6th Edition 15.15

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

beckyolmedo1G
Posts: 34
Joined: Fri Sep 28, 2018 12:21 am

6th Edition 15.15

For a reaction, when OH conc. was doubled, the rate doubled. When the CH3OH concentration alone was increased by a factor of 1.2, the rate increased by a factor of 1.2. Write the rate law for the reaction.

The answer key says "rate is 1st order in both reactants." But I thought it would be a 2nd order reaction. Since it would look like K[CH3Br][OH-]. Can someone explain this to me? Do these essentially mean the same thing?

Sean Reyes 1J
Posts: 67
Joined: Fri Sep 28, 2018 12:24 am

Re: 6th Edition 15.15

The overall reaction order would be 2, yes. But in terms of each reactant, the rate is first order. The rate law you wrote down is correct, but because the order of each reaction is 1, then the overall reaction order is 2.

chaggard
Posts: 37
Joined: Fri Sep 28, 2018 12:19 am

Re: 6th Edition 15.15

The answer is saying that each individual reactant is first order, which is shown through the doubling directly doubling the rate. However when combined they do produce a second order reaction with respect to the product.