## Problem 15.27 vs. 15.35

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

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Tiffany Wu 1K
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Joined: Wed Sep 21, 2016 2:56 pm

### Problem 15.27 vs. 15.35

For questions 27 and 35 in Chapter 15 in the textbook, they both ask for the time needed for the concentration of a half-life reaction to decrease the concentration of A to a specific amount. However, question 27 is in first-order while questions 35 is in second-order.

For first-order, if it asked for one eight of its original concentration, I solved it like... (1/2)^n=1/8 then I solved for n and multiplied it by its half life. Why can't I do the same method for second-order?

stephanieyang_3F
Posts: 62
Joined: Wed Sep 21, 2016 2:55 pm

### Re: Problem 15.27 vs. 15.35

For the half life of a first order, the formula is $t_{1/2}=\frac{.693}{k}$. However for the half life of second order, the formula is $t_{1/2}=\frac{1}{k[R]_{o}}$. Notice that for first order reactions, the half time is constant while for second order reactions the time it takes for the substance to half increases over time. (For zero order, the half time decreases). Here's a link that explains it visually: https://www.chem.purdue.edu/gchelp/howt ... flife.html

Basically the reason why you can use your first method is because the time it takes for the substance to half the first time is directly proportional to the second, third, fourth, etc. time the substance is halved. Hope this helps!

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