JEE Advanced levelSingle correct optionDifficulty 3 of 3ck092
The hydrolysis of an ester in a large excess of dilute acid is followed by titrating equal samples with alkali. The titre volumes are V0=12.0mL at t=0, Vt=16.0mL at t=40min and V∞=24.0mL at completion. Use 2.303, log2=0.3010 and log3=0.4771. The pseudo first order rate constant is
A0.02747min−1
B0.00719min−1
C0.01733min−1
D0.01014min−1
Answer: D
The acid produced is proportional to Vt−V0, so k=(2.303/t)logV∞−VtV∞−V0=(2.303/40)log8.012.0=(2.303/40)(log3−log2)=(2.303/40)(0.1761)=0.01014min−1. The other options use incorrect volume combinations.