Topic: predicting equilibrium molar ratio of N to O with kh values and T (Read 5693 times)

Sodium-Potassium Pump · « **on:** February 18, 2007, 11:21:09 AM »

I don't want an answer this problem but I just don't even know where to start so if someone could give me a basic idea of how to go about solving this I would really appreciate it. Thanks!

Air contains 78% N₂ and 21% O₂ (mole percentages).
Predict the equilibrium molar ratio of nitrogen to oxygen ([N₂]/[O₂]) dissolved in water which is exposed to air at 298 K.
N₂(g) over water: k_H= 8.57×104 atm
O₂(g) over water: k_H = 4.34×104 atm
(Assume ideal gas behaviour.)

Borek · « **Reply #1 on:** February 18, 2007, 12:16:22 PM »

Henry's Law

Sodium-Potassium Pump · « **Reply #2 on:** February 18, 2007, 01:27:47 PM »

OK, if K is supposed to be solubility (mols/L) why is it given as a pressure? I am so confused.

Borek · « **Reply #3 on:** February 18, 2007, 02:16:44 PM »

IMHO k is supposed to be Henry's Law constant, L*atm/mol or something.

Topic: predicting equilibrium molar ratio of N to O with kh values and T (Read 5693 times)

Sodium-Potassium Pump

predicting equilibrium molar ratio of N to O with kh values and T

Borek

Re: predicting equilibrium molar ratio of N to O with kh values and T

Sodium-Potassium Pump

Re: predicting equilibrium molar ratio of N to O with kh values and T

Borek

Re: predicting equilibrium molar ratio of N to O with kh values and T

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