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chris2009317 2013-7-14 03:37 AM

chem 問題

1,
0.001 mol  of PCI5(g) was introduced into a 250cm3 container and allowed to reach equilibrium with PCI3(g) and CI2(g) at a certain temperature. 9.65 X 10[size=1]-4[/size] mol of CI2(g) was found in the equilibrium mixture
A).write a balanced chrmical equation for the reaction
B). write the expression for the equilibrium constant for the reaction
C).Calculate the equilibrium constant for reaction

2)
At 460C, H2(g)+I2(g)-->2HI(g)
2.0g of H2(g) was mixed with 508g of I2(g) in a 5.0dm3 container and the mixyure was allowed to reach equilibrium.
a) Calculate the initial number of moles of H2(g) and I2(g) respectively
b) Calculate the equilibrium concentration of H2(g),I(g) and HI(g) in the mixture

3)

XO(g)+O2(g)--->XO3(g)

The value of the equilibrium constant for this reaction at 398K is 0.0001 mol-1dm-3. if 1.00 mol of XO(g) and 2.00 mol of O2(g) are placed in a 2.00 dm3 vessel and are allowed to attain equilibrium at 398K ,find the equilibrium concentration of each chemical species

chungkin81 2013-7-14 12:44 PM

1,
0.001 mol  of PCl5(g) was introduced into a 250cm3 container and allowed to reach equilibrium with PCl3(g) and Cl2(g) at a certain temperature. 9.65 X 10-4 mol of Cl2(g) was found in the equilibrium mixture
A).write a balanced chemical equation for the reaction
B). write the expression for the equilibrium constant for the reaction
C).Calculate the equilibrium constant for reaction

A). PCl5(g) <==> PCl3 (g) + Cl2 (g)
B)  Kc = [PCl3(g)] [Cl2(g)] / [PCl5(g)]
C)                  PCl5(g)        <==>         PCl3(g)           + Cl2(g)
At start (mol)   0.001                           0                          0
At eqm (mol)   0.001 - 9.65 x 10^-4    9.65 x 10^-4          9.65 x 10^-4

At eqm,
[PCl5 (g)] = (0.001 - 9.65 x 10^-4) / (250 / 1000) = 1.4 x 10^-4 mol dm-3
[PCl3 (g)] = 9.65 x 10^-4 / (250 / 1000) = 3.86 x 10^-3 mol dm-3
[Cl2 (g) ] = 9.65 x 10^-4 / (250 / 1000) = 3.86 x 10^-3 mol dm-3

Kc = (3.86 x 10^-3) (3.86 x 10^-3) / (1.4 x 10^-4)
    = 0.106 mol dm-3

chungkin81 2013-7-14 01:10 PM

2)
At 460C, H2(g)+I2(g)-->2HI(g)
2.0g of H2(g) was mixed with 508g of I2(g) in a 5.0dm3 container and the mixture was allowed to reach equilibrium.
a) Calculate the initial number of moles of H2(g) and I2(g) respectively
b) Calculate the equilibrium concentration of H2(g),I2(g) and HI(g) in the mixture

a)  Initial number of moles of H2 = 2 / (1 x 2) = 1 mol
     Initial number of moles of I2 = 508 / (126.9 x 2) = 2.00 mol

b) Assume Kc = 48.9
    Let no. of moles of HI formed at equilibrium as x,
    At eqm,
    No. of moles of H2(g) = 1 - x
    No. of moles of I2(g) = 2 - x
    No. of moles of HI(g) = x
   
    [H2(g)] = (1 - x) / 5
    [I2 (g)] = (2 - x) / 5
    [HI (g)] = x / 5

     Kc = [HI(g)]^2 / ([H2 (g)] [I2 (g)])
         = (x / 5)^2 / {[(1 - x) / 5] [(2 - x) / 5]}
          = ( x^2 / 25) / [(1 - x) (2 - x) / 25]
          = x^2 / [(1 - x) (2 - x)
   49.8 = x^2 / [(1 - x) (2 - x)]
  49.8 (1 - x) (2 - x) = x^2
  49.8 (x^2 - 3x + 2) = x^2
48.8 x^2 - 149.4 x + 99.6 = 0
  x = 2.0804 (rejected) or x = 0.9810

[H2 (g)] = (1 - 0.9810) / 5 = 3.8 x 10^-3 mol dm-3
[I2 (g)] = (2 - 0.9810) / 5 = 0.2038 mol dm-3
[HI (g)] = 0.9810 / 5 = 0.1962 mol dm-3

[[i] 本帖最後由 chungkin81 於 2013-7-14 01:12 PM 編輯 [/i]]

chungkin81 2013-7-14 01:52 PM

3)

XO(g)+O2(g)--->XO3(g)

The value of the equilibrium constant for this reaction at 398K is 0.0001 mol-1dm3. if 1.00 mol of XO(g) and 2.00 mol of O2(g) are placed in a 2.00 dm3 vessel and are allowed to attain equilibrium at 398K ,find the equilibrium concentration of each chemical species.

3.                       XO(g) + O2(g) --> XO3 (g)
    At start (mol)    1           2               0
    At eqm (mol)   1 - x      2 - x            x

At eqm,
[XO (g)] = (1 - x) / 2
[O2 (g)] = (2 - x) / 2
[XO3 (g)] = x / 2

Kc = [XO3 (g)] / {[XO (g)] [O2 (g)]}
0.0001 = (x / 2) / {[(1 - x) / 2] [ (2 - x) / 2]
0.0001 = (x / 2)  / [(1 - x) (2 - x) / 4]
2x = 0.0001 (1- x) (2 - x)
2x = 0.0001 (2 - 2x - x + x^2)
2x = 0.0002 - 0.0003x + 0.0001x^2
0 = 0.0002 - 2.0003x + 0.0001x^2
x = 20003 (rejected) or x = 1.00 x 10^-4

[XO(g)] = (1 - 1.00 x 10^-4) / 2 = 0.500 mol dm-3
[O2(g)] = (2 - 1.00 x 10^-4) / 2 = 1.000 mol dm-3
[XO3(g)] = (1.00 x 10^-4) / 2 = 5 x 10^-5 mol dm-3
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