp.
• UltimaOnline

  21 Aug 15, 18:05

  Gohby asked : Hi UltimaOnline,
  

  I have a question on organic synthesis.

  I understand that P will react with aqueous sodium hydroxide (nucleophilic substitution), resulting to the substitution of the 3 Cl groups to 3 OH groups. However I do not understand how/why the ketone in Q is formed.

  
  
  UltimaOnline replied : Hi Gohby,
  
  Due to close proximity of the 2 OH groups in geminal diols (in the intermediate between P and Q), intramolecular proton transfer (can be catalyzed by acid or base) followed by elimination of a H2O molecule with thermodynamically favourable positive entropy change can readily occur.
  
  As predicted by Gibbs free energy formula, (positive entropy change) is encouraged by heating (as specified in the question), which also provides the activation energy for the proton transfer and elimination to occur (subsequent to the SN2 and SN1 hydrolysis nucleophilic substitutions).
  
  Furthermore, heating also removes H2O (in gaseous state) from the solvated reaction mixture, and concordantly as predicted by Le Chatelier's principle, shifts the position of equilibrium over to the aldehyde/ketone side.
  
  If large amounts of water are added at lower temperatures, the position of equilibrium can be made to shift back to the germinal diol side.
  
  Additional info : 2 other factors that can affect the position of equilibrium between geminal diols and ketones/aldehydes, are : presence of strongly electron-withdrawing by induction groups (eg. ninhydrin and chloral hydrate) increasing the electrophilicity of the carbonyl C atom for nucleophilic hydrolysis thus favouring the hydrolysis product the geminal diol ; and ring strain due to angle strain (eg. cyclopropanone and its geminal diol hydrate) favouring the less severely strained tetrahedral sp3 gemial diol species, over the more severely strained trigonal planar sp2 ketone/aldehyde species.
  
  While this reaction isn't specified in the basic H2 syllabus, but with sufficient clues given in the question (ie. the molecular formula, with which you can work out the degree of unsaturation to clue you into the structure of the final product, yet another useful tool which H3 / Olympiad / BedokFunland JC students can use as an advantage over Singapore JC H2 Chem students), such questions can, and has indeed (eg. in recent Singapore-Cambridge A level H2 Chem papers) been asked of H2 Chem students.
  
  Bonus qn for your students : what would you get if you started with a geminal tri-halide (instead of a geminal di-halide)?
• UltimaOnline

  27 Aug 15, 02:10

  Originally posted by Light5:

  Q1.)

  (a) State the colour you would expect for the interhalogen compound BrCl

  (b.) Predict the equation of the reaction between BrCl (interhalogen) and AgNO3.

  (c.) Describe what you would observe when the products of (a) are added to Dilute NH3 in a test tube.

  Q2.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

     Please explain how to solve Q13 and Q24 from the above paper. 

  Mark Scheme :         http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_ms_11.pdf

  Thank You !

  
  Where did you get the interhalogen qn from? Is it a CIE 'A' level Chem qn, or a school's exam qn (if so, which country is the school based in)?
  
  Q1a) Somewhere between the colours of Cl2 (yellow-green) and Br2 (reddish-brown).
  
  Q1b) Although AgBr is less soluble than AgCl, but due to the greater electronegativity of Cl over Br, we have Br+ (not Br-), hence the products would be AgCl(s) and (covalent) BrNO3. BrCl is a strong oxidizing agent, but neither Ag+ nor NO3- (with N already in it's most positive OS) can be readily oxidized (it is difficult to oxidize Ag+ and O atoms having OS of -2).
  
  Q1c) The AgCl ppt would dissolve in NH3(aq), as the soluble ionic diamminesilver(I) coordination complex is generated. The Br atom of BrNO3 would also be nucleophilically attacked by NH3 and H2O, but the reaction products are not readily observable (hence irrelevant to the question).
  
  Q13. This qn tests you on the thermal stabilities of Group 2 carbonates (determined by cationic charge densities). As a fire retardant, you would want two things : greater moles of CO2(g) generated per mole of retardant, and lower thermal decomposition temperature to release the CO2(g) more readily.
  
  Q24. The mechanism involves 2 x nucleophilic substitution (first SN2 on the less sterically hindered terminal C atom, followed by deprotonation, then intramolecular SN2 or SN1 attack on the 5th C atom, followed by deprotonation) of the Br atoms on the 1st and 5th C atoms, by the NH3 nucleophile, to generate the required product Coniine.
• UltimaOnline

  27 Aug 15, 04:18

  Originally posted by Light5:

  Sorry if the interhalogen question was vague....i actually constructed the question by myself because cambridge has a knack of asking weird questions from time to time. They have already tested displacement reactions of interhalogen compounds so i wondered if they could test a question like this as well.

  In q1(c) , we would obviously have a colourless test tube as all nitrates are soluble and AgCl is also soluble but what would be the products of the nucleophilic attack of NH3 and H2O on BrNO3?

  In q13(from the CIE paper), how can i predict the temperature required for decomposition of the 3 fire retardants since there are 2 Group 2 elements in one fire retardant.

  Thanks Again.

  
  No prob.
  
  Q1c) NH2Br + HNO3, and BrOH + HNO3, all of which are colourless.
  
  Q13. Each metal carbonate will decompose according to their own temperature, and you can treat each mineral as a mixture of two separate metal carbonates. Eg. For Mg3Ca(CO3)4, magnesium carbonate decomposes first, followed by calcium carbonate. Hence BaCa(CO3)2 is the worst fire retardant because Ba2+ has the lowest cationic charge density and thus requires the highest temperature to decompose, so by the time sufficient CO2(g) is released, your house would have burnt down completely.
• UltimaOnline

  27 Aug 15, 14:58

  Originally posted by Mrworry:

  Hi again :D

  For Organic chem made easy page 258,

  Example 8.4, part b)

  Why cannot it be:

  1. Add conc H2SO4(aq) 170C

  2.Cl2 (aq), heat

  3. Add Ethanolic KCN, heat

  4. HCl (aq), reflux

  xie2=3

  
  
  Your proposed synthesis pathway will score some marks, but not full marks.
  
  You'll be penalized *mostly* because your proposed synthesis pathway isn't the shortest possible (ie. George Chong's answer). Always give the shortest possible pathway, because using a longer-than-necessary synthesis pathway will result in a lower atom economy and/or equilibrium yield, and hence will always be penalized by Cambridge. (But hey, if you can't think of any shorter pathway, getting some marks is certainly preferable to getting no marks, so go ahead and just write the shortest pathway you can think of, even if it's 1 or 2 steps longer than specified by the qn).
  
  In addition (this is a minor point), although your step 2 will indeed obtain the required intermediate (used in your step 3) as the major product, but nonetheless there will still be a mixture of products, which should still be avoided in synthesis qns whenever possible (it's perfectly fine to use it when there's no other choice, but in this qn, there *is* another, even shorter pathway, ie. George Chong's answer).
  
  But I emphasize that the main problem with your answer is that it isn't the shortest possible synthesis pathway. If it had the same no. of steps as the shortest possible pathway, then Cambridge (in most cases) wouldn't penalize you for using electrophilic addition of aqueous halogen. No worries.
  
  On this note, using halogenation via free radical substitution will always get you penalized for synthesis qns (as it generates a variety of products, hence low overall atom economy which means wasteful on resources), unless there is absolutely no other choice for that qn.