## Question 15.19 on 6th edition hw

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

Marina Gollas 1A
Posts: 36
Joined: Fri Sep 28, 2018 12:20 am

### Question 15.19 on 6th edition hw

In Q: 15.19 on the 6th Edition Hw, how come the reaction of B is in second order? The concentration 2 of B was 3.02 and the concentration 1 of B was 1.25. The rate 2 is 50.8 and the rate 1 is 17.4. I end up getting 2.416^n=2.919, so n is about around the 1.2 area. Why did the solutions manual round it up to 2, making the reaction 2nd order, when I thought it would be first because of the 1.2. Did I do something incorrectly?

705192887
Posts: 77
Joined: Fri Sep 28, 2018 12:18 am

### Re: Question 15.19 on 6th edition hw

So to find the reaction of B, you divide reaction 2 by reaction 4. This cancels out K, A, and C, leaving just the rate in relation to B. (8.7/50.8)=(1.25)^y/(3.02)^y. By solving this equation, you get (1/5.8)=(1/2.4)^y. This simplifies to (2.4)^2=(2.4)^y. Therefore, y, which is the exponent on [B] and B's reaction rate, is 2. This means that B is a second order reaction. Hope this helps!