LN 8 IDLE MIND SOLUTIONS 1 T x 10 4

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1 LN 8 IDLE MIND SOLUTIONS. k Ae RT k 2 3k Ae RT 2 3 ln 3 R T T 2 e R T T 2 R x ln x kjmole 2. (a) k Ae RT.7 x 0 4 xe 2.5x05 8.3x s (b) Requires knowledge of k 600 : k x 0 4 e 2.5x x c/c o = e kt = e 0.84x600 =.3 x c = 0: reaction complete to 00% (c) k 3xk 3 x 28.8 Ae RT x ln 86.4 ln(.7 x 0 4 ) RT x T x R[ln86.4 ln(.7 x 0 4 )] T x = 063K = 790 o C

2 3. c o c c = c o e kt.4 ek x65 ln.4 = k x 65 k = ln.4/65 = 5.8 x 0 3 s e 5.8 x 03 x M 4. (a) A plot of ln c vs time places all points on a straight line, indicative for a first order reaction. c ln c t ln c k t (b) We may obtain the rate constant from points () and (4). k x 0 2 m k = 4.4 x 0 2 m (You may check the correctness of the results by applying the exponential form of the rate equation, c/c o = e kt.)

3 5. D D o e RT x ln D o D RT x T x Rxln D o D 3.5 x x ln.7 5x0 563K 290 o C 6. c o /c = e kt (c = 0.06 x c o ) c o ln kt 0.06 c x o ln 0.06 = kt x t x ln 0.06 ln 2 t 2 ln years 7. If 28% Ag* decays in.52 hours, then c = (0.72 c o ) c o c o e kt ln 0.72 = kt k ln (hrs ) t hours (2 Ag *)

4 8. Decay is a st order reaction: dc dt kc 2.3 x 0 6 ats k 3.5 x 06 g x 6.02 x g k x 8.85 x 0 5 k 2.3 x x x 00 s 9. ln 88 ln kt k x 02 (h ) t 2 ln 2 k hours 5.34 x T( C) /T(/ K) k ln k x x x x x x x x ln k /000T (continued)

5 0. Continued. Notice: the first and fourth points are on the slope. (Knowing this) we could forego the plotting: 2.44 x x 0 3 Ae RT Ae RT 2 e R T T 2 R T T 2 ln(2.44 x 0 3 ) ln(3.5 x 0 3 ) R (2.47 x 0 3 ) (2.3 x 0 3 ) = 33 kj/mole. If the reaction is first order, a plot of ln c vs time will yield a straight line. The shape will give k, which allows calculation of the time required for 70% completion. c ln c t 5.00x x x x x ln c The plot indicates a first order reaction. However, one data point is erratic (presumably an error in recording since the last point is again on the straight line). We may take points () and (5) for the determination of the slope. (continued) k t

6 . Continued. k x 0 2 k = 4.4 x 0 2 min t 2 ln 2 k 5.8 min 70% completion of the reaction means: c/c o = 0.3; accordingly: ln c c o kt x t x ln 0.3 k 27.4 min 2. (a) Knowing the half life, we can determine the rate (decay) constant, k: k ln 2 t x 0 2 min According to C = C o e kt : C C o e 3.46 x 02 x3x60.97 x 0 3 C =.97 x 0 3 x C o = 0.2% of original amount of C remains after 3 hours (b) Since the half life is independent of concentration (for first order reactions), the answer must be the same as in (a):.97 x % remains after 3 hours

7 3. In this question we are given a concentration of product (or rather, a volume proportional to a product). We want to convert the information given into quantities proportional to the original reactant (to be able to treat the problem in the familiar manner). We recognize that the volume of alkali for t = (29.7 ml) should be proportional to C o, the concentration of the reactant at t = 0. The volume at t = 27 min, proportional to the remaining reactant concentration, is ( ) ml and the volume at t = 60 min, proportional to the remaining reactant concentration, is ( ) ml. We can now plot the data in terms proportional to reactant concentration vs time. time (min) V (ml) prop. to product V (ml) prop. to reactant 29.7 ( ) ( ) 0 ln V (prop. to reactant) lnc o t(min) From the graph, the reaction is shown to be first order. For the slope determination, we may use t = 0 and t = 60 min. k t = 3.47 x 0 2 min ln 2 20 min 3.47 x k k 2.02 e R T 2 T ln.02 R T 2 T R T T 2 T xt 2 R T T 2 ln.02 R T T 2 T RxT2 x ln x 0 K 0.K The required temperature stability is beyond the capabilities of conventional thermostats.

8 5. () ax = by x b/a x b a y c a y > (2) x = Ae y/rt ln x RT y ln A ln x /(RT) y > (3) xy = ax + by xy ax = by x(y a) = by y a by x b a by x /x a/b /y > x a by b 6. k ln 2 t a 5.54 x 02 a c c o e kt 0.4 C o C o ln 0.4 = kt t ln years 5.54 x 02

9 7. D D e R Rln kjmole D 00 D 200 e R D 200o C 05 e x 0 5 cm 2 sec 8. T = 600K /T = 6.25 x 0 4 From the data in the problem: D 600K (In > Si) = 8 x 0 2 cm 2 /sec C C 2 erfc x 2 Dt C = /2 C 2 X = 0 3 cm erfc x Dt erf x Dt erf (Z) = 0.5 ; from tables, Z = x Dt x2 4Dt t (03 ) x 4D.37 x 05 sec.59 days

10 9. (a) For the first order reactions: c o c o /2 ln c straight line (b) at 25 o C, k = hr > t /2 = constant (time) t 25 C 2 ln c 3 4 k = hr t (hr) at 50 o C, k = min 50 C 2 ln c 3 4 k = min t (min) (c) t 2 ln 2 k ln min

11 9. Continued. (d) k A exp E a RT We need to determine A (and E a ): k 25 k 50 ln k 25 k 50 exp E a R T 25 E a E a R T 25 T 50 Rln k 25 k 50 T 25 T 50 T ln8.88 x x 0 2 k hr x = 8.88 x 0 4 min hr 60 min x 03 Jmole At 25 o C: k = A exp (E a /RT) A k exp RT (8.88 x 04 ) exp 8.77 x x 298 A = 2.22 x 0 2 m So, at 70 o C (= 343K): k A exp E a k = 9.28 x 0 2 min RT (2.22 x 02 ) exp 8.77 x x 343 and (finally!): t 2 ln 2 k ln x min min (continued)

12 9. (e) ln c c o c = c o e kt kt t ln c c o k If reaction is 42% complete, then 58% of reactants remain. Therefore, c = 0.58 c o. ln0.58 c o c o 0.22 hr 0 hr, 3 min, 2. sec hr 20. t /2 = (ln 2)/k = 590 years k = 4.36 x 0 4 years c/c o = e kt c/c o = exp (4.36 x 0 4 )(5) = Hence, fraction decayed = = E a = 50 x 0 3 J/mole k 25 k 50 k A exp E a RT exp E a R k 25 k ; k 0 o k 25 o If we consider a half life for tires, t /2 = (ln 2)/k: at 0 o C, t 2 0 o ln 2 t 0 ln k t 2 25 o at 25 o C, t 2 25 o ln 2 k 25 We have increased the life of the rubber by a factor of 2.92!

13 22. (a) Plot ln c vs t and /c vs t. If ln c vs t yields a straight line, then the reaction is first order and k equals the slope of the line. If /c vs t yields a straight line, then the reaction is second order and k equals the slope of the line. (b) k x 0 4 sec k = 6. x 0 4 sec See following graph. Realize: these data are same data as that of problem ln c st order!! t (sec)

14 23. For a first order reaction we have: k ln 2 t 2 and k ln 2 t 2 Ae RT (a) k Ae RT ln A ln k RT T R ln A k R ln A ln 2 t 2 ln k ln A RT ln A k RT 04 x ln 5x = 347 o K = o C (b) T R ln A ln 2 t 2 04 x ln 5x03 ln 2 30 x 24 x 3600 = 268 o K = 5.0 o C 24. k k 3 3 Ae RT Ae RT 2 ln 3 R T T 2 R e T T 2 ln 3 ln 3 R T T 2 ln 3 x R T T 2 (a) ln 3 x x 03 Jmole

15 24. (b) ln 3 x x 03 Jmole 25. SO 2 Cl 2 SO 2 + Cl 2 : k = 2.2 x 0 5 sec (at T = 593 o K) ln C C o kt C C o e kt e 2.2 x 05 x 2 x or: 4.7% has decomposed 26. k 2x x 03 e 8.34 x x 0 3 sec t 2 ln x sec at 700o K 7.72 min at 700 o K 27. At T x, k A k B or : 0 5 e 25x03 RT x x 03 e RT x ln x 03 RT x ln x 03 RT x ln RT x (25 x 0 3 ) (83.6 x 0 3 ) T x (25 x 03 ) (83.6 x 0 3 ) Rxln x o K 808 o C 8.34 x ln

16 28. k ln 2 t 2 k k 300o C k 2 k 500o C k k 2 ln x 03 sec (T 573 o K) ln x 02 sec (T o K) E 6.9 x e R a T T x 02 ln 0.97 E a R T T 2 E a ln 0.97 x R T T x 0 3 Jmole 29. (a) Plot ln c vs t and /c vs t. If ln c vs t yields a straight line, then the reaction is first order and k equals the slope of the line. If /c vs t yields a straight line, then the reaction is second order and k equals the slope of the line. ln c k According to the graph, the reaction is st order!! t (sec) k x 0 4 sec (b) k k = 6. x 0 4 sec (continued)

17 29. Continued. (c) C C o e kt (for a first order reaction) C 2.33 x e 6. x 04 x 4500 C = 0.5 moles/ltr 30. In this question the rate of disintegration (decay) is given ( dc/dt =.4 x 0 5 /min). To determine t /2 (or k), we look at the rate equation: dc kc dt We need to know the concentration (c) at t = 0; it is given through the data provided: µg = 0 6 g Pu 239 The isotope Pu 239 has the atomic weight 239: Thus: g 0 6 gx 6.02 x 023 atoms mole 239 g mole 2.52 x 0 5 atoms dc dt kc and k dc dt c.4 x x x 0 min t 2 = 2.92 x 0 5 years , 700 years 2.92 x 05

18 3. (a) A look at the data shows that after three minutes the concentration drops from 6.5 to 3.25 m moles/ltr, and after 6.5 min c has decreased to.45 m moles a typical first order reaction behavior. (We could forego plotting the data and take t /2 = 3 min, test the validity of this value, and calculate the requested data.) time conc ln c (min) ln c 7 (b) k = min t (min) k min (c) t 2 ln 2 k min (d) c = c o e kt = 6.5 x 0 3 e x = x 0 3 m moles/ltr 32. First order reaction: k ln 2 t 2 t /2 = 30d = 30 x 24 x 3600 s = x 0 6 s k ln x x07 s k Ae ERT x ; lnk ln A E RT x T x E R(ln A ln k) 324K

19 33. We may solve the problem graphically: plot ln k t vs /T, which yields E/R as the slope which will permit us to determine the pre exponential, A. Alternately we can take T and T 4 (which are on the curve) and solve numerically: ln k ln A E RT ln k 2 ln A E RT 2 ln k k 2 E R T T 2 for T = 273K T 2 = 38K E ln k k 2 xr T T 2 E = 9 x 0 4 J/mole A = k x e E/RT =.8 x 0 2 s 34. ln k k 2 ln 2 E R T T 2 E ln 0.5 x R E = 53.6 kj/mole for T = 300K T 2 = 30K 35. (a) c co e kt c e =.x0 2 moles/l (continued)

20 35. (b) e t In.65 = 4.8x0 4 x t t = In.65 = 043 sec (a) In c In c /c time (min) The graph indicates, that the In c vs time plot yields almost a straight line, the /c plot does yield a straight line. This identifies the reaction as a 2nd order reaction. (b) The rate constant (k) is given by the slope of the /c vs time plot: k mole min (c) The half life of a second order reaction (given in class) can be simply obtained from the second order rate equation. It is: 2 k C o = 2.5 min

21 37. k A e RT E k T A ert e k T2 A e E RT 2 E R T In 3 In 3 R T T 2 T 2 kt 3 kt 3 R In kj mole 38. (a) k 350 k 400 e R x x [In 74.2] R kj mole (b) First we find A, then we determine (k) A e R400 A e R k e sec

THEORY OF REACTION RATES

3.091 Introduction to Solid State Chemistry Lecture Notes No. 8 THEORY OF REACTION RATES * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Sources for Further