Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H₃O⁺ by OH^-, which has a second-order rate constant of 1.3⨉10¹¹ M^-1 s^-1 at 25 °C. (a) If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously, how much time is required for 99.999% of the acid to be neutralized?

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Hello. Everyone in this video we're calculating for the time that is required for 95.0% of H. 02 to react. And the problem with being told that we have a second order reaction. So Gayle had utilized the second order integrated rate law so that the equation is one over the final concentration of the molecule that we're interested be going to cave are constant times T. Plus one over the initial concentration of the molecule that were interested in. All right. So here we're going to first start off the calculations by doing M one V. One. You're going to M. Two V. Two. So this is just a dilution reaction or equation. So we're being told the problem that V. two. So the final volume here is twice its initial volume. So that's just two times v. one. So then we get if we replace that to be M. One V. One equaling two M. Two. And the V. One is just two V. One or two times V. One. And here this problem we're solving for M. Two was rearranging this by dividing both sides by two V. One. So M. T equals two M. One V one Over two times v one. Alright, so for the volume, we go ahead. Just assume that we have one leader and then for M. One that is given to us to be 1.50 molars. Again, we're just swimming one here two times one. And that just gives us the M. Two. So the final concentration to be 0.75 molars. And that just means that our final concentration of the H. 02. So what we're interested Then the show value of that is 0.75 more years. The concentration of the H. O. to sleep molecule of interest as well. We have $0.75. We need go ahead and take into account the $0. times are 0.95 value. Every scroll up to the problem. That's just the percentage. And that gives us the concentration to be zero 0375. And of course units just being smaller for the concentration and this is for our final value. So it should probably change that. This is our Nutrition of H. 0. 2. All right. So now that we have both values here we go ahead and use that second order integrated rate law. Again that is one over the concentration of the molecule that were interested in. Equal to K. Times T. Plus one over the initial concentration of the molecule that were interested. So parking on parking in those values. The final concentration we calculate to spread above and read that 0.0375 molars. And that equals to K. is 1.4 times 10 to the nine means being moller times seconds. And this negative one here is just well one over M. Or this is just one over S. Alright. And we're multiplying this by T. Which are known. And that's what we're solving for here in this problem over one over the concentration of our initial our initial concentration of what cares about and that's just right above 0.75. All right, So we can go ahead and do some simplifications and that's just 26.6666. And to the negative one you go into 1. times 10 to the nine. That doesn't change here And then Ryan this with 1. M. to the -1. Alright. And I'm gonna go ahead and take this. I'm subtracted by both sides to basically isolate this right here. That goes as a value of 25.3333 M to the negative one equaling to 1.4 times 10 to the nine. Again just saying things. This doesn't change. Alright. And last step is to isolate R. T. Which is what we're selling for. We go ahead and divide both sides by this. So that goes as a value that T. Is equal to the numerical value is 1.81 times to the negative eight. And then our units is just seconds. So our final answer then is just This right here, 1.8, 1 times 10 to the -8 seconds.

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Key Concepts

Diffusion-Controlled Reactions

Second-Order Kinetics

Rate Constant and Reaction Time