The average rate of the reaction is 0.00405 M/s, and the amount of Br2 formed in the first 9 seconds is 0.07146 moles.
The question is asking to determine the average rate of the reaction for the decomposition of HBr and the amount of Br2 formed in a given time interval. To find the average rate, we can use the change in concentration of HBr over the time period. The concentration of HBr decreases from 0.663 M to 0.574 M in 22 seconds, which gives us:
Rate (M/s) = ∆[Reactant] / ∆Time
Rate = (0.574 M - 0.663 M) / 22 s
Rate = -0.089 M / 22 s
Rate = -0.00405 M/s
Note that the rate of a chemical reaction is positive, so we take the absolute value of the rate:
Average rate = 0.00405 M/s
For the second part, since the reaction vessel volume is 1.96 L, and we know the rate, we can calculate the moles of Br2 produced in the first 9 seconds:
Moles Br2 = Rate × Volume × Time
Moles Br2 = 0.00405 M/s × 1.96 L × 9 s
Moles Br2 = 0.07146 moles
Complete question is:
Consider the reaction:
2 HBr(aq) -> H2(g)
a. In the first 22 s of this reaction, the concentration of HBr dropped from 0.663 M to 0.574 M. What is the average rate of the reaction during this time interval? (Remember to normalize the rate of the reaction for all reactants and products.)
rate (M/s) = number (rtol=0.03, atol=1e-08)
b. If the volume of the reaction vessel was 1.96 L, what amount of Br2 (in moles) was formed during the first 9 s of the reaction?
moles = number (rtol=0.03, atol=1e-08)