When 75% reactant A decompose in 10 sec according to zero order reaction A-2B then rate constant of the reaction is ( assume initial concentration of A as 0.1 M)
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When 75% reactant A decompose in 10 sec according to zero order reaction A-2B then rate constant of the reaction is ( assume initial concentration of A as 0.1 M) is 0.0025 [M] s^-1.
According to zero order kinetics,
[A]' = [A]° - kt where ...(1)
[A]° is the initial concentration and,
[A]' is the concentration of the reactant at time 't'
k is the rate constant.
Given, from the question, time 't' = 10sec.
Initial concentration [A]° = 0.1[M]
Also, it is given that after 10sec., the reactant is 75% dissociated,
so remaining reactant's concentration = .75(initial concentration)[M]
= 0.75(0.1) [M]
So, [A]' = 0.075[M]
So, putting the value of [A]' = 0.075[M] and value of [A]° and t in (1), we get,
0.075 = 0.1 - k(10)
k(10) = 0.025
k = 0.0025 [M] s^-1.
Thus, rate constant is 0.0025 [M] s^-1.
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