## Question 15.13 Part B (Sixth Edition)

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

Steve Magana 2I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Question 15.13 Part B (Sixth Edition)

Question: When 0.52 g of H2 and 0.19 g of I2 are confined to a 750.-mL reaction vessel and heated to 700. K, they react by a second-order process (first order in each reactant), with k = 0.063 L.mol^-1.s^-1 in the rate law (for the rate of formation of HI). (a) What is the initial reaction rate? (b) By what factor does the reaction rate increase if the concentration of H2 present in the mixture is doubled?

How do we determine the answer for part b? Thank you!

Eshwar Venkat 1F
Posts: 32
Joined: Fri Sep 28, 2018 12:22 am

### Re: Question 15.13 Part B (Sixth Edition)

Since each reactant has a first-order process, rate is directly proportional to the concentration. Thus, if you double the concentration of H2, the rate will double.

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Re: Question 15.13 Part B (Sixth Edition)

This is not a question that requires calculation as much as it is a question that requires conceptual understanding. In a 1st order reaction, the rate of the reaction varies proportionally with the concentration of reactant. But in a 2nd order reaction, the rate will quadruple if the concentration of a reactant doubles.