# Goals: Dependence of rate on conc. Distinguish different rates How T affects rate How catalysts affect rate Determine simple mechanisms Suggested Problems:

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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

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.

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

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

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

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?

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

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)?

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

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

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

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

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?

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

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

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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