If the rate constant for a base catalyzed ester hydrolysis reaction is 0.20 L mol⁻¹ s⁻¹, half-life (in s) of the ester (Given [ester]₀ = [base]₀ = 0.05 mol.L⁻¹) would be closest to

Concept:

The rate law for a base-catalyzed ester hydrolysis reaction is:

rate = k[ester][OH^-]

where k is the rate constant, [ester] is the concentration of the ester, and [OH^-] is the concentration of the hydroxide ion.

We can use the half-life formula for a first-order reaction to relate the rate constant to the half-life:

t_½ =\(\frac{ ln(2) }{k}\)

where ln(2) is the natural logarithm of 2, which is approximately 0.693.

Explanation:

RCOOR' + OH^- → RCOO^- + R'OH

where R and R' are organic groups.

The rate law for this reaction is:

rate = k[ester][OH^-]

where k is the rate constant, and [ester] and [OH^-] are the concentrations of the ester and hydroxide ion, respectively.

At the start of the reaction, the concentrations of the ester and hydroxide ion are both 0.05 mol/L, as given in the question.

To find the half-life of the reaction, we can use the first-order rate law equation:

ln([ester]_t/[ester]₀) = -kt

Assuming that the base completely dissociates to form hydroxide ions, the initial concentration of [OH^-] is also 0.05 mol/L.

Therefore, the half-life is:

t_½ = ln(2) / k

= ln(2) / (0.20 L mol^-1 s^-1 x 0.05 mol/L x 0.05 mol/L)

t½ ≈ 98 s.

Conclusion:
Therefore, the half-life of the ester is closest to 100 s.