Methyl acetate is hydrolyzed with 0.1 N HCl at 25^(@)C. 5mL of the reacting mixture is withdrawn at various time intervals and is quickly titrated is withdrawn NaOH. The volume of NaOH consumed are as follows: |{:("Time (s)",339,1242,2745,4546,oo),("vol of NaOH (mL)",26.34,27.80,29.70,31.81,39.81):}| Show that the hydrolyiss is a first order reaction. Also find the rate constant (k).

Methyl acetate is hydrolyzed with 0.1 N HCl at 25^(@)C. 5mL of the rea

1.	Methyl acetate is hydrolyzed with 0.1 N HCl at 25^(@)C. 5mL of the reacting mixture is withdrawn at various time intervals and is quickly titrated is withdrawn NaOH. The volume of NaOH consumed are as follows: \|{:("Time (s)",339,1242,2745,4546,oo),("vol of NaOH (mL)",26.34,27.80,29.70,31.81,39.81):}\| Show that the hydrolyiss is a first order reaction. Also find the rate constant (k).
Answer» Solution :For ester hydrolyiss, equation`k = (2.303)/(t)LOG.(V_(oo)-V_(0))/(V_(oo)-V_(t))` Since `V_(0)` is not given Let `V_(oo)-V_(0) = c_(0)` and `V_(oo) - V_(t) = c_(t)`. When `t = t_(1)` `:. k = (2.303)/(t_(1))log.(c_(0))/(c_(t_(1)))`...(i) ismilarly, when `t = t_(2)` `k = (2.303)/(t_(2))log.(c_(0))/(c_(t_(2)))`...(ii) Oprating EQ.(i) Eq. (ii), we GET `k(t_(1)-t_(2)) = 2.303log.(c_(t_(2)))/(c_(t_(1)))` or `k = (2.303)/((t_(1)-t_(2)))log.(c_(t_(2)))/(c_(t_(1)))` ...(iii) Now find `c_(t)`. and `t_(1) - t_(2) = 339 - 1242 = -903 s` `t_(1)-t_(3) = 339 - 2745 = -2406 s` `t_(1)-t_(4) = 339-4546=-4207 s` Uisng relation given in Eq. (iii), `:. k_(1) = (2.303)/(-903)log.(12.0)/(13.47) = -0.0025 XX log(0.89)` `= -0.0025 xx -0.0506 = 1.27 xx 10^(-4) s^(-1)` `k_(2) = (2.303)/(-2406)log.(10.11)/(13.47) = -0.00095 xx log(0.75)` `= -0.00095 xx -0.1249 = 1.19 xx 10^(-4) s^(-1)` `k_(3) = (2.303)/(-4207)log.(8.0)/(13.47) = -0.00054 xx log(0.59)` `= -0.00054 xx -0.2291 = 1.23 xx 10^(-4) s^(-1)` `k_("avergae") = 1.23 xx 10^(-4) s^(-1)`