Calculate the time at which ester concentration reduce to $0.25 M [ln 2 = 0.7]$

(Given density of pure H₂O = 1 gm/mL)

Question

Calculate the time at which ester concentration reduce to 0.25 M [ ln 2 = 0.7 ] (Given density of pure H₂O = 1 gm/mL)

TestHub · Accepted Answer

Question Explanation: Find time at which ester concentration reduce to 0.25 M

Concept : This question is based on pseudo first order reaction.

Solution: The rate law is $r = K [H^{+}] [H_{2} O] [es t er] = k^{'} [es t er]$ , where $k^{'}$ is the pseudo-first-order rate constant.

Given: $[H^{+}] = 0.1 M$ (from $0.1 M H Cl$ catalyst).

$[H_{2} O] = \frac{d e n s i t y}{M o l a r ma ss} = \frac{1000 \frac{g}{L}}{18 \frac{g}{m o l}} = \frac{1000}{18} M$

$k^{'} = K [H^{+}] [H_{2} O] = (1.8 \times 1 0^{- 3}) \times 0.1 \times \frac{1000}{18} = \frac{0.18}{18} = 0.01 se c^{- 1}$ .

Initial concentration $[es t er]_{0} = 1 M$ , final concentration $[es t er]_{t} = 0.25 M$ .

For first-order kinetics, the integrated rate law is:

$t = \frac{1}{k ^{'}} ln (\frac{[ es t er ] _{0}}{[ es t er ] _{t}})$

$t = \frac{1}{0.01} ln (\frac{1}{0.25})$

$t = 100 ln (4)$

$t = 100 \times 2 ln (2)$

$t = 200 \times 0.693$

$t = 138.6 sec$

Rounding to one significant figure, $t \approx 140 sec$ .

Chemistry - Chemical Kinetics Question with Solution | TestHub