After a certain pesticide compound is applied to crops, its decomposition is a first-order reaction with a half-life of 56 days. What is the rate constant, k, for the decomposition reaction?

The rate constant for the decomposition reaction has been 0.0124/days.

The rate constant has been given by;

log [Final concentration] = kt + log [Initial concentration]

k has been the rate constant, and t has been the time.

The initial concentration has been 1, after the half life period the concentration  has been half i.e. 0.5 .

The given half life has been 56 days.

The rate constant can be calculated as:

log 1 = k(56) + log 0.5

k(56) = 0.693

k = 0.0124/days.

The rate constant for the decomposition reaction has been 0.0124/days.

For more information about the rate constant, refer to the link:

link

The rate constant, k, for the decomposition reaction :  k = 0.0124 / days

Given

The half-life of 56 days

Required

The rate constant, k

Solution

For first-order, rate law : ln[A]=−kt+ln[A]o

The half-life :  the time required to reduce to half of its initial value.

The half life :

t1/2 = (ln 2) / k

k = (ln 2) / t1/2

k = 0.693 / 56 days

k = 0.0124 / days

one state would change   like louisiana to texas. thats a change

explanation:

yes

ya te ayud espera