A first-order reaction is 75.0% complete in 320. s. a. What are the first and second half-lives for this reaction?
Answer: The first and second half-lives for The following formula can be used to calculate this reaction: equation: t1/2 = 0.693/k, where k is the rate constant.
Using a rate constant of 0.0022 s-1, the first and second half-lives would be 313.6 s and 627.2 s, respectively. To calculate the time it takes for 90.0% completion, we can use the equation:
t = t1/2 * ln(0.1/C), where C is the initial concentration of the reactant.
Using an initial concentration of 1 and the first half-life of 313.6 s, it would take 577 s for 90.0% completion.
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