Answer :
a. the average rate of the reaction is (0.337 M) / (24 s) = 0.01404 M/s.
b. amount of Br2 (in moles) was formed during the first 11 s of the reaction 0.13040 moles of Br2.
a. To calculate the average rate of the reaction during the first 24 seconds, we can use the formula: average rate = (change in concentration) / (change in time).
The concentration of HBr dropped from 0.792 M to 0.455 M, so the change in concentration is 0.792 - 0.455 = 0.337 M. The change in time is 24 seconds. Therefore, the average rate of the reaction is (0.337 M) / (24 s) = 0.01404 M/s.
b. To determine the amount of Br2 formed during the first 11 seconds of the reaction, we first need to find the change in HBr concentration during this time.
Using the average rate we calculated, we can find the change in HBr concentration during 11 seconds: (0.01404 M/s) * (11 s) = 0.15444 M.
Since the reaction ratio is 2 HBr to 1 Br2, the change in Br2 concentration is half of the change in HBr concentration: 0.15444 M / 2 = 0.07722 M. Now, multiply the change in Br2 concentration by the volume of the reaction vessel
(1.69 L) to find the amount of Br2 formed in moles: 0.07722 M * 1.69 L = 0.13040 moles of Br2.
To know more about reaction refer here: https://brainly.com/question/17434463#
#SPJ11
