BedokFunland JC's A Level H2 Chemistry Qns (Part 1)
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'O' Level Qn.
Question :
A sample of nitrogen gas contains the same number of atoms as found in 4.0g of methane gas. What is the mass of the sample of nitrogen gas?
Answer :
(2)(x/28) = (4/16)(5)
x = 17.5g -
'O' level and 'A' level Qn.
Draw the Dot-&-Cross diagram of Urea (Chemistry 'O' Level Qn), given that urea is CO(NH2)2.
Solution :
Wikipedia Image of Urea displayed structural formula
So your diagram should have :
between O and C, double bond, so "xoxo"
between C and each N, single bond, so "xo"
between N and each H, single bond, so "xo"
make sure your O has two lone pairs, ie. "xx" and "xx" (not bonded with C)
make sure your N has one lone pair, ie. "xx" (not bonded with H or C)
check that total no. of valence electrons (both bond pairs and lone pairs) = 12 pairs = 24 valence electrons
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'O' and 'A' level Qn.
Methane was burnt in an incorrectly adjusted burner. The methane was converted into a mixture of carbon dioxide and carbon monoxide in the ratio of 99:1, together with water vapour. What will be the volume of oxygen consumed when 'y' dm3 of methane is burnt?
Solution :
Write and balance the equations :
Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2OCarbon monoxide
CH4 + 3/2 O2 --> CO + 2H2OBased on stoichiometry and info given in qn, we find
Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2O
0.99y 2(0.99y) 0.99y (don't care)Carbon monoxide
CH4 + 3/2 O2 --> CO + 2H2O
0.01y 3/2(0.01y) 0.01y (don't care)Total vol. of oxygen required =
vol of O2 needed for 1st equation + vol of O2 needed for 2nd equation
= 2(0.99y) + 3/2(0.01y)
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'O' level & 'A' level Qn.
For instance, given that average molar mass of Cl is 35.5, and that there are 2 isotopes, Cl-35 and Cl-37, find the relative frequency of each isotope.
Solution :
(x/100)(35) + ((100-x)/100)(37) = 35.5
x = 75
Hence 75% of all chlorine atoms have a molar mass of 35g, and 25% of all chlorine atoms have a molar mass of 37g.
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Soln to Q8.
Q8) Using the following information, draw the mechanism for the reduction of an amide to an amine, using LiAlH4 in dry ether.
[Description : after the hydride ion attacks the δ+ carbon and the pi bond electron density gets kicked up to become a lone pair on the oxygen; the oxygen then attacks AlH3 (again because Al in AlH3 lacks a full octet), and subsequently becomes a leaving group as the lone pair on the nitrogen forms a pi bond with the δ+ carbon; finally, another hydride ion attacks the δ+ carbon, and the pi bond electron density shifts back to become a lone pair on the nitrogen.]
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'A' level Qn :
How is it that transition metals can function as catalysts?
Solution :
Transition metals and their compounds function as catalysts either because of their capacity for variable oxidation states (hence allowing them to readily transfer electrons; this provides an alternative pathway with lower activation energy (as compared to the uncatalyzed pathway), hence increasing the rate of reaction) which is usually the modality of homogenous catalysts; or because of their capacity to adsorb/adsorp other substances (specifically, the reactants) on to their surface, and activate them in the process (transition metals have partially filled d-orbitals and hence can have different number of bonds, and can bond to a wide variety of ions and molecules; and so the catalyst can form temporary bonds with the reactant molecules, which weakens the remaning bonds within the reactant molecules (eg. from double to single bonds), eventually causing them to dissociate into highly reactive atoms, and/or by holding the reactant molecules in closer proximity with each other on the catalyst surface as compared to in free gaseous state, all of which lowers activation energy as well as increasing surface area concentration of reactants available for reaction), which is usually the modality of heterogenous catalysts.
A case of the former (homogenous catalyst; partially filled d-orbitals with capacity for variable oxidation states) would be the redox reaction (catalyzed by Fe2+/Fe3+ ions) between peroxydisulphate(VI) ions S2O82-, and iodide ions I-; to generate sulfate(VI) ions SO42- and iodine molecules I2.
Overall Redox : 2I- + S2O8 2- --> I2 + 2SO4 2-
Step 1 : S2O8 2- + 2e- <---> 2SO4 2-
Reduction potential = +2.01V
Step 2 : Fe3+ + e- <---> Fe2+
Reduction potential = +0.77V
Step 3 : I2 + 2e- <---> 2I-
Reduction potential = +0.54V
The uncatalyzed reaction (steps 1 and 3) is energetically feasible (according to redox potential values); however, the reaction is slow because the negatively charged S2O8 2- and I- anions repel each other.
In the presence of Fe2+/Fe3+ ions, a alternative pathway (that is also energetically feasible, but in addition) with much lower activation energy (since the reactants are oppositely charged ions), and with a much faster rate of reaction, can now occur :
S2O8 2- + 2Fe2+ --> 2SO4 2- + 2Fe3+
2Fe3+ + 2I- --> I2 + 2Fe2+
Two examples of the latter (heterogenous catalyst) would be nickel acting as a catalyst for the hydrogenation reaction between ethene and hydrogen in the presence of a nickel catalyst; as well as iron acting a catalyst for the synthesis of ammonia in the Haber process.
For heterogenous solid metal catalysts in the hydrogenation reaction of alkenes, the mechanism is as follows :
- Binding of the unsaturated bond (ie. adsorbtion/adsorption)
- hydrogen dissociation into atomic hydrogen onto the catalyst
- Addition of one atom of hydrogen
- Addition of the second atom
- Alkane leaves the surface of the nickel catalyst
A similar mechanism occurs for the Haber process :
First the H2 and N2 molecules form temporary bonds with the surface of the iron catalyst (ie. adsorbtion/adsorption); this interaction with the catalyst weakens the covalent bonds within the molecules and eventually causing the H2 and N2 molecules to dissociate into atomic hydrogen and nitrogen; which are highly reactive, and rapidly combine with each other to form NH3 molecules, which then leave the surface of the iron catalyst.
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'A' level Qn.
hydrogen iodide, hydrogen bromide, hydrogen chloride, hydrogen fluoride.
Arrange (and explain your arrangement) the above 4 hydrogen halides according to
a) their boiling points, from lowest to highest
b) their acidity when dissolved in water, from strongest to weakest
Solution :
The lowest to the highest boiling points :
Hydrogen chloride (-85.1°C) , hydrogen bromide (–66.38°C), hydrogen iodide (–35.36 °C), hydrogen fluoride (19.54 °C).
Hydrogen fluoride has the highest boiling point due to strong hydrogen bonding between molecules (fluorine is the most electronegative element in the periodic table, hence is capable of the strongest hydrogen bonding).
From hydrogen chloride down the Group to hydrogen iodide, the boiling points increase due to the increasing numbers of electrons in the molecules that in turn leads to an increase in the frequency and strength of (induced dipole - induced dipole) van der Waals forces.
The strongest to weakest acids :
HI > HBr > HCl > HF
This is because the bond energies for the hydrogen halides further down the group are lower, so they dissociate more easily (see Data booklet values below). Furthermore, the conjugate base halide ions increase in stability down the Group, due to the negative charge being spread out over a larger ionic radius (directly related to number of electron shells).
Bond energies
H-F 562 kJ/mol
H-Cl 431 kJ/mol
H-Br 366 kJ/mol
H-I 299 kJ/mol
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Solution to Qn 9.
Q9) When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product.
Solution :
1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.
2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.
3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.
4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 highly electronegative and hence strongly electron-withdrawing Cl groups.
5) However, carbanions are still more unstable compared to alkoxide or alkonoate ions, hence the next step is
6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H+); bond pair between proton (H+) and oxygen becomes lone pair on oxygen), forming
7) trichloromethane and sodium methanoate (negative charge of HCOO- counterbalanced by Na+; note that these ions are aqueous meaning dissolved in water, having a hydration shell, ion-dipole interactions with polar water molecules).
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'A' levels.
Qn.
Describe the bond enthalpies of the halogen molecules down the periods. Explain any anomaly present in the trend.
Ans.
(Quote Data Booklet values here.)
Down the group, due to increasing number of electron shells, the bond lengths increase and hence bond strengths and bond enthalpies decrease.
An anomaly presents itself in that the bond enthalpy of F-F is significantly lower than the bond enthalpies of Cl-Cl and even Br-Br.
The reason for this anomaly, is that due to the very short bond length of F-F, the two F atoms are brought exceedingly close together, that there is significant electrostatic repulsion between the lone pairs of both F atoms; consequently decreasing the effective bond enthalpy of F-F.
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'A' Level 2008 Qn.
Qn. In another separate experiment, a new alkane F, C6H14, was produced. When reacted with bromine under ultraviolet light, F produced only two isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H, explaining your reasoning.
Solution :
The only C6H14 alkane that when monobrominated would yield only 2 isomers one of which is a chiral compound, is 2,3-dimethylbutane.
Notice that 2,3-dimethylbutane has a mirror plane of symmetry. Substituting away H on the 4 terminal C atoms (1st, 4th, 2 methyl and 3 methyl carbons) are equivalent. That leaves the 2nd and 3rd C atoms, which are also equivalent thanks to this mirror plane.
To obtain the chiral compound, substitute away H on (any of the 4) terminal carbons. The 2nd (or 3rd) C atom would then be the chiral carbon.
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H2 'A' level 2008 paper :
List and explain the following alkyl halides (halogenoalkanes) according to their boiling points :
Alkyl bromide, alkyl iodide, alkyl chloride.
Solution :
There are two opposing factors, which you have to describe and explain in the exam to score full marks.
Electronegativity and hence strength of permanent dipole-dipole interactions would point to (highest boiling point) alkyl chloride, alkyl bromide, alkyl iodide (lowest boiling point).
However, no. of electrons and hence strength of induced dipole-dipole van der Waals interactions would point to (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).
As it turns out, based on experimental evidence, van der Waals interactions trumps/owns/pwnz/outweighs permanent dipole-dipole interactions (specifically in THIS context of alkyl iodide versus alkyl bromide versus alkyl chloride; because we're talking about a lot of electrons here).
Hence, the correct answer is (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).
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'A' level 2008 Qn
The molecules of compond P, C7H15Br, are chiral. On treatment with NaOH(aq), P produces alcohol Q, C7H15OH, which does not react with hot, acidified Na2Cr2O7(aq). The elimination of HBr from compound P produces a mixture of 4 different isomeric alkenes with the formula C7H14, only two of which are geometrical isomers of each other. Suggest the structural formulae of compound P and the 4 alkanes.
Solution :
P is 3-bromo-2,3-dimethyl-pentane.
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'A' level Qn :
Explain how chlorine has a bleaching effect.
Solution :
Did you know...
The bleaching effect of chlorine, is actually due to the hypochlorous acid (formed when chlorine reacts with water), rather than the chlorine itself?
The bleaching effect is really a result of oxidation; and hypochlorous acid is a stronger oxidizing agent than chlorine itself.
Household bleach is the sodium salt of hypochlorous acid, sodium hypochlorite, of which the ClO- ion (hypochlorite ion aka chlorate(I) ion) undergoes disproportionation at higher temperatures to generate Cl- ion (chloride ion) and ClO3- ion (chlorate ion aka chlorate(V) ion) which is an even stronger oxidizing agent (compare the oxidation states of chlorine in Cl2 vs ClO- vs ClO3-).
>>> How Bleach Works
http://chemistry.about.com/od/chemistryfaqs/f/bleach.htm
An oxidizing bleach works by breaking the chemical bonds of a chromophore (part of a molecule that has color). This changes the molecule so that it either has no color or else reflects color outside the visible spectrum.
A reducing bleach works by changing the double bonds of a chromophore into single bonds. This alters the optical properties of the molecule, making it colorless.
In addition to chemicals, energy can disrupt chemical bonds to bleach out color. For example, the high energy photons in sunlight (e.g., ultraviolet rays) can disrupt the bonds in chromophores to decolorize them. <<<
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'A' Level Qn.
Many students, as taught by the JCs, do not understand many aspects of Chemistry, they just blindly memorize. As a teacher and tuition teacher, I always I always encourage students to understand the *meaningfulness* behind anything they're required to memorize. Here, students should appreciate the *purpose* of hybridization, and why the mathematical formula to determine hybridization works.
Consider Ethyne and Ammonia.
1) Ethyne.
To minimize electron repulsion, the VSEPR molecular geometry is linear, and the VSEPR electron geometry is also linear (since there are no lone pairs). To achieve linear VSEPR electron geometry, hybridization of C orbitals must be sp (without hybridization, the bond angles would result in significant electron repulsion and an unstable geometry). This leaves 2 p orbitals (s s p p --> s p p p --> sp sp p p) for overlap (with the other C's 2p orbitals) to form the 2 pi bonds (over the sigma bond; hence triple bond between the 2 Cs).
2) Ammonia.
The molecular VSEPR geometry is trigonal pyramidal, but the electron geometry is tetrahedral. Hence hybridization is sp3 (to achieve tetrahedral electron VSEPR geometry, to minimize electron repulsion).
A related question to this :Identify and explain whether PH3 is a stronger or weaker nucleophile and base, as compared to NH3.
The solution, is to recognize that P is one electron shell larger than N; consequently the electron repulsion (between electron pairs, regardless lone or bond) is less in PH3 as compared to NH3; consequently there is less need and therefore less extent of hybridization of electron orbitals; consequently the bond angles in PH3 are almost orthogonal as compared to (slightly less than, due to greater repulsion between lone pair and bond pairs) tetrahedral bond angles in NH3; which implies that the lone pair in PH3 is largely the s orbital as compared to sp3 orbital in NH3; consequently the s orbital lone pair of PH3 is less available for nucleophilic attack and/or accepting a proton, as compared to NH3. -
(drinking chocolate silently...)
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'A' Level Qn.
Someone posted :
>>> What is the major organic product of the reaction of 2,3-Dimethyl-1,3-butadiene with Br2? <<<Solution / Comments :
For the mechanism, the first step is nucleophilic attack resulting in electrophilic addition (where the pi bond acts as the nucleophile); the second step is nucleophilic attack resulting in nucleophilic addition (where the Br- (or OH- if aqueous conditions are used) acts as the nucleophilie).Four points to consider :
1) The solvent used matters. With a non-polar, inert solvent, you'll get the addition of bromine only. Using aqueous bromine, you have a competing hydroxide ion nucleophile that will attack the cyclic bromonium carbocation intermediate, in effect adding not just Br, but also OH, across the double bonds.
2) If we regard the cyclic bromonium ion as a non-cyclic carbocation, (eg. imagine you're adding H-Br instead of Br-Br), then the more stable carbocation would be the one with more electron donating (non-electronegative-substituted) alkyl groups. If both double bonds are added across simultaneously, and based on this argument alone, you would end up with 1,4-dibromo-2,3-dihydroxy-2,3-dimethyl-butane.
3) However, because oxygen is significantly more electronegative than bromine, the resulting electronic instability of 1,4-dibromo-2,3-dihydroxy-2,3-dimethyl-butane (with the two partial +ve Cs being adjacent), may see the emergence of 2 other slightly more stable isomers (at least in regard to the two partial +ve adjacent Cs), 1,3-dibromo-2,4-dihydroxy-2,3-dimethyl-butane and 2,3-dibromo-1,4-dihydroxy-2,3-dimethyl-butane.4) Let's consider sterics. Br is huge, but OH is small. But Br has to be added on first, since the pi-bond acts as a nucleophile (and attacks the +ve charged Br). So the next step of OH- nucleophile attacking the carbocation (or bromonium carbocation), might preferentially add on to the terminal C's, to minimize steric repulsion (or van der Waals repulsion), so this argument tends towards the isomer 2,3-dibromo-1,4-dihydroxy-2,3-dimethyl-butane.
The answer (single major product) isn't directly provided by this post, but hopefully these 4 points above might have possibly helped you understand the processes or considerations involved a little better. -
'O' Level Chem.
anpanman posted :
>>> We encounter questions regarding the reduction of metal oxides by a reducing agent, such as carbon. The "standard" equation goes something like this
Metal oxide + carbon ---(heat)---> metal + carbon dioxide
However, i discovered that there were some reactions such as reduction of zinc oxide and lead(II) oxide which produce CO instead of CO2. I would appreciate if someone could explain. Thanks.
<<<I replied :
Zinc is a relatively (everything's relative... just ask Einstein) reactive metal, which means that it won't be that easy to grab away zinc's girlfriend ("oxygen"). Carbon can grab away oxygen from zinc, but it's harder for carbon monoxide to do so.
Carbon : "Eh gimmie ur oxygen gf, or else I call my gang whack u!"
Zinc Oxide : "Ok ok, I give u face this time... but I'll remember ur face... u be careful..."
Carbon (now in monoxide form, ie. Carbon monoxide) to Zinc (who has lost his oxygen) : "Muhahaha! Ok, now ur brudder's turn... tell your younger didi zinc oxide to give up his oxygen too!"
Zinc & his brudder Zinc Oxide : "Oi! U already got one oxygen liao, don't be too greedy hor! Don't see we all quiet quiet good to bully, think we sick kitten izzit! Knn u nvr die b4 izzit, I tell you..."
Carbon monoxide : "Ok lah! ok lah! Kao peh lah!"
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'O' & 'A' level Qn.
(inspired by a cold rainy morning on 24 Dec 08)
Qn : Why are rainy days so cold?
Ans : 3 reasons.
Firstly, when it rains, there is cloud cover which blocks out the sun (and infra red waves from solar radiation results in heat, or an increase in temperature).
Secondly, when it rains, the rain is often accompanied by strong winds, or in increase in wind velocities. This enhances the evaporating rate, which brings us the the third reason.
Thirdly, during and after a rain, the abundance of liquid water in the environment evaporates (a continuous process) to gaseous water into the atmosphere. This is an endothermic process, as energy (in the form of heat) from the environment is absorbed by the liquid water, in order to overcome the strong hydrogen bonding (for 'O' level students, say "intermolecular forces of attraction") between water molecules in close proximity in the liquid state, so that as the water molecules gain sufficient energy and vibration to overcome the hydrogen bonding attractive forces to move further away (from the rest of the water molecules in the liquid state), it is able to locate itself as an individual particle, ie. it is now in the gaseous state, as water vapour.
This continuous endothermic process of liquid water evaporating draws heat away from the environment, and hence exerts a cooling effect on the environment. (Think of your hand as the thermometer; in an exothermic process such as combustion a.k.a. burning, your hand feels hot, conversely an endothermic process such as water or alcohol on your skin evaporating makes your skin feel cold.)
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'O' & 'A' Level Qn.
Compound A liberates carbon dioxide gas with calcium carbonate, but produces only a small percentage of carbon dioxde gas (when reacted with calcium carbonate) as compared with hydrochloric acid (of the same molarity as an aqueous solution of A). Compound B decolourizes acidified potassium manganate(VII). Both compounds A and B have a molar mass of 74g each, and both compounds contain only 3 elements. When an equimolar mixture of A and B is boiled (together with a few drops of sulfuric acid as catalyst), an equilibrium mixture is obtained, which contains a new compound C, which is sweet smelling. Draw compounds A, B and C.
Answer :
A - CH3CH2COOHB - CH3CH2CH2CH2OH
C - CH3CH2COOCH2CH2CH2CH3
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'A' Level Qn.
>>> i know that enthalpy change of soln= enthalpy change of hydration - L.E. but how do i represent it in this diagram..? <<<
That's a lousy formula, which most JC students dogmatically memorize without understanding.
A better formula, which makes obvious sense, and one which would immediately self-explain how to draw the Hess Law energetics cycle diagram, would be :
Solution Enthalpy = Lattice Dissociation Enthalpy* + Solvation Enthalpy (when the solvent is water, solvation enthalpy is called "Hydration Enthalpy", for obvious reasons).
Afterall (ie. the reason why this formula is superior and intuitive), the 1st step is the endothermic process of investing energy to overcome the electrostatic attraction between the oppositely charged ions, in order to produce the gaseous cations and anions. Thereafter, the 2nd step is the exothermic process in which ion-dipole interactions or 'bonds' are formed, between the ions and solvent molecules (which are usually polar water molecules). Hence the formula makes perfect sense, and the Hess Law energetics cycle diagram becomes self-explanatory.
(*Lattice Dissociation Enthalpy is endothermic; Lattice (association) Enthalpy is exothermic. Same magnitude, since we're talking about the same process of overcoming-vs-forming a lattice formation, but in reverse to each other.)
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'O' & 'A' Level Qn.
>>> Explain why application of pressure to ice at 0 degree Celsius makes it melt. Why do most other solids not show the same behaviour? <<<
Why ice melts under pressure :
Water in solid state exists as a fixed crystalline lattice structure, thanks to hydrogen bonds between water molecules that result in fixed (and precise) geometries with fixed (and precise) hydrogen bond lengths and intermolecular distances. Putting pressure would disrupt such a fixed crystalline lattice structure (that we perceive as a solid state), and hence the chaotic disorderly structure of water that results, is what we perceive as the liquid state.
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