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.

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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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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