15.35

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Chew 2H
Posts: 34
Joined: Fri Sep 29, 2017 7:05 am

15.35

For 15.35 part a, the half life for the second order reaction of a substance A is 50.5s when [A]initial is 0.84M. it then asks us to calculate the time needed for the concentration of A to decrease to one sixteenth of its original value. For this question why can't we just multiply 50.5s by 4 like we did for question 15.27?

Thank you in advance!

Hubert Tang-1H
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

Re: 15.35

The reason is because the reaction is a second-order reaction, meaning we must use the equation

1/[A]=kt+1/[A_0]. We notice that that for 16*[A]=[A_0], we have
16/[A_0]=kt+1/[A_0]
kt=15/[A_0].

Compared to the half -life 2[A]=[A_0]
2/[A_0]=kt+1/[A_0]
kt=1/[A_0].

Thus, we see that for second-order reactions, because of the order, we can't just multiply by 4.

Michelle Nguyen 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Re: 15.35

Question 15.27 asks you to use the half life for a first order reaction, t(1/2)= ln2/k, which does not depend on the concentration of reactants at all. That means the half life will always be the same length of time no matter how far the reaction proceeds or what the initial concentration is, so you can just multiple t(1/2) by a constant. For 15.35, you have to use the half life reaction of a second order reaction, t(1/2)= 1/k[A] initial, which does depend on initial reactant concentration. That means each consequent half life you need to calculate is a different length of time, because each time the initial concentration is half of the last initial concentration. Thus, you need to compute multiple half lives based on different initial concentrations, then add them together to get the total amount of time it takes for that amount of decomposition of reactant to occur.