The rate constant for this zero‑order reaction is 0.0240 M ⋅ s − 1 0.0240 M·s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long (in seconds) would it take for the concentration of A A to decrease from 0.900 M 0.900 M to 0.210 M?

Zero order reaction : A reaction is said to be of zero order if the rate is independent of the concentration of the reactants, that means the rate is directly proportional to the zeroth power of the concentration of the reactants.

Expression for the zero order kinetics:

[tex][A]=-kt+[A]_o[/tex]

where [A] = concentration left after time t = 0.210 M

[tex][A]_o[/tex] = initial concentration = 0.900 M

k= rate constant =[tex]0.0240Ms^{-1}[/tex]

t = time for reaction = ?

[tex]0.210=-0.0240\times t+0.900[/tex]

[tex]t=28.75s[/tex]

Thus it will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s

andromache andromache · Answer 2 · 2022-03-01T14:42:32+01:00

It will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s

Calculation of the time:

A reaction is said to be of zero-order when the rate should be independent of the concentration of the reactants, which represents the rate should be directly proportional to the zeroth power of the concentration of the reactants.

So, the time taken should be

[A]= -kt +A_o

Here,

[A] should be concentration left after time t i.e. 0.210 M

A_o = initial concentration = 0.900 M

k= rate constant =0.0240 M

So,

0.210 = -0.0240 * t + 0.900

t = 28.75s

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