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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

Respuesta :

MissPhiladelphia
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.
meerkat18
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.

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