For more information, please see full course syllabus of General Chemistry

For more information, please see full course syllabus of General Chemistry

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### Chemical Kinetics

- Kinetics studies the factors that can influence the rate of a chemical reaction.
- The method of initial rates and the use of integrated rate laws can help solve for the rate orders and for the rate constant.
- The slowest step of a reaction mechanism dictates the overall reaction rate.
- The Arrhrenius equation relates the temperature of a reaction directly to its reaction rate.

### Chemical Kinetics

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Lesson Overview
- Introduction
- Introduction cont'd
- Rate Laws
- Rate Laws cont'd
- Methods of Initial Rates
- Integrated Rate Laws
- Reaction Mechanisms
- Reaction Rates and Temperatures
- Catalysis
- Summary
- Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed
- Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction

- Intro 0:00
- Lesson Overview 0:06
- Introduction 1:09
- Chemical Kinetics and the Rate of a Reaction
- Factors Influencing Rate
- Introduction cont'd 2:27
- How a Reaction Progresses Through Time
- Rate of Change Equation
- Rate Laws 7:06
- Definition of Rate Laws
- General Form of Rate Laws
- Rate Laws cont'd 11:07
- Rate Orders With Respect to Reactant and Concentration
- Methods of Initial Rates 13:38
- Methods of Initial Rates
- Integrated Rate Laws 17:57
- Integrated Rate Laws
- Graphically Determine the Rate Constant k
- Reaction Mechanisms 21:05
- Step 1: Reversible
- Step 2: Rate-limiting Step
- Rate Law for the Reaction
- Reaction Rates and Temperatures 26:16
- Reaction Rates and Temperatures
- The Arrhenius Equation
- Catalysis 30:31
- Catalyst
- Summary 32:02
- Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed 32:54
- Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction 35:24

### General Chemistry Online Course

I. Basic Concepts & Measurement of Chemistry | ||
---|---|---|

Basic Concepts of Chemistry | 16:26 | |

Tools in Quantitative Chemistry | 29:22 | |

II. Atoms, Molecules, and Ions | ||

Atoms, Molecules, and Ions | 52:18 | |

III. Chemical Reactions | ||

Chemical Reactions | 43:24 | |

Chemical Reactions II | 55:40 | |

IV. Stoichiometry | ||

Stoichiometry I | 42:10 | |

Stoichiometry II | 42:38 | |

V. Thermochemistry | ||

Energy & Chemical Reactions | 55:28 | |

VI. Quantum Theory of Atoms | ||

Structure of Atoms | 42:33 | |

VII. Electron Configurations and Periodicity | ||

Periodic Trends | 38:50 | |

VIII. Molecular Geometry & Bonding Theory | ||

Bonding & Molecular Structure | 52:39 | |

Advanced Bonding Theories | 1:11:41 | |

IX. Gases, Solids, & Liquids | ||

Gases | 35:06 | |

Intermolecular Forces & Liquids | 33:47 | |

The Chemistry of Solids | 25:13 | |

X. Solutions, Rates of Reaction, & Equilibrium | ||

Solutions & Their Behavior | 38:06 | |

Chemical Kinetics | 37:45 | |

Principles of Chemical Equilibrium | 34:09 | |

XI. Acids & Bases Chemistry | ||

Acid-Base Chemistry | 43:44 | |

Applications of Aqueous Equilibria | 55:26 | |

XII. Thermodynamics & Electrochemistry | ||

Entropy & Free Energy | 36:13 | |

Electrochemistry | 41:16 | |

XIII. Transition Elements & Coordination Compounds | ||

The Chemistry of The Transition Metals | 39:03 | |

XIV. Nuclear Chemistry | ||

Nuclear Chemistry | 16:39 |

### Transcription: Chemical Kinetics

*Hi, welcome back to Educator.com.*0000

*Today's lecture from general chemistry is going to be on chemical kinetics.*0002

*Here is our brief overview of the lesson today.*0008

*As always we will go ahead and start off with our introduction.*0010

*Then we are going to get into what is really at the heart of chemical kinetics.*0013

*That is the rate of chemical reactions; that is how fast does a reaction occur?*0017

*We are then going to see what are the mathematical equations which actually attempt to quantify the rate of an equation.*0021

*That is what we call a rate law.*0029

*Our objective is to then try to derive these equations for any given reaction.*0031

*There is two sets of experiments that you can do to perform this.*0037

*Number one is what we call the method of initial rates.*0040

*Number two is what we call the integrated rate laws.*0043

*We will then jump into how a reaction proceeds, the step by step process which is*0045

*called a reaction mechanism, followed by the effect that temperature has on reaction rates.*0051

*We will go ahead and wrap up the lesson with a brief overview of*0059

*catalysis followed by our summary and a pair of sample problems.*0062

*Chemical kinetics basically is the area of chemistry that examines the factors which can influence the rate of a reaction.*0070

*What do we mean by factors that can influence the rate?*0081

*There are several--temperature, pressure, the reactant concentration, the addition of a catalyst, and mechanical force.*0084

*We are going to see that basically as temperature goes up, the rate goes up.*0095

*We are going to see that in general as pressure goes up, the rate goes up.*0104

*We are going to see that as the concentration of reactant goes up, the rate goes up.*0112

*When we add a catalyst, we are going to define what that exactly means.*0120

*But this also increases the rate.*0123

*Finally mechanical force, this is what you probably do in lab.*0126

*All we mean by mechanical force is something as simple as stirring or shaking.*0129

*Of course this will increase the rate of the reaction.*0137

*This is what we mean by factors which can influence the rate of a reaction.*0142

*Let's take a look though how a typical reaction progresses through time.*0150

*Consider the reaction A plus 2B goes to C.*0154

*This is telling me that for every one of A, two Bs are required.*0158

*One C is going to be produced.*0165

*Remember we are only looking here at the forward direction only.*0167

*When we go in the forward direction, we say that reactants are consumed which means concentration of reactant go down.*0176

*The products are made which means that the concentration of product goes up.*0190

*It looks like because twice the amount of B is going to be consumed as A,*0201

*the rate of consumption of B is going to be double that of A.*0206

*It is very typical for us in chemical kinetics to graph this.*0221

*We can go ahead and graph the concentration of the reactant with respect to time and the concentration of product too.*0225

*At time zero, I have no product; I only have reactants.*0237

*Down here right at the origin, this is the concentration of C initial.*0240

*The way that concentrations change with respect to time is not linear.*0247

*But instead it is going to be characterized by a simple curve just like that.*0251

*You notice that at some time t, the concentration of C stops changing.*0260

*We are going to plateau; it is not going to increase forever.*0264

*How about the concentration of A and B?*0269

*The concentration of A let's say is here and the concentration of B.*0272

*Let's say we had equal amounts.*0278

*Let me go ahead and draw now a curve that represents the concentration of A changing with respect to time.*0280

*It is going to be a mirror image of the concentration of C changing because they are both 1:1 ratio.*0287

*But how is the concentration of B changing?*0293

*The concentration of B, it is going to drop at double the rate.*0296

*The concentration of B is going to drop much much much... quicker than the concentration of A.*0302

*You notice that for all of the reactions, there reaches a point where the curves plateau*0315

*which means that the concentration of reactants and the concentration of products no longer changes.*0325

*We are going to talk about this region in another lecture.*0337

*This is what we call the equilibrium region.*0340

*In chemical kinetics, what we are interested in is really the start of the reaction, the early part*0346

*of the reaction where we actually can monitor change in the concentration of reactants and products.*0351

*That is the representation graphically; how about mathematically?*0360

*It turns out that the rate of change is going to be equal to the following.*0364

*The change in the concentration of A, Δ[A]Δt, this is going to be equal to*0368

*1/2 Δ[B]Δt which is equal to the change in concentration of C Δ[C]Δt.*0382

*We have to put a negative sign in front of the reactants such that the overall number is*0392

*going to be positive because Δ[A]Δt is going to be slope of the curve.*0401

*This is going to be slope of the curve you see.*0408

*This is going to be the slope of that curve too.*0411

*This gives us a general equation for relating the rates of change of each reactant and product concentration with respect to time.*0416

*Now that we have gone through a brief introduction, let's go now into*0429

*what is at the core of chemical kinetics--that is the rate law.*0434

*A rate law is going to be a mathematical equation which relates the rate of reaction to the concentration of the reactants.*0439

*Basically the rate law tells us that the rate of a reaction is proportional to the concentration of reactants typically.*0447

*When we write out the rate law, all rate laws have the following equation*0458

*Of the form rate which is going to be usually in molarity per second...*0461

*is equal to some constant k times the concentration of A raised to*0467

*some power times the concentration of the B raised to some power, etc.*0472

*Let's go ahead and define what each of these mean.*0478

*k is what we call the rate constant.*0484

*The rate constant is going to be unique to a reaction at a certain temperature.*0491

*In addition, we are going to see that the rate constant, the units of k vary.*0508

*We are going to see that very soon.*0519

*x and y are what we call rate orders.*0522

*Once again x and y are what we call rate orders.*0529

*If x is equal to 1, we say reaction is first order with respect to the concentration of A.*0537

*If x is equal to 2, we say the reaction is second order with respect to the concentration of A.*0554

*In addition, if the sum of the rate orders x plus y is equal to 0, we say zero order overall.*0564

*If the sum of the orders is equal to 1, we say first order overall.*0577

*If x plus y is equal to 2, we say second order overall.*0585

*This is just some terminology that we want to introduce and clarify.*0592

*Let's go ahead and get back to the units of k.*0598

*The units of k will vary.*0601

*For zero order, the units of k is just going to be reciprocal seconds.*0604

*For first order, the units of k is going to... excuse me.*0619

*For zero order.. my apologies.*0627

*For zero order, the units of k is going to be molarity per second.*0629

*For first order, the units of k is just reciprocal seconds.*0636

*For second order overall, the units of k is going to be inverse molarity inverse second.*0640

*You can easily plug that back into the equation for k and see that the units will cancel.*0649

*This is the general rate law.*0657

*Rate is equal to some constant k times the concentration of A raised to*0659

*the x power times the concentration of B raised to the y power.*0662

*But let's go ahead now and see what the significance of the rate orders are.*0666

*Consider the following reaction: A plus 2B goes to C.*0673

*The rate law is given to us to be k times A squared times the concentration of B.*0678

*Let's go ahead and study what happens if we change the concentration of one of these reactants.*0686

*If the concentration of A doubles holding B constant, then we see that the rate is going to increase by 4.*0692

*It is going to quadruple.*0710

*If the concentration of A triples holding B constant, we see that the rate is going to increase by a factor of 9.*0714

*What if the concentration of B doubles holding concentration of A constant?*0728

*If that happens, the rate is just going to increase by 2.*0739

*If the concentration of B triples holding A constant, we see that the rate is going to increase by a factor of 3.*0746

*You see what the significance of the rate order is.*0756

*The rate order is really proportional to the sensitivity of a reaction rate on*0760

*the concentration of a specific reactant molecule, on the concentration of a specific reactant.*0778

*Pretty much we see that the larger the value of these rate orders, the more sensitive*0788

*your reaction is to a change in concentration of a specific reactant.*0795

*Basically what the kinetics deals with is focusing on this rate law and solving for x, y, and k.*0802

*In other words, our goal is to solve for the rate constant and rate orders.*0811

*We can do this in one of two ways.*0820

*The first experimental way to derive a rate law is what we call the method of initial rates.*0822

*In the method of initial rates, you basically do what we just did.*0828

*You change one reactant concentration holding all else constant and seeing how the rate varies.*0834

*Let's go ahead and take a look at the following.*0857

*You are usually given some table of data.*0859

*The table of data is going to list different concentrations, different molarities of each reactant and the rate that was measured.*0865

*For example, in experiment number one, this concentration of iodide was found.*0873

*This concentration of thiosulfate was found; this is the initial rate.*0881

*In experiment number two, we see that the concentration of thiosulfate was fixed.*0886

*The concentration of iodide was tripled; we see that the rate also tripled.*0894

*What does that mean?--basically as the concentration of iodide tripled, the rate tripled.*0901

*This is a 1:1 correspondence.*0913

*Anytime you have this 1:1 correspondence, it is a rate order of 1; rate order of 1.*0917

*Let's now see what the rate order is for S _{2}O_{8}^{2-}, thiosulfate.*0927

*For thiosulfate, we are going to look at experiments two and three.*0933

*You see that as the concentration of thiosulfate tripled... I'm sorry.*0941

*We have to do experiments one and three, not two and three.*0953

*Here as the concentration of thiosulfate tripled, iodide was held constant.*0956

*What happened to the rate?--the rate tripled.*0966

*As S _{2}O_{8}^{2-} tripled, the rate tripled holding everything else constant.*0968

*This was also a 1:1 correspondence; the rate order is also 1.*0975

*Therefore we have the following rate law.*0986

*The rate of this reaction is equal to some constant k times the concentration of I ^{-}*0989

*raised to the first power and the concentration of S _{2}O_{8}^{2-} raised to the first power.*0994

*The next thing now that we have the rate orders, we can go ahead and now solve for k.*1002

*We can solve for k by simply plugging in any experiment--one, two, or three--directly into the equation.*1011

*Solve for k using any experiment number.*1019

*For example, let's use experiment one; the rate was 0.044 molarity per second.*1029

*That is going to be equal to the rate constant k times the concentration of I ^{-} which is*1037

*0.125 molar times the concentration of thiosulfate, 0.150 molar, all raised to the first power.*1041

*Then you can just use your algebra to go ahead and solve for the value of k for this second order overall reaction.*1050

*Here we are going to get units of inverse molarity inverse seconds.*1063

*That is how we use the method of initial rates--very straightforward experiment to determine the rate law.*1070

*The second way of determining a rate law is to use what we call integrated rate laws.*1078

*Integrated rate laws, they quantify the relationship between the reactant concentration and time; and time.*1085

*Basically these are all derived mathematically.*1095

*You should always ask your instructor if you need to know how to derive it or not.*1100

*For zero order rate law, the rate is equal to the rate constant k.*1104

*The integrated rate law is the following.*1108

*The concentration of A at any given time is equal to the initial concentration of A minus kt.*1111

*For the first order overall, the integrated rate law is natural log of A _{0} minus natural log of A equal to kt.*1116

*Finally the second order integrated rate law is 1 over A minus 1 over A _{0} is equal to kt.*1125

*The nice thing about these equations is that they are all linear.*1133

*If several concentrations are determined at different times, you can get the rate constant graphically.*1139

*For a zero order overall reaction, we are basically going to graph the concentration of A as a function of t.*1146

*That is going to give us a nice straight line with slope equal to ?k.*1156

*For first order overall, we can go ahead and graph the natural log of the concentration of A versus t.*1162

*We are also going to get a straight line whose slope is equal to ?k.*1174

*For second order overall, we are going to plot 1 over the concentration of A which is equal to time.*1179

*Here we are going to get a nice straight line with a positive slope equal to k.*1189

*Once again this is a graphical determination of the rate constant anytime*1194

*you have data from several different time intervals and measured reactant concentration.*1198

*Another thing we like to talk about too is the half-life.*1210

*For zero order, the half-life is equal to the initial concentration of A over 2k.*1214

*For first order, the half-life is equal to the natural log of 2 over k.*1221

*For second order, the half-life is equal to 1 over k times the concentration of A _{0}.*1228

*Once again you should ask your instructor to make sure if you have to know the derivation or not.*1235

*Basically the half-life is very important because it tells us the time required to reach 1/2 of the initial concentration of A.*1241

*Again that is what we call integrated rate laws.*1261

*The next thing we are going to look at is what we call a reaction mechanism.*1267

*A reaction mechanism basically represents the step by step reactions which when combined give you the overall net reaction.*1271

*For example, let's say we had the following given, step one.*1278

*Step one was 2NO gas going on to form N _{2}O_{2} gas.*1285

*This tells us that the reaction is occurring both in the forward and reverse directions.*1293

*This is what we call the reversible reaction; this is usually very very fast.*1299

*Step number two is going to be O _{2} gas plus N_{2}O_{2} gas going on to form 2NO_{2} gas.*1305

*You are told that this reaction is very very slow.*1318

*The overall reaction is going to be 2NO gas plus O _{2} gas going on to form 2NO_{2} gas.*1323

*You notice that N _{2}O_{2} gets cancelled out.*1336

*Any item in a chemical reaction mechanism that gets cancelled is what we call an intermediate.*1342

*That is it is both formed and consumed during the course of a chemical reaction; formed and consumed during a reaction.*1351

*The reason why we care about what the slow step is is because of the following.*1370

*The slow step, which is in this case step two, is like the weakest link in your chain.*1376

*It is like the slowest person on your track and field relay team.*1382

*The slow step determines, it limits how fast a reaction can go.*1387

*We call this the rate limiting step.*1392

*The rate limiting step is equal to the rate of the overall reaction.*1399

*If we were to write out the rate law for this, we would get the following.*1408

*The rate is equal to the rate constant k _{2}...*1412

*This is k _{2}, of the second step.*1417

*Times the concentration of O _{2} raised to the first power*1419

*times the concentration of N _{2}O_{2} raised to the first power.*1422

*We can always use the coefficients as the orders if the reaction you are*1427

*looking at is a part of the mechanism, what we call an elementary reaction.*1438

*Coefficient is equal to the rate orders for elementary steps only.*1444

*Otherwise you would have to go through integrated rate laws or method of initial rates*1451

*to go through the whole process again to find what x and y are.*1454

*When we look at this, we have a problem.*1458

*We have N _{2}O_{2} appearing in this rate law.*1460

*This is the intermediate.*1463

*You can never have an intermediate appearing in the rate law.*1465

*We have to do something about this; what we do is the following.*1468

*We use what is called a steady state approximation.*1473

*You utilize the fast equilibrium step where the k _{1} times the concentration of NO squared*1481

*equals to k _{-1} times the concentration of N_{2}O_{2} where k_{1} represents the forward.*1492

*k _{-1} represents the reverse.*1502

*What we do then is we solve for the intermediate.*1505

*Concentration of N _{2}O_{2} is equal to k_{1} over k_{-1} times the concentration of NO_{2} squared.*1507

*We then plug this back into our experimental rate law.*1518

*Rate is equal to k _{2} times the concentration of O_{2}*1522

*times k _{1} over k_{-1} times the concentration of NO squared.*1526

*You can collect all of the constants together and just call that what we call k _{observed}.*1534

*We get left with O _{2} times the concentration of NO squared.*1539

*Here we have our final rate law that is going to be the rate law for the overall reaction.*1546

*It is just by coincidence here that the rate orders are the coefficients.*1555

*That is not usually the case.*1559

*But that is how we solve for it where k _{observed} is equal to k_{2}k_{1} over k_{-1}.*1560

*Once again this is how you deal with reaction mechanism problems.*1571

*Let's now move on to the next topic.*1577

*This is the relationship between temperature and reaction rate.*1579

*Basically in general, as temperature increases, so does the reaction rate; just think about this.*1583

*You know the tea bag is going to brew faster in warm water than cold water.*1588

*You can see that visually happening right before you.*1594

*In order for a reaction to occur, reactant molecules must do two things.*1597

*They must collide; they must collide with sufficient energy.*1602

*They must collide in the proper orientation; sufficient energy and proper orientation.*1606

*Basically we can look at a sample here.*1613

*Let the y-axis be fraction of sample; let the x-axis be temperature.*1616

*At any given temperature, I am going to have a bell curve distribution of molecules just like that.*1628

*Let's call this T _{1}.*1639

*What happens to T _{2}?--what happens when we have a hotter temperature?*1641

*When I have a hotter temperature, my bell curve is going to shift just like that.*1649

*I call this T _{2}; T_{2} is greater than T_{1}.*1652

*Let's say that in order for the sufficient energy, in order for the*1658

*reaction to occur, let's go ahead and make that as a dotted line.*1663

*I am going to call this dotted line E _{A}; E_{A} is equal to activation energy.*1669

*What this is, it is the minimum energy required for the reaction to proceed, for collisions to occur.*1677

*Basically anything below E _{A}, anything less than E_{A}, you get zero collisions and no reaction.*1691

*Anything greater than or equal to E _{A}, you get collisions; therefore a reaction will occur.*1704

*Basically you see that as you go from T _{1} to T_{2},*1712

*the fractional molecules with an energy greater than E _{A} significantly increases.*1717

*At T _{2}, larger percent of molecules with an energy greater than or equal to E_{A}.*1723

*This is why as temperature goes up, so does the rate of a reaction in general.*1739

*We have a nice equation which can actually quantify this.*1749

*This is called the Arrhenius equation.*1754

*The natural log of k _{1} over k_{2} equals to E_{A} over R times 1 over T_{2} minus*1756

*1 over T _{1} where R is our universal gas constant in terms of energy, 8.314 joules per K times mole.*1766

*Temperatures T _{1} and T_{2} are kelvin temperatures; what this basically says is the following.*1779

*If I do a series of reactions at different temperatures and I calculate the*1789

*rate constant, I can then do graphical determination of the activation energy.*1794

*If I graph natural log of k _{1} over k_{2} as a function of 1 over temperature,*1800

*I am going to get a nice straight line with slope equal to ?E _{A} over R.*1809

*Again the Arrhenius equation is very useful because it gives us graphical approximation of the activation energy for a reaction.*1815

*Again this is the relationship between rate and temperature.*1829

*Finally the last factor we are going to study is a catalyst.*1833

*Basically a catalyst's job is to do the following.*1837

*A catalyst assists reactant molecules to be in the proper orientation for proper collision to occur.*1840

*What that does is that the activation energy is lower.*1847

*If this reaction represents without a catalyst, the activation energy is going be basically right here.*1853

*This is the energy that you must overcome for the reaction to go form products.*1867

*I can then proceed and draw another curve where this is a catalyst now.*1873

*With the catalyst, you see that the activation energy E _{A} is much lower.*1878

*Activation energy catalyst is going to be always much less than the activation energy no catalyst.*1884

*Because of that, with the lower activation energy, that means a faster reaction.*1892

*A catalyst again speeds up a reaction by lowering the activation energy.*1900

*We see that of course the nice about catalysts is that they can be reused over and*1907

*over again because during the course of a reaction, they are recovered; they are recovered.*1913

*That is catalysis.*1921

*Let's now get into our summary before we jump into our sample problems.*1923

*Kinetic studies the factors that can influence the rate of a chemical reaction.*1927

*We saw that we can have two main experiments to help us determine*1932

*the rate law--the method of initial rates and the integrated rate law.*1936

*We found that in a reaction mechanism, that the slowest step dictates the overall reaction.*1940

*That is what we call the rate limiting step.*1946

*Finally we introduced the Arrhenius equation which gives a mathematical relationship between the reaction rate and the temperature.*1952

*The nice thing about this equation again, this gave us graphical estimate of E _{A}.*1963

*Let's now get into a pair of sample problems.*1975

*A certain first order reaction has a half-life of twenty minutes.*1977

*Calculate the rate constant; that is part A.*1980

*Part B, how much time is required for this reaction to be 75 percent complete?*1984

*Let's go ahead; you are told that the reaction is first order.*1989

*For a first order reaction, T _{1/2} is equal to the natural log of 2 over the rate constant k.*1992

*You are told that the rate constant is 20.0 minutes.*2001

*Here the rate constant is simply going to be equal to the natural log of 2 divided by 20.0 minutes.*2007

*That gives us our answer in units of reciprocal minutes; that is part A.*2014

*Let's go ahead and do part B now.*2024

*How much time is required for this reaction to be 75 percent complete?*2026

*As soon as you see the word time, you should immediately, immediately, think integrated rate law because*2029

*method of initial rates does not have time in it; only integrated rate laws does.*2036

*For the first order integrated rate law, it is the natural log of the initial concentration*2041

*of A minus the natural log of the concentration of A is equal to kt.*2047

*We know what k already is because we already solved for that in part A.*2056

*We are good to go on that.*2060

*The question is asking for how much time is required.*2062

*This is what we are trying to solve for.*2065

*All that matters is is what is the identity of A _{0} and what is the identity of A?*2068

*You are told that how much time is required for this reaction to be 75 percent complete?*2074

*Let's say A _{0} is going to be 100.*2079

*If the reaction is 75 percent complete, that means only 25 percent of A is remaining.*2083

*A is going to be 25.*2090

*Again we now have enough information to solve for t.*2092

*We are going to get our final answer of t in units of minutes.*2097

*How do you know if you have done something wrong?*2105

*You should always check your answer because if you get a ?t, again that just doesn't make physical sense.*2107

*You know you have done something mathematically wrong.*2119

*Always check your answer for a negative time.*2121

*That is sample problem number one.*2125

*Let's go ahead and move on to sample problem number two.*2127

*Consider the following reaction.*2130

*2N _{2}O_{5} gas goes on to form 4NO_{2} gas plus O_{2} gas.*2132

*Here we are given several rate constants that were found at several temperatures.*2137

*The only equation that we know that deals with this is the Arrhenius equation.*2145

*The natural log of k _{1} over k_{2} is equal to*2150

*E _{A} over R times 1 over T_{2} minus 1 over T _{1}.*2154

*We know that we are going to be using this equation quite easily.*2161

*Here we can calculate E _{A} from there.*2165

*The nice thing about this is because this is going to give us a nice straight line,*2170

*we can use any pair of k and T data points to go ahead and solve for E _{A}.*2174

*Once again this is going to be using the Arrhenius equation to solve for E _{A}.*2193

*Our units of E _{A} is going to be in kilojoules per mole.*2199

*Next one is what is the order of the reaction.*2206

*This is kind of a trick question; this is something I have asked students before.*2208

*This is something I have seen asked by other instructors before.*2213

*The order of the reaction, you don't have to do any work for that.*2217

*The reason is because they already give you the units for k.*2219

*The units of the rate constant tell us the rate order.*2226

*It is only first order where the units of k is reciprocal time; first order overall.*2234

*Again just watch out for that when you do problems.*2248

*Again the units of k tell us a great deal of information without doing any work.*2251

*That is our lecture from general chemistry concerning kinetics.*2257

*I want to thank you for your time.*2262

*I will see you next time on Educator.com.*2263

0 answers

Post by Nasser Fiture on January 12 at 01:55:24 AM

If a proposed reaction mechanism fits the experimental rate law and reaction stoichiometry, then the mechanism is known to be completely correct.

True/False

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Post by Nasser Fiture on January 12 at 01:54:20 AM

all reactions which show first-order kinetics are also known as unimolecular reactions.

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Post by Nasser Fiture on January 12 at 01:53:31 AM

the rate determining step of a chemical reaction is the step which has the transition state highest in potential energy.

True/False

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Post by Jarrah alharbi on September 18, 2016

is the ----> t1/2=In1/2 / k

or t1/2=In2/k

2 answers

Last reply by: Professor Franklin Ow

Mon Jun 20, 2016 11:35 PM

Post by Parth Shorey on June 20, 2016

I don't understand why you used the negatives in front of the recants? You said because of the slopes? I still don't understand.

1 answer

Last reply by: Professor Franklin Ow

Thu Jun 16, 2016 3:29 PM

Post by Parth Shorey on June 13, 2016

Is the Q&A active?

1 answer

Last reply by: Professor Franklin Ow

Thu May 28, 2015 12:28 PM

Post by BRAD POOLE on May 7, 2015

At about the 17 min mark you said that the example was a 2nd order reaction. How did you come up with this? Are you just going by whatever your units are for "k"? I always thought you used the exponents of the reactants to figure out what order it was, then again could be why I can't seem to get these kinetics problems.

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Post by Saadman Elman on January 14, 2015

Great lecture as usual!

0 answers

Post by Saadman Elman on January 14, 2015

What you mean by we can write co-effecient as rate order ONLY for elementary steps. What you mean by elementary steps exactly?

2 answers

Last reply by: David Gonzalez

Thu Jul 31, 2014 12:25 PM

Post by David Gonzalez on July 31, 2014

Hi Professor Ow. First of all, great lecture. Although, there is one problem that I have.

In the example (around the 16-minute mark), when determining the order for S2O8, you mentioned that there was a 3x increase in initial rate from 0.015 to 0.044. I'm confused, because 0.015 x 3 is 0.045 - don't these problems need to be exact? Or can they sometimes be slightly off?

Thanks.

1 answer

Last reply by: Professor Franklin Ow

Tue Jun 24, 2014 1:29 PM

Post by brandon joyner on June 24, 2014

Also why for Iodine did you only do the first experiment for the second experiment you went from experiment 3 to 1.

4 answers

Last reply by: brandon joyner

Fri Jun 27, 2014 1:46 PM

Post by brandon joyner on June 24, 2014

For Iodine how is it a triple jump when all you had to do was go up 2?

1 answer

Last reply by: Professor Franklin Ow

Wed May 21, 2014 1:54 AM

Post by Ashley Gwemende on May 20, 2014

Where do I find a lecture of how to read a potential energy diagram ?