Need HELP on CIE Chemistry AS/A2 level

[The SMA]

hi, need help n explanation on nov 2008 p1,
q2,10,15,24,30,39 and 40.

[tmisterr]

q2 is C. Volume of gas after burning...use mole ratios, it was burned with 50cm³ of oxygen but only 30 reacted so 20 remains. It forms 10 of CO2 and 20 of SO2, total is therefore 50. After addition if NaOH which is alkaline, the acidic gases, CO2 and SO2 will react leaving behind only the 20cm³ of O2

q10 D. you should know that Kc and Kp are not affected by concentration or pressure changes or by catalyst! a catalyst will only reduce the time at which equilibrium is reached, increasing concentration of O2 will also reduce the time in which equilibrium is reached but it will be reached at the same Kp values as before. same with increasing pressure. Temperature will however affect the Kp value since the forward reaction is exothermic, if you increase the temp, the extra energy will favour the backward reaction and vice versa.

q15 C, RCl will break down heterolytically forming the carbacation and the chloride ion which has a full octet, which bonds with AlCl3 with a dative bond to Al since Al has an incomplete octed of electrons

q24 C, the reaction takes place in two parts, the first part is slow meaning it has a high activation energy, then the second part is fast showing it has a low activation energy. Look at the reaction pathway of C, it shows this, the first "bump" is high then the second "bump" is lower

q30 C, Find the theoretical mass of ester that should be obtained. 30 grams of ethanol is 5/6 moles and 30 grams of ethanoic acid is 0.5 mole so ethanol is in excess. therefore Use moles of ethanoic acid for calculations. 0.5 moles of ester are formed, which is 44 grams of ester. 22g were however formed so yield is (22/44)*100 =50%

q39. B. so a possible termination step in this reaction is when two propyl free radicals form one hexane molecule. possible products are therefore hexane (C6H14) or some of its isomers can form, one and two can be formed but three cannot because of the position of the methyl branch.

q40 B. it is soluble in water because hydrogen bonds can form between water and the OH groups of the molecule. It has a carbon-carbon double bond so it can de colourise bromine water but it cannot react with fehling's reagent since it is a ketone not a aldehyde.

[The SMA]

q2 is C. Volume of gas after burning...use mole ratios, it was burned with 50cm³ of oxygen but only 30 reacted so 20 remains. It forms 10 of CO2 and 20 of SO2, total is therefore 50. After addition if NaOH which is alkaline, the acidic gases, CO2 and SO2 will react leaving behind only the 20cm³ of O2

q10 D. you should know that Kc and Kp are not affected by concentration or pressure changes or by catalyst! a catalyst will only reduce the time at which equilibrium is reached, increasing concentration of O2 will also reduce the time in which equilibrium is reached but it will be reached at the same Kp values as before. same with increasing pressure. Temperature will however affect the Kp value since the forward reaction is exothermic, if you increase the temp, the extra energy will favour the backward reaction and vice versa.

q15 C, RCl will break down heterolytically forming the carbacation and the chloride ion which has a full octet, which bonds with AlCl3 with a dative bond to Al since Al has an incomplete octed of electrons

q24 C, the reaction takes place in two parts, the first part is slow meaning it has a high activation energy, then the second part is fast showing it has a low activation energy. Look at the reaction pathway of C, it shows this, the first "bump" is high then the second "bump" is lower

q30 C, Find the theoretical mass of ester that should be obtained. 30 grams of ethanol is 5/6 moles and 30 grams of ethanoic acid is 0.5 mole so ethanol is in excess. therefore Use moles of ethanoic acid for calculations. 0.5 moles of ester are formed, which is 44 grams of ester. 22g were however formed so yield is (22/44)*100 =50%

q39. B. so a possible termination step in this reaction is when two propyl free radicals form one hexane molecule. possible products are therefore hexane (C6H14) or some of its isomers can form, one and two can be formed but three cannot because of the position of the methyl branch.

q40 B. it is soluble in water because hydrogen bonds can form between water and the OH groups of the molecule. It has a carbon-carbon double bond so it can de colourise bromine water but it cannot react with fehling's reagent since it is a ketone not a aldehyde.

Many thanks tmisterr for the help in explaining it all.
i have little doubts on ur explanation q10 n q40

in q10,since Kp and Kc are only affected by Temperature
how do u differentiate whether the reaction is Exothermic
or Endothermic if its not given in the question?

in q40,just a little curiousity, i couldn't manage to see/identify where is the ketone group
instead i see only alcohols groups [CH(OH), C(OH) & -CH2OH],
alkene group (C=C) and ester group (-CO2-).
im not quite sure though but are u referring to the carbonyl group where -CO2-
is attached? please correct me if im wrong.

Your help is very much appreciated. Thankyou soo much

[The SMA]

Need more help in O/N 2008 p1
q8,14,17,22,31,35,38

[tmisterr]

Many thanks tmisterr for the help in explaining it all.
i have little doubts on ur explanation q10 n q40

in q10,since Kp and Kc are only affected by Temperature
how do u differentiate whether the reaction is Exothermic
or Endothermic if its not given in the question?

in q40,just a little curiousity, i couldn't manage to see/identify where is the ketone group
instead i see only alcohols groups [CH(OH), C(OH) & -CH2OH],
alkene group (C=C) and ester group (-CO2-).
im not quite sure though but are u referring to the carbonyl group where -CO2-
is attached? please correct me if im wrong.

Your help is very much appreciated. Thankyou soo much

Ah sorry, for 40 u are right its ester not ketone, was a bit in a hurry but the point is  there is no aldehyde group so it will not react with fehling's reagent.
for q10 forward reaction is combustion, all combustions are exothermic. but its just an example. Kc and Kp are only affected by temp, nothing else. if was an endothermic reaction go the right then increase in temp increases products so Kp rises e.Take Care. Kp and Kc are used to give you an idea on whether the reaction is moving right or left.

[The SMA]

Ah sorry, for 40 u are right its ester not ketone, was a bit in a hurry but the point is  there is no aldehyde group so it will not react with fehling's reagent.
for q10 forward reaction is combustion, all combustions are exothermic. but its just an example. Kc and Kp are only affected by temp, nothing else. if was an endothermic reaction go the right then increase in temp increases products so Kp rises e.Take Care. Kp and Kc are used to give you an idea on whether the reaction is moving right or left.

Its ok. I understood everything now. Thanks again for making it more clear 

[The SMA]

Need more help in O/N 2008 p1
q8,14,17,22,31,35,38

could someone help me in explaining these questions please?

[nid404]

yes, when i log in next

[The SMA]

ok nid 

[nid404]

14) C SO2 always first convert to SO3...so basically even when you do reaction with water, it converts to sulphurous acid first, sulphuric later

17) D because if you notice the reaction with iodide. I2 as vapour is formed with H2SO4...ox state 0  whereas with H3PO4, it is HI that forms..ox state of I is -1 So higher ox state with sulphuric acid, which shows it oxidises better...

Q22) A & B ruled out at the very start....no chiral possible, no isomerism either
        C not possible cause you can't have a chiral...try substituting an  H with a chlorine...try all possible subs ....no chiral..Hence D (second from left substituted, you get chiral)

I have to leave urgently....I'll get back from school and answer

[The SMA]

14) C SO2 always first convert to SO3...so basically even when you do reaction with water, it converts to sulphurous acid first, sulphuric later

17) D because if you notice the reaction with iodide. I2 as vapour is formed with H2SO4...ox state 0  whereas with H3PO4, it is HI that forms..ox state of I is -1 So higher ox state with sulphuric acid, which shows it oxidises better...

Q22) A & B ruled out at the very start....no chiral possible, no isomerism either
        C not possible cause you can't have a chiral...try substituting an  H with a chlorine...try all possible subs ....no chiral..Hence D (second from left substituted, you get chiral)

I have to leave urgently....I'll get back from school and answer

allright Nid, i'll wait for the other explanations later  

By the way, I understood q14 & 17 now.
but i'm a lil bit confused on q22), do you mean substitute the CH3- group on the left with Cl
so ClCH=CHCH2CH3
but this is the confusing part..i can't identify which one is the Chiral carbon atom..
please correct me if i'm wrong

[nid404]

naah..... I meant
CH2=CHC*H(Cl)CH3  (*=chiral centre)

[nid404]

Q31)Can't really think how to explain 31...let me look up for something

Q35) It can't be 1 because halogen-halogen bond energy is related to atomization.. .They always exists as diatomic molecules. when you've eliminated the 1st option, obv ans is 2&3(C)
The weaker the van der waals forces, more volatile it is.

Q38) has to be D. 2 and 3 are much obviously flammable.

[The SMA]

naah..... I meant
CH2=CHC*H(Cl)CH3  (*=chiral centre)

got it now. Thanks! ^^

[The SMA]

ok, take your time..
I tends to score low in section B  i don't find it easy to ans questions from section B. :S

oh yeah and for Q8 of the same year, why is B is not the answer?
the ans is actually D though..