Posted by **andy** on Thursday, April 8, 2010 at 12:23am.

I also had a problem with this type of questionthe initial pressure of NOCl(g) is 4.329 atm, calculate the % of NOCl(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kp at 400.0 °C is 1.99. The initial pressure of the reaction products is 0 atm.

2NOCl(g) = 2NO(g)+Cl2(g)

- chemistry -
**DrBob222**, Thursday, April 8, 2010 at 12:54am
initial:

NOCl = 4.329

NO = 0

Cl2 = 0

change:

NOCl = -2x

NO = +2x

Cl2 = x

equilibrium:

NOCl = 4.329 - 2x

NO = 2x

Cl2 = x

Substitute into Kp and solve for x, multiply by 2 and subtract from original. That gives pressure of NOCl at equilibrium. I don't know if percent is to be based on pressure or not. I would think it would be based on grams but I don't see a way to quickly convert to grams.One way might be to convert Kp to Kc and find concns at equilibrium, then tak a liter.

- chemistry -
**Hmm**, Thursday, April 8, 2010 at 11:35pm
1.99 = (3x)/(4.329-2x)

8.615 - 3.98x = 3x

8.615 = 6.98x

x = 1.23

NO = 2.46 atm

Cl2 = 1.23 atm

The question now is, how to find the % NOCl left over?

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