The observed first-order rate constant of an ester hydrolysis follows k_obs = k_0 + k_H[H+]. It is 0.95 x 10^-4 s^-1 at [H+] = 0.050 M and 3.50 x 10^-4 s^-1 at 0.200 M. Find the half-life of the ester, in seconds to the nearest integer, at [H+] = 0.400 M. Use ln 2 = 0.693.

JEE Advanced levelNumerical answerDifficulty 3 of 3sc101

The observed first-order rate constant of an ester hydrolysis follows $k_{o b s} = k_{0} + k_{H} [H X^{+}]$ . It is $0.95 \times 1 0^{- 4} s^{- 1}$ at $[H X^{+}] = 0.050 M$ and $3.50 \times 1 0^{- 4} s^{- 1}$ at $0.200 M$ . Find the half-life of the ester, in seconds to the nearest integer, at $[H X^{+}] = 0.400 M$ . Use $ln 2 = 0.693$ .

Numerical answer type, answer in s, tolerance 1.

Subtracting the two readings gives

k_{H} = (0.000350 - 0.000095) /0.150 = 0.00170 M^{- 1} s^{- 1}

, and then

k_{0} = 1.00 e - 05 s^{- 1}

. At 0.400 M,

k_{o b s} = 6.900 e - 04 s^{- 1}

. So

t_{1/2} = 0.693/ k_{o b s} = 1004.3

s, i.e. 1004 s.