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 February 26th, 2018, 12:52 AM #1 Newbie   Joined: Feb 2018 From: Afghanistan Posts: 17 Thanks: 0 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?
 February 26th, 2018, 01:26 AM #2 Member   Joined: Feb 2018 From: Iran Posts: 30 Thanks: 3 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 Thanks from studiot and Marva
February 26th, 2018, 05:00 AM   #3
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 Originally Posted by Elize 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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