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February 25th, 2018, 11:52 PM   #1
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Chemical equilbrium

H2(g) + I2(g)⇋2Hl(g) Kc = 49
The above reaction reaches equilibrium in a 1.12 L
container at 0 °C.
What will be the total pressure at equilibrium, if the
system starts with 0.9 Μ H2 and 0.9 Μ I2 in the container?
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February 26th, 2018, 12:26 AM   #2
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H2+I2⇋2HI
0.9M 0.9M ---
-x -x +2x
----------------------------
(0.9-x) (0.9-x) (2x)
Kc=49=(2x)^2/(0.9-x)^2
X=0.7 SO we have 0.2 M H2 and I2 and 1.4 mole HI
Now convert each of them to P using this formula p=CRT
And find sum of them
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February 26th, 2018, 04:00 AM   #3
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Quote:
Originally Posted by Elize View Post
H2+I2⇋2HI
0.9M 0.9M ---
-x -x +2x
----------------------------
(0.9-x) (0.9-x) (2x)
Kc=49=(2x)^2/(0.9-x)^2
X=0.7 SO we have 0.2 M H2 and I2 and 1.4 mole HI
Now convert each of them to P using this formula p=CRT
And find sum of them
Good job, Elize, but there is a short cut.

Since the total number of gas moles does not change as the reaction proceeds, the pressure does not change either (Avogadro).

So all that is needed is to use the gas law to calculate the pressure of 1.8 moles of ideal gas in 1.12L , at the temperature concerned. The molar volume is 22.4 L at STP (0 degrees C and 1 Atm)
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