The first-order decomposition of cyclopropane has a rate constant of 6.7 x 10-4 s-1. if the initial concentration of cyclopropane is 1.33 m, what is the concentration of cyclopropane after 644 s?

a. 0.43 m

b. 0.15 m

c. 0.94 m

d. 0.86 m

e. 0.67 m

The first order rate law has the form: - d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:

ln [A] = - kt + ln [A]_o, where,

k is the rate constant

t is the time of the reaction

[A] is the concentration of the species at the given time

[A]_o is the initial concentration of the species

For this problem, we simply substitute the known values to the equation as in:

ln[A] = - (6.7 x 10⁻⁴ s⁻¹) (644 s) + ln (1.33 M)

We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.

For a first-order reaction, the equation is written below:

lnA = lnA₀ - kt

where

A₀ is the original concentration

A is the concentration left after time t

k is the rate constant

Substituting the values:

lnA = ln (1.33 M) - (6.7*10⁻⁴ s⁻¹) (644 s)

Solving for A,

A = 0.863 M

Therefore, the answer is letter D.