Cyclopropane is converted to propene in a first-order process. The rate constant is 5.4 × 10-2 hr-1. If
the initial concentration of cyclopropane is 0.150 M, what will its concentration be after 22.0 hours?
.0457
To determine the concentration of cyclopropane after 22.0 hours, we can use the first-order rate equation:
ln(c/c₀) = -kt
Where:
- c is the concentration of cyclopropane at a given time
- c₀ is the initial concentration of cyclopropane
- k is the rate constant
- t is the time
We can rearrange the equation to solve for c:
c = c₀ * e^(-kt)
Substituting the given values:
- c₀ = 0.150 M (initial concentration)
- k = 5.4 × 10^(-2) hr^(-1) (rate constant)
- t = 22.0 hours (time)
c = 0.150 M * e^(-5.4 × 10^(-2) hr^(-1) * 22.0 hours)
Now, we can calculate the concentration of cyclopropane after 22.0 hours.
To determine the concentration of cyclopropane after 22.0 hours, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where:
[A]t is the concentration of cyclopropane at time t
[A]0 is the initial concentration of cyclopropane
k is the rate constant
t is the time interval
Rearranging the equation, we can solve for [A]t:
[A]t = [A]0 * e^(-kt)
Given:
[A]0 = 0.150 M (initial concentration)
k = 5.4 × 10^(-2) hr^(-1) (rate constant)
t = 22.0 hours (time interval)
Substituting the given values into the equation, we have:
[A]t = 0.150 * e^(-5.4 × 10^(-2) * 22)
Now we can calculate the concentration of cyclopropane after 22.0 hours using this equation.
k=5.4x10^-2
t=22.0 hours
initial=0.150 M
final=?
ln(x/0.150) = - (5.4x10^-2)(22.0)
x= 0.127