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Goals: Dependence of rate on conc. Distinguish different rates How T affects rate How catalysts affect rate Determine simple mechanisms Suggested Problems: 1, 7, 9, 11, 13, 15, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 41, 45, 47, 49, 53, 55, 57, 59, 61, 65

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Relative Rates of Change Your Turn: Show the relative rates of change for the reaction: N 2 O 5(solvent) --> 2 NO 2(solvent) + ½ O 2(g) Hint: First, convert to whole number stoichiometric coefficients.

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Relative Rates of Change Problem: Show the relative rates of change for the reaction: N 2 O 5(solvent) --> 2 NO 2(solvent) + ½ O 2(g) 2 N 2 O 5(solvent) --> 4 NO 2(solvent) + O 2(g) Not Given

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Deriving Rate Laws Your Turn: Derive rate law and k for CH 3 CHO(g) ---> CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO: Rate of rxn = - k [CH 3 CHO] ? Expt. [CH 3 CHO]Disappear of CH 3 CHO (mol/L)RATE (mol/Lsec) 10.100.020 20.200.081 30.300.182 40.400.318 C/ t

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Deriving Rate Laws Rate of rxn = k [CH 3 CHO] 2 Here the rate goes up by FOUR when initial conc. doubles. Therefore, we say this reaction is Second order. Now determine the value of k. Use expt. #3 data— 0.182 mol/Ls = - k (0.30 mol/L) 2 k = - 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T. C/ t Not Given

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Deriving Rate Laws C/ t Note: In this example the rate of removal depends on the conc. in soln. Increase conc. increase removal Go back to the ethanol problem, rate is independent of the presence of ethanol. So what will happen?

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Your Turn: Determine the rate order expression for the following data set:

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Concentration/Time Relations Tylenol dose is 650 mg Human body has 5L of blood. Thus, Initial [Tylenol] = C o = 8.61 x 10 -4 M (confirm this!) Rate of removal of Tylenol is k = 0.21 hr -1 What is conc. when you need to take another dose in 4 hours? C/ t Acetaminophen (Tylenol) C 8 H 9 O 2 N How long will it take to drop the conc. by 90% of C o (10% is left or 8.61 x 10 -5 M)?

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Concentration/Time Relationships Rate of removal of Tylenol = k [Tylenol], k = 0.21 hr -1. If initial [Tylenol] = 8.61 x 10 -4 M: What is the blood conc. when you take another dose in 4 hrs; How long will it take to drop the blood conc. by 90% of C o or to 8.61 x 10 -5 M? Use the first order integrated rate law C/ t

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Concentration/Time Relationships Rate of removal of Tylenol = k [Tylenol], k = 0.21 hr -1. If initial [Tylenol] = 8.61 x 10 -4 M: What is the blood conc. when you take another dose in 4 hrs; How long will it take to drop the blood conc. by 90% of C o or to 8.61 x 10 -5 M? C/ t Use the first order integrated rate law

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Your Turn If k = 0.1 min -1 what is half-life? If half-live = 1000 years, what is k? THERE IS NO ANSWER TO THIS PROBLEM ON THE WEB

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Your Turn: You are working as a technician for the State crime lab and you receive a urine sample for cocaine analysis. The sample is from a driver who fled the scene of an accident and was apprehended 3 hours later. It has been 5 hours after the accident when you obtain the sample and you find a concentration of 6.04 x 10 -7 M. The average removal rate of cocaine from the human body is 0.8 hr -1. The court decides that a concentration above 3.3 x 10 -6 M will impair the driver. What was the cocaine concentration in the driver at the time of the accident? Pharmacology / Forensic Science

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Your Turn: You are working as a technician for the State crime lab and you receive a urine sample for cocaine analysis. The sample is from a driver who fled the scene of an accident and was apprehended 3 hours later. It has been 5 hours after the accident when you obtain the sample and you find a concentration of 6.04 x 10 -7 M. The average removal rate of cocaine from the human body is 0.8 hr -1. The court decides that a concentration above 3.3 x 10 -6 M will impair the driver. What was the cocaine concentration in the driver at the time of the accident?

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Natural Formation of Radioactive Elements: 14 6 C N 2 in the atmosphere is bombarded by cosmic radiation. One type of radiation is neutrons. 14 7 N + 1 0 n ===> 14 6 C + 1 1 H 14 6 C formed in this reaction, goes on to form radioactive CO 2 …which is taken up by plants….which is eaten by animals….which are eaten by more animals. With time 14 6 C decays. 14 6 C ===> 0 -1 + 14 7 N

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Half-Life Problem You are working as an anthropologist, and you find a well-preserved burial ground containing wooden artifacts. You have some of these dated using a 14 C dating technique and the results show that 35% of the original 14 C remains. Given that the half-live of 14 C is 5730 years, what is the age of the burial site? ln([A] t /[A] o )=-kt Hint: calc. k using t ½, then calc. age

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Half-Life Problem Soln: ln([A] t /[A] o )=-kt ln(0.5)=-k (5735yr) 1.21 x 10 -4 /yr= k ln(0.35) = - (1.21 x 10 -4 /yr)t age = t = 8686 years

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1Chemistry 2C Lecture 20: May 17 th, 2010 1)Introduction to Kinetics 2)Rate Laws 3)Orders and Reaction Constants 4)Initial Slopes 5)Zero th order reactions.

1Chemistry 2C Lecture 20: May 17 th, 2010 1)Introduction to Kinetics 2)Rate Laws 3)Orders and Reaction Constants 4)Initial Slopes 5)Zero th order reactions.

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