Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.
| t/min | 0 | 30 | 60 | 90 |
| C/mol L–1 | 0.8500 | 0.8004 | 0.7538 | 0.7096 |
Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant (55 mol L–1), during the course of the reaction. What is the value of k' in this equation?
Rate = k' [CH3COOCH3] [H2O]
Dear student..
Given, the concentration of H2O remains nearly constant. Therefore, the reaction should be first order with respect to ester(methyl acetate). Now, the rate constant K for a pseudo first order reaction is
From the above data we get the following values,
|
t/min |
C/mol L-1 |
K/min-1 |
|
0 |
0.8500 |
--- |
|
30 |
0.8004 |
2004 ×10-3 |
|
60 |
0.7538 |
2002 ×10-3 |
|
90 |
0.7096 |
2005 ×10-3 |
It can be seen that K = K’[H2O] is constant and nearly equal to 2.004 × 10-3 min–1 and hence, it is pseudo first order reaction.
Now,
K’ [H2O] = 2.004 × 10–3 min–1
K’ [55 mol L–1] = 2.004 × 10–3 min–1
K’= 3.64 × 10–5 mol–1 L min–1
Hope it is clear
Regards