A reaction was experimentally determined to follow the rate law, Rate = k[A]2 where k = 0.456 s-1M-1. Starting with [A]o = 0.500 M, how many seconds will it take for [A]t = 0.250 M?
If the rate law is rate = k[A]^2 you know it is second order. The integratedrate law for a second order reaction is 1/(A) - 1/(A)o = kt
1/(0.250) - 1/(0.500) = 0.456t. Splve for t in seconds. Post your work if you get stuck.
Oh boy, let me grab my lab coat and my funny-shaped glasses for this one! So, according to the rate law, we have Rate = k[A]^2. Now, we want to find out how many seconds it will take for [A] to reach 0.250 M, starting with [A]o = 0.500 M.
Well, since we have the rate constant k = 0.456 s^-1M^-1, we can use it to solve this problem. Since the rate is proportional to [A]^2, we can write:
Rate = k[A]^2
Now, let's plug in the values we know:
0.456 s^-1M^-1 = 0.500 M^2
Since 0.500 M^2 is equal to 0.250 M, we can rewrite the equation as:
0.456 s^-1M^-1 = 0.250 M
Now, let's solve for time (t) using a bit of mathematical magic:
0.456 s^-1M^-1 = (0.250 M) / t
t = (0.250 M) / 0.456 s^-1M^-1
t = 0.548 s
So, it will take approximately 0.548 seconds for [A] to reach 0.250 M. Hope that answers your question and brings a smile to your face!
To determine the time it takes for the concentration of reactant A to decrease from [A]o = 0.500 M to [A]t = 0.250 M, we can use the integrated rate law for a second-order reaction.
The integrated rate law for a second-order reaction is:
1/[A]t - 1/[A]o = kt
Rearranging this equation, we can solve for t:
t = (1/kt) * ([A]t - [A]o)
Given that k = 0.456 s^(-1) M^(-1), [A]o = 0.500 M, and [A]t = 0.250 M, we can substitute these values into the equation to calculate the time:
t = (1/(0.456 s^(-1) M^(-1))) * (0.250 M - 0.500 M)
t = (1/(-0.456 s^(-1) M^(-1))) * (-0.250 M)
t = (1/0.114) * 0.250 s
t ≈ 2.19 s
Therefore, it will take approximately 2.19 seconds for the concentration of reactant A to decrease from [A]o = 0.500 M to [A]t = 0.250 M.
To find out how many seconds it will take for [A]t to reach 0.250 M, we can use the integrated rate law equation for a second-order reaction. The integrated rate law equation for a second-order reaction is:
1/[A]t - 1/[A]o = kt
Where [A]t is the concentration of A at time t, [A]o is the initial concentration of A, k is the rate constant, and t is the time.
In this case, the rate law for the reaction is:
Rate = k[A]2
Therefore, we can rearrange the rate law equation to get:
[A]2 = Rate/k
Substituting the given values, we have:
0.250 M = 0.456 s-1M-1 * [A]2
Solving for [A], we take the square root of both sides of the equation:
[A] = sqrt(0.250 M / 0.456 s-1M-1)
[A] = 0.516 M
Now, we can use the integrated rate law equation to find the time it takes to reach [A]t = 0.250 M.
1/[A]t - 1/[A]o = kt
Substituting the known values:
1/0.250 - 1/0.500 = 0.456 s-1M-1 * t
Simplifying the equation:
4 - 2 = 0.456 s-1M-1 * t
2 = 0.456 s-1M-1 * t
Now, we can solve for t:
t = 2 / (0.456 s-1M-1)
t ≈ 4.39 seconds
Therefore, it will take approximately 4.39 seconds for [A]t to reach 0.250 M.