Textbook Question 7B 13

[Ian_Lee_1E]

For this question,

I used half life equation in order to find the time it would take for the concentration to decrease to one-sixteenth of its original, which would be (1/2)^4, making the time 4 x 50.5 seconds.

However, the answer used the Second order integrated equation, making the answer equal to 7.4 x 10^2 s

How do I know when to use integrated or half -life equation and why are they different?

[Rich Luong 1D]

Honestly, what I did was a whole different method, yet it was able to produce the desired results. I first use the 2nd order half life equation to find the value of k. Afterwards, I just plug everything into the 2nd order integrated rate law equation: 1/[A] = kt + 1/[A]0. Note that when you plug in the value for [A], it must be the desired fraction of the concentration 0.84. After plugging in everything, do a bit of algebra and it should get you the needed time in seconds. I hope this helps!

[Eliot Kagan 2G]

I think it has to do with the rate flatting out as the reaction goes on in a 2nd order reaction. The 2nd order graph has a decaying exponential line. Correct me if I'm wrong but this is my guess.