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CHM 1120 – General Chemistry II Paul Stein – S2208B – pstein@css.edu - 6065

CHM 1120 – General Chemistry II Paul Stein – S2208B – pstein@css.edu - 6065 MWF 10:30 – 11:30 TR 1:00 – 1:50 Or drop in any time . Text: Chang – Chemistry 10 th Ed. Lab manual from bookstore iClickers – bring this Friday. CHM 1120 ― Supplemental Instruction (SI).

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CHM 1120 – General Chemistry II Paul Stein – S2208B – pstein@css.edu - 6065

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  1. CHM 1120 – General Chemistry II Paul Stein – S2208B – pstein@css.edu - 6065 MWF 10:30 – 11:30 TR 1:00 – 1:50 Or drop in any time Text: Chang – Chemistry 10th Ed. Lab manual from bookstore iClickers – bring this Friday

  2. CHM 1120 ― Supplemental Instruction (SI) Every Monday and Wednesday Evening Beginning Monday, January 23 4:30 – 6:30 Room TBA

  3. A B rate = D[A] D[B] rate = - Dt Dt Chemical Kinetics Thermodynamics – Which way does a reaction go & how far? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

  4. A B rate = D[A] D[B] rate = - Dt Dt Molarity or

  5. 2A B aA + bB cC + dD rate = - = = rate = - = - D[B] D[C] D[B] D[A] D[A] D[D] rate = 1 1 1 1 1 Dt Dt Dt Dt Dt Dt d 2 c b a Reaction Rates and Stoichiometry Two moles of A disappear for each mole of B that is formed. It makes no difference which reactant or product you “follow”

  6. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) time 393 nm Detector light red-brown t1< t2 < t3 D[Br2] aD Absorption

  7. Br2 = 3.10 g ml-1. A highly volatile, reddish brown liquid. What volume of Br2 is needed to make 100. ml of 0.120 M Br2. Br2 = 159.8 g mol-1. Add to 100 ml volumetric flask and dilute top mark with water. 0.120 mol• 0.100 L • 159.8 g • 1.00 ml = 0.619 ml or 619 ml L mol 3.10 g

  8. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial initial rate = (0.0120 – 0.0072)/100 ~ 4.8 x 10-5 M s-1. = 3.0 x 10-5 M s-1 instantaneous rate = rate for specific instance in time

  9. rate k = [Br2] rate a [Br2] rate = k [Br2] = rate constant = 3.50 x 10-3 s-1

  10. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) Instantaneous rate at t = 0s … or initial rate [Br] (M) t (s) What is the value of the initial rate? a) 2.5 x 104 M s-1 b) 4.0 x 10-5M s-1 b) 2.3 x 10-5 M s-1

  11. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) Rate Laws and Reaction Order A + B → C + D Rate Law: rate = k [A]a [B]b k = rate constant - A and B are reactants a/b= partial reaction order with respect to A, B. Reaction order (n) = a + b the overall order of the reaction (n) is the sum of all the partial orders rate = k [Br2] rate = k [Br2]1 [HCOOH]0 This is a first order reaction (n = 1)

  12. Most Common Reaction Orders linear rate equations 1storder ln([A]/[Ao]) = -kt ln[A] = -kt + ln[Ao] plot ln[A] vs. t (slope = -k) rate = k [A] r = k [A] = - d[A]/dt 2ndorder 1/[A] – 1/[Ao] = kt 1/[A] = kt + 1/[Ao] plot 1/[A] vs. t (slope = k) rate = k [A]2 or.... rate = k [A][B]

  13. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) rate = k [Br2] 1st: ln ([A]/[Ao]) = -kt ln [A] = -kt + ln[Ao] plot ln[A] vs. t (slope = -k) ln [Br2] 1/[Br2] 2nd: 1/[A] – 1/[Ao] = kt 1/[A] = kt + 1/[Ao] plot 1/[A] vs. t (slope = k) Time (s)

  14. Determining the Rate Law Initial rate method ― A + B → Products Determine the rate with a fixed [A] and [B] Change (typically double) [A] while holding [B] constant – measure rate - determine the partial order with respect to A. Repeat experiment for [B] while holding [A] constant - determine the partial order with respect to B. Write the rate law for the reaction. Calculate k from rate law. Rate (M/s)[A] (M)[B] (M) Trial 1 ??? 0.01 0.01 Trial 2 x•??? 0.02 0.01 Trial 3 y•??? 0.01 0.02 If x = 1 …. Order = 0; 20 = 1 If x = 2 …. Order = 1; 21 = 2 If x = 4 …. Order = 2; 22 = 4

  15. F2(g) + 2ClO2(g) 2FClO2(g) Determining the Rate Law X = 1 rate = k [F2]x[ClO2]y r2 = k•(2[F2])x•[ClO2]y = k•2x•[F2]x•[ClO2]y = 2x = 2 r1 k•[F2]x•[ClO2]y k• [F2]x•[ClO2]y r2 = k•[F2]x•(4•[ClO2])y = k•[F2]x•4y•[ClO2]y = 4y = 4 r1 k•[F2]x•[ClO2]y k•[F2]x• [ClO2]y y = 1 rate = k [F2][ClO2] ― Note that partial orders ≠ reaction coefficients. Problem 13.70

  16. Problem 13.70 r1/r2= (c1/c2)x. Choose correct trials Solve for x 2H2 + 2NO → 2H2O + N2 rate = k[H2]1[NO]2 rate = k[H2]x[NO]y k = 2.4 x 10-6/((.01 • .0252) = 0.38 M-2 s-1 What is x (the partial order with respect to H2)? a) 0 b) 1 c) 2 d)3 k = ? What is y (the partial order with respect to NO)? a) 0 b) 1 c) 2 d)3 What is the overall reaction order? a) 0 b) 1 c) 2 d)3

  17. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) 1st: ln ([A]/[Ao]) = -kt ln [A] = -kt + ln[Ao] plot ln[A] vs. t (slope = -k) rate = k [Br2] [Br2]0 = 0.0120 M How much Br2 left after 400 s ? -slope = k = 0.0035 s-1 ln ([Br2]/0.012) = -0.0035 s-1• 400 s ln ([Br2]/0.012) = -1.40 [Br2] = 0.0030 M ([Br2]/0.012) = 0.247 ln [Br2] Time (s)

  18. Integrated Rate Laws 1st Order 2nd Order Rate law: r = k[A] Rate law: r = k[A]2 ln [A] = -kt + ln [Ao] 1/[A] = kt + 1/[Ao] Half-life: The time it takes for ½ of a reactant to be depleted or …. t½ = t such that [A] = [A0]/2 ln [A] - ln [Ao] = -kt and… ln ([A]/[Ao]) = -kt 1/[A] – 1/[Ao] = kt 1/([A0]/2) - 1/[Ao] = kt½ and … ln (([A0]/2)/[Ao]) = -kt½ ln (½) = - kt½ and ….. t½ = 0.693/k 2/[A0] – 1/[Ao] = kt½ 2/[A0] – 1/[Ao] = kt½ = 1/[A0] t½ = 1/(k[A0])

  19. Integrated Rate Laws 1st Order 2nd Order Rate law: r = k[A] Rate law: r = k[A]2 ln [A] = -kt + ln [Ao] or ln ([A]/[Ao]) = -kt 1/[A] = kt + 1/[Ao] t½ = 0.693/k t½ = 1/(k[A0])

  20. Theoretical - rate of loss of Reactant 1st order: t½ = 0.693/k as [A0]↓ t½ stays the same 2nd order 2nd order: t½ = 1/(k[A0]) as [A0]↓ t½ increases [A] t1/2 1st order t1/2 t1/2 t

  21. Radiocarbon dating The natural abundance of 14Cis very low compared to 12C. It remains constant because it is constantly produced by cosmic ray neutrons according to the following …. 147N + 10n  146C + 11H Living organisms maintain 14C at a constant level. When the organism dies the 14C begins to decay ….. 14C  14N + 0-1b This leads to the depletion of 14C over time. The more time that passes the less the amount of 14C in material that came from a once living source.

  22. Radioactive decay is 1st Order ln([A0]/[A]) = kt & t½= 0.693/k or k = 0.693/t1/2 14C t½= 5730 years – What is k? 14C decay: k = 0.693/5730 = 1.21 x 10-4 years? A vendor tries to cell you a wooden tool that he claims is from a Neanderthal settlement in Europe. You have the wood tested and find that it contains 0.420 times the 14C activity as fresh wood. Is the artifact a fraud? ln(1/0.420) = 1.21 x 10-4 • t & t = 0.868/1.21 x 10-4 ~ 7200 yrs old. Neanderthals became extinct more than 20,000 years ago.

  23. ‡ A + B AB C + D + Exothermic Reaction Endothermic Reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction. The rate of a reaction depends upon …. 1) The frequency of collisions 2) The energy of each collision (what fraction of collisions will succeed?)

  24. ‡ A + B AB C + D + The rate of a reaction depends upon …. 1) The frequency of collisions • The energy of each collision • (what fraction of collisions will succeed?) The rate law for a 2nd order reaction is …. r = k [A] [B] Increasing the [A] will …. a) increase the frequency of collisions b) increase the energy of collisions c) decrease the energy of collisions A higher value of k means …. a) the frequency of collisions is greater b) the activation energy is greater c) the activation energy is lower Increasing T will …. a) increase the frequency of collisions. b) increase the energy of each collision. c) increase k d) all of the above

  25. ‡ A + B AB C + D + Temperature Dependence of the Rate Constant (Arrhenius equation) Thermal energy available Collision energy required rate constant Arrhenius constant frequency factor The rate of a reaction depends upon …. 1) The frequency of collisions 2) The energy of each collision (what fraction of collisions will succeed?)

  26. ‡ A + B AB C + D + Temperature Dependence of the Rate Constant (Arrhenius equation) ln k = ln (A • e(-Ea/RT)) ln k = ln A + ln (e(-Ea/RT)) ln k = ln A – Ea/RT or … ln k = -Ea/R • 1/T + ln A y m x b

  27. cyclopropaneÞpropene ln k 1/T ln k = -Ea/R  1/T + ln A slope = (rise/run) = D(ln k)/D(1/T) = -Ea/R ln (k2/k1) = - Ea/R (1/T2 – 1/T1) At what T will rate double? compared to 600K What is Ea using only 600 and 650K data?

  28. cyclopropaneÞpropene ln k = -Ea/R •1/T + lnA What is Ea using only 600 and 650K data? slope = (rise/run) = D(ln k)/D(1/T) = (ln k2 – ln k1)/(1/T2 – 1/T1) = -Ea/R = ln (k2/k1)/(1/T2 – 1/T1) = - Ea/R Ea = -8.314 • ln (2.19 x 10-7/3.30 x 10-9)/(1/650 – 1/600) = 272,000 J or 272 kJ At what T will rate double (compared to 600K)? Hint: (if rate doubles, k doubles) a) 608K b) 616K c) 670K d) 590k Ea = 272,000 = -8.314 • ln(2)/(1/T – 1/600) T = 608K A rule of thumb suggests that the rate of a reaction doubles for every 10°C increase in T.

  29. ABc‡ Transition state Transition state Eact(with catalyst added) Eact Energy (G) A + B DG˚ C + D Catalyst – Species that is part of the reaction but regenerated - reusable functions by lowering the activation energy (c). A catalyst increases the rate by which of the following: a) increasing T b) increasing Ea c) decreasing Ea d) increasing collisions A + B  C + D + free energy (G) c AB‡ + c + c

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