ALL CIE CHEMISTRY DOUBTS HERE !!

[Deadly_king]

I have attached June 2001 as well as November 2001.It seems that these exams are difficult to find so i thought these might help you practice more questions. Best wishes

Thank you very much for thinking about us dude.

Really appreciated.

+rep

[moon]

Anytime ...

[moon]

here are the markskemes for june & Nov.2001

[moon]

pls I need help in Nov.06 Q9, 11 12 31, 32.
June 01 Q6,11,37
thanks for ur help.

[Hypernova]

plzzzzzzzzz help me in these questions. June 07 Q 1, 9
                        
                        June 01 Q6,11,37
                      
                        June 2002 Q 38

I would really apreciate any help. thanx a lot.

Was waiting for the marking scheme..i don't post untill im 100% sure

June 01 Q6

See the attached file

Using Hess's law, the enthalpy change of a reaction is the same, irrespective of the route taken

So we choose the route I drew.
If we move in the direction of the reaction we add the enthalpy change, if we move in the opposite direction to the reaction, we subract the enthalpy change, so
the equation looks like this

 -109 - -111 + -395

=-393 KJ/mol

Ans is C

11)

Its a basice math question that tests your ability to read of the graph.

They want the fraction of molecules higher than the activation energy

That is: the total number of molecules above Ea
                total number of molecules

so the total number of molecules above Ea is reprisented by the areas shaded R and Q
and
total number of molecules is represented by the areas shaded P and Q

which is: Q + R
           P + Q




37)

Can't do it, its got something to do with oxinumbers

thanks for the papers
+Reputation

[Deadly_king]

pls I need help in Nov.06 Q9, 11 12 31, 32.
thanks for ur help.

Nov 06 p1

9. Reaction : CO_2(g) + H_2(g) -----> CO_(g) + H₂O_(g)

So we are converting CO₂ to CO while H₂ is being oxidised to H₂O. However both H₂ and H₂O are in gaseous state.

Conversion of CO to CO₂ is given as -283 KJmol^-1. Since we're converting CO₂ to CO, it will be +283 KJmol^-1.

Converting H₂ to water in liquid is given -286 KJmol^-1. But we have to convert it back to gaseous state, hence net energy will be (-286 + 44) = -242 KJmol^-1

Hence Change in enthalpy of reaction : (+283 - 242) = +41 KJmol^-1

Answer turns out to be C

11. For the reaction K_c = [CH₃CO₂C₂H₅][H₂O] / [C₂H₅OH][CH₃CO₂H]

Since from the reaction mole ratio is 1:1 for every reactant and products, it becomes easier.

                    [C₂H₅OH]        [CH₃CO₂H]      [CH₃CO₂C₂H₅]      [H₂O]
Initially              1.0                1.0                 0.0                 0.0

At equilibrium      1-             1-                                    

Therefore Kc = ()() / (1-)(1-) = 4
It also implies [/(1-)]² = 4

Now you simplify and you'll get = 2/3
Hence number of moles of ethyl ethanoate formed is 2/3

Answer is B

12. At low temperature, less molecules will have have reached activation energy to undergo the reaction. Hence the graph with lower peak is to be used.

Energy of such a graph is found on the x-axis.

Hence answer is D

31. The chlorine oxide radical is ClO^..

It is made up of a covalent bond between the Cl and O atom. Since the oxygen atom carries only one bond, it has one more free electron available for bonding.

Hence the oxygen atom contains an odd number of electrons. However no dative bond is present in this molecule.

Hence answer is B

32. Strong acid imply that it dissociates rapidly to form the corresponding ions while a weak acid does not dissociate much.

Hence sulfuric acid is a strong one which implies it will form much H⁺ and HSO₄^-. The latter being a weak acid will not dissociate much such that SO₄^- will only be present in minimal amount.

Answer is D.

[haris94]

state 2 reasons why noble gases behave less ideally as you go down the group 8?

pls explain
thanx

[Deadly_king]

state 2 reasons why noble gases behave less ideally as you go down the group 8?

pls explain
thanx

1. As you go down the group, the atoms increase in size such that their volume become large enough. Hence their volume are no more negligible.

2. As you go down the group the atoms of the noble gases become larger due to increase in number of shells. Hence the outermost electron is further away from the nucleus. Therefore the force of attraction decreases such that it is easier to remove an electron from it. It becomes less stable and it is easier to form attraction.

Ideal gases [How gases should behave but don’t]-

   1. No attraction between molecules/atoms
   2. Molecules have a negligible volume
   3. Collisions are elastic
   4. Particle movement is random

Real gases VERY RARELY BEHAVE LIKE IDEAL GASES since

         1. There IS an attraction between particle (van der Waals)
         2. The volume of particles are NOT negligible, esp. at low temps & high-pressure since atoms/molecules are close together

***HYDROGEN and HELIUM are the most IDEAL gases.*** Also, Diatomic molecules and nonsymmetrical molecules & noble gases act the most ideal. THE SMALLER THEY ARE THE MORE IDEAL THEY BEHAVE.

Hope it help

[birchy33]

10. At a total pressure of 1.0 atm, dinitrogen tetraoxide is 50% dissociated at a temperature of 60 C,
according to the following equation.
 
N2O4  2NO2
 
What is the value of the equilibrium constant, Kp, for this reaction at 60oC?
 
 A 1/3 atm
 B 2/3 atm
 C 4/3 atm
 D 2 atm

Please explain. Its from MJ/06 P1 thank-you. Oh and please can you held with 18 and 20 from the same exam.

[moon]

pls can u help in Nov.08 Q3,30,39 and Nov. 09 Varient 1 Q5,21,22,28,29,31. Thanx in advance.

[Master_Key]

Nov 08
Q3 - Ans:- D

One electron in S orbital means that it is in S-block of periodic table but the first group.

[elemis]

Nov. 09 Variant 1

Question 5

A and D are eliminated as they are symmetrical and hence have no overall dipole moments.

In C Chlorine is very electronegative and hence tugs hard at the electrons. Oxygen which is even more electronegative tugs even harder so the net dipole moment is very small.

 In B we see that hydrogen can offer little resistance to the effect of the highly electronegative oxygen causing a large dipole moment in the direction of oxygen

Ans = b

[Deadly_king]

pls can u help in Nov.08 Q3,30,39

Nov 08 p1

30. First you need to use the mass of the respective molecules given to find the number of moles of each.

No of moles = Mass / Mr

No of moles of ethanol : 30/46 = 0.652
No of moles of ethanoic acid : 30/60 = 0.50
Ratio of moles of ethanol to ethanoic acid was supposed to be 1:1. Hence we can deduce that ethanol is in excess. Only 0.5 mole of each reactant will react.

No of moles of ethyl ethanoate : 22/88 = 0.25.
Hence it can be noted that only 0.25 mole of ester is formed when 0.5 mole of reactants react together.

% Yield is therefore : 0.25/0.5 x 100 = 50%

Answer is C

39. From the information given above, you can note that for methane which contains only one carbon atom, ethane with two carbon atoms can be obtained as product. In other words the number of carbon atoms double.

So for propane which contains 3 carbon atoms, the resulting product may contain maximum 6 carbon atoms.

Hence we can reject the last suggestion leaving us with only 1 and 2 as potential answers.

Answer is B

[Deadly_king]

10. At a total pressure of 1.0 atm, dinitrogen tetraoxide is 50% dissociated at a temperature of 60 C,
according to the following equation.
 
N2O4  2NO2
 
What is the value of the equilibrium constant, Kp, for this reaction at 60oC?
 
 A 1/3 atm
 B 2/3 atm
 C 4/3 atm
 D 2 atm

Please explain. Its from MJ/06 P1 thank-you. Oh and please can you held with 18 and 20 from the same exam.

Jun 06 p1

10. Kp = [(NO₂)]² / (N₂O₄)

At equilibrium, pressure of N₂O₄ is while that of NO² is 2. Since total pressure is given to be 1.0 atm, this implies that + 2 = 1.0

Therefore = 1/3, i.e pressure of N₂O₄ is 1/3 while that of NO² is 2/3.

Kp = (2/3)² / (1/3) = 4/3

Answer is C.

18. C is the most appropriate solution since it is cheaper and the reaction is not exothermic.

20. First reaction is removal of hydrogen atom, i.e oxidation of an alcohol to an aldehyde.


Second reaction is a nucleophilic reaction as an aldehyde is made to react with an alcohol to form a ketone.

Third reaction is again oxidation.

Answer is C.

[birchy33]

Jun 06 p1

10. Kp = [(NO₂)]² / (N₂O₄)

At equilibrium, pressure of N₂O₄ is while that of NO² is 2. Since total pressure is given to be 1.0 atm, this implies that + 2 = 1.0

Therefore = 1/3, i.e pressure of N₂O₄ is 1/3 while that of NO² is 2/3.

Kp = (2/3)² / (1/3) = 4/3


Answer is C.

18. C is the most appropriate solution since it is cheaper and the reaction is not exothermic.

20. First reaction is removal of hydrogen atom, i.e oxidation of an alcohol to an aldehyde.


Second reaction is a nucleophilic reaction as an aldehyde is made to react with an alcohol to form a ketone.

Third reaction is again oxidation.

Answer is C.

Cheers mate, I'm just a bit unsure about the 2x + x = 1 bit. Secondly There's another equilibrium question: Question 9 from MJ/07. Your help would be much appreciated. Cheers + rep.