Consider two reactions, both with equilibrium constants of 1.5 x 10 -2 . One rea

Question

Consider two reactions, both with equilibrium constants of 1.5 x 10-2 . One reaction, 2A ---> B is first-order, with a rate constant of 1.5 x 102 sec-1; the other, A+B--->C, is second-order with a rate constant of 1.5 x 102M-1sec-1. Suppose you start with 1M concentrations of each of the starting materials. For each reaction, which will happen first: Establishment of equilibrium, or one half-life? How long will it take for the first-order reaction to reach equilibrium? Can you explain it clearly please ?

Explanation / Answer

for the first reaction , -d[A]/dt = k [A]
=> [A] = e^(-t)

half life of the reaction = ln(2)/k = 0.00462 s.
[B] = (1-e^(-t))/2

for equilibrium , ke = [B]/[A]^2
=> [A]^2 = [B]/ke = { (1-e^(-t))/2 } / ke = { e^(-t) }^2
=> e^(-t) = 0.971675.
=> t =0.0287s.

hence half life takes place in 1st reaction.

second reaction:
-d[A]/dt = [A][B] ; -d[B]/dt = [A][B].
now , d[A]/dt = d[B]/dt => [A] = [B] ------------>[since initial concentrations are same]
=>-d[A]/dt = [A]^2
=> [A] = 1/(t+1).
hence half life = t =1.
for equilibrium , [C] = ke [A][B].
=>[C] = ke[A]^2
=>1 - [A] = ke[A]^2.
=> [A] = 0.98543 = 1/(t+1)
=> t = 0.014785s
hence equilibrium takes place first.

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