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General Chemistry 2 (CHEM 215) Week 1 Class Notes

by: snufkin

General Chemistry 2 (CHEM 215) Week 1 Class Notes CHEM 216

Marketplace > San Francisco State University > Chemistry > CHEM 216 > General Chemistry 2 CHEM 215 Week 1 Class Notes
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Week 1 lecture notes from CHEM 215 (General Chemistry 2 course) going over the rate expressions and rate law.
General Chemistry 2 Laboratory
Kent Lau
Class Notes
Chemistry, General Chemistry, rateexpressions, ratelaw




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This 4 page Class Notes was uploaded by snufkin on Monday September 5, 2016. The Class Notes belongs to CHEM 216 at San Francisco State University taught by Kent Lau in Fall 2016. Since its upload, it has received 104 views. For similar materials see General Chemistry 2 Laboratory in Chemistry at San Francisco State University.

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Date Created: 09/05/16
Chemistry Notes: Week 1    ❈ Derive rate expressions from balanced chemical reactions and visa  versa    ❈ Determine experimental rate law with reaction orders and rate constant  from initial rates data    Kinetics­ ​  how reactions progress with time  *can reveal reaction mechanism  *predicts yield of a chemical at a given time      H​  + I​ →  2HI  2​ 2      Which concentration changes faster?  answer →  HI    ­rates must be defined to quantify reaction progress    Δquantity rate=  Δtime          speed=  ΔdΔtimece   Δconcentration reaction rate=    Δtime   m mol mol M units:  s , s ​ L s ​ min    rise Δ[c] slope=  run ​ Δt = rate    Rate expressions:  Δ[H2] rate of appearance HI=  Δt   −Δ[H2] −Δ[I2] rate of disappearance H​ =  2​ Δt = Δt     −Δ[H2] 1 ΔHI −Δ[I2] Δt =  2 Δt = Δt     In general: aA + bB →  cC + dD    rate= ​­ 1 ΔA = ​­ 1 ΔB =  1 ΔC ​  1 ΔD   a Δt b Δt c Δt d Δt   Rate Laws:  ­not the same as rate expressions  ­describes how rate depends on concentration  ­reactions rely on collisions  ­higher concentration leads to higher rate  ***must be determined by experiment    In general: aA + bB +cC →  xX + yY +zZ    m​ n​ p  rate = k [A]​ [B]​ [C]​   (k= rate constant; m DOES NOT EQUAL a)        Ex. 2H​ O2​ 2​2H​ O + 2​ 2    −Δ[c1] rate​1​ Δt   −Δ[c2] rate​2​ Δt   −Δ[c3] rate​3​ Δt     rate= k [H​ 2​  2​   How to Calculate Rate Law:    NO(g) + O​  →3​O​ (g) + 2​ (g)  2​   Total  [NO] mol/L  [O​3​mol/L  Rate mol/L/s  1  1.00  3.00  0.660  2  1.00  6.00  1.31  3  1.00  9.00  1.98  4  2.00  9.00  3.96  5  3.00  9.00  5.94    Solving m:  Pick a concentration two rows that have the same number, for  ­­­­­­­​example rows 1 and 2 for [NO] are the same. Then, divide the rates,  rate2=1.31 ­­­­­­­​in case of rows 1 and 2 in column [NO]:  rate1=0.660 .  or  In this case, we can solve for m without calculating:  The rate doubles when NO doubles: linear; m=1      Solving n:  Same as solving for m, pick a concentration (a different one than from  ­­­­­­­­​m) two rows that have the same number and divide the rates.  rate1 =  k[NO]1[O2]n = 1.31m/s = [6.00M]n   rate2 k[NO]1[03]n 0.660m/s [3.00M]n n 1.98= (2.00)​   take the natural log of both sides  ln(1.98)= (n) ln(2.00)  n= ln(1.98) ≈ 1  ln(2.00)   Solving for k:    rate 0.660M/s 1 k= [NO][O3] ​ 1.00M×3.00M ​ 0.220  Ms    


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