Iodine atoms combine to form I₂ in liquid hexane solvent with a rate constant of 1.5⨉10¹⁰ L/mols. The reaction is second order in I. Since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instantaneously, usually by photochemical decomposition of I₂. Suppose a flash of light creates an initial [I] concentration of 0.0100 M. How long will it take for 95% of the newly created iodine atoms to recombine to form I₂?

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Hey everyone in this example, we're told that we have a second order reaction with a rate constant of 2.3 times 10 to the 10th power, inverse polarity times inverse seconds. We need to assume an initial concentration of 100.200 moller and calculate the time that it takes for 80% of our eight atoms to convert the product A. Two. So what we're going to recognize is that because 80% of our eight atoms convert to the product, that means we have 20% left of our reactant. A. And so we would say that therefore our initial concentration for our reactant is equal to zero point or 20% as a decimal being 0.20 moller, which again is given in the prompt with an extra trailing zero as 00.200 moller. We also want to recognize that our concentration of our reactant a after the reaction occurs and produces a two as a product is going to be 20. multiplied by the initial concentration of A. And sorry, that should be outside of the brackets. So this is going to be equal to .2, multiplied by .2. And this is going to give us a value in our calculators equal to 0.04 Mueller as our concentration of a. After it converts to the product a two. So we also should recognize that we're given the rate constant in the prompt. And we're told that we have a second order reaction. So we should recall our integrated rate law for a second order reaction where we take one over our concentration of our reactant. After the reaction occurs equal to the rate constant. K. multiplied by time and then added to one over our initial concentration of our react in a. So plugging in what we know from the prompt, we would have won over our concentration of a. After the reaction occurs which is 0.4 moller. As we stated above, this is going to be equal to our rate constant. K. 2.3 times 10 to the 10th power inverse polarity times inverse seconds multiplied by time And then added to one over our initial concentration of a. Which above the stated is .20 Molar. And so next we're going to simplify this by from both sides subtracting 1/0 0.20 moller. And so what we would be able to simplify to is one sorry, 1/0 10.4 moller subtracted from 1/ 0.20 moller is equal to On the right hand side. We would just have 2.3 times 10 to the 10th power universe similarity times inverse seconds. And this is still going to be multiplied by time. T. Which is what we are solving for here. So, to simplify this in our calculators, we're going to get for this difference a value equal to 20 in verse similarity. Now, since we're in the numerator with these units. And this is set equal to the right hand side where we have our rate constant. So we want to continue to isolate for time and sorry, that is second, their inverse seconds. So did it continue to isolate for time? We're gonna divide both sides by our rate constant. K. So we divide both sides by 2.3 times 10 to the 10th power, inverse polarity times inverse seconds. And what we would isolate for is for time of our reaction or of our react in a to convert to the product a two is equal to a value of 8.7 times 10 to the negative 10th power. And now we would just be able to cancel our units of inverse similarity, leaving us with inverse seconds. However, now that we're no longer in the denominator, our units are just seconds and this is going to complete this example here as our final answer. So I hope that everything I reviewed was clear. But if you have any questions, just leave them down below and I will see everyone in the next practice video

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Key Concepts

Second-Order Reactions

Rate Constant (k)

Integrated Rate Law