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ALL CIE CHEMISTRY DOUBTS HERE !!

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Hypernova:
Lets first find the number of moles of CH3C00H at equilibrium. It reacts with NaOH in 1:1 ration.

number of moles=concentration*volume
               =1.5 * (72.5/1000)
               =0.10875 moles of CH3C00H at equilibrium

When 1 mole of CH3C00H reacts with 1 mole of C2H5OH 1 mole of CH3COOC2H5 and H2O is produced.
So if 0.09125 moles of CH3C00H reacts, then 0.09125 of CH3COOC2H5 and H2O is formed.


                      CH3C00H      +     C2H5OH <->     CH3COOC2H5 +      H2O

Before equilibrium      0.2                0.11                     0              0
At equilibrium         0.10875           0.01875            0.09125        0.09125


  KC= 0.09125 * 0.09125
             0.10875 * 0.01875

    
    =4.08

    
Note: the concentration should be used to find Kc, but since they all in the same container, the volumes is the same and will cancel out in this case, so i didn't need to put in volumes.


For the second part, you need to find enthalpy change of reaction. enthalpy change = bonds broken - bonds made
but thats long.

$H00t!N& $t@r:
Thanks alot for the detailed answer!
+rep  :)

Greed444:

--- Quote from: Deadly_king on November 05, 2010, 06:05:24 am ---
Conditions                                   Product at anode                          Product at cathode

ZnCl2(l)                                        Cholrine                                      Zinc

ZnCl2(concentrated aqueous)            Chlorine                                    Hydrogen

ZnCl2(diltue aqueous)                       Oxygen                                    Hydrogen



NOTE : Positive ions get discharged at the cathode while negative ones are discharged at the anode.

i) Zinc Chloride in liquid form is composed of Zn2+ and Cl- ions which are discharged at the electrodes forming the respective compounds.

ii) Concentrated aqueous Zinc Chloride contains the following ions : Zn2+, Cl-, H+ and OH-.
For cations, you should compare them from the reactivity series. The lower the cation is in the reactivity series the earlier it will be discharged in preference to the most reactive cation.
This is why hydrogen gets preferentially discharged at the expense of Zn2+

For anions, it's about the same thing but you should note that the second part is concentrated aqueous, which means that Cl- is abundant while OH- is present in very small amount.

Hence oxygen will be formed in both cases but it in so minimal amount that you won't even notice and Cl- will immediately be discharged after that.


--- End quote ---
Thanks Bro! i Completely understand now :)

Greed444:
does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)

in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S

moon:

--- Quote from: Greed444 on November 06, 2010, 02:46:00 pm ---does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)

in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S

--- End quote ---


Yes I am confused too . I would really appreciate any help Thanks in advance

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