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- 1. The Rate Law: How Concentration Affects Rate • The rate law expresses the rate of any reaction at a fixed temperature on the reactant concentration , AND the order of reaction. Rate law is determined from experimental data . - For a general reaction, aA + bB + … cC + dD … etc. - The rate law has the form: Rate = k [A] m [B] n … etc. Where k is the rate constant, and m and n are the reaction orders (could be the number 0, 1, or 2) - The overall order of the reaction is m + n +…etc
- 2. • Explaining Reaction Order - Using the simple general reaction of A B The rate law would be: Rate = k [A] n - Depending on the nature of the reaction, the order with respect to [A] can be 0, 1, or 2. The order (n) as well as the k value must be experimentally determined. - In the rate law equation, the order represents the power of [A] (the reactant concentration). Mathematically, the order tells you to what extent the concentration affects the rate.
- 3. - From the equation, you can see that any concentration of [A] 0 equals 1. - The rate of a zero order reaction is solely determined by the k constant, which has units of mol . L -1. time -1 The reaction would always proceed at a constant rate. A plot of [A] vs. time would be linear. • If m = 0, or a zero order reaction , the concentration has no affect on rate of the reaction The rate law equation would be: Rate = k [A] 0
- 4. • If m = 1, it is a first reaction order . - The rate of reaction will be directly proportional to the concentration of the reactant [A]. The higher the concentration, the faster the rate - The k constant for a 1 st order reaction has units of time -1 The rate of a 1 st order reaction slows down linearly as reactant is used up. A plot of [A] vs. time would be curved. The rate law equation would be: Rate = k [A] 1
- 5. • If m = 2, it is a second reaction order . - The rate of reaction will be proportional to the concentration of the reactant [A] squared . The higher the concentration, the faster the rate - The k constant for a 2 nd order reaction has units of L . mol -1. time -1 The rate of a 2 nd order reaction would slow down exponentially as reactant is used up. A plot of [A] vs. time would be curved The rate law equation would be: Rate = k [A] 2
- 6. For the reaction A B
- 7. Integrated Rate Law: How Concentration Changes Over Time - The rate expression defined rate as: For the general reaction of A B or - The rate law is defined rate as: k [A] n • Equating the two terms produces what is called the differential rate law . - Integrating this equation produces integrated rate law equations depending on the order of reaction (n value) - The integrated rate law equation simply relates how the concentration of reactant(s) changes over time (according to the rate constant) in a LINEAR fashion Δ [A] Δ t d [A] d t Δ [A] Δ t k [A] n =
- 8. - Through methods of calculus, the integrated rate law equation for zero order reactions is: [A] t – [A] o = - k t Concentration of A at time t Concentration of A originally Rearranging: [A] t = - k t + [A] o y = m x + b Therefore, in a plot of [A] vs t, slope = - k Zero Order: • Zero Order Integrated Rate Law
- 9. If we observed the change in concentration of A over time for a zero order reaction [A] (mol/L) Time (unit) Slope = - k - Zero order reactions are relatively easy to identify straight away since a plot of reactant concentration over time is linear .
- 10. - 1st order reactions are more difficult to interpret since the plot of [reactant] vs time would be curved . - Through methods of calculus, the integrated rate law equation for a first order reaction is: Rearranging: ln [A] t = - k t + ln[A] o y = m x + b Therefore, in a plot of ln[A] vs t, slope = - k [A] o [A] t ln = k t First Order: • First Order Integrated Rate Law
- 11. [A] (mol/L) Time (units) ln [A] Time (units) Slope = - k If the reaction is first order , transforming the plot of [A] vs t, to ln[A] vs t will yield a linear plot where the slope = - k
- 12.
- 13. - 2nd order reactions are also hard to interpret since the plot of [reactant] vs. time is curved, and looks very similar to first order plots - Through methods of calculus, the integrated rate law equation for a second order reaction is: Rearranging: • Second Order Integrated Rate Law [A] o [A] t = k t 1 - 1 y = m x + b [A] t 1 = k t + [A] o 1 Therefore, in a plot of 1/[A] vs. t, slope = k
- 14. [A] (mol/L) 1/[ A] Time (units) If the reaction is second order , transforming the plot of [A] vs t, to 1/[A] vs t will yield a linear plot where the slope = k Slope = k Time (units)
- 15. N 2 O 5 (g) N 2 O 4 (g) + ½ O 2 (g) • Example: Decomposition of dinitrogen pentoxide Rate = k[ N 2 O 5 ] a What is the order of the reaction?
- 16. Since it fits the ln plot, it is a 1 st order Reaction with respect to [N 2 O 5 ] Original data transformed mathematically Data will fit a straight line equation for only ONE of the plots.
- 17. Reaction Half-Life • The half-life (t 1/2 ) is the time required for the reactant concentration to reach half its initial value. At t 1/2 , [A] t = ½[A] o - By substituting ½[A] o into the integrated rate law equations, we get: t 1/2 Zero Order: t 1/2 t 1/2 = [A] o 2 k = ln 2 k = 1 k [A] o First Order: Second Order:
- 18. Carbon 14’s half-life is 5, 700 years. Taking the ratio of C-12 to remaining C-14 gives an estimate of an objects age, good for about 60,000 years. Method can date only living/organic artifacts.
- 19. • Technique of Swamping - Kinetics analysis of reactions becomes more complicated when there are more than 1 reactant Eg. A + B Products Rate = k[A] m [B] n - As the reaction goes, both [A] and [B] will decrease over time and BOTH are affecting the rate (as long as the orders with respect to [A] and [B] are not 0) - In “swamping”, we purposely make the starting concentration of one of the reactants so high compared to the other, we assume the reactant of high concentration remains the same concentration throughout the experiment.
- 20. • Initial Rates Method For Determining Order Reaction: O 2 (g) + 2 NO(g) 2 NO 2 (g) Rate = k [O 2 ] m [NO] n - Multiple experimental runs are required for this method. [NO 2 ] time Trial #1 [NO 2 ] time Trial #2 (Concentration of reactant changed) Initial rate changes
- 21. Rate = k [O 2 ] m [NO] n
- 22. Effect of Temperature on Reaction Rate • A plot of rate constant versus temperature (in K) shows an exponential increase. The relationship between k and T is defined by the Arrhenius equation : k = Ae -Ea/RT A is related to the colliding molecules, E a is the activation energy, R is the universal gas constant (J/mol . K), and T is the temperature in Kelvin.
- 23. • Taking the natural log of the Arrhenius equation gives a linear relationship of k and T. y = b + m x Activation energy can be calculated from a minimum of two experiments done at different temperatures. - Two-Point form of Arrhenius equation ln k = ln A - E a R 1 T ln k 2 k 1 = E a R 1 T 1 1 T 2 -
- 24. Reaction Mechanisms: Multiple Steps in the Overall Reaction • Many reactions do not occur in one single step. Rather, multiple reaction steps occur to eventually convert the reactant(s) to the product(s) • Each reaction step in an overall reaction is called an elementary step Rare • Elementary reaction steps have predictable rate law equations, where the order with respect to a reactant is the coefficient of the reactant
- 25. • A proposed mechanism must meet 3 criteria: 1. The elementary steps must add up to the overall balanced equation. 2. The elementary steps must by physically reasonable. This means that they should either be unimolecular or bimolecular. Termolecular is very unlikely. 3. The mechanism must correlate with the actual rate law for the overall reaction. Rate-determining step: The slowest elementary step. Since it limits how fast the overall reaction can proceed, the rate law for the rate-determining step defines the rate law for the overall reaction.
- 26. • Reaction mechanisms are hypothetical , but are based on chemical evidence of intermediates and data from rate experiments 2 NO 2 (g) + F 2 (g) 2NO 2 F(g ) Example: - Experimentally, it has been determined that the rate law is first order in both reactants. (1) NO 2 (g) + F 2 (g) NO 2 F(g) + F(g) (slow; rate determining) (2) NO 2 (g) + F(g) NO 2 F(g) (fast) • The mechanism provides insight into how and why any given reaction has its characteristic kinetics Rate = k [NO 2 ][F 2 ] The proposed mechanism: Is it reasonable?
- 27. Reaction profile for: 2 NO 2 (g) + F 2 (g) 2NO 2 F(g ) The rate determining step is the transition state with the highest energy, not the step with the greatest E a At the transition state , the molecular entity Is believed to look like a combination of both reactant and product.
- 28. Reaction Coordinate I II III IV 1. How many transition states? 2. How many intermediates 3. Which is the rate determining step?
- 29. • If the first elementary step is not the slowest (rate determining) step, determination of the rate law equation is more complicated Observed rate law: Rate = k[H 2 ][NO] 2 - Proposed mechanism is 3 elementary steps, where the 2 nd step is the slowest. - Does the mechanism fit the observed rate law? 2 H 2 (g) + 2 NO(g) 2 H 2 O(g) + N 2 (g) Eg.
- 30. Riddle: 4 men went camping in the woods and became lost for days. They have no more food or water and some are seriously injured. They will all die within an hour, unless they can cross a bridge that leads them to salvation. A sign on the bridge says that as soon as it is stepped on, it will collapse in 17 minutes. It’s dark and since they have only 1 flashlight, only 2 men at most can cross at any given time. The time it takes for 2 men to cross is limited by the slowest man. Once 2 have crossed, one person must come back to bring back the flashlight before more can cross. Man 1 is fine and can cross in 1 minutes. Man 2 has blisters on his feet and can cross in 2 minutes. Man 3 has a sprained ankle and can cross in 5 minutes. Man 4 has a broken leg and can crawl across in 10 minutes. How can they all make it across in 17 minutes?
- 31. • Example of a 1-step overall reaction: CH 3 Br + OH - CH 3 OH + Br - The topmost point of energy is called the transition state , where the molecular entity looks like a combination of both reactant and product. Rate = k [ CH 3 Br ][ OH - ] Observed rate law:
- 32. Where R-X represents CH 3 Br • Example of a multi-step overall reaction: CH 3 Br + H 2 O CH 3 OH + HBr The reaction step of highest energy is the rate- determining step Observed rate law: Rate = k[ CH 3 Br ]

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