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General Chemistry 2 (CHEM 215) Week 2 Class Notes

by: snufkin

General Chemistry 2 (CHEM 215) Week 2 Class Notes CHEM 215

Marketplace > San Francisco State University > Chemistry > CHEM 215 > General Chemistry 2 CHEM 215 Week 2 Class Notes
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About this Document

Week 2 lecture notes from CHEM 215 (General Chemistry 2 course) going into depth about rate law and introducing reaction mechanism.
General Chemistry 2
Russel Jensen
Class Notes
Chemistry, General Chemistry, reactionmechanism, ratelaw, PV=nRT




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This 5 page Class Notes was uploaded by snufkin on Thursday September 15, 2016. The Class Notes belongs to CHEM 215 at San Francisco State University taught by Russel Jensen in Fall 2016. Since its upload, it has received 48 views. For similar materials see General Chemistry 2 in Chemistry at San Francisco State University.


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Date Created: 09/15/16
Chemistry Notes: Week 2      Ex. Rate Expression:  Ozone decomposes to oxygen:  2O​ (g)→3O​ (g)  3​ 2​ What is the rate reaction for this reaction in terms of the  ­­­­­­­​ isappearance of O​ (g)?  2​   aA→bB  rate= ­ 1 Δ[A] ​ ­1 Δ[B] = 1 Δ[C] = 1 Δ[D]   a Δt b Δt c Δt d Δt   Ex. The rate of a reaction depends on what?  Answer: The concentration of reactants, temperature, and the  ­­­­­­­­­­­­­­­­­­­­​structure of the reactants.    Reaction Orders:    Reaction order­ the sum of orders for each reactant in the rate law    Ex. The rate law for the reaction of chlorine and nitric acid can be  expressed as rate= k[Cl​ ][NO]​ . Determine the units for the rate constant.  2​   rate M/s 1 k= [A][B]2 = M 3 = 2   M s   Ex. An experiment shows that the reaction of nitrogen dioxide with carbon  ­­­­­​monoxide.  NO​ 2​) + CO(g)  →  NO(g) + CO​ (g)  2​ ­­­­­​It is in second order in NO​ (g) and zeroth order in CP at 100°C. What is  2​ ­­­­­​ he rate law?  rate=k[NO​ ]​     2​ note: the zeroth power does not contribute to the rate order at all.      ***this type of problem will be on the test:  Ex. Acetaldehyde decomposes when heated to yield methane and carbon  ­­­­­​monoxide  CH​ 3​O(g) → CH​ (g) + CO4​)    Determine the rate law using the experimental data:    Δ[CH CHO] Trial:  [CH​ 3​O](mol/L)  ­ 3 ( mol )  Δt Ls −3 −11 1  1.75× 10   2.06× 10   2  3.51× 10 −3  8.24× 10 −11   −3 −10 3  7.00× 10   3.30× 10     rate2=k[C 2 n 8.24×10 −11M/s k(3.50×10 M) n rate =k[C ] n =  −11 ​ −3 n   1 1 2.06×10 M/s k(1.75×10 M)   8.24×10 −11 3.50×10 −3 n 2.06×10 −11= ( 1.75×10 −3 )​     4=(2)​   ln(4) = n ln(2)  ln(4) n= ln(2) = 2    ❈Predict the rate determining step (RDS) in a reaction mechanism from an  experimental rate law, or visa versa including when RDS is not first.    ­RDS is the slow step  Rate Laws:  ­must be determined by experiment  ­can’t be predicted from chemical equations with multiple steps  ­can be predicted from single step elementary reactions    Elementary Reactions:  ­single step  ­rate law follows stoichiometry  ­combine elementary steps to describe multistep reactions  ­most common types:  ● unimolecular: A→products; rate=k[A]  2​ ● bimolecular: A+A→products; rate=k[A]​  or rate=k[A][B]  2​ ● termolecular (rare): 2A+B→products; rate=k[A]​ [B]      Reaction Mechanism:  NO​ (g) + CO(g)  →  NO(g) + CO​ (g)  2​ 2​    ­uses elementary steps to describe    ­­­​ ultistep reactions  2 Observed rate law: rate=k[NO​ ]​   2​    ­can contain intermediates      Intermediates:          (NOT stoichiometrically correct)            ­must be consumed          ­hard to measure  NO​  2​NO​    2​ NO​  + NO (3​ow)  2         ­therefore can’t be in experimental       rate=k[NO​ ]​ 2​ ­­­­­­­​rate law  NO​  3​CO  →  NO​  + CO​  2​ast)  2​      rate=k[NO​ ][CO]  3​   Combine equations:  2NO​  +2​NO​  + 3​  →  N ​ O​3​+ NO + ​NO​  + C2​    2      Intermediates: NO​  , NO​    2​ 3   NO​  2​CO → NO + CO​    2     Rate of the entire process ≈  rate of the slowest step    Rate determining step:  ­slowest step in a reaction mechanism        Ex. Given a reaction mechanism, identify:  1)Chemical equation  2)Intermediates  3)Elementary rate laws  4)Observed rate law    (fast)  O​ (g) ⇋ O​ (g) + O(g) rate​ = k​ [ O​ ]; rate​ = k​ [O​ ][O]  3​ 2​ 1 ​ 1​ 3​ ­1 ​ ­1​ 2​ (slow) O​ (g) + O(g) →  O​ (g) rate​ = k​ [ O​ ][O]  3​ 2​ 2 ​ 2​ 3​ __________________________________________________________    2O​ (3​ + ​O(g)  ​ → 3O​ (g)2​ O ​ (g) intermediates: O(g)    2O​ (g)→ 3O​ (g)   2​ 2​   Fast equilibrium:  ­use when slow step is not first  ­rate forward = rate reverse    rate​ = rate​  → k​ [ O​ ] = k​ [O​ ][O]  1​ ­1​ 1​ 3​ ­1​ 2​ k1[ O ]3 [O] =  k [O ]   −1 2   k [O ] rate​  = k​ [ O​ ] × 1 3   2​ 2​ 3​ k −1 ] 2   k2 1[O ]3 2​ ­1 rate​  2​ k [O ] ​ k[O​ ]3​[O​ ]​2​  −1 2 rate​  = k[O​ ]​ [O​ ]​   (this is the observed rate law)  2​ 3​ 2​   Ex. Ammonia gas reacts with dioxygen gas to produce nitric oxide and  water. 956 liters per second of dioxygen is consumed when the reaction  runs at 257°C and 0.88 atm. What is the rare that NO is produced in kg/s?    Steps:  1) find the chemical equation  2) use PV=nRT to get mols of O​   2 3) use mol to mol ratio to relate O​  to N2​ 4) convert mol NO to kg    NH​ 3​) + O​ (2​ → NO(g) + H​ O(g)  2​   Balance equation:  4NH​ (3​ + 5O​ (g2​→ 4NO(g) + 6H​ O(g)  2​   5:4 mol ratio of O​ :2​O    n= PV    RT   What we know:  P= 0.77atm , V= 956L/s , T= 257°C + 273.15= 530.15K ,   R= 0.08205 L×atm/ mol × K    n=  0.88aL×atm956L/s   ​ 19.24 mol/s of O​   2 0.0805 mol×K ×530.15K   19.24 molO /2 4 mol NO 1 ×  5 mol O = 15.47 mol/s of NO  2   15.47 molNO/s 30.0061 g/mol 1 kg 1    1 ×  1000g = 0.46 kg/s og NO 


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