Chemistry: Post your doubts here!

[AbbbbY]

Here's a question which i tried but remained unable to solve. So i would really be grateful if anyone attempts to give it a try...
Ques: Two moles of compound P were placed in a vessel. The vessel was heated and compound P
was partly decomposed to produce Q and R. A dynamic equilibrium between chemicals P, Q and
R was established.
At equilibrium x moles of R were present and the total number of moles present was (2 +2/x ).
What is the equation for this equilibrium reaction? [Key: --> is the equilibrium sign]
A P --> 2Q + R
B 2P --> 2Q + R
C 2P --> Q + R
D 2P --> Q + 2R

2-x : x/2 : x

In essence, it's a pretty simple question if you ask me. People tend to get confused seeing variables instead of constants.

Anyhow.

There are two approaches to this.

One is quicker but much more confusing. If you want, let me know and I'll enlist that too.

If not, simply take the equations and manipulate.

A- Inital: 2 : 0 : 0
Final : 2-x: 2x : x
Total moles: 2-2+2x+x = 3x so A is incorrect.

B-
Initial: 2 : 0 : 0
Final: 2-2x : 2x : x
Total moles: 2-2x+2x+x = 2+x so B is incorrect

C-
Initial: 2 : 0 : 0
Final 2-2x:x:x
Total = 2-2x+x+x = 2

D-
Initial: 2 : 0 : 0
Final: 2-x : x/2 : x [x = 2R]
Total moles = 2-x+x/2+x =2+x/2 mols!

 

[Mairaxo]

yah thats what the questions says
how did u now 1 will give 2 ketone ?? on which basis did u rely ...
what i now ..if there are 2 alkyl gp .. the ketone will be produced by the oxidation of alkenes..here its cyclic and difficult to now it ?

okay does this help? just break the double bonds and form a ketone or acid depending on whether its secondary or tertiary C atom 2 has only one product and 3 produces 2 out of which one is an acid

 

[AbbbbY]

umm sorry for saying sis maybe changing your avatar will be good
rarely i get correct answer is hess law is there any tip which i can keep in mind while doing this questions

I simply use the formulae for MCQs. I personally don't make the whole Hess Diagram. You can, if you're comfortable that way and think you'll make a mistake using the formulae.

i.e :
ΔHc = ΣΔHr - ΣΔHp
ΔHf = ΣΔHp - ΣΔHr


The way I remember these is using the word CRAP.

CRAP == Cr-p
And for formation, the other way round. Works for me. Every time.

[For Born-Haber cycle related questions, ΣΔHf = ΣΔEVERYTHING else!. Everything else = Total atomization energies, electron affinities, ionization energies and lattice enthalpy. This is P4 only though.]

In addition to that, I've uploaded all the handouts covering AS and A2 energetics. You might want to give them a read. I studied from them alone and get it right 90% of the time. You just gotta focus on what it's asking.

(https://www.xtremepapers.com/commun...st-your-doubts-here.9859/page-286#post-573574)

 

[ahmed abdulla]

okay does this help? just break the double bonds and form a ketone or acid depending on whether its secondary or tertiary C atom 2 has only one product and 3 produces 2 out of which one is an acid

for a cyclic chain .. how do we now it its primary / secondary ..? there is no "OH" or C=O ... how to now that ?

 

[Mairaxo]

for a cyclic chain .. how do we now it its primary / secondary ..? there is no "OH" or C=O ... how to now that ?

if u get confused then just show the C atoms and bonds yourself and it will be much more simple.

 

[ahmed abdulla]

if u get confused then just show the C atoms and bonds yourself and it will be much more simple.

still not getting into my brain
anyway thanks
if as u said ... then ALL should be either secondary and tertiary and there will be no primary

 

[Mairaxo]

still not getting into my brain
anyway thanks
if as u said ... then ALL should be either secondary and tertiary and there will be no primary

yupp cz in cyclic every C has atleast 2 other C bonded!

 

[ahmed abdulla]

yupp cz in cyclic every C has atleast 2 other C bonded!

then from where did the carboylic acid come in the 2nd case if its not primary !

 

[Mairaxo]

then from where did the carboylic acid come in the 2nd case if its not primary !

it forms an aldehyde 1st and then gets converted into an acid!

 

[Mairaxo]

then from where did the carboylic acid come in the 2nd case if its not primary !

dont confuse this with alcohols! we are talking about C atoms only! there are no OH groups secondary alcohols dont form acids on oxidation but here its a double bond breaking to form an aldehyde which then forms an acid

 

[KWIKIW]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Do you know how to do Qs 33 , 32, 20, 16, 11 in 01/M/J/2003?
I am a bit confused about the answers to those Qs.

 

[Ahmedraza73]

dont confuse this with alcohols! we are talking about C atoms only! there are no OH groups secondary alcohols dont form acids on oxidation but here its a double bond breaking to form an aldehyde which then forms an acid

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_ms_1.pdf
Help me with Q:38,39,40

 

[snowbrood]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_ms_1.pdf
Help me with Q:38,39,40

for Q38
here X is the halide concentration depends upon the halide for 1 we have two Cl atoms making it double the reaction .
br- and I- are weaker than Cl- bond so weaker bonds breaks more quickly than a stronger one
for q39
problem with dehydration here is that there would be 4 double bonds which are not possible with one carbon atom one carbon atom cannot form 4 double bonds so it cant happen 3 option there are five carbon atoms so again not possible

 

[snowbrood]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_ms_1.pdf
Help me with Q:38,39,40

however it would react with sodium like any alcohol does

 

[snowbrood]

for Q40 primary or secondary alcohol oxidize to form aldehyde and ketone respectively. u reverse the process that is called reduction

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Do you know how to do Qs 33 , 32, 20, 16, 11 in 01/M/J/2003?
I am a bit confused about the answers to those Qs.

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Do you know how to do Qs 33 , 32, 20, 16, 11 in 01/M/J/2003?
I am a bit confused about the answers to those Qs.

for Q11
lone pairs make compounds polar more lone pairs means more polar so ammonia can not be more polar than water.
stronger base will pull those H+ protons with greater ease.
for Q16
for option A u get a complex ion where Cl has oxidation number of -1
for option B u get HCl again -1
for option C u get NaClO here u get +1
for option D u get NaClO3 here u get +5 so this is the correct answer.
for q32
right hand side of the molecule is polar as the oxygen atom has lone pairs any compound that has lone pair is considered polar.
like gets like which means polar molecule dissolves in polar and non polar dissolves in nonpolar. so polar cant dissolve in non polar oil u know oil does not dissolve in water.
for q33
refer to this diagram
when u break the bonds in graphite and diamond u get carbon gas carbon gas would be at the same energy that is the hess law.

 

[Raiyan3]

View attachment 28143

Ahaa Thank you so much! That helped alot bro!

 

[Rahma Abdelrahman]

Ahaa Thank you so much! That helped alot bro!

bro??!!!
U r wlcm anyways

 

[ahmed abdulla]

it forms an aldehyde 1st and then gets converted into an acid!

sorry if i am annoying you !
y not a ketone .... how u knew it gonna be an acid