In a pseudo first order hydrolysis of ester in water the following result6s were obtained:
`{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}`
(i) Calculate the average rate of reaction between the time interval `30` to `60` seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Options

A. `1.91 xx 10^(-2) s^(-1)`
B. `4.67 xx 10^(-3) mol L^(-1) s^(-1)`
C. `1.98 xx 10^(-3) s^(-1)`
D. `2.07 xx 10^(-2) s^(1)`

Answer

B

Explanation

Average rate during the time interval 30-60s.
Rate `= - (C_(2) - C_(1))/(t_(2) - t_(1)) = - ((0.17 - 0.31))/(60 - 30) = (0.14)/(30)`
`= 4.67 xx 10^(-3) mol L^(-1) s^(-1)`