Qualification > Queries

ANY DOUBTS HERE!!!

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Saladin:
Hey u guys, post all ur doubts here, and I will do my best to answer them form u.

Just trying out the tex editor.

Sue T:
i hope chemistry doubts go here as well
past paper says:

 The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic
acid, C2H5OH + CH3CO2H CH3CO2C2H5 + H2O, at 60(degree)C is 4.00.
When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60(degree)C, what
is the number of moles of ethyl ethanoate formed?
A 1/3
B 2/3
C 1/4
D 3/4

3ishakay:

--- Quote from: Sue T on April 06, 2010, 04:55:15 pm ---i hope chemistry doubts go here as well
past paper says:

 The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic
acid, C2H5OH + CH3CO2H CH3CO2C2H5 + H2O, at 60(degree)C is 4.00.
When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60(degree)C, what
is the number of moles of ethyl ethanoate formed?
A 1/3
B 2/3
C 1/4
D 3/4

--- End quote ---



hey ... dyu noe the anser to tht?
wht paper ist frm?

im thinkn the anser is B = 2/3

this is how i did it ..(any1 tell me if i got it wrong .. plz!!)


                      ethanol   +     ethanoic acid    ----->   ethylethanoate   +     water
                      C2H5OH   +    CH3CO2H         ------->  CH3CO2C2H5     +     H2O

b4 eq.                  1                   1                                   0                      0

aftr eq.                1-x                1-x                                 x                       x




thn Kc= [ethylethanoate][water]
            -----------------------
            [ethanol][ethanoic acid]


so its like  


          [ x]^2           =  4
        --------
          [1-x]^2


thn u ju solv it like in maths ..

and thn u get 2/3 =D

rechek the smae crap..


go


           ( 2/3)^2              =    4!!   TADA!!! :D:D:D
          ----------
            (1/3)^2




nid404:
hey Thanks :)

3ishakay:

--- Quote from: nid404 on April 06, 2010, 06:05:15 pm ---hey Thanks :)

--- End quote ---

not a prbm  8)   ;)

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