(1 )4.21 moles of S2Cl4 are introduced into a 2L vessel.
S2Cl4<-----> 2SCL2
At equilibrim,1.25 moles of S2CL4 are found to remain in the container .
calculate Kc for this reaction.
(2) Calculate the concentration of malonate ion (C3H2O4 2-) in a 0.200 M solution of malonic acid (C3H4O4). The malonic acid, Ka1 = 1.4 x10^(-3), Ka2 = 2.0 x 10^(-6)
急!!!化學題目求解 請求化學高手
2013-06-08 5:29 am
回答 (3)
2013-06-08 6:42 am
✔ 最佳答案
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參考: pingshek
2013-06-08 6:55 am
Actually, H2A partially dissociates to give HA- and H+, and then HA- partially dissociates to give A-- and H+.
Hence, [H2A] = (0.200 - x) M, [HA-] = (x - y) M, [A--] = y M and [H+] = (x + y) M
In 洛雲殤's answer, it is assumed that [HA-] = 0 M, i.e. x = y. This is incorrect.
Hence, [H2A] = (0.200 - x) M, [HA-] = (x - y) M, [A--] = y M and [H+] = (x + y) M
In 洛雲殤's answer, it is assumed that [HA-] = 0 M, i.e. x = y. This is incorrect.
2013-06-08 5:57 am
1.
initial concentration S2Cl4 = 4.21/ 2.0= 2.11 M
concentration S2Cl2 at equilibrium = 1.25/2.0 = 0.625 M
S2Cl4 <------> 2 SCl2
initial concentration
2.11
change
-x . . .. . . . . . . . . +2x
2.11 - x = 0.625
x = 1.49
2x = [SCl2]= 2.98
K = [SCl2]^2/ [S2Cl4] = (2.98)^2/ 0.625 =14.2
2.
Given two equilibria:
H2A ↔ HA- + H+ (Ka1)
HA- ↔ A-- + H+ (Ka2)
The overall reaction and equilibrium constant is:
H2A ↔ A-- + 2H+ (K = Ka1Ka2 = 2.8x10^-9)
K = [A--] [H+]² / [H2A]
Initially [H2A] = 0.200 M and [A--] & [H+] = 0
In going to equilibrium, [H2A] loses x, [A--] gains x, and [H+] gains 2x
So at equilibrium,[H2A] = 0.200-x, [A--] = x, and [H+] = 2x.
Substituting,
2.8x10^-9 = (2x)² (x) / (0.200 - x).
x will be very small, so we ignore it in the denominator. The equation then easily solves to:
x = [A--] = 5.2x10-4 M.
Note that this equilibrium approach will NOT give total solution pH, just malonate dianion concentration.
Hoping to help you!
initial concentration S2Cl4 = 4.21/ 2.0= 2.11 M
concentration S2Cl2 at equilibrium = 1.25/2.0 = 0.625 M
S2Cl4 <------> 2 SCl2
initial concentration
2.11
change
-x . . .. . . . . . . . . +2x
2.11 - x = 0.625
x = 1.49
2x = [SCl2]= 2.98
K = [SCl2]^2/ [S2Cl4] = (2.98)^2/ 0.625 =14.2
2.
Given two equilibria:
H2A ↔ HA- + H+ (Ka1)
HA- ↔ A-- + H+ (Ka2)
The overall reaction and equilibrium constant is:
H2A ↔ A-- + 2H+ (K = Ka1Ka2 = 2.8x10^-9)
K = [A--] [H+]² / [H2A]
Initially [H2A] = 0.200 M and [A--] & [H+] = 0
In going to equilibrium, [H2A] loses x, [A--] gains x, and [H+] gains 2x
So at equilibrium,[H2A] = 0.200-x, [A--] = x, and [H+] = 2x.
Substituting,
2.8x10^-9 = (2x)² (x) / (0.200 - x).
x will be very small, so we ignore it in the denominator. The equation then easily solves to:
x = [A--] = 5.2x10-4 M.
Note that this equilibrium approach will NOT give total solution pH, just malonate dianion concentration.
Hoping to help you!
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