- Tfnh oxi boa cua SO 2: SO 2+ 2H2S^ 3S +2H2O Tinh khft cua SO2 :
TOC DO PHAN LTNG VA CAN BANG HOA HOC
VA CAN BANG HOA HOC
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TOO DO P H A N Q N G H O A HOC
7.1 a) Ndng do chd't B la 0,98 mol/l.
At
n 78 — 0 80
Tinh theo ndng dd chd't A : v = —-——-^— = 0,001 (mol//.ph).
Tinh theo ndng dd chd't B : v = - ^ ^ ^ ^ ^ ^ = 0,001 (mol//.ph).
Nhu vdy tfnh theo ndng dd chd't A hay chat B, tdc dd phan ftng cung vdn nhu nhau.
7.2 a) Ndng do chd't B la 3,99 mol/l. Ndng do chd't C la 0,03 mol/l.
b) Tdc dd trung binh cua phan ftng tinh theo ndng do chat A trong khoang thdi gian dd la :
L O O - 1 , 0 1
V =
20 = 0,0005 (mol//.ph).
7.3 Gia sft ndng do ban ddu cua chat A la a mol/l, ciia chdt B la b mol/l, tdc
2
dd ban ddu cfta phan ftng la : Vj = k.a.b .
a) Khi ndng dd chat B tang 3 ldn va ndng do chat A khdng ddi thi tdc dd phan ftng la : V2 = k.a.(3b)^ = 9kab
V 9kab^
Nhu vdy • — = TT = 9, nghia la tdc dd phan ftng tang 9 ldn.
Vj kab^
b) Khi dp sud't cua he tang 2 ldn thi ndng do mdi chd't diu tang 2 ldn ; tdc dd phan ftng luc dd la :
V 8kab
Nhu vdy : — = ^ = 8, nghia la tdc dd phan iing tang 8 ldn. 7.4 B.
7.5 Ndng dd ban ddu cfta N2 la 2,5 mol/l; cfta H2 la 6 mol/l.
75-25
7.6 Tdc dd phan ftng tdng 2 10 = 2^ = 32 (ldn).
7.7 Phan ftng phai dugc thgc hien d 70''C.
70-40
7.8 Tdc dd {)han ftng giam 4 10 = 4^ = 64 (ldn). 7.9 A. Ndng dd oxi tang len.
B. Chd't xuc tac. C. Nhiet dd. D. Kfch thudc hat.
7.10 a) Tft thf nghidm (1) va (2), ta thd'y khi ndng dd CO giam 10 ldn (ndng dd
CI2 khdng ddi), td'e dd phan ftng cung giam 10 ldn. Nhu vdy, tdc dd phan ftng ti le thudn vdi ndng dd eua CO. Tft thf nghidm (3) vd (4), ta thdy khi ndng dd CI2 giam 100 ldn (ndng dd CO khdng ddi), td'e dd phan ftng cung giam 100 ldn. Nhu vdy, tdc dd phan ftng ti le thudn vdi ndng dd cua CI2. Tdc dd phan ftng ti Id thudn vdi ndng dd eua CO va cfta CI2.
Vivdy :v = k[C0].[Cl2]
b) Gid tri trung binh eua k = 1,31.10" .
7.11*Khi nfyitt dd tdng them 40 - 20 = 20 (°C), thdi gian phan ftng giam di
27
— = 9 ldn, nghia la tdc dd phan ftng tdng 9 ldn. Nhu vdy, mdi khi nhiet dd tdng thdm 10°C, tdc dd phan ftng da tang 3 ldn (3^ = 9).
Nhu vdy khi tang nhiet dd them 55 - 40 = 15 (°C) thi td'e dd phan ftng tang 15
3io=3^'5ldn.
Do dd, thdi gian dl hod tan hoan todn mdu kem-dd d 55°C la : ^•^^•=34,64 (gidy).
3I.5 152 152
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CAN BANG HOA HOC
7.12. Cdc phdn ftng a, b vd c Id phan ftng mdt ehilu. Phan ftng d la phan ftng
thudn nghich. Br2 (/) + H2O (/) < » HBr (dd) + HBrO (dd). 7.13. D. 7.14. A. 7.15 C. 7.16 B. l2 m A<2 7.11. K= ^ ^ " 3 ] - ^ (0,4)- ^ ^ [N2].[H2f 0,01.(2,0)3
Ndng dd ban ddu eua nito la 0,21 mol/l, cua hidro la 2,6 mol/l.
7.18. Lugng chd't C khi cdn bdng la 1,5 mol. Do dd, lugng chd't D la 1,5 mol ;
lugng chd't A = lugng chd't B = 0,5 mol. Hdng sd cdn bdng K bdng 9.
7.19. Ddt ndng dd cdn bang cua CO2 la x mol/l thi ndng do cdn bdng cua H2 la
X mol/l, ciia CO la (0,1 - x) mol/l va cua H2O la (0,4 - x) mol/l.
„ [C02].[H2]_ x^ _ . x - 0 n 8 [CO].[H20] " (0,1 - x).(0,4 - X) - ' "^ '^ - "'"^•
Vdy ndng dd cdn bdng cfta CO2 la 0,08 mol/l; cua H2 la 0,08 mol/l; cua CO la 0,02 mol/l vd cfta H2O la 0,32 mol/l.
7.20. Phan ftng b cd hieu sud't cao nhd't va phan ftng c cd hieu sud't thdp nhd't. 7.21. a) Cdn bdng chuyin dich sang trai, 7.21. a) Cdn bdng chuyin dich sang trai,
b) Cdn bdng chuyin dich sang phai, c) Cdn bdng khdng thay ddi,
d) Cdn bdng chuyin dich sang phai, d) Cdn bdng chuyin dich sang phai.
7.22. Khdng. Sg thay ddi dp sud't chi gdy ra su chuyin dich cdn bdng dd'i vdi
cdc phan img thudn nghich cd mat chdt khf va sd mol khf d hai vd eua phuang trinh khdc nhau.
7.23. a) Sai; b) Dung ; c) Dftng ; d) Dung.
7.24. ' ^ .Phdn dng .Phdn dng 1 2 3 4
Cdn bang chuyen dich the'ndo khi a) tdng dp sudt sang trai sang phai sang phai sang trdi b) tdng nhiet dd sang phai sang trdi sang trai sang phai
7.25*. Ap sud't tdng len vi hai If do : nhiet do tdng va sd mol khi tdng. Nhidt dd tft 0°C (tfte 273K) tdng len 546°C (tfte 819K) nghia la nhiet dd tuyet dd'i
3,3
tdng 3 ldn. Nhu vdy sd mol khf chi tdng : --^ = 1,1 ldn.
Gia sft ban ddu trong binh chfta n mol khf NH3 va x mol chd't dd da hi phdn buy : 2NHq 13 <- n n - X N2 0 , 0,5x +
Sdmol khi ban ddu Sdmol khi liic cdn bdng
Tdng sd mol khf luc cdn bang : n - x + 0,5x + l,5x = l,ln x = 0 , l n Ndng do luc cdn bdng : [N2] = ^ i M i l E ^ o,05 (mol//).
[H2] = l i ^ ^ = 0,15 (mol//). [NH3] = ILzAl!! = 0,9 (mol//). Luu y : — = ndng do ban ddu cua NH3 ^ — = 1.
u- -' - K^ V [N2].[H2f 0,05.(0,15)^ oAQin-
Hang so can bang : K = ^ = ^—^^— = 2,08.10
3H2 0 l,5x
[NH3r (0,9)^ 154
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