Edexcel CHEMISTRY DOUBTS!!!!

[~ Miss Relina ~]

I'm glad you managed to find the answer, Relina, but for the sake of other's benefit I'll clear it out.  

2SO₃(g) <----> 2SO₂(g)+ O₂(g)

Moles at the start of the experiment: 2 (for SO3) 0 (for SO2) 0 (for O2)

Moles at equilibrium: 1.5 (for SO2 ---> "75% of SO₃", so you have to do 75% of the initial 2moles of SO₃), O.5 (for SO₃, because 75% has been disassociated; subtract the moles of initial moles of SO3 with the equilibrium moles of SO2) and 0.5 moles of 0₂ (you did notice in the equation there's only one mole of O₂, so it's half of SO₂ in the equilibrium.)

and than you have to find out the partial pressure, I guess y'all know how to do that.

P.S - Sorry for the late reply.

I was also going to post the answer  but mine was  a different one as ur answer is like the mark sheme and what I didn't undersatnd from it is from where did  he get  the initial moles ???from equation 

[~ Miss Relina ~]

Thanks fer the help 

really sorry for late reply I'll post now the way I found it so that u would use it too:

At a particular temperature, 75% of the sulphur trioxide is dissociated, producing a
pressure of 10 atm. Calculate the value of Kp at this temperature paying, attention to its units.

OK You are given the equation:

2SO3 --> 2SO2 + O2

If 75% of the SO3 dissociates it leaves 25% undissociated.

.. and as 1 mole of SO3 produces 1 mole of SO2 there will be 75% of the original moles of SO3 made into SO2 (and half that of O2 )

And, you are told that the overall pressure is 10atm.

Well, total pressure = sum of the partial pressures.

Let initial moles of SO3 = x

Final moles of SO3 = 0.25x
Final moles of SO2 = 0.75x
Final moles of O2 = 0.375x

Sum of moles = 1.375 = 10 atm

PP of SO3 = 0.25/1.375 x 10 = 1.818
PP of SO2 = 0.75/1.375 x 10 = 5.455
PP of O2 = 0.375/1.375 x 10 = 2.727

kp = (5.455)(5.455)(2.727)/(1.818 x 1.818) = 24.55 atm


hope u get it    and accept my apologise 

[Banana]

Let initial moles of SO3 = x

Final moles of SO3 = 0.25x
Final moles of SO2 = 0.75x
Final moles of O2 = 0.375x

hope u get it    and accept my apologise 

Thanks a lot But when you found the final moles aren't you supposed to multiply the SO3 moles by 2 as per the equation? 

[~ Miss Relina ~]

Thanks a lot But when you found the final moles aren't you supposed to multiply the SO3 moles by 2 as per the equation? 

no 0.25x is already multiplied but for for equation u have O₂ half  SO₂ that's why dividing 0.75x/2= 0.375x

[Banana]

no 0.25x is already multiplied but for for equation u have O₂ half  SO₂ that's why dividing 0.75x/2= 0.375x

Aha I see...that's simpler to understand  Thanks a lot, yo...this'll help me a lot in unit 4 

[~ Miss Relina ~]

Aha I see...that's simpler to understand  Thanks a lot, yo...this'll help me a lot in unit 4 

u're welcome 

[**RoRo**]

A sample of 0.025 mol of the chloride of an element was dissolved in distilled water & the solution made upto 500 cm^3. 12.5 cm^3 of this solution reacted with 25 cm^3 of 0.1 mol dm^-3 silver nitrate solution. What is the formula of the chloride?

[Deadly_king]

I know am extremely late but still try these

http://www.chemguide.co.uk/basicorg/isomerism/optical.html

http://chemistry.umeche.maine.edu/CHY252/stereo.html

http://webhost.bridgew.edu/fgorga/Stereochem/chiral.htm

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/sterism2.htm

http://www.elmhurst.edu/~chm/vchembook/209optical.html

[~ Miss Relina ~]

how bacteria can cause a sample of 2-hyroxypropanoic acid that contains equal amounts if opyical isomers to cause the rotationof plane =[polarised light


Thanks

and other one too what is the effect of increasing chain length on Pka of carboxylic acids
and why ?

[SZM]

dear iluvme,

do u understand the concept based on- How fast? Rates of chemical change??

If so, can u explain me in details with simple english in bullet point form on
1. techniques to measure rate of reaction
2. rate equations, rate constants and the order of a reaction
3. determining the order of a reaction and the rate equation from experimental data.
4.graphical represnentation of kinetic measurements.
5. activation energy and types of catalyst
6. investigating the activation energy of a reaction
7. relating a mechanism to the rate-determining step
8. the mechanism of the reaction of iodine with propane.

pls help me dear....

i need ur reply urgenlty.
these questions are from edexcel A2 chemistry in unit 4///

[SZM]

dear iluvme, where is ur reply???/

i was waiting 4 ur reply.

pls send ur reply very soon...

[iluvme]

SZM, I myself haven't yet started Reaction Kinetics and Equilibria and besides I do CIE not Edexcel.

So I suppose I couldn't help you much, though take a look here, it might help you immensely InshaAllah

http://www.s-cool.co.uk/a-level/chemistry/reaction-kinetics

[The Golden Girl =D]

I hope someone helps me with this cuz it's H.W. and I'm dead meat if I don't do it 

several grams(an excess) of magnesium hydroxide are added to 1 dm³ of water at 289 K (room temperature). The solution becomes Saturated when 10.31*10^-4 moles of magnesium hydroxide had dissolved.

Calculate the pH of this solution ,given that Kw = 1*10^-14 mol²dm^-6 at 289 K.

That the only thing given in the Question as well as the following  Equation;


Mg(OH)₂ (s) +aq <-> Mg^-2 (aq) +2OH^- (aq)

Thanks in Advance =D

[The Golden Girl =D]

^ Anyone ? =[

[Amelia]

I hope someone helps me with this cuz it's H.W. and I'm dead meat if I don't do it 

several grams(an excess) of magnesium hydroxide are added to 1 dm³ of water at 289 K (room temperature). The solution becomes Saturated when 10.31*10^-4 moles of magnesium hydroxide had dissolved.

Calculate the pH of this solution ,given that Kw = 1*10^-14 mol²dm^-6 at 289 K.

That the only thing given in the Question as well as the following  Equation;


Mg(OH)₂ (s) +aq <-> Mg^-2 (aq) +2OH^- (aq)

Thanks in Advance =D

Dont blame me, if it's wrong. Not my work, anyway. 


"you need to work out how many OH- ions are in solution.

10.31 x 10^-4 moles of Mg(OH)2 means twice as many OH- ions

 You know how many moles of OH- are in that dm³ and you know that [H+] x [OH-] = 1 x 10^-14 so divide 1x10^-14 by the [OH-] to get the H+ concentration. Then use -log of that to get the pH."