AS Chemistry P2 Prep.

[saudha]

Paired electrons repel, so they have higher energy and are thus more easily removed. Phosphorus has its 3p orbital just half-filled, which is relatively stable. Sulphur has one paired electrons in 3p. The paired electrons shall have slightly higher energy, so it requires less energy to remove the outermost electron from sulphur.

thankyou

 

[Saad (سعد)]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_23.pdf

Help needed in Q5 part d ...

That's a very tricky question, and difficult to explain on here, but I'll try insha'Allah.

X, as you know in part (c)(i), is OHCH2CH2COOH - it has an Alcohol group at one end, and a Carboxylic acid at the other. Basically, when you warm X with a little conc. H2SO4(aq) you get an Esterification reaction; X undergoes an esterification with itself (i.e. one molecule of X with another molecule of X); forming a cyclic compound. The -COOH group on one end of the first molecule of X, esterifies with the -OH group of another molecule of X, while the -OH group of that other molecule esterifies with the -COOH group of the first molecule of X.

So the final product is the cyclic compound shown in the Mark Scheme. And in (d)(ii), the answer is Esterification.

I hope that helped. If it didn't, stare really hard at the answer in the Mark Scheme, for a couple minutes.

 

[jsg]

That's a very tricky question, and difficult to explain on here, but I'll try insha'Allah.

X, as you know in part (c)(i), is OHCH2CH2COOH - it has an Alcohol group at one end, and a Carboxylic acid at the other. Basically, when you warm X with a little conc. H2SO4(aq) you get an Esterification reaction; X undergoes an esterification with itself (i.e. one molecule of X with another molecule of X); forming a cyclic compound. The -COOH group on one end of the first molecule of X, esterifies with the -OH group of another molecule of X, while the -OH group of that other molecule esterifies with the -COOH group of the first molecule of X.

So the final product is the cyclic compound shown in the Mark Scheme. And in (d)(ii), the answer is Esterification.

I hope that helped. If it didn't, stare really hard at the answer in the Mark Scheme, for a couple minutes.

Thanks alot !!

 

[Arpit17]

That's a very tricky question, and difficult to explain on here, but I'll try insha'Allah.

X, as you know in part (c)(i), is OHCH2CH2COOH - it has an Alcohol group at one end, and a Carboxylic acid at the other. Basically, when you warm X with a little conc. H2SO4(aq) you get an Esterification reaction; X undergoes an esterification with itself (i.e. one molecule of X with another molecule of X); forming a cyclic compound. The -COOH group on one end of the first molecule of X, esterifies with the -OH group of another molecule of X, while the -OH group of that other molecule esterifies with the -COOH group of the first molecule of X.

So the final product is the cyclic compound shown in the Mark Scheme. And in (d)(ii), the answer is Esterification.

I hope that helped. If it didn't, stare really hard at the answer in the Mark Scheme, for a couple minutes.

is this part of the syllabus?!

 

[angelgirl:)]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_21.pdf​

For question 3 d, how do you draw the Hess Law diagram? It doesn't seem to work out when I try using both diagrams in the coursebook for enthalpy change of reaction and combustion.​

wheres the answer for this Q...i saw some1 had answered it...i have doubt in it?

 

[angelgirl:)]

is this part of the syllabus?!

of course...

 

[Saad (سعد)]

is this part of the syllabus?!

Don't fret yourself, its pretty easy, once you've figured it out. You've been doing this since O' Levels; just think, how does an ester form?

And use your head.

 

[AngelColdplay]

wheres the answer for this Q...i saw some1 had answered it...i have doubt in it?

Yeah I still don't get it because the reply wasn't specific to this question. Would you kindly point out the answer for me?

 

[Saad (سعد)]

Yeah I still don't get it because the reply wasn't specific to this question. Would you kindly point out the answer for me?

I would like someone to help me understand that entire question; 3 (b) onwards. o-o;

 

[Saad (سعد)]

How do you determine how many isomers there are for any given mol. formula? Is there any trick method or something, or a list of rules to follow to get them all down?

I always seem to miss out one or two isomers in the questions when they tell you to draw all possible isomers for a given formula. >.<

 

[littlecloud11]

No one?

Also,

Paper: Click Me
Mark Scheme: Yeah, Click Him, Why Dontcha'?

Question 3; the whole question. Someone please explain it to me in detail. Jazak Allah Khayr.

OK, here goes nothing.
3b i) metal carbonate + acid = salt + water + CO2
K2CO3 + 2HCl = 2KCl + H2O + CO2
just complete the neutralization reaction and balance.

ii) use the formula Q= m*c* delT where, m is the mass of water in the solution. c is the specific heat capacity (4.18) of water and delT is the change in temp. [ for the mass of water, mass= density* volm. As the density of water is 1gcm^-3 the mass of water is the same as it's volume in the question]
so Q = 30* 4.18 * (26.2-21) = 652.08 J

iii) now divide the previous answer by the moles of K2CO3 to find the energy per mole. 652.08/.02= 32604 J or 32.6 KJ
As the temperature rose during the reaction, this is an exothermic reaction and the sign would be negative.

iv) excess HCl is used to ensure that all the K2CO3 have reacted and has been completely neutralized.

ci) metal hydrogen carbonate + acid = salt + H2O + CO2
KHCO3 + HCl = KCl + H2O +CO2

ii) use the same formula as in bii to calculate the delH. answer is 463.98 J

iii) divide answer to cii by the number of moles of KHCO3 to calculate the energy per mole. 463.98/.02 = 23199J or 23.2 KJ
As the temp dropped during the reaction this is an endothermic reaction and the sign is positive.

d) you know the enthalpy change of neutralization for KHCO3 and K2CO3 from the previous answers. Neither H2O nor CO2 is neutralized by HCl so you will not have to be bothered by them while calculating the delH for this reaction.
delH = 32.6 - (-23.2 *2) = +79 kJ/mol


[multiply 23.3 with 2 as 2 moles of KHCO3 is neutralized]

Hope this helped!

 

[AngelColdplay]

OK, here goes nothing.
3b i) metal carbonate + acid = salt + water + CO2
K2CO3 + 2HCl = 2KCl + H2O + CO2
just complete the neutralization reaction and balance.

ii) use the formula Q= m*c* delT where, m is the mass of water in the solution. c is the specific heat capacity (4.18) of water and delT is the change in temp. [ for the mass of water, mass= density* volm. As the density of water is 1gcm^-3 the mass of water is the same as it's volume in the question]
so Q = 30* 4.18 * (26.2-21) = 652.08 J

iii) now divide the previous answer by the moles of K2CO3 to find the energy per mole. 652.08/.02= 32604 J or 32.6 KJ
As the temperature rose during the reaction, this is an exothermic reaction and the sign would be negative.

iv) excess HCl is used to ensure that all the K2CO3 have reacted and has been completely neutralized.

ci) metal hydrogen carbonate + acid = salt + H2O + CO2
KHCO3 + HCl = KCl + H2O +CO2

ii) use the same formula as in bii to calculate the delH. answer is 463.98 J

iii) divide answer to cii by the number of moles of KHCO3 to calculate the energy per mole. 463.98/.02 = 23199J or 23.2 KJ
As the temp dropped during the reaction this is an endothermic reaction and the sign is positive.

d) you know the enthalpy change of neutralization for KHCO3 and K2CO3 from the previous answers. Neither H2O nor CO2 is neutralized by HCl so you will not have to be bothered by them while calculating the delH for this reaction.
delH = 32.6 - (-23.2 *2) = +79 kJ/mol


[multiply 23.3 with 2 as 2 moles of KHCO3 is neutralized]

Hope this helped!

Thank you so much! But i have a question, why did you reverse the signs for the enthalpy energies you calculated? Why +23.3 became -23.3?

 

[littlecloud11]

Thank you so much! But i have a question, why did you reverse the signs for the enthalpy energies you calculated? Why +23.3 became -23.3?

The direction of the arrows are also reversed. so the signs have to be reversed too.

when K2CO3 and KHCO3 react with HCl the signs are + and - respectively. but in the diagram i combined the products to produce the reactants so the signs have to be reversed.

 

[RGBM211]

can someone explain me the Boltzmann distribution of molecular energies ?

 

[AngelColdplay]

The direction of the arrows are also reversed. so the signs have to be reversed too.

when K2CO3 and KHCO3 react with HCl the signs are + and - respectively. but in the diagram i combined the products to produce the reactants so the signs have to be reversed.

I'm eternally grateful to you. God bless you! Thanks a bunch again.

 

[AngelColdplay]

can someone explain me the Boltzmann distribution of molecular energies ?

The area under the graph (Boltzmann distribution) represents the amount of molecules which either possess enough energy to overcome the activation or not. The amount of molecules to the left of the activation energy is the amount that HAS the required activation energy.
Uhm... Hope that helps.

 

[littlecloud11]

I'm eternally grateful to you. God bless you! Thanks a bunch again.

Haha. You're very much welcome. And all the very best of luck for your exam.

 

[Arpit17]

Don't fret yourself, its pretty easy, once you've figured it out. You've been doing this since O' Levels; just think, how does an ester form?

And use your head.

yeah i know about ester formation,but the cyclo thingy?! i havent ever done it before!

of course...

 

[JulyMei]

hi.. I wanna ask a question abt organic chem..

Can alkane react with NaOH(aq) to form Alcohol??

 

[littlecloud11]

The area under the graph (Boltzmann distribution) represents the amount of molecules which either possess enough energy to overcome the activation or not. The amount of molecules to the left of the activation energy is the amount that HAS the required activation energy.
Uhm... Hope that helps.

you mean to say the no of molecules to the RIGHT of Ea has enough energy to react, i think.