# 005 M/s = k (001 M) (001 M) 2 k = (005 M/s)(001 M) (001 M) 4 2 k = 5 10 /M s in C# Recognizer Data Matrix 2d barcode in C# 005 M/s = k (001 M) (001 M) 2 k = (005 M/s)(001 M) (001 M) 4 2 k = 5 10 /M s

2 005 M/s = k (001 M) (001 M) 2 k = (005 M/s)(001 M) (001 M) 4 2 k = 5 10 /M s
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Sometimes, because of the numbers complexity, you must set up the equations mathematically The ratio of the rate expressions of two experiments will be used in determining the reaction orders The equations will be chosen so that the concentration of only one reactant has changed while the others remain constant In the example above, the ratio of experiments 1 and 2 will be used to determine the effect of a change of the concentration of NO on the rate, and then experiments 1 and 3 will be used to determine the effect of O2 Experiments 2 and 3 cannot be used, because both chemical species have changed concentration Remember: In choosing experiments to compare, choose two in which the concentration of only one reactant has changed while the others have remained constant Comparing experiments 1 and 2: 0 05 M/s = k [0 01]m [0 01]n 0 20 M/s = [0 02]m [0 01]n
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n Canceling the rate constants and the [001] and simplifying:
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1 1 = 4 2 Comparing experiments 1 and 3:
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m = 2 (use logarithms to solve for m) 0 05 M/s = k [0 01]m [0 01]n 0 10 M/s = [0 01]m [0 02]n Canceling the rate constants and the [001]n and simplifying: n 1 1 = 2 2 n =1 Writing the rate equation:
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2 Rate = k[NO] [O2]
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Again, the rate constant k could be determined by choosing any of the three experiments, substituting the concentrations, rate, and orders into the rate expression, and then solving for k
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Thus far, only cases in which instantaneous data are used in the rate expression have been shown These expressions allow us to answer questions concerning the speed of the reaction at a particular moment, but not questions about how long it might take to use up a certain reactant, etc If changes in the concentration of reactants or products over time are taken into account, as in the integrated rate laws, these questions can be answered Consider the following reaction: A B Assuming that this reaction is first order, then the rate of reaction can be expressed as the change in concentration of reactant A with time: Rate = and also as the rate law: Rate = k[A ] Setting these terms equal to each other gives: and integrating over time gives: ln[A ]t ln[A ]0 = kt where ln is the natural logarithm, [A ]0 is the concentration of reactant A at time = 0, and [A]t is the concentration of reactant A at some time t If the reaction is second order in A, then the following equation can be derived using the same procedure: 1 1 = kt [A ]t [ A ]0 Consider the following problem: Hydrogen iodide, HI, decomposes through a secondorder process to the elements The rate constant is 240 10 21/M s at 25 C How long will it take for the concentration of HI to drop from 0200 M to 0190 M at 25 C Answer: 20 21 110 10 s In this problem, k = 240 10 /M s, [A ]0 = 0200 M, and [A]1 = 0190 M You can simply insert the values and solve for t, or you first can rearrange the equation to give t = [1/[A ]t 1/[A ]0]/k You will get the same answer in either case If you get a negative answer, you interchanged [A ]t and [A]0 A common mistake is to use the first-order equation instead of the second-order equation The problem will always give you the information needed to determine whether the first-order or second-order equation is required The order of reaction can be determined graphically through the use of the integrated rate law If a plot of the ln[A] versus time yields a straight line, then the reaction is first order 1 with respect to reactant A If a plot of [A] versus time yields a straight line, then the reaction is second order with respect to reactant A The reaction half-life, t1/2, is the amount of time that it takes for a reactant concentration to decrease to one-half its initial concentration For a first-order reaction, the half-life [ A ] = k [A ] t [A ] t