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Chapter 14 and Chapter 21 Notes

by: Ryan Nannetti

Chapter 14 and Chapter 21 Notes CHEM 132

Marketplace > Towson University > Chemistry > CHEM 132 > Chapter 14 and Chapter 21 Notes
Ryan Nannetti
GPA 2.6
General Chemistry II
Dr. Richard Preisler

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These are the notes from Chapter 14 and parts of 21 on Nuclear Chemistry and Reaction Rates.
General Chemistry II
Dr. Richard Preisler
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This 11 page Bundle was uploaded by Ryan Nannetti on Friday October 9, 2015. The Bundle belongs to CHEM 132 at Towson University taught by Dr. Richard Preisler in Fall 2015. Since its upload, it has received 28 views. For similar materials see General Chemistry II in Chemistry at Towson University.

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Date Created: 10/09/15
Chapter 14 Notes Note used to represent concentration A used to represent the change in concentration andor time 141 The Rate of 23 Reaction Chemical Kinetics the branch of chemistry that looks at the speed or rate at Which a chemical reaction occurs 0 Kinetic energy is the energy available because of an object s motion Reaction rate the change in concentration of a reactant or a product over time Usually in units of molarity per secondMs 0 Because A decreases over time and thus the B increases over time AAAt is negative and ABAt is positive 0 Rate is given in terms of this generic form aA bB cC dD 0 Which yields 1 MA 1 MB 1 MC 1 0 Rate a At b At c At d MD At Example 141 Write the rate expressions of the following reactions in terms of the disappearance of the reactants and the appearance of the products a 139 aq 0C139 aq D Cl39 aq 0139 aq b 302 g I 203 g C 4NH3 g 502 g I 4N0 g 6H20 g 2 3 CH 01 2 2 AlOCll z z At a At Z Ar Ar Z Z Z 1 M02 1 A03 b 3 At 2 At 1 ANH3 1 M02 1 AN0 i M1120 C39 4 At 5 At 4 At 6 At Example 142 Consider the reaction 4N02 g 02 g D 2N205 g Suppose that at a particular moment during the reaction molecular oxygen is reacting at the rate of 0037 Ms a at What rate is N205 being formed b At What rate is N02 reacting 1 First we need to write the stoichiometric rate equation 1 AN02 A02 1AN205 Rate 4 At At 2 At 2 We were given the concentration of 02 over time so A 02 A t 0037 Ms decreases with time 3 Using this information we can find the other concentrations M02 1 A102 M a At 5 3f and At 0037 Ms so A t 2 0037 0074 Ms product is appearing thus the result is positive 1AN02 M02 AN02 b I At At so At 4 0037 015 Ms reactant is disappearing thus the result is negative Ideal Gas Eguation n n PV nRT or P 7 RT V is equivalent to molarity M 2 The Rate Laws Rate constant k a proportionality constant between the reaction rate and the concentrations of the reactants Rate law an expression relating the rate of a reaction to the rate constant and the concentrations of the reactants rate AXBy Overall reaction order the sum of the powers to which all reactant concentrations appearing in the rate law are raised the X y in the example above Experimental Determination of Rate Laws If a reaction involves only one reactant the rate law can be readily determined by measuring the initial rate of the reaction as a function of the reactant s concentration 0 If the rate doubles when the reactant concentration doubles then the reaction is first order IN THE REACTANT If the rate quadruples when the concentration doubles the reaction is second order IN THE REACTANT If a reaction involves more than one reactant the rate law can be determined one at a time using concentrations and initial rates see example 143 Example 143 The reaction of nitric oxide with hydrogen at 1280 C is 2N0 g 2H2 g I N2 g 2H20 g From the following data collected at the this temperature determine a the rate law b the rate constant k and c the rate of the reaction when N0 120 10393 M and H2 60 10393 M Expt NO M H2 M Initial Rate Ms 1 50 10393 2 10393 13 10395 2 10 10393 2 10393 5 10395 3 10 10393 4 10393 10 10395 The rate law in this case will take the form of 0 Rate k NOquotH2y The first thing we need to do is find the order of the reactants Then we need to calculate k and then we can find what the rate is at a certain concentration a Experiments 1 amp 2 show that when we double the NO at constant H2 the rate quadruples Using the ratio rate2 5105Ms k10103M gtXlt ratel 1310 5Ms 4 k510 3M x 210 3My 2103My x This shows how they got the quadrupled rate The two y values cancel out here so we are left with 1010 3 x 5103 x 2X and we know 2quot 4 so x 2 NO is second order Experiments 2 and 3 indicate that doubling H2 at constant NO doubles the rate rate3 10105Ms k1010 3M x 410 3y rate2 510 5Ms 2 k1010 3 x 210 3y The two X values cancel out so we are left with 4 10 3y 553v Dylwy1 So the rate law for this reaction is rate kNO2H2 and xy 21 3 overall reaction rate b Now we can use the date from experiment 2 to calculate k for the experiment rate 5105Ms k W 10103M2210 3M c We now know the rate law and the rate constant so we can find the rate under particular conditions Rate 25102 Mzs 1210 32610393 143 Relation between Reactant Concentrations and Time FirstOrder Reactions Firstorder reaction a reaction whose rate depends on the reactant concentration raised to the first power 0 A I products MA 0 Rate is At A A 0 From the rate law rate kA thus At kA We can determine the units of first order rate constants by transposition AAM1 M i 1 o k Tam ms so M s s 9 Air Eauation 143 In W kt Eauation 144 In At kt 1n A0 The fact that the points in this form are in a straight line shows that this is a first order reaction Example 144 The conversion of cyclopropane to propene in the gas phase is a firstorder reaction With a rate constant of 6710394 s391 at 500 C a If the initial concentration of cyclopropane is 025 M What is the concentration after 88 minutes b How long in minutes Will it take for the concentration of cyclopropane to decrease from 025 M to 015 M c How long in minutes Will it take to convert 74 of the starting material a Given A0 025M we need At after 88 minutes First convert 88 minutes into seconds 88 minutes60 seconds in 1 minute 528 seconds So 1n At kt 1n A0 1n At 6710394 s39l 528 s In 025 M 1n At 174 so At b From equation 143 W E In W ktln 025 6710 4t t76102s c Using equation 143 again 026 In 100 6710394t t2103s HalfLife HalfLife tug the time required for the concentration of a reactant to decrease to half of its initial concentration Equation 145 t12 0693k 0 Shows that the halflife of a first order reaction is independent of the initial of the reactant It takes the same time for the reactant to decrease from 1 M to 05 M as it would to decrease in from 010 M to 005 M Example 145 The decomposition of ethane C2H6 to methyl radicals is a firstorder reaction with a rate constant of 53610 4 s391 at 700 C C2H6 g gt 2 CH3 g Calculate the halflife of the reaction in minutes From Equation 145 t12 0693k O69353610394 129103 s Second Order Reactions Second order reaction a reaction whose rate depends on the concentration of one reactant raised to the second power or on the concentrations of two different reactants each raised to the first power L Equationl46 Alt Am kt 1 Halflife t12 HA 0 equation 147 Example 146 Iodine atoms combine to form molecular iodine in the gas phase I g 1 g I 12 g This reaction follows second order kinetics and has the high rate constant 7 109Ms a If the initial concentration ofI was 0068 M calculate the after 35 minutes b Calculate the half life of the reaction if the initial concentration OH is 053 M and if it is 039 M a To calculate the concentration of a species at a later time of a second order reaction we need the initial concentration and the rate constant both given in problem Applying equation 146 1 At 1ltt A 1 1 0 D At 7109Ms2108 0068M L l 1 Alt 1471012 so A 14710121 low 1 due to high initial 1 b Now we use equation 147 For I0 053 M l 1 9 t12 kA0 710 s 053M For I0 039 M 1 l 9 t12 kA0 710 5 039M ZeroOrder Reactions The rate of a zero order reaction is a constant independent of the reactant concentration At kt A0 Equation 148 A10 7 For halflife t12 Equation 149 Most KNOWN zeroorder reactions take place on a metal surface Table 142 Summary of the Kinetics of ZeroOrder FirstOrder and SecondOrder Reactions Order Rate Law ConcentrationTime Equation HalfLife 0 rate k At kt A0 A02k 1 rate k A ln AtA0 kt 0693k 2 rate k A2 1At kt 1A0 1kA0 144 Activation Ener2v and Temperature Dependence of Rate Constants Reaction rates increase with increasing temperature with very few exceptions Kinetic Molecular Theory of Gases states that gas molecules frequently collide with one another Collision theory of chemical kinetics we expect the rate of a reaction to be directly proportional to the number of molecular collisions per second or to the frequency of molecular collisions 0 Not all collisions lead to reactions Activation energy E the minimum amount of energy required to initiate a chemical reaction Activated complextransition state the species temporarily formed by the reactant molecules as a result of the collision before the product forms If products are more stable than reactants heat is released exothermic If products are less stable than reactants heat is absorbed endothermic 0 In both cases we plot the potential energy of the reacting system vs the progress of the reaction Arrhenius Eaum The dependence of the rate constant of a reaction on temperature is shown in this equation 0 k Ae39E RT Where E1 is the activation energy of the reaction usually in kJmol R is the gas constant 8314 JKOmol T is the absolute temperature A is the collision frequency aka the frequency factor Because of the minus sign in the exponent EaRT the rate constant decreases With increasing activation energy and increases With increasing temperature Ea Equation 1411 ln k ln A E Ea 1 Can also take the linear form In k f T In A Equation 1412 Thus a plot of In k vs 1T gives a straight line Whose slope m is equal to the EaR and Whose intercept b has the ordinate yaxis in In A k1 Ea T1 T2 Equation 1413ln E W T1gtIltT2 Example 148 The rate constant of a first order reaction is 46810 2 s391 at 298 K What is the rate constant at 375 K if the activation energy for the reaction is 331 kJmol Given k1 468810 2 5391 E1 331 kJmol or 33100 Jmol T1 298 K R 8314 JKmol T2 375 K k2 Use equation 1413 k1 Ea T 1 T2 468102s1 33100 Jmol 1n k 2 W T1gtIltT2 lZIln k2 8314JKmol 298K 375K 46810 2 298KgtIlt375K D 1n 2 274 46810 2 46810 2 k 2 e3923974 so k2 00646 which makes 145 Reaction Mechanisms Elementarv stepsreactions a series of simple reactions that represent the progress of the overall reaction at the molecular level Reaction mechanism the sequence of elementary steps that lead to product formation 0 A net chemical equation which represents the overall change is given by the sum of the elementary steps Intermediates appear in the mechanism of the reaction that is the elementary steps but not in the overall balanced equation I always formed in an early elementary step and consumed in a later elementary step Molecularitv of a reaction the number of molecules reacting in an elementary step Bimolecular reaction involves two molecules Unimolecular reaction involves one molecule Termolecular reaction involves three molecules extremely rare The reaction order for each reactant in an elementary reaction is equal to its stoichiometric coefficient in the chemical equation for that step Ratedetermining step the slowed step in the sequence of steps leading to product formation Elementary steps must satisfy two requirements 1 The sum of the elementary steps must give the overall balanced equation for the reaction 2 The ratedetermining step should predict the same rate law as is determined experimentally Example 149 The gasphase decomposition of nitrous oxide N20 is believed to occur via two elementary steps Step 1 N20 gt N2 0 k1 Step 2 N20 O gt N2 02 k2 Experimentally the rate law is found to be rate kNzO a Write the equation for the overall reaction b Identify the intermediates c What can you say about the relative rates of steps 1 and 2 a By adding the two equations together and cancelling out elements that are common on each side we get an overall equation 2N20 gt 2N2 02 b Since the O atom is produced in the first step but does not appear in the overall equation it is the only intermediate c If we assume the ratedetermining step is step 1 that is if k2gtgtllt1 then the rate of the overall reaction is Rate k1 N20 and k k1 146 Catalysis Catalyst a substance that increases the rate of a reaction by lowering the activation energy 0 Speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence Three tVDes of Catalvsis 1 Heterogeneous the reactants and the catalyst are in different phases Usually the catalyst is a solid and the reactants are either gases or liquids a The Haber synthesis of ammonia i In 1905 after testing literally hundreds of compounds at various temperatures and pressures the German chemist Fritz Haber discovered that iron plus a few percent of oxides of potassium and aluminum catalyze the reaction of hydrogen with nitrogen to yield ammonia at 500 C Haber process 2 Homogeneous the reactants and the catalyst are dispersed in a single phase a The reaction of ethyl acetate with water to form acetic acid and ethanol normally occurs too slowly to be measured b In the absence of the catalyst water the rate law is rate kCH3COOC2H5 ethyl acetate c Reaction can be catalyzed by an acid In the presence of hydrochloric acid the rate is rate kCCH3COOC2H5H 3 Enzyme catalysis a Enzymes are biological catalysts b Substrates are another name for reactants c Enzyme is typically a large protein molecule that contains one or more active sites where interactions with substrates take place winter 21 Nuclear Chemistrv onlv 211 and 213 g11 The Nature of Nuclear Reactions 0 With the exception of hydrogen all nuclei contain two kinds of fundamental particles protons and neutrons 0 Radioactivity nuclei emitting particles andor electromagnetic radiation spontaneously 0 Nuclear transmutation results from the bombardment of nuclei by neutrons protons or other nuclei 0 Elementary particles Protons neutrons electrons positrons and alpha particles ble 211 ComDaLison of Chemical Reactions and Nuclear Reactions Chemical Reactions 1 2 3 4 Atoms are rearranged by the breaking and forming of chemical bonds Only electrons in atomic or molecular orbitals are involved in the breaking and forming of bonds Reactions are accompanied by absorption or release of relatively small amounts of energy Rates of reaction are in uenced by temperature pressure concentration and catalysts Nuclear Reactions 1 Elements or isotopes of the same elements are converted from one to another Protons neutrons electrons and other elementary particles may be involved Reactions are accompanied by absorption or release of tremendous amounts of energy Rates of reaction normally are not affected by temperature pressure and catalysts Positron has the same mass as the electron but bears a 1 charge 0 In balancing any nuclear equation we observe these rules I The total number of protons plus neutrons in the products and in the reactants must be the same conservation of mass number I The total number of nuclear charges in the products and in the reactants must be the same conservation of atomic number 213 Natural Radioactivity Radioactive decgv series a sequence of nuclear reactions that ultimately result in the formation of a stable isotope 0 All radioactive decays obey first order kinetics Therefore the rate of radioactive decay at time t is O t AN where 0 A is the first order rate constant 0 N is the number of radioactive nuclei present at time t Using Equation 143 and substituting we can get the number of radioactive nuclei at time zero N0 and time t N Nl ln m t and using equation 145 and substituting we get the halflife of a reaction 0693 t12 A Radiocarbon dating 0 The carbon14 isotope is produced when atmospheric nitrogen is bombarded with energetic neutrons produced by cosmic rays 0 The decreasing ration of carbon14 to carbon12 can be used to estimate the age of a specimen


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