A reaction of the form aA---> Products is second order with a rate constant of 0.169 L/(mol*s). If the intial concentration of A is 0.159 mol/L, how many seconds would it taje for the concentration of A to decrease to 6.07X10^-3 mol/L
I think the equation for a second order reaction is
(1/A) -(1/Ao) = kt.
You know A, Ao, and k. Calculate time.
To solve this problem, we can use the second-order rate equation:
Rate = k[A]^2
Where:
Rate is the rate of reaction (change in concentration per unit time),
k is the rate constant, and
[A] is the concentration of A.
We can rearrange the equation to solve for time. Let's call the initial concentration of A [A]₀, and the concentration of A at a later time [A]₁:
Rate = k[A]^2
k[A]^2 = ( [A]₁ - [A]₀ ) / t
Here, [A]₀ = 0.159 mol/L and [A]₁ = 6.07x10^-3 mol/L.
Substituting the given values, we find:
0.169 L/(mol*s) * (0.159 mol/L)^2 = (6.07x10^-3 mol/L - 0.159 mol/L) / t
Simplifying the equation by canceling units:
0.169 * (0.159)^2 = (6.07x10^-3 - 0.159) / t
Now we can solve for t:
t = (6.07x10^-3 - 0.159) / (0.169 * (0.159)^2)
Calculating this expression, we find:
t ≈ 4.536 seconds
Therefore, it will take approximately 4.536 seconds for the concentration of A to decrease to 6.07x10^-3 mol/L.
To solve this problem, we can use the integrated rate equation for a second-order reaction:
1/[A]t - 1/[A]0 = kt
Where:
[A]t is the concentration of A at time t
[A]0 is the initial concentration of A
k is the rate constant
t is the time
First, we need to rearrange the equation to solve for t:
1/[A]t - 1/[A]0 = kt
Multiply both sides by [A]t[A]0:
[A]0 - [A]t = kt[A]0[A]t
Rearrange the equation:
kt[A]t[A]0 = [A]0 - [A]t
Now, we can substitute the given values into the equation:
k = 0.169 L/(mol*s)
[A]0 = 0.159 mol/L
[A]t = 6.07 × 10^(-3) mol/L
0.169 * t * (6.07 × 10^(-3)) * (0.159) = 0.159 - 6.07 × 10^(-3)
0.0102593t = 0.158943
Now, divide both sides by 0.0102593:
t = 0.158943 / 0.0102593
t ≈ 15.48 seconds
Therefore, it would take approximately 15.48 seconds for the concentration of A to decrease to 6.07 × 10^(-3) mol/L.