I need help with parts i and iiScreenshot 2025-01-08 193033.pngScreenshot 2025-01-08 193042.png
This reaction follows first-order kinetics because the half-life is constant throughout the process. By analyzing the graph, we can observe that the time required for the concentration to decrease from 1.60 M to 0.80 M is 8 ms. Similarly, the time required for the concentration to decrease from 0.80 M to 0.40 M is also 8 ms.
This constancy of half-life is a defining characteristic of first-order reactions. In first-order kinetics, the half-life is independent of the initial concentration, as expressed by the formula:
𝑡1/2= 0.693/k
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k is the rate constant. Thus, the observation that the half-life remains unchanged at different concentration levels confirms that the reaction is first order.
Similar question we did in the class.
Now it is first order kinetics so we can use
ln[A]t=-kt+ln[A]0
You can rearrange this equation as
t = 1/k x ln([A]0/[A]t
I discussed this format also.
[A]0= 100 (assume because percent is given)
[A]t= 1 (100-99)
k = 0.693/t1/2 This we calculated in part i as 8ms
Just plug in all values and work out the answer as ms.