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Ch.14 - Chemical Kinetics

Chapter 14, Problem 61

The reaction 2NO1g2 + 2 H21g2S N21g2 + 2 H2O1g2 is first order in H2 and second order in NO. Write the rate law, and specify the units of the rate constant.

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Hello in this problem, we are told if the order of the reaction of nitric oxide and ozone reacting for nitrogen dioxide and oxygen is second in nitric oxide and first in ozone. What is the rate law? And what are the units for the reaction rate constant? So recall that the general form of the rate law we have, the rate is equal to the reaction rate constant, sums are concentrations of our reactant to some order X and Y. It's okay again, is our reaction rate constant? And be our our reactant and X and Y are orders of the reaction. And they have to be determined experimentally. So in this problem then we start off with our rate again is equal to some reaction rate constant. We have our reactant nitric oxide and ozone. We're told that it is second order in nitric oxide. And it is first order in ozone. So the ones are not explicitly written, the ones are understood. So we don't put a superscript one um around the ozone. So this is our rate law expression or the reaction of nitric oxide and ozone where it's second order in nitric oxide. And first order in ozone. Now we are asked to find the reaction rate constant units. So we are going to solve for a reaction rate constant. It's okay then we'll be equal to the rate divided by the concentration of nitric oxide squared by by the concentration of ozone. And so these are our units recall that rate is always going to have some change in concentration over a period of time. So we have a concentration over a period of time. Those are always going to be our units for the rate and then we follow our units for the concentrations and then raise them to the powers indicated by the reaction rate law. So we have the concentration of nitric oxide is squared and that is then times the concentration of ozone which is to the one power. And now we can simplify this so we can cancel out one of the concentration units. And what we are left with then is one over the concentration squared and seconds. We can rewrite this And put the units in the numerator there in the denominator. Then the power is now minus. We bring it up to the numerator. So concentration to the -2 Seconds to the -1. And so our units for the reaction rate constant Are again concentration to the -2 seconds to the -1. Keep in mind that we know what the units for the rate has to be. There always have to be some concentration per unit time. And then we have concentrations which have units of polarity and then we pay attention to the order of the reaction to determine whether they are squared as in this example or first order is also shown in this example. Thanks for watching. Hope this help
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Textbook Question

From a plot of the concentration–time data in Worked Example 14.9, estimate: (b) the initial rate of decomposition of NO2.

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Textbook Question
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Textbook Question
Chlorite is reduced by bromide in acidic solution according to the following balanced equation: ClO2 -1aq2 + 4 Br-1aq2 + 4 H+1aq2S Cl-1aq2 + 2 Br21aq2 + 2 H2O1l2 (a) If Δ3Br24>Δt = 4.8 * 10-6 M>s, what is the value of Δ3ClO2 -4>Δt during the same time interval?
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Textbook Question
Initial rate data at 25 °C are listed in the table for the reaction NH4 +1aq2 + NO2 -1aq2S N21g2 + 2 H2O1l2 (b) What is the value of the rate constant?

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Textbook Question
Trimethylamine and chlorine dioxide react in water in an electron transfer reaction to form the trimethylamine cation and chlorite ion: 1CH323 N1aq2 + ClO21aq2 + H2O1l2S 1CH323 NH+1aq2 + ClO2 -1aq2 + OH-1aq2 Initial rate data obtained at 23 °C are listed in the following table. (b) What would be the initial rate in an experiment with initial concentrations 31CH323 N4 = 4.2 * 10-2 M and 3ClO24 = 3.4 * 10-2 M?
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Textbook Question
What is the half-life (in minutes) of the reaction in Problem 14.74 when the initial C4H6 concentration is 0.0200 M? How many minutes does it take for the concentration of C4H6 to drop from 0.0100 M to 0.0050 M?
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