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Ch.14 - Chemical Kinetics

Chapter 14, Problem 131

The reaction 2 NO1g2 + O21g2S 2 NO21g2 has the thirdorder rate law rate = k3NO423O24, where k = 25 M-2 s-1. Under the condition that 3NO4 = 2 3O24, the integrated rate law is 13O242 = 8 kt +113O24022 What are the concentrations of NO, O2, and NO2 after 100.0 s if the initial concentrations are 3NO4 = 0.0200 M and 3O24 = 0.0100 M?

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Hello. In this problem, we are told the reaction between nitric oxide and chlorine gas obeys an overall third order rate law with the rate constant of three per concentration squared per second. We are told that when the concentration of nitric oxide and chlorine are equal, the integrated rate law is given by the equation below. Were asked if the initial concentration of nitric oxide in chlorine are 0.3 moller and 0.150 molar respectfully whether the concentrations of nitric oxide, chlorine and nitrogen chloride after 50 seconds. Looking at the information provided in the problem statement, when the concentration of nitric oxide and chlorine are equal, then we can write the great law as the rate is equal to the reaction rate constant times the concentration of foreign cubed. Or we could write the rate is equal to the reaction rate constant times the concentration of nitric oxide cubed. Given that our integrated rate law is in terms of chlorine, that means then that the rate law that was used to derive the integrated rate law is that it involves chlorine that that this is a third order reaction integrated rate law. Probably more familiar with integrated rate laws for a zero first or second order reaction In this problem, we're going to begin then by making use of this integrated rate law to find the concentration of chlorine after 50 seconds. So making use of our integrated rate law, we have concentration of corinne After 50 seconds and that squared is equal to two times our reaction rate constant times time plus one over the concentration of chlorine time zero squared from the problem statement, we are told that the reaction rate constant is equal to three. Her concentration squared per second. That our time is 50 seconds. And that our initial concentration of chlorine is $0.150. So we'll plug this information into our integrated rate law and we get one over Concentration of chlorine at seconds squared Equal two times our reaction rate constant Times are time of 50 seconds Plus one over our initial concentration of boring not squared. So we can simplify what's on the right hand side. Note that our units of seconds will cancel. And so we'll be left with units of concentration the -2. This works out to 344.44. Hilarity The -2. So we'll invert both sides and then we get concentration of chlorine Squared at 50 seconds Is equal to. Then our concentration squared over 344. Is equal 0.00 - 903. And our concentration is squared will now take the square root of both sides. And we get the concentration of chlorine, then at 50 seconds Is equal to 0.0538, Reporting that to the correct numbers and figures a concentration of pouring, 50 seconds is equal to 0.0539 Mueller. And now we'll make use of that concentration in our second step, find the concentration of nitric oxide and chloride after 50 seconds. So we'll begin by writing our reaction. So we have two moles of nitric oxide reacts with one mole of chlorine to form two moles of nitrogen chloride. So initially we were told we have 0.300 moller nitric oxide, 0.150 Mueller chlorine. And initially we have no nitrous l chloride. Our change then is minus two X. We have two moles of nitric oxide minus X one mold chlorine and plus two X. We have two moles of nitric fluoride. So combining these two, we get 0.300 minus two X 0.150 minus X and two X. So we found that this will represent the time of This row of 50 seconds. So we know that the concentration pouring in at 50 seconds, which is 0.15 -X is equal to 0.05388. And so X. is equal to 0.09612. Our concentration then of nitric oxide after 50 seconds is equal to 0.300 -2 x. Which works out to 0.108 Mueller. And our concentration of nitrous il chloride After 50 seconds is equal to two x. That is equal to 0.19, So the concentration then of chlorine after 50 seconds is 0.539 moller. Our concentration of nitric oxide after 50 seconds is 0. moller, and our concentration of nitrogen chloride after 50 seconds is 0.192 Moeller. Thanks for watching. Hope this helps.
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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. (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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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 following experimental data were obtained in a study of the reaction 2 HI1g2S H21g2 + I21g2. Predict the concentration of HI that would give a rate of 1.0 * 10-5 M>s at 650 K.

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Textbook Question

Values of Ea = 6.3 kJ/mol and A = 6.0⨉108/(M s) have been measured for the bimolecular reaction: NO(g) + F2(g) → NOF(g) + F(g) (a) Calculate the rate constant at 25 °C.

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Textbook Question

Values of Ea = 6.3 kJ/mol and A = 6.0⨉108/(M s) have been measured for the bimolecular reaction: NO(g) + F2(g) → NOF(g) + F(g) (d) Why does the reaction have such a low activation energy?

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