Question

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s	0	30	60	90
Ester/mol L−1	0.55	0.31	0.17	0.085

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer 1

i. Average rate during the interval 30 to 60 seconds

= `"Change (decrease) in concentration"/"Time interval involved in the change"`

= `- (0.17 - 0.31)/(60 - 30)`

= `- (-14)/30`

= `14/30`

= 4.67 × 10⁻³ mol L⁻¹ s⁻¹

ii. For a first-order reaction,

k = `2.303/t log_10  [A]_0/([A])`

In the present case, [A]₀ = 0.55 M

When t = 30 s, k = `2.303/30 log_10  0.55/0.31` = 1.91 × 10⁻² s⁻¹

When t = 60 s, k = `2.303/60 log_10  0.55/0.17` = 1.96 × 10⁻² s⁻¹

When t = 90 s, k = `2.303/30 log_10  0.55/0.085` = 2.07 × 10⁻² s⁻¹

∴ Average value of k = `(1.91 xx 10^-2 + 1.96 xx 10^-2 + 0.07 xx 10^-2)/3`

= 1.98 × 10⁻² s⁻¹