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 H 3 O + by OH - , which has a second-order rate constant of 1.3⨉10 11 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?

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.

Ch.14 - Chemical Kinetics

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

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McMurry 8th Edition

Ch.14 - Chemical Kinetics

Problem 130a

Chapter 14, Problem 130a

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?

Verified step by step guidance

Identify the reaction: HCl + NaOH -> NaCl + H2O. This is a neutralization reaction where H3O+ is neutralized by OH-.

Since the reaction is diffusion-controlled, use the second-order rate equation: $ \frac{1}{[A]} - \frac{1}{[A]_0} = kt $, where $[A]_0$ is the initial concentration and $[A]$ is the concentration at time $t$.

Calculate the initial concentration of H3O+ and OH- after mixing. Since equal volumes of 2.0 M solutions are mixed, the concentration of each will be halved: $[H3O^+]_0 = 1.0\, M$ and $[OH^-]_0 = 1.0\, M$.

Determine the concentration of H3O+ when 99.999% is neutralized. This means only 0.001% remains: $[H3O^+] = 0.00001 \times [H3O^+]_0 = 0.00001 \times 1.0\, M$.

Substitute $[A]_0$, $[A]$, and $k = 1.3 \times 10^{11} \text{ M}^{-1} \text{s}^{-1}$ into the rate equation to solve for $t$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Diffusion-Controlled Reactions

Diffusion-controlled reactions occur when the rate of reaction is limited by the rate at which reactants collide in solution. In these cases, the reaction proceeds as quickly as the molecules can move and encounter each other, leading to very high rate constants. This concept is crucial for understanding how quickly certain reactions can occur, especially in dilute solutions.

Second-Order Kinetics

Second-order reactions depend on the concentration of two reactants or the square of the concentration of one reactant. The rate law for a second-order reaction can be expressed as rate = k[A][B], where k is the rate constant. In the context of the given question, the second-order rate constant indicates how the reaction rate changes with varying concentrations of H3O+ and OH-.

Rate Constant and Reaction Time

The rate constant (k) is a proportionality factor in the rate law that quantifies the speed of a reaction at a given temperature. For the neutralization reaction mentioned, the rate constant is 1.3⨉1011 M-1 s-1. To determine the time required for a specific percentage of reactants to react, one can use integrated rate laws that relate concentration changes over time to the rate constant.

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Textbook Question

Consider the reversible, first-order interconversion of two molecules A and B: where k_f = 3.0⨉10-3 s^-1 is the rate constant for the forward reaction and kr = 1.0⨉10^-3 s^-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants k_f and k_r have comparable values.

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Textbook Question

Assume that you are studying the first-order conversion ofa reactant X to products in a reaction vessel with a constantvolume of 1.000 L. At 1 p.m., you start the reactionat 25 °C with 1.000 mol of X. At 2 p.m., you find that0.600 mol of X remains, and you immediately increase thetemperature of the reaction mixture to 35 °C. At 3 p.m.,you discover that 0.200 mol of X is still present. You wantto finish the reaction by 4 p.m. but need to continue it untilonly 0.010 mol of X remains, so you decide to increase thetemperature once again. What is the minimum temperaturerequired to convert all but 0.010 mol of X to productsby 4 p.m.?

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Textbook Question

The half-life for the first-order decomposition of N2O4 is 1.3 * 10-5 s. N2O41g2S 2 NO21g2 If N2O4 is introduced into an evacuated flask at a pressure of 17.0 mm Hg, how many seconds are required for the pressure of NO2 to reach 1.3 mm Hg?

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Textbook Question

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. (b) Under normal laboratory conditions, would you expect the rate of the acid–base neutralization to be limited by the rate of the reaction or by the speed of mixing?

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Textbook Question

The reaction 2 NO1g2 + O21g2S 2 NO21g2 has the thirdorderrate law rate = k3NO423O24, where k = 25 M-2 s-1.Under the condition that 3NO4 = 2 3O24, the integratedrate law is13O242 = 8 kt +113O24022What are the concentrations of NO, O2, and NO2 after100.0 s if the initial concentrations are 3NO4 = 0.0200 Mand 3O24 = 0.0100 M?

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