Bài tập Home work II

Student’s name: DO THI KIM HUE Class: K52 Advanced Chemistry ********** Home work II Problem 7.1 a. SSA states that d[O]/dt = 0 => r 1 = r 2  r = 3 []dO dt  3 []dO dt  = r = 2r 1 =2k 1 [O 3 ][M] or 2 []dO dt  = 3k 1 [O 3 ][M] b. Cannot assume Quasi-Equilibrated Adsorption (No Langmuir isotherm) r =   2 d CO dt ; at steady state, so r 1 = r 2 , Site Balance: L =   SO + [S]           1 1 2 2 2 2 [] k N O S k CO S O S O N O [] k S k CO                  1 2 2 2 1 2 2 Lk k CO N O r k CO S O k N O k CO     c. Step 1 is a quasi equilibrated, but step 2 and 3 isn’t, so no Langmuir isotherm, then we use SSA to determine surface concentration. We also assume single-site adsorption and C 2 H 2(ad) is MARI. We assume: d[CH 2(ad) ]/dt = 0 → d[CH 2 ]/dt = 2k 1 [C 2 H 2 ][H 2 ] - k 2 [CH 2(ad) ][H 2 ] = 0 → [CH 2(ad) ] = 2k 1 [C 2 H 2 ] /k 2 → 2k 1 = k 2 (1) r = d[CH 4 ]/dt = k 2 [CH 2(ad) ][H 2 ] (2) Site balance for C 2 H 2(ad) is MARI: L = [C 2 H 2(ad) ] + [S] or 1 = θ C2H2 + θ v → [S] = L - [C 2 H 2(ad) ] (3) From eq. QE 1: K = [CH 2(ad )][H2]2 [C2H6][S] → [C 2 H 2(ad) ] =  C2H6][S] [H2]2 (4) Replace (3) into (4), we obtain: [C 2 H 2(ad) ] =  [C2H6] [H2]2 ( L - [C 2 H 2(ad) ]) → [C 2 H 2(ad) ] =  [C2H6] (1+[26/[2]2])  H2  2 (5) Replace (5) into (2), we have the rate: r = d[CH 4 ]/dt = k 2 [CH 2(ad) ][H 2 ] = k2 [C2H6]   26  2   +  H2  Problem 7.2   3 [ ] [ ] r k B S d A d B dt dt      12 [ ] [ ] K , K [ ][ ] [ ] A S B S A S A S    Site balance: L=[S] +[A-S] + [B-S] a) If [B-S] is the MARI, then L= [s] + [B-S] and                2 1 2 12 L S K A S S K K A S S 1 K K A L         r= k 3 K 2 [A-S]= [K 1 K 2 k 3 [A][S] =       3 1 2 12 Lk K K A 1 K K A b) If [A-S] is the MARI, then L= [s] + [A-S] and L= [S] + K 1 [A][S] ] =>       1 S 1 K A L   r= k 3 K 2 [A-S]= [K 1 K 2 k 3 [A][S] =       3 1 2 1 Lk K K A 1 K A the mathematical forms are identical. Problem 7.4 r = -       2 2 2 d H O d H d CO dt dt dt  r = k 3 [H 2 O*] = k 4 [O*][CO*] and K 1 =     co CO* P* L = [*] + [CO*] + [H 2 O] + [O*] (from site balance) Steady – state approximations: On H 2 O* : k 2 P H2O [*]= k 3 [H 2 O*] On O* : k 3 [H 2 O] = k 4 [O*][CO+]         2 3 2 22 34 H O* ( )P & O* H O* / CO* k HO k k k            3 2 1 CO H2O 2 34 L K P * P * * H O / CO* k k kk     the last term can be negligible, so   2 2 1 CO H O 3 L K P P 1 * k k             22 2 3 2 2 H O 32 3 1 CO H O . Lk P r k H O* * 1 K P k'P HO k k P k     Problem 7.5 a. A + S 1 1 k k    A S reversible RDS (For an ideal gas) B + S 2 K   B S A S + B S 3 K   C S + D S C S 4 K   C + S 5 K D S D S    A + B   C + D The rate of reaction r = k 1 P A [S] – k -1 [A S] From site balance: L = [S] + [A S] + [B S] + [C S] + [D S] Moreover     2 4 5 [ ] [ ] [] [] B C D B S K P S PS CS K PS DS K       and        3 C S D S AS K B S      K = CD AB PP PP = K 1 K 2 K 3 K 4 K 5 [A S] =   CD 5 4 3 2 B P P S K K K- K P So we can obtain: L = [S] CD 2 3 4 5 B 2 B C D 45 1 P P 11 K K K K P K P P P KK     Thus r =   1 C D 1A 2 3 4 5 B Lk P P Lk P K K K K P S      = 1 C D 1A 2 3 4 5 B CD 2 3 4 5 B 2 B C D 45 Lk P P Lk P K K K K P 1 P P 11 K K K K P K P P P KK          b. A + S 1 K   A S B + S 2 K   B S A S + B S 3 K   C S + D S C S 4 K   C + S 5 5 k k D S D S      Reversible RDS A + B   C + D The rate of the reaction: r = k 5 [D S] – k -5 P D [S] Using site balance: L = [A S] + [B S] + [C S] + [D S] + [S], we also have some relations:     2 1 4 [ ] [ ] [] [] B A D B S K P S A S K P S PS DS K       and       3 C S D S K A S B S    So [D S] =        3 1 2 3 4 A B C K K K K P P S P K A S B S CS    L =   C 1 2 3 4 A B 1 A 2 B 4C P K K K K P P 1 K P K P KP S        Thus r = 5 1 2 3 4 A B 5D C C 1 2 3 4 A B 1 A 2 B 4C Lk- K K K K P P – Lk P P P K K K K P P 1 K P K P KP      In this case, K = K 1 K 2 K 3 K 4 K 5 and K 1 = K A , K 2 = K B , 4 1 K = K C , 5 1 K = K D = 5 5 k k  , K 1 K 2 K 3 K 4 = 5 55 kK K Kk   Substituting on the rate equation, we get: r = 5 A B 5D C AB 1 A 2 B C Lk- KP P – Lk P P KK P P 1 K P K P P K P D CC      c. A 2 + 2S 1 1 k k    2A S reversible RDS 2[B + S 2 K   B S] 2[A S + B S 3 K   C S + D S] 2[ C S 4 K   C + S] A 2 + 2B   2C The rate equation is: r = k 1 P A2 [S] 2 – k -1 [A S] 2 , Applying site balance: L = [A S] + [B S] + [C S] + [S]   2 4 [ ] [ ] [] B C B S K P S PS CS K     and       3 C S S K A S B S     [A S] =        3 * * * C S S AS K B S  , K = K 1 K 2 2 K 3 2 K 4 2 or K 1/2 = K 1 1/2 K 2 K 3 K 4 Then, [A S] =   2 3 4 B K K K P C PS . And L =   2 2 3 4 4 1 CC B B PP S K P K K K P K       So we can calculate that: r =   2 2 22 11 2 3 4 B 2 2 2 3 4 4 22 K K K P 1 C A CC B B PS ZZ L k P L k LL PP KP K K K P K           with 2 Z L = site-pair probability and Z = coordination number Moreover, K 2 K 3 K 4 = K 1/2 /K 1 1/2 then: r = 2 2 ' 2 1 2 2 2 3 4 4 1 C A B CC B B P Lk P KP PP KP K K K P K           d. A 2 + 2S 1 K   2A S 2[B + S 2 K   B S] 2[A S + B S 3 K   C S + D S] 4 4 2 k k C S C S        reversible RDS A 2 + 2B 2C The rate equation is: r = k 4 [C S] – k -4 P C [S] Also we have: [A S] 2 =   2 2 1A K P S ; [B S] = K 2 P B [S], and       3 C S S K A S B S    [C S] =       2 11 22 3 1 2 3 [A ] K AB S B S K K K P P S S   From site balance: L = [S] + [A S] + [B S] + [C S] =     1/2 1/2 1/2 1/2 1 A2 2 B 1 2 3 A2 B 1 K P K P K K K P PS    Moreover, we also have: K = K 1 K 2 2 K 3 2 K 4 2 K 1 1/2 2 1 2 3 4 K K K K K  So 2 1/2 1/2 2 4 2 3 A B 4 C 1/2 1/2 1/2 1/2 1 A2 2 B 1 2 3 A2 B L k K K K P P L k P 22 1 K P K P K K K P P ZZ LL r       With K 1 = K A2 , K 2 = K B , 4 1 K = K C  K = K A K B 2 K 3 2 K 4 2 So   1/2 1/2 4 A2 B C 1/2 1/2 1/2 A2 A2 B B C A2 B L k K P P – P 2 1 K P K P KK P P Z r      Problem 7.6 a. The rate equation is: r = 2 4 2 26 3 [] C H H d C H Lk P dt   2 4 2 25 C H H 2 C H H P P k    Applying steady-state approximation on total surface carbon atoms, we obtain another equation for 25 CH  , dθ C(total) /dt = 0. Additionally, because steps 2 and 4 are very rapid:  2 6 2 5 2 4 2 C(total) 1 C H 1 C H 3 C H H d k P – Lk – Lk P 0 dt     2 6 2 4 2 5 1 C H 3 C H H2 1 C H H k P – Lk P Lk        2 5 2 4 25 1 C H 3 C H H2 C H 1 k P – Lk P Lk H        2 6 2 4 2 24 2 1 C H 3 C H 2 CH 1 H H H H k P Lk P K P Lk          = 2 6 2 2 4 2 1 C H 1 H 2 3 C H 1 K k P / Lk P – LK k / Lk   26 24 2 1 2 C H 23 CH 1 1 H k K P Kk 1 k Lk P             2 6 2 24 12 C H H 1 CH 2 3 1 kK P / P Lk 1 K k k         and 2 2 6 2 26 3 H 1 2 C H 1H 1 2 3 C H C2H6 2 3 2 3 11 Lk P k K P Lk P k K k P r k’P K k K k 1 1 kk                      but θ C2H4 can be rewritten to give:     2 6 2 25 1 C H H 3 C H 1 2 3 k P / P k L 1 k / K k        so that 26 1 C H 1 23 k P r k 1 Kk       b. C 2 H 6 + 2S 1 1 k k    C 2 H 5 S + H S C 2 H 5 S + H S 2 K   C 2 H 4 S + S + H 2 C 2 H 4 S + H 2 + S 3 k  2CH 3 S 4 3 2 4 2 2 2 K CH S H CH S    C 2 H 6 + H 2  2CH 4 The rate of reaction can be defined as: r =   26 d C H dt  = k 3 [C 2 H 4 S][S] 2 H P K 2 = [C 2 H 4 S][S] 2 H P / [C 2 H 5 S][H S] Steady-state approximation on all surface C atoms gives: k 1 26 CH P [S] 2 – k -1 [C 2 H 5 S][H S] – k 3 [C 2 H 4 S][S] 2 H P = 0 k 1 P C2H6 [S] 2 – k 3 [C 2 H 4 S][S] 2 H P = k -1 [C 2 H 5 S][H S] and [C 2 H 5 S] =        26 2 2 1 C H 3 2 4 11 k P S k [H ] H k C H S S P S k H S     [C 2 H 4 S] =           26 2 2 1 3 2 4 2 1 CH H k P S k C H S K H S S P k H S     26 2 1 2 C H 23 24 1 1H k K P Kk [C H ] 1 k kP S          2 6 2 12 C H H 1 24 23 1 kK ( )P / P k [C H ] Kk 1 k S      Now, site balance to get [S] is: L = [S] + [H S] +[C 2 H 5 S] + [C 2 H 4 S] + [CH 3 S] (a) is obtained only if [S]~ L, it means the surface is essentially free of all adsorbed species or θ H , θ C2H5 , θ C2H4 , θ CH3 << 1. This is a questionable assumption. Problem 7.7 Plotting ln rate vs. lnP i using a power rate law gives the following reaction orders: T(K) Reaction Order N 2 O O 2 N 2 623 0.08 -0.31 0 653 0.24 -0.12 0 673 0.31 -0.07 0 The simplest L-H model would be for unimolecular decomposition: (1) 2[N 2 O + * 2 NO K   N 2 O *] (2) 2[N 2 O* k  N 2 + O*] (3) 2O* 2 1/ O K   O 2 +2* 2N 2 O ==> 2N 2 + O 2 r m = 1 m dN N 2 dt = 1 m   2   = k[N 2 O*] From(1) : K N2O = [N 2 O  ]   2  [] , so [N 2 O*] = K N2O P N2O [*] From(3) : K O2 = [] 2   2 [] 2 , so [O*] = K O2 1/2 P O2 1/2 [*] Site balance gives: L=[*] +[N 2 O*] +[O*] thus L =[*] +  2    2  [*] +   2 1/2   2 1/2 [*] , and [*] =  (1 +   2    2  +   2 1/2   2 1/2 ) , consequently, r = k K N2O P N2O [*] =   2    2  1+   2    2  +   2 1/2   2 1/2 Arrhenius plots of the fitting parameters listed in Table 2 provide the following values : For K N2O :∆H ad o = -17 kcal mole -1 and ∆S ad o - 21 cal mole -1 K -1 (e.u.) For K O2 :∆H ad o = -25 kcal 1 and ∆S ad o - 35 e.u. For k : E RDS = 57 kcal mole -1 The enthalpy and entropy values for adsorption fulfill all the guidelines in Table 6.9, thus they are consistent. From either a linear extrapolation of the high-P portions of the two isotherms in Figure 1 or using the difference between the two at 100 Torr CO pressure, the irreversible uptake is 580 µmole CO g cat -1 . The dispersion of Cu is : D Cu = Cu s / Cu tot , and with CO ad /Cu s = 1, D Cu = 580 µmole Cu s g cat 1  0.0456 g Cu g cat 1   mole Cu 63.55 g Cu  (10 6 µmole mole = 0.81 Under differential reaction conditions, P O2 O and can be ignored; therefore, an easy way is to choose a known differential rate and correct for temperature , for example :  823   673  =  823  673 =  36200 cal /mole /(1.987cal /mole .K )(823K )  36200 cal /mole /(1.987cal /mole .K)(673K ) = 139 thus TOF 823K =  823    = (12.6 µmole /s.g) (13 atm 1)(0.0666 atm )(139) [1+ (13 atm 1)(0.0666 atm )](580 µmole Cu s/g) = 1.4 s - Problem 7.8 Step 2 defines the rate: r m = 1 m d N 2 dt = 1 m (- d N 2 dt ) = k[N 2 O*] Step 1 gives : K N2O = [N2O] PN2O [] , so [N 2 O*] = K N2O P N2O [*] Assuming all surface species are included, a site balance gives L =[*] +[N 2 O*] + [O*] To remove the unknown [O*], the SSA must be used: d [O ] dt = k[N 2 O*] + k -1 P O2 [*] 2 – k 1 [O*] 2 = 0 And [O*] =( k[N2O] + k 1 PO2 [] 2 k1 ) 1 2 . Student’s name: DO THI KIM HUE Class: K52 Advanced Chemistry ********** Home work II Problem 7.1 a. SSA states that d[O]/dt = 0 => r 1 = r 2  r = 3 [ ]dO dt  3 [ ]dO dt  = r =. atoms:  [N*] >> [*] and [N*] ~ L So 2 22 2 Z r k L ZLk L  In this case, over a La 2 O 3 catalyst at 923K, this reaction exhibited a reaction order on NO of about 1.2 with no O 2 . B 5D C C 1 2 3 4 A B 1 A 2 B 4C Lk- K K K K P P – Lk P P P K K K K P P 1 K P K P KP      In this case, K = K 1 K 2 K 3 K 4 K 5 and K 1 = K A , K 2 = K B , 4 1 K = K C , 5 1 K = K D