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

last SG · 2010-07-26T23:40:35+00:00

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'A' Level Qn : Acid-Base Equilibria

Calculate the molarity of aqueous ammonia required to initiate the precipitation of iron(II) hydroxide from a 0.003 mol/dm³ aqueous iron(II) chloride. pKa of NH₄⁺ = 9.2553, solubility of Fe(OH)₂ = 1.4255 x 10^-3 g/dm³.

Answer :

molarity of aqueous ammonia needs to be greater than 2.60 x 10^-6 mol/dm3

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'A' Level Qn : Acid-Base Equilibria

Q6. Which of the following is true of pure water at (i) 274 K? (iii) 299 K?

a) pH = 7 b) pH < 7 c) pH > 7

d) pOH = 7 e) pOH < 7 f) pOH > 7

Answer :

(i) c & f

(ii) b & e

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'A' Level Qn : Organic Chemistry

A molecule contains a primary alcohol group, an aldehyde group, an alkene group, a carboxylic acid group, and an ester group. How would you carry out the following reactions?

a) Oxidize the primary alcohol group without oxidizing the aldehyde group?

b) Oxidize the aldehyde group without oxidizing the primary alcohol group?

c) Simultaneously reduce the aldehyde group and the ester group without reducing the alkene group?

d) Reduce the aldehyde group without reducing the ester group?

e) Reduce the alkene group and the aldehyde group without reducing the carboxylic acid and ester group?

Answers :

a) Pyridinium chlorochromate

b) Fehling's / Benedict's / Tollen's

c) LiAlH4 (in dry ether), followed by protonation or hydrolysis

d) NaBH4 (in dry ether), followed by protonation or hydrolysis

e) Hydrogen gas and nickel / platinum / palladium catalyst

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'A' Level Qn : Organic Chemistry

With relevant diagrams and/or equations, explain the chemistry behind “rebonding” and “perming” hair.

Answers :

http://www.humantouchofchemistry.net/hairstyling

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'A' Level Qn : Chemical Structure & Bonding

Cubane is a hydrocarbon with the following cube-like structure :

http://upload.wikimedia.org/wikipedia/commons/7/75/Cuban.svg

Explain why cubane is unstable and hence reactive.

Answer :

Each C atom in cubane has 4 bond pairs and 0 lone pairs, which ideally means it would be sp3 hybridized and have a tetrahedral geometry with bond angles of 109.5 degrees, to minimize inter-electron bond pair repulsions. But the C-C-C bond angles in cubane are forced to be orthogonal (ie.of 90 degrees), and consequently there is considerable angle strain due to inter-electron bond pair repulsions.

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H2 Chem Equilibria Qn.

0.2 mols of SbCl5 is placed in a beaker fixed with a piston, pressure maintained at 0.2bar, temp maintained at 448 degrees. Calculate the volume of the equilibrium mixture. Assume all gases are ideal.

SbCl5 <==> SbCl3 + Cl2

Kp = 1.48 at 448 degrees

Solution :

Our ICE will be in moles.

SbCl5 <==> SbCl3 + Cl2
I | 0.2 | 0 | 0
C | -x | +x | +x
E | 0.2 - x | x | x

Total no of moles of gas = (0.2 + x)

P V = n R T
(0.2 x 10^5 Pa) (Vm3) = (0.2 + x) (8.31) (746)
Make V the subject, in terms of x.

P(SbCl3) = mole fraction x total pressure = (x / (0.2 + x)) x (0.2 x 10^5 Pa)
P(Cl2) = mole fraction x total pressure = (x / (0.2 + x)) x (0.2 x 10^5 Pa)
P(SbCl5) = mole fraction x total pressure = ((0.2 - x) / (0.2 + x)) x (0.2 x 10^5 Pa)

Kp = ( P(SbCl3) x P(Cl2) ) / P(SbCl5) = 1.48
==> ( (x / (0.2 + x)) x (0.2 x 10^5 Pa) x (x / (0.2 + x)) x (0.2 x 10^5 Pa) ) / ( ((0.2 - x) / (0.2 + x)) x (0.2 x 10^5 Pa) ) = 1.48

Solve for x.

Substitute into expression for V, and solve for V.

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hesaniga posted :

"Is the lone pair of electrons on a double bonded O atom ever available? As in can it act as nucleophile or as ligand?"

ThunderFbolt replied :

"yeap, if not oxygen won't be able to bind to hemoglobin."

Ultima Online added :

UltimaOnline SG

H2 Chem

Inorganic Chem Trends Qn / Organic Chem Mechanisms Qn

Longchamp3 asked :

"The trichlorides of nitrogen and phosphorus each react with water to give two products but they react in different ways. Phosphorus trichloride, PCl3 reacts to give HCl as the only chlorine containing product. NCl3, produces HOCl as the only chlorine containing product. In each case, predict the other product and write an equation for its production. Suggest an explanation for this difference in behaviour. I know that the other products are H3PO3 and NH3, but im unable to explain the difference in behaviour.
Once again, help is greatly appreciated!"

UltimaOnline replied :

The different behaviours of the two Grp V chlorides are the consequence of the different electronegativities of N and P versus Cl, resulting in NCl3 being nucleophilic/basic, and PCl3 being electrophilic/acidic.

One other factor would be that P has vacant, low lying, energetically accessible 3d orbitals to expand its octet. Notice that P has an expanded octet in H3PO3. This is not possible for N.

The best way to answer such a question would be to draw the mechanisms for the hydrolysis in each case, which will clearly explain and account for their different behaviours.

Water nucleophile attacks P of PCl3, Cl atom leaves, abstracts a proton from +ve formal charged O atom of H2O just added, repeat, repeat, one of the O atoms forms pi bond with P, +ve formal charged O atom loses proton, -ve formal charged P atom acceps proton; we have H3PO3 and 3 HCl.

N atom of NCl3 abstracts a proton from H2O, OH- nucleophile thusly generated attacks Cl atom which cleaves away from +ve formal charged N atom, repeat, repeat; we have NH3 and 3 HOCl (hypocholorous acid aka chloric(I) acid).

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H2/H1 Chem

Acid-Base Equilibria Qn.

Animizer asked :

"10mL of 0.10 H2SO3 is titrated with 0.10M KOH. Determine the pH at the 1st equivalence point. Ka1 = 1.2 x 10^-2"

UltimaOnline replied :

Moles of HSO3- at equivalence point = 1 x 10^-3
Volume of solution at equivalence point = 20cm3
[HSO3-] at equivalence point = 0.05

Since Ka2 not given, we assume Kb1 > Ka2. In other words, we assume salt is basic rather than acidic.

(Note : If Ka2 is given, we have to compare Kb1 vs Ka2, to determine if HSO3- is more acidic or basic, and calculate pH accordingly.)

On the assumption that HSO3- is basic rather than acidic (ie. Kb1 > Ka2), HSO3- will undergo hydrolysis and generate OH-

After doing the ICE table at equivalence point, for the hydrolysis of HSO3- acting as a base, we have :

Kb = 8.333 x 10^-13 = [OH-]^2 / [HSO3-]
[OH-]^2 = 4.167 x 10^-14
[OH-] = 2.041 x 10^-7

Since pOH = 6.69, hence pH = 7.31

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Two vessels containing 4dm^3 of argon and 10dm^3 of water vapour respectively at 150 degree celsius and 101 kPa were connected together.The connected vessels were allowed to cool to room temperature.What was the final pressure in the connected vessel?

Ans:20.3kPa

i used p1v1 + p2v2 = p3v3,couldn't get the answer..

Solution (method) :

Use PV=nRT to find the number of moles of argon only (before connecting the vessels).

After connecting the vessels (ie. new combined volume), apply PV=nRT with n = number of moles of argon only.

Solve for P.

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H2 Organic Chem

Qn : What is the reaction that occurs when benzene is mixed with benzoyl chloride? Draw the mechanism.

Ans : Electrophilic (aromatic) substitution and nucleophilic (acyl) substitution.

(Note : the words in brackets above are not required for the H2 syllabus, but it's better to include them.)

Mechanism :

1. Benzene nucleophile attacks acyl carbon; acyl/carbonyl pi bond shifts up to become a lone pair on O.

2. Loss of proton H+ from benzene intermediate restores aromaticity; lone pair on the +ve formal charged O atom shifts back down to restore the acyl/carbonyl pi bond; Cl leaves as Cl- ion.

Product is diphenylmethanone aka diphenyl ketone aka benzoylbenzene aka benzophenone (main product) and hydrogen chloride (byproduct).

UltimaOnline SG

H2 Chemistry Organic Chem Qns.

greenapplejellybean asked :

hello everyone, i'm taking a levels this year (more specifically next week), i have a question for chem...
LiAlH4 in dry ether, and H2 with Ni catalyst, are both reducing agents. But from what I know, LiAlH4 can't reduce C=C double bonds and H2 with Ni catalyst can't reduce carboxylic acids and esters. Can someone clarify why this is so? And can H2 with Ni catalyst reduce things like aldehydes, ketones etc? Thanks in advance for any help rendered.

UltimaOnline replied :

The reason why LiAlH4 cannot reduce alkenes is because the way LiAlH4 works as a reducing agent is that it functions as a source of H- hydride ion nucleophiles. Alkenes are also nucleophilic due to their localized pi bonds. Hence both being nucleophiles will naturally repel away each other (think of it as : guys only like girls, guys do not like guys; nucleophiles only like electrophiles, nucleophiles do not like nucleophiles).

Yes, H2 with nickel catalyst (at temperatures above 140 deg C) can reduce alkenes, aldehydes, ketones and amides. Under usual conditions (ie. H2 syllabus) it cannot readily reduce carboxylic acids though.

FacadesofLight asked :

hi guys, hope u all can help me clarify these concepts.

1.) In Benzene ring, if the -OH group is attached to 1 and 4 position, upon addition with aq Bromine where would the bromine be directed.
We all know OH is electron donating and activates the ring and put it at 2 4 6 position. But isnt it relative concept?
If you add br with respect to position 1 OH, the bR will be 2 substituted. but if you "invert" the paper, it will be 3 substitution. can any chemistry expert help me clarify this? thanks.

2.) Which one would produce the monosubstitution or tri 2 4 6 substitution upon reaction with Phenol. Is it aq bromine or liquid bromine. Because i remember my chem teacher in school said that addition of aq bromine is mono sub and liquid br is 2 4 6 sub, but the chem notes im reading say the other way round. can anyone help me to clarify these concepts? thanks

UltimaOnline replied :

The phenolic OH group is electron donating by resonance, and is hence both an activator and ortho-para directing. Therefore, with aqueous bromine, all 4 available positions (ie. ortho & para with respect to both OH groups) will under electrophilic aromatic substitution of Br onto the benzene ring.

The strongly polar aqueous (ie. water) solvent will induce a stronger temporary dipole in the non-polar Br2 molecule, allowing for electrophiilc aromatic substitution to occur more readily (ie. the benzene ring nucleophile finds it easier to attack the Br2 electrophile because it has a larger magnitude of separation of partial charges); and consequently tri-substitution (onto the ortho, ortho and para positions) of the halogen occurs onto the phenolic benzene ring.

With a non-polar solvent such as CCl4, there is only a weak temporary instantaneous dipole on the Br2 molecule (ie. the benzene ring nucleophile finds it harder to attack the Br2 electrophile because it has a smaller magnitude of separation of partial charges), and hence only mono-substitution will occur (onto either the ortho or para position).

UltimaOnline SG

2010 Cambridge H2 Chem P3 Qns :

[QUOTE=midoforce]what's the bond angle between NO2 and O3, must be exactly accurate, not a range[/QUOTE]

NO2 : 134.3 deg

O3 : 116.8 deg

If (assuming) the Cambridge Mark Scheme is reasonable and fair, they will accept any reasonable angle approximately 120 deg (eg. for NO2, based on the trigonal planar electron geometry), provided you gave valid rationale (eg. "the single unpaired electron on the N atom takes up less space than a lone pair, hence the bond angle will be expected to be greater than 120 deg, eg. 130 deg").

Glutamic acid: HOOCCH2CH2CH(NH2)COOH (they gave skeletal but i cant draw it here)

There are 3 pKa values associated with glutamic acid: 2.1, 4.1, 9.5

Make use of these pKa values to suggest major species present in solutions of glutamic acid with the following pH values

pH1

pH3

pH7

pH11

pH 1

HOOCCH2CH2CH(NH3+)COOH

pH 3

HOOCCH2CH2CH(NH3+)COO-

pH 7

-OOCCH2CH2CH(NH3+)COO-

pH 11

-OOCCH2CH2CH(NH2)COO-

The idea is to, as you go from acidic environment to alkaline environment, deprotonate step-by-step one at a time (initially in very acidic conditions all groups are protonated; next deprotonate 1st the alpha carboxylic acid, then 2ndly the R group carboxylic acid, then 3rdly the ammonium group).

The alpha carboxylic acid is stronger than the R group carboxylic acid, due to the electron-withdrawing by induction effect of the (+vely charged) amine group.

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

Structure of NO2 :

NO2 has an unpaired electron (free radical) on N, which is doubly bonded to an O atom, and singly datively bonded to another O atom. The N atom has a +ve charge, the singly bonded O atom a -ve charge, and the doubly bonded O atom has no charge. N lacks a stable octet in NO2, and the molecule is a free radical.

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

Why FO2 doesn't exist :

Reason #1

The problem with FO2, is that F (being in period 2), cannot expand its octet (since it doesn't have vacant, energetically accessible 3d orbitals to use). Hence, it cannot form a similar structure as ClO2, which has expanded its octet (4 bond pairs + 2.5 lone pairs on Cl = 11 electrons in terms of an expanded octet).

Reason #2

For FO2 to have single bonds between the F and two O atoms, F will gain a 2+ charge and each O atom with gain a -ve charge, and this separation of charge is destabilizing, since F is more electronegative than O.

Reason #3

In addition, in such a model, F atom would have an unpaired electron, making the molecule an unstable free radical.

Reason #4

And furthermore, the F atom wouldn't have a stable octet in such a model, which defeats the purpose of forming the structure in the 1st place.

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

CuO reacting with NH3 :

The oxide anion (ie. the base) abstracts a proton from NH3 molecule (ie. the acid), generating first the OH- ion and NH2- ion. These two anions, together with four other NH3 molecules, act as ligands to donate dative bonds to the Cu2+ ion, generating an octahedral complex ion.

Another possibility is that the OH- abstracts a further proton from another NH3 molecule (since liquid ammonia used means NH3 is present in large excess), generating H2O and another NH2- ligand. In such an octahedra complex ion, two ligands are NH2-, one is the H2O generated, and 3 are NH3 molecules.

The resulting complex ion should still be deep blue in colour, or a slight variant* of the deep blue colour of the usual tetraaminecopper(II) complex ion, since the Cu2+ ion remains as Cu2+, and the ligands are similar*.

(* Wikipedia out "spectrochemical series" for "strong field vs weak field" ligands)

CuO + 6NH3 --> Cu(NH2)(OH)(NH3)4

CuO + 6NH3 --> Cu(NH2)2(H2O)(NH3)3

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'A' Level H2 Qn on Benzene :

wintersolstice posted :

1. The following shows the hydrogenation of cyclohexene.
C6H10 + H2 ---> C6H12
a)Calculate the enthalpy change for the hydrogenation of cyclohexene using data booklet.[2]

[I]my answer is -124 kJ mol^(-1)....
[/I]
b) In 1865, Friedrich Kekule described benzene as a ring of alternating single and double bonds.
From experimental determination, the enthalpy change of hydrogenation of benzene was found to be -205 kJ mol^(-1).
Explain, in terms of structure of benzene, why this value is not three times your answer in (a). Use an energy level diagram to illustrate your answer. [3]
[I]
i was thinking.....is it becoz the bond energy values provided in the data booklet is for gaseous compound ......but cyclohexene is a liquid.......so enthalpy change for vapourisation was not included.......so the value differ..????[/I]

any help is much appreciated.
thanks.

Ultima Online replied :

Your reason may not be applicable, because the question did not specify the states for both species (benzene and cyclohexene), for which the values are as given. Even if it were applicable, it would be a minor reason, and you would have missed the point of the question, which is about RESONANCE or AROMATICITY.

(Nonetheless, your point about liquid vs gaseous state is indeed the correct answer for many similar questions on "why the discrepancy?", but just not (the correct answer) for this particular question.)

Now, the concepts of RESONANCE or AROMATICITY are very important fundamental concepts in Organic Chemistry, but are poorly dealt with in the H2 syllabus. Usually only H3 Chem students and/or Chem Olympiad students have a better understanding of it. Students taking Chem at solely the H2 or H1 level generally have a much more simplified understanding of RESONANCE (in some JCs the teachers don't even introduce the term "RESONANCE" to H2 students), if at all.

If you (ie. all JC students reading this) are serious about taking Science in the University (specifically Chemistry-based courses, eg. Medicine, Life Sciences, etc), I recommend you read up on RESONANCE, eg. on Wikipedia, from recommended textbooks, or ask your school/private tutor to teach you about it.

Back to your question, the correct answer involves a brief description and explanation on how the pi electrons of benzene are DELOCALIZED by RESONANCE throughout the ring (ie. the benzene ring is AROMATIC) ; include a diagram showing the overlapping unhybridized 2pz orbitals to form a continuous ring above and below the plane of the C-C sigma bonds ; state that all C-C bonds in benzene are of equal length and strength (with enthalpy/energy values between that of single and double bonds) ; draw the 2 resonance contributors (ie. both with alternating single and double bonds) and the resonance hybrid of benzene (ie. benzene with the circle in the middle) ; state that the capacity for delocalization of the pi electrons around the benzene ring by RESONANCE (aka its AROMATICITY) confers a greater stability upon the benzene ring (since it's pi electrons are now delocalized instead of being concentrated like alkenes; consequently alkenes are stronger nucleophiles and undergo electrophilic addition while benzene is a weaker nucleophile and undergoes electrophilic aromatic substitution instead); consequently with greater stability benzene has less energy and therefore has a less exothermic enthalpy of hydrogenation (compared to the mythical cyclohex-1,3,5-triene; ie. the non-resonance model of benzene, which some JCs' teachers erroneously label as "the kekule structure/model of benzene" ) .

An energy level diagram (such as this one, on Jim Clarke's webpage) will illustrate the last point nicely. Note that your question specifies that such an energy level diagram is required, to gain full credit/marks.

(Ok, you're not required to write all the points above for just 3 marks, but other H2 Chem exam questions on benzene may require some of the other points I've made above; so if you're serious about scoring a distinction in Chemistry, and/or studying Science in the University, then you shoud read up on RESONANCE and related concepts and do your best to understand all the points I've made).

Jim Clarke's 'A' Level Chemistry website on Benzene :
http://www.chemguide.co.uk/basicorg/bonding/benzene1.html
http://www.chemguide.co.uk/basicorg/bonding/benzene2.html

Rod Beavon's (yeah, the UK lecturer now famous in Singapore because his question was used in the Singapore-Cambridge 2010 'A' Level H2 Chemistry Exam question on Planning) 'A' Level Chemistry website on Benzene :
http://www.rod.beavon.clara.net/benzene.htm

Wikipedia on Benzene :
http://en.wikipedia.org/wiki/Benzene

I'll now leave you with the above webpages for your perusal and enjoyment. Have fun!

UltimaOnline SG

'A' Levels H2 Chemistry

midoforce posted :

did my chem As recently just want to ask the following questions:
1) is skeletal drawings accepted as "structural formula"
eg. qn 1d) in chem p3, i put all my answers in skeletal form.. will minus mark?

2) is state symbols required when they ask to suggest a balanced equation? i know for certain processes such as to show combustion or solution etc you need to state, what about inorganic questions? must u state without them asking for it? (state symbols)

1 question says "suggest a balanced equation between barium ozonide and water" and in the description says barium ozonide is a solid, reacting with water to form oxygen to form alkaline solution, must i state the state symbols in my equation?

3) is writing [cu(nh3)4]2+ acceptable? or must it be [cu(nh3)4(h20)2]2+ no matter what

4) for dot and cross dative bond, i drew dative bonds as :-> (electron pair with an arrow) is it wrong?

1) Unless the question specified "full" or "displayed" structural formula, skeletal drawings are accepted. Don't worry.

2) State symbols are not required unless specified by the question. Don't worry.

3) The abbreviated form [Cu(NH3)4]2+ is accepted. Don't worry.

4) Technically it is wrong (this is one of many areas which different teachers, even those within a same JC, teach differently and some teach wrongly). It is wrong because an electron pair is either a lone pair or a bond pair, not both at once. The (misguided) teachers who teach the wrong way you described (ie. dative bonds drawn as :--->) imagine the straight arrow functions as a curved arrow, which is used in Organic Chemistry to show the electron-flow process or mechanism of a chemical reaction. In Physical Chemistry (by contrast), the lone pair has already become a dative bond, and should no longer be shown as a lone pair (or else the donor atom, if it is an element of period 2, may be misinterpreted to imply its octet has been violated).

But don't worry, Cambridge will accept it and won't penalize you for it. Usually Cambridge will ask you to show dative bonds in the form of a dot-&-cross diagram, in which case the dative bond must be shown as a dot-dot (or cross-cross).

UltimaOnline SG

H2 Chemistry - Topic : Electrochemistry

In this topic, the formula for Ecell is Ecell= E(reduction half cell) - E (oxidation half cell). So usually we look at the standard electrode potential, the one with the more +ve value undergoes reduction, fit into formula to get ans. Besides questions asking us to determine whether a rxn is feasible (that one dont have to see which one is more +ve or -ve, we look at the reaction equation to see which rxtant undergoes oxid. or reduction), I chanced upon this question which wasnt properly explained to us in school and which I dont understand why we DONT have to see which eqn is MORE +ve or -ve to determine whether they undergo oxid or reduc. Here goes:

When aq solution of Copper II sulfate and potassium iodide are mixed, a brown ppt is formed. On addition of aq sodium thiosulfate to this mixture, the brown colour disappears and a white ppt REMAINS (means it is formed previously and stays there or the white ppt has just formed and stayed there)?

What is the std cell potential of the rxn of ppotassium iodide with copper II sulfate?

During lecture we werent given much explanation regarding this. The teacher just said well, I2 was formed and Cu+ was formed...

then write this equation

2I- ---> I2 + 2e- +0.54V

Cu2+ +e --->Cu+ +0.15V

Ecell = (+0.15) - (+0.54) = -0.39V

From the 2 equations of cuz i can see that Cu2+ undergoes reduc and I- undergoes oxidation. But why in this case we look at the equation to determine which one undergoes oxid or reduc? Why not the numbers? (quite silly question but still want to ask)

If the brown colour I2 disappears eventually, why do we still use the equation 2I- ---> I2 + 2e- ?? shouldnt it be the opposite way round?

If possible, can someone write for me the entire equations of this whole reaction? (for better understanding) thanks

UltimaOnline replied :

In the 1st reaction :

Cu2+ reduced to Cu+, and I- oxidized to I2

Ppt formed is copper(I) iodide, which is white; but because iodine is brown, the result appears to be a brown ppt (which is really a mixture of two separate species, one white one brown).

In the 2nd reaction :

S2O3 2- oxidized to S4O6 2-, and I2 reduced to I-.

Hence the brown colour dissapears (I2 reduced to colourless I-), while the white ppt (copper(I) iodide) remains.

The following calculations assume standard molarities for all species (per molar quantities specified by equation), which of course isn't actually the case (based on the stoichiometry of the balanced redox eqns; furthermore the following reaction would not be feasible or spontaneous in the forward direction with a negative cell potential). So the following is "just for argument's sake".

For 1st reaction :

2I- ---> I2 + 2e- -0.54V oxidation potential at anode

Cu2+ + e- ---> Cu+ +0.15V reduction potential at cathode

E cell = Reduction potential at Cathode + Oxidation potential at Anode

= (+0.15) + (-0.54) = -0.39V

For 2nd reaction :

2S2O3 2- ---> S4O6 2- + 2e- -0.09V oxidation potential at anode

I2 + 2e- ---> 2I- +0.54V reduction potential at cathode

E cell = Reduction potential at Cathode + Oxidation potential at Anode

= (+0.54) + (-0.09) = +0.45V

UltimaOnline SG

H2 Chem Qns.

Some questions to ask.

1. Use of the data booklet is relevant to this question

The standard enthalpy change of formation of methane is given in the equation below.

C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol

what is the enthalpy change of atomisation of graphite?

+693 +843 +1129 +2587

tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.

2. The decomposition of H2O2 is known to be first order reaction

2H2O2 -> 2H2O + O2

the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?

2.0 mol/dm3

1.0 mol/dm3

0.5 mol/dm3

0.25 mol/dm3

3. is SiCl4 a simple molecular compound? thanks everyone

Q1.

Atomization enthalpy of graphite + atomization enthalpy of H2 + 4 x (bond formation enthalpy of C-H) = -75kJ/mol

Atomization enthalpy of H2 given in data booklet.

Bond formation enthalpy of C-H given in data booklet.

Make Atomization enthalpy of graphite the subject.

Q2.

From rate constant, calculate half-life.

Then apply the formula :

Molarity (desired) / Molarity (initial) = (1/2)^n
where n = no. of half-lives.

UltimaOnline SG

H2 Chem Qns.

Some questions to ask.

1. Use of the data booklet is relevant to this question

The standard enthalpy change of formation of methane is given in the equation below.

C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol

what is the enthalpy change of atomisation of graphite?

+693 +843 +1129 +2587

tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.

2. The decomposition of H2O2 is known to be first order reaction

2H2O2 -> 2H2O + O2

the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?

2.0 mol/dm3

1.0 mol/dm3

0.5 mol/dm3

0.25 mol/dm3

3. is SiCl4 a simple molecular compound? thanks everyone

Q1.

Atomization enthalpy of graphite + atomization enthalpy of H2 + 4 x (bond formation enthalpy of C-H) = -75kJ/mol

Atomization enthalpy of H2 given in data booklet.

Bond formation enthalpy of C-H given in data booklet.

Make Atomization enthalpy of graphite the subject.

Q2.

From rate constant, calculate half-life.

Then apply the formula :

Molarity (desired) / Molarity (initial) = (1/2)^n
where n = no. of half-lives.

UltimaOnline SG

H2 Chem qns

Daniel asked :

When 0.192g one of the following oxobromides, SiO2Br8, Si2OBr6, Si3O2Br8, Si4O3Br10, was completely reacted with water, all its bromine was converted to bromide ions and produced 0.392g of a cream precipitate when an excess of silver nitrate was added. Which oxobromide was reacted?

2) What is the difference in structure between AlF3 and AlCl3?

3)Hydrogen Sulphide reacts with water accordin to the following eQuilibrium eQuation

H2S(aQ) + 2H20(l) ---> 2 H3O+(aQ) + S2-(aQ)

Which of the following would not cause the position of eQuilibrium to shift when added to the eQuilibrium mixture?
A)Water
B)Ammonium Sulphate
C)Sodium Carbonate
D)Lead(II) Nitrate

4)A cell is set up by connecting a Cu2+/Cu half-cell and an acidified Cr2O72-/Cr3+ half-cell

Which of the following correctly describe the effect of the change to emf of the cell?

A Replacing copper with an alloy of copper and silver increase emf
B Adding H2SO4(aQ) to reduction half-cell decreases emf
C Adding NaOH(aQ) to oxidation half-cell decrease emf
D Adding KNO3(aQ) to reduction half-cell does not affect emf

Ans to Q1. Find the no. of moles of each silicon oxobromide (assuming the stated sample mass). Find the no. of moles of the AgBr ppt. Apply stoichiometry (of the formulae of all 4 oxobromides) to see which fits.

Ans to Q2. The former has a giant ionic lattice structure (large magnitude of electronegativity difference), the latter exists as simple covalent molecules (small magnitude of electronegativity difference).

Ans to Q3.
A)Water, doesn't shift position of equilibrium since it's already present in large excess (since the state symbols of other reactants and products are aqueous).
B)Ammonium Sulphate, acidic hence donates protons to water to generate H3O+, consequently shifting position of equilibrium to the LHS.
C)Sodium Carbonate, basic hence accepts protons from H3O+ to generate H2CO3 which exists in equilibrium with and hence decomposes into CO2 and H2O; consequently lowering the molarity of H3O+ and thereby shifting position of equilibrium to the RHS.
D)Lead(II) Nitrate, increases the ionic product of PbS until it exceeds solubility product of PbS, thereby causing S2- to precipitate out as PbS and thereby lowering the molarity of S2- ions; consequently shifting position of equilibrium to the RHS.

Ans to Q4.
A) This change makes the average oxidation potential at the anodic compartment less positive (because silver, being even less reactive than copper, has a oxidation potential even less positive than copper), consequently emf decreases. Hence wrong option.
B) This change shifts the position of equilibrium to the RHS (in the cathodic compartment), making the reduction potential at the cathode more positive; consequently emf increases. Hence wrong option.
C) This change causes Cu2+ to be precipitated out as Cu(OH)2, thereby lowering the molarity of Cu2+(aq), shifting the position of equilibrium to the RHS (in the anodic compartment) making the oxidation potential at the anode more positive; consequently emf increases. Hence wrong option.
D) While K+ and NO3- are inert, the additional water added lowers the molarity of both Cr2O7 2- and Cr3+, but to an equal extent (because both are aqueous species); hence reduction potential at cathode does not change; consequently emf does not change. Hence correct option.

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