作者shujeff (好累)
站內NTUCE-96
標題[情報]普化期中解答
時間Fri Apr 2 01:03:03 2004
不想看的...
想直接找我印的... 可以直接出去了....XD
1.
[Ag]o = 4.29 * 10^-4 M [(S2O3)2-]o = 2.86 M
Ag+ + 2 S2O3 2- --> Ag((S2O3)2)3-
初 4.29 * 10^-4 2.86 0
末 ~0 ~2.86 4.29 * 10^-4
所以
[Ag((S2O3)2)3-] = 4.29 * 10^-4 M
[Ag((S2O3)2)3-]
又 k2 = -------------------------
[Ag(S2O3)-] [(S2O3)2-]
所以
[Ag(S2O3)-] = 3.8 * 10^-9 M
從 K1 可以得到
[Ag+]= 1.8 * 10^-18 M
2.
滴定Asprin得NaOH = 50 * 0.5 - 31.92 * 0.289 = 15.8 (m mole)
所以 15.8
------ * 180.2 = 1420 (mg) = 1.42 (g)
2
1.42 / 1.427 = 0.995 =
99.5%
3.
Cr3+ + H2EDTA --> CrEDTA- + 2 H+
初 0.001 0.05 0 10^-6
末 0 0.049 0.001 10^-6
平衡 x 0.049 + x 0.001 - x 10^-6
(0.01 - x ) * (10^-6)^2
Kf = --------------------------- 其中..-x +x 可忽略 因x太小
( 0.049 + x ) * x
x = [Cr3+] =
2 * 10^-37 M
4.
NH4+ <--> NH3 + H+
CN- + H2O <--> HCN + OH-
+) 2 H2O <--> H3O+ + OH-
-----------------------------------
NH4+ + CN- <--> HCN + NH3
[HCN] [NH3] [H+] [NH3] [HCN] 1
K = --------------- = ------------ * ------------ = Ka(NH4+) * --------
[NH4+] [CN-] [NH4+] [H+] [CN-] Ka(HCN)
5.6 * 10^-10 x^2
= -------------- = 0.90 = ---------------
6.2 * 10^-10 ( 0.1 - x )^2
所以 x = [NH3] = [HCN] =
4.9 * 10^-2 M
[H+] = Ka * [HCN] / [CN-] = 6 * 10^-10 M
所以 pH =
9.22
5.
(a)
x^2
--------- = 6.2 * 10^-10
1 - x
所以 x = [H+] =
2.5 * 10^-5 M
(b)
x^2
------------------ = 6.2 * 10^-10
1.0 * 10^-4 - x
x = 2.5 * 10^-7 M 因為極稀溶液 所以須考慮水的解離
1
所以 6.2 * 10^-10 = Ka = ---------------------------
10^-14 - [H+]^2 - 10^-14
-----------------
[H+]
[H+] 設為 3.0 * 10^-7 帶入..
可以解得 [H+] = 2.68 * 10^-7 在帶回原式
得[H+] = 2.7 * 10^-7 所以 pH =
6.57
6.
(a)
ΔG°= 3 * 191.2 - 78.2 =
495.4 (KJ)
ΔH°= 3 * 241.3 - 132.8 =
591.1 (KJ)
ΔH° - ΔG°
ΔS°= ------------- =
321 (J/K)
T
(b)
ΔG° = -R * T * LnK 所以 K =
1.47 * 10^-87
(c)
ΔH° 和 ΔG° are temperature independent
ΔG° = 591.1 (KJ) - 3000 * 0.321 (KJ/K) = -372(KJ)
- ΔG°(3000K)
所以 Ln K = ---------------- = 14.9 K =
3 * 10^6
R * T
7.
ΔG° = -561 + 2 * -109 - (-797) = 18(KJ)
ΔG° = -R * T * Ln Ksp
所以 Ksp =
7 * 10^-4
8.
Kp = Pco2 所以 Pco2 要大於 Ksp 才能防止 Ag2CO3 分解
K2 ΔH° 1 1
由 Ln ---- = ------ * ( ----- - ----- )
K1 R 298 383
K = 7.5(torr) Pco2 > 7.5 torr
9.
3 O2 --> 2 O3
ΔH° = 2 * 143 = 286 (KJ)
ΔG° = 2 * 163 = 326 (KJ)
所以 Ln K = - ΔG° / ( R * T ) 可以得到 K = 7.22 * 10^-58
求在 230K 下的 K 值
K2 ΔH° 1 1
Ln ---- = ------ * ( ---- - ---- )
K1 R T1 T2
7.22 * 10^-8 286 * 10^3 1 1
Ln ---------------- = ------------ * ( ----- - ----- ) = 34.13
K(230K) 8.314 230 298
可得 K(230K) = 1.1 * 10^-72
Po3^2 Po3^2
又 K(230K) = ------- = ------------ 可得 Po3 =
3.3 * 10^-41
Po2^3 1.0 * 10^3
從 PV = nRT ， V = 9.5 * 10^-17 (L) --> 一個 O3 分子所佔的體積
所以兩個 O3 在 9.5 * 10^-17 升的體積中時，反應並不平衡，碰撞也不會發生
Thus the couceutration of ozone is not large enough to
maintained equilibrium
10.
(a)
Isothermal:
ΔE = 0, ΔH = 0
nRT nRT
w = - P*ΔV = - 2.45 * 10^3 * ( ------------- - -------------- ) * 101.3
2.45 * 10^-3 2.45 * 10^-2
=
-2230(J)
q = -w =
2230(J)
P1 2.45 * 10^-2
ΔS = n * R * Ln(----) = 1 * 8.314 * -------------- =
19.1 (J/K)
P2 2.45 * 10^-3
ΔG = ΔH - T * ΔS = 0 - 298 * 19.1 =
-5.69(KJ)
(b)
ΔE = 0, ΔH = 0, ΔS = 19.1(J/K), ΔG = -5.69(KJ) <-- State Funtion
q rev
ΔS = ------- 可得
q rev = 5.69 (KJ)
T
w = -q rev =
-5.69(KJ)
(c)
ΔE = 0, ΔH = 0, ΔS = -19.1(J/K), ΔG = 5.69(KJ) <--符號相反
1 1
w = -2.45 * 10^-2 *(-------------- - --------------)* 101.3 =
22.3(KJ)
2.45 * 10^-2 2.45 * 10^-3
q = -w =
-22.3(KJ)
(d)
^
__________ (a)
|
__________ (b)
|
__________ (c)
|
2.45*10^-2 | |.¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯|
|
| . |
|
| . |
|
| . |
|
| . |
|
| . |
|
| . |
|
| . |
2.45*10^-3 | |___________________.|
|
|
|________|____________________|_______________
V1 V2
(e)
-q actual
ΔSsurr = -----------
T
- 2230
所以
ΔSsurr,a = -------- =
-7.48(J/K)
298
ΔSsurr,b = - ΔS =
-7.48(J/K)
22300
ΔSsurr,c = -------- =
74.8(J/K)
298
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