[急]Equalibrium, find volume

[匿名]

2009-01-28 7:57 pm

For the dissociation of at about 1200 degree Celcius,
I2 (g) -> 2I (g), Kc= 1.1* 10^(-2)

Note: The reaction is reversible

What volume flask should we use if we want 0.33mol I to be present for every 1.00mol I2 at equilibrium?

It's very URGENT!!!!!!!!!!!!!!!!!!
Thank you very much for your kind help!!!!

回答 (1)

Uncle Michael

2009-01-28 8:34 pm

✔ 最佳答案

I2(g) ≒ 2I(g) **Kc = 1.1 x 10-2

Let the volume of the flask be V dm3.

At eqm, mole ratio I2 : I = 1 : 0.33
No. of moles of I2 = y mol
No. of moles of I = 0.33y mol
[I2] = y/V
[I] = 0.33y/V

Kc = (0.33y/V)2/(y/V) = 1.1 x 10-2
0.11y/V = 1.1 x 10-2
V = 10y

The volume of the flask depends on the amount of I2 at equilibrium.
When there are y mol of I2 at eqm, volume of the flask is 10y dm3.
For every 1 mol of I2 at eqm, flask volume is 10 dm3.
=

2009-01-28 12:36:58 補充：
The units of [I2] and [I] are both mol dm^-3.

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