## 7th Edition 7A.11 2nd Order RXN

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

abbydouglas1K
Posts: 65
Joined: Fri Sep 28, 2018 12:26 am

### 7th Edition 7A.11 2nd Order RXN

In this problem it says that these reactants react in a second order process so that k[H2][I2] is the rate. However I thought that a second order process was k[A]^2 . Does A just represent reactants and the square mean there is two reactants hence a second order process?

Schuyler_Howell_4D
Posts: 66
Joined: Fri Sep 28, 2018 12:28 am
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### Re: 7th Edition 7A.11 2nd Order RXN

I think so. If you expanded it to be k[A][A] then you would be able to find that A could be two different reactants.

Gracie Ge 2E
Posts: 28
Joined: Wed Nov 21, 2018 12:19 am

### Re: 7th Edition 7A.11 2nd Order RXN

for a reaction with rate k[A]^a[B]^b, the order is simply a+b.
which means k[H2][I2] and a general k[A]^2 are both second order.