Asked by andy
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)
2NOCl(g) = 2NO(g)+Cl2(g)
Answers
Answered by
DrBob222
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.
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.
Answered by
Hmm
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?
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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