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Chemistry: Post your doubts here!

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darknessinme

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afsah shaikh said:
Two moles of compound P were placed in a vessel. The compound P was partly decomposed by
heating. A dynamic equilibrium between chemicals P, Q and R was established.
At equilibrium, x mol of R were present and the total number of moles present was (2 + x).
What is the equation for this equilibrium?
A P ---> 2Q + R
B 2P---> 2Q + R
C 2P---> Q + R
D 2P----> Q + 2R

Can someone help with this, i've tried a lot but i cant get it right :)
Click to expand...

You just have to try each option to see if it works, using the given stoichiometric equation.
Going straight to B: If you had two moles of P originally and then x mol of R were present at equilibrium, then you'd expect 2x mol of Q to be formed at equilibrium. Furthermore you'd use up 2x mol of P, so that there is 2-2x mol of P at equlibrium. If you add the total: 2-2x+2x+x=2+x
 
snowbrood

snowbrood

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OHCH2CONH2 PRODUCES NH3 WHEN HEATED WITH NAOH CAN ANYONE TELL ME HOW AND WHAT GROUPS ARE RESPONSIBLE FOR THAT
 
Rahma Abdelrahman

Rahma Abdelrahman

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Hassan Ali Abid said:
hydroden halides are Reducing agents .....and power of reducing agent increases down the group ..... HCl<HBr<HI.......
p.s the non metals (helogens cl,br,I) are not reducing agents ...they are O.A .....metals are R.A...

and lastly i didnt get what you were saying above but i tried to explain what i have understood :(
Click to expand...
no problems about the other one...
I asked my teacher 2day n she said that Oxidising ability decreases down the gp 7, so iodine is a reducing agent :)
Thanks anyway... :D
 
afsah shaikh

afsah shaikh

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darknessinme said:
You just have to try each option to see if it works, using the given stoichiometric equation.
Going straight to B: If you had two moles of P originally and then x mol of R were present at equilibrium, then you'd expect 2x mol of Q to be formed at equilibrium. Furthermore you'd use up 2x mol of P, so that there is 2-2x mol of P at equlibrium. If you add the total: 2-2x+2x+x=2+x
Click to expand...
are u sure wouldn't every term in the equation just cancel out each other tbh i tried it that way too....:(
 
Hassan Ali Abid

Hassan Ali Abid

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Rahma Abdelrahman said:
no problems about the other one...
I asked my teacher 2day n she said that Oxidising ability decreases down the gp 7, so iodine is a reducing agent :)
Thanks anyway... :D
Click to expand...

u are saying it wrong .....

look if u go down the group ,the oxidising ability increases .........just confirm this once again :)
 
P

PhyZac

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Hassan Ali Abid said:
u are saying it wrong .....

look if u go down the group ,the oxidising ability increases .........just confirm this once again :)
Click to expand...
Rahma Abdelrahman said:
no problems about the other one...
I asked my teacher 2day n she said that Oxidising ability decreases down the gp 7, so iodine is a reducing agent :)
Thanks anyway... :D
Click to expand...

Down the group the oxidising ability increase, thus down the group the better reducing agents.
 
snowbrood

snowbrood

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P

PhyZac

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Kyusam said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q-9 stuck bad......help please :D
Click to expand...
The Kc expression for reaction I is
Kc(1)= [X2Y]^2 / [X2]^2 + [Y2]

And for reaction two

Kc(2)= [X2] + [Y2]^1/2 / [X2Y]

Now we should rearrange the first expression in order to get a form similar to second expression.

( [X2]^2 + [Y2]) Kc(1) = [X2Y]^2
( [X2]^2 + [Y2]) Kc(1) / [X2Y]^2 = 1
[X2]^2 + [Y2] / [X2Y]^2 = 1/Kc(1)
now square root all terms to get
[X2] + [Y2]^1/2 / [X2Y] = 1/(Kc(1))^1/2 (see the power of half is a square root)

we can see that

Kc(2) = 1/[Kc(1)]^1/2
= 1/(2)^1/2
 
scouserlfc

scouserlfc

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snowbrood said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf
can anyone solve q 28 i dont get it please
Click to expand...

First remove C6H5 from the molecular formula as a phenyl is attached to the alkyl grps and...... now u get C4H9O
Now look at what the second requirement says not reactive to mild oxi agents which means it was a tertiary alcohol ! Now u can easily throw A and B out as they wont ever form a tertiary alcohol and C can also be thrown out as it shows a double bond at a Carbon where only a secondary alcohol can form event though u have that methyl at the end ! remember that when dehydration takes place the OH obviously is removed plus an H from the next door carbon to the carbon on which the OH was attached is also removed and between these carbons a double bond is formed this is what happens in alcohol dehydration !
 
afsah shaikh

afsah shaikh

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saad12 said:
http://www.freeexampapers.com/index.php/directory/download?location=A Level/Chemistry/CIE/2002 Jun/9701_s02_qp_1.pdf plz explain q34
Click to expand...
answer should be C
the first one says the reaction is exothermic in the forward direction, the graph shows increase in kp while the temperature increases, according to le chatier's principle if a reaction is exothermic then then on a increase in temperature the reaction should proceed in the backward direction, hence this statement is incorrect
so if the first one is incorrect then according to our options we have only one choice left
if u still need further explanation i suggest visit www.chemguide.co.uk and learn about le chatier's principal
 
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