Question

2 moles each of CH3OH & CH3COOH are taken and heated in the presence of conc.H2SO4 so that
equilibrium is established. If Kc for esterification process is 4, moles of ester present at equilibrium is
1) 1.33
2) 1.6
3) 0.67
4) 0.33 ​

Answer 1

Answer:

Dear Student,

CH3OH      +     CH3COOH    −→−−−−−conc.H2SO4    CH3COOCH3       +      H2O

2moles               2moles                                    0 mole                   0 mole   (at t=0)

  (2-x)                   (2-x)                                          x                             x           (no. of moles at t= equilibrium)

 

Kc = −→−−−−−−−−−−−−[CH3OH][CH3COOH][CH3COOCH3][H2O]   

    = −→−−−−−−(2−x)*(2−x)x*x

4 = x2(2−x)2

Taking square root both side

2  = x(2−x)

4 - 2x  =  x

3x  = 4

x = 43

Thus 4/3 moles of ester (CH3COOCH3) is formed.

Molar mass of ester is = 74g/mol

mass = mole * molar mass

= 43*74

=98.6g

hope this helps u

pls mark it as the brainliest

#BAL

#ANSWER with QUALITY

Answer 2

CH3OH      +     CH3COOH    −→−−−−−conc.H2SO4    CH3COOCH3       +      H2O

2moles               2moles                                    0 mole                   0 mole   (at t=0)

 (2-x)                   (2-x)                                          x                             x           (no. of moles at t= equilibrium)

 

Kc = −→−−−−−−−−−−−−[CH3OH][CH3COOH][CH3COOCH3][H2O]    

   = −→−−−−−−(2−x)*(2−x)x*x

4 = x2(2−x)2

Taking square root both side

2  = x(2−x)

4 - 2x  =  x

3x  = 4

x = 43

Thus 4/3 moles of ester (CH3COOCH3) is formed.

Molar mass of ester is = 74g/mol

mass = 98.4

moles = 98.6 / 74 = 1.33