[docnet]

Kp = 5.3E⁵ for the reaction N_{2 (g)} + 3H_{2 (g)}  2NH_{3 (g)}

when a certain partial pressure of ammonia gas is put into an otherwise empty rigid vessel at the same temperature, equilibrium is reached when 50 percent of the original ammonia has decomposed. what was the original partial pressure of ammonia before any decomposition occurred?

I wasn't sure how to approach the problem, so I set up an equilibrium expression and tried to express all partial pressures in terms of P_NH3.

I made the following assumptions:

-The partial pressure of nitrogen gas must be half of the equilibrium partial pressure of ammonia
-The partial pressure of hydrogen gas must be one and a half times the equilibrium partial pressure of ammonia
-let x equal to the equilibrium partial pressure of ammonia

5.3E⁵ = x²/[(x/2)(3x/2)³]

the pre-equilibrium pressure of ammonia using this assumption is 2.24 E^-6, which is far from the answer given which is 2.1E^-3 ATM. what's the correct way to approach this type of problem?
Thanks in advance for any input.

[Borek]

Check your math.

BTW, it is not E^-6. It is either 10^-6 or E-6.

[docnet]

thank you Borek!! (I got the right answer this time)

And also thanks for saying something about the E.

[docnet]

My thing with the E⁶ started when I started writing in small and dense letters when solving equilibrium problems. On paper and in pen it makes the number that's being powered easier to visually distinguish when written as a superscript. I write so small that after a while all the E's and the numbers look the same to me. I always knew that E-6 means 10^^-6 (which is the scientific notation)

A few times in class I read out loud from my calculator "E to the power of". Everyone including the teacher thought I was actually saying "2.718 to the power of..."  and that's when I knew I had a problem.

[docnet]

E-6 means 10^^-6 (which is the scientific notation)

EDIT: The symbol ^ shouldn't be there.