Ryan Ong Meng How's answer to Angelica's Junior College 1 H2 Maths question.
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First thing you need to note is that the ratio of partial pressure is not affected by the molecular mass of the compounds. It's only affected by the number of atoms.
Thus we can effectively say that every 2atm of NH3 would mean a loss of 1atm N2 and 3atm H2.
So at the start from the ratio there is 200atm and the partial pressure of both gases are..
N2:H2 > 1:3 > 50:150
Then the lost of N2 = (27/2)*1=13.5
And the lost of H2 = (27/2)*3=40.5
So the Kp=(NH3)/(N2)(H2)
=(27)/(50-13.5)(150-40.5)
=6.755*10^-3
Thus we can effectively say that every 2atm of NH3 would mean a loss of 1atm N2 and 3atm H2.
So at the start from the ratio there is 200atm and the partial pressure of both gases are..
N2:H2 > 1:3 > 50:150
Then the lost of N2 = (27/2)*1=13.5
And the lost of H2 = (27/2)*3=40.5
So the Kp=(NH3)/(N2)(H2)
=(27)/(50-13.5)(150-40.5)
=6.755*10^-3
Date Posted:
6 years ago
Oops I made a mistake
Kp=(NH3)^2/(N2)(H2)^3
Thus it is actually equals to...
Kp=27^2/(50-13.5)(150-40.5)^3
=1.52*10^-5
Kp=(NH3)^2/(N2)(H2)^3
Thus it is actually equals to...
Kp=27^2/(50-13.5)(150-40.5)^3
=1.52*10^-5
Thanks alot!!!
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