# U Step 4 Review the Knowledge You Need to Score High in C# Decode Data Matrix in C# U Step 4 Review the Knowledge You Need to Score High

202 U Step 4 Review the Knowledge You Need to Score High
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is a constant, independent of reactant concentration, and can be shown to have the following mathematical relationship: ln 2 0 693 t 1/2 = = k k For second-order reactions, the half-life does depend on the reactant concentration and can be calculated using the following formula: 1 t 1/2 = k [A ]0 This means that as a second-order reaction proceeds, the half-life increases Radioactive decay is a first-order process, and the half-lives of the radioisotopes are well documented (see the chapter on Nuclear Chemistry for a discussion of half-lives with respect to nuclear reactions) Consider the following problem: The rate constant for the radioactive decay of 11 thorium-232 is 50 10 /year Determine the half-life of thorium-232 Answer: 14 1010 yr This is a radioactive decay process Radioactive decay follows first-order kinetics The solution to the problem simply requires the substitution of the k-value into the appropriate equation: 11 1 10 t l/2 = 0693/k = 0693/50 10 yr = 1386 10 yr which rounds (correct significant figures) to the answer reported Consider another case: Hydrogen iodide, HI, decomposes through a second-order 21 process to the elements The rate constant is 240 10 /M s at 25 C What is the halflife for this decomposition for a 0200 M of HI at 25 C 21 Answer: 208 10 s The problem specifies that this is a second-order process Thus, you must simply enter the appropriate values into the second-order half-life equation: t1/2 = 1/k[A ]0 = 1/(240 10
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/M s)(0200 M) = 208333 1021 seconds
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which rounds to the answer reported If you are unsure about your work in either of these problems, just follow your units You are asked for time, so your answer must have time units only and no other units
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Activation Energy
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A change in the temperature at which a reaction is taking place affects the rate constant k As the temperature increases, the value of the rate constant increases and the reaction is faster The Swedish scientist Arrhenius derived a relationship in 1889 that related the rate Ea /RT where k is the constant and temperature The Arrhenius equation has the form: k = Ae rate constant, A is a term called the frequency factor that accounts for molecular orienta 1 tion, e is the natural logarithm base, R is the universal gas constant 8314 J mol K , T is the Kelvin temperature, and Ea is the activation energy, the minimum amount of energy that is needed to initiate or start a chemical reaction The Arrhenius equation is most commonly used to calculate the activation energy of a reaction One way this can be done is to plot the ln k versus 1/T This gives a straight line whose slope is Ea /R Knowing the value of R allows the calculation of the value of Ea Normally, high activation energies are associated with slow reactions Anything that can be done to lower the activation energy of a reaction will tend to speed up the reaction
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Kinetics 203
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In the introduction to this chapter we discussed how chemical reactions occurred Recall that before a reaction can occur there must be a collision between one reactant with the proper orientation at the reactive site of another reactant that transfers enough energy to provide the activation energy However, many reactions do not take place in quite this simple a way Many reactions proceed from reactants to products through a sequence of reactions This sequence of reactions is called the reaction mechanism For example, consider the reaction A + 2B E + F Most likely, E and F are not formed from the simple collision of an A and two B molecules This reaction might follow this reaction sequence: A+B C C+B D D E+F If you add together the three equations above, you will get the overall equation A + 2B E + F C and D are called reaction intermediates, chemical species that are produced and consumed during the reaction, but that do not appear in the overall reaction Each individual reaction in the mechanism is called an elementary step or elementary reaction Each reaction step has its own rate of reaction One of the reaction steps is slower than the rest and is the rate-determining step The rate-determining step limits how fast the overall reaction can occur Therefore, the rate law of the rate-determining step is the rate law of the overall reaction The rate equation for an elementary step can be determined from the reaction stoichiometry, unlike the overall reaction The reactant coefficients in the elementary step become the reaction orders in the rate equation for that elementary step Many times a study of the kinetics of a reaction gives clues to the reaction mechanism For example, consider the following reaction: NO2(g) + CO(g) NO(g) + CO2(g) It has been determined experimentally that the rate law for this reaction is: Rate = 2 k[NO2] This rate law indicates that the reaction does not occur with a simple collision between NO2 and CO A simple collision of this type would have a rate law of Rate = k[NO2][CO] The following mechanism has been proposed for this reaction: NO2(g) + NO2(g) NO3(g) + NO(g) NO3(g) + CO(g) NO2(g) + CO2(g) Notice that if you add these two steps together, you get the overall reaction The first step has been shown to be the slow step in the mechanism, the rate-determining step If we 2 write the rate law for this elementary step it is: Rate = k[NO2] , which is identical to the experimentally determined rate law for the overall reaction Also note that both of the steps in the mechanism are bimolecular reactions, reactions that involve the collision of two chemical species In unimolecular reactions a single chemical species decomposes or rearranges Both bimolecular and unimolecular reactions are common, but the collision of three or more chemical species is quite rare Therefore, in developing or assessing a mechanism, it is best to consider only unimolecular or bimolecular elementary steps
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