[ReVeR]

Hello.
I got the following querstion:
We have a reaction
2NCl3<->N2 +3Cl2
We have Kc=3.3e-12.
We also have 4.6e-4 mol/l of N2 present at equalibrium. THe question is how much NCl3 was in the flask initialy.
I did the following:
3.3e-12=(27 * (4.6e-4))/y^2  
After i solve for y i get 1.2, the answer is 6e-5.
Any ideas where i made the mistake?

[dolphinsiu]

What is e ?

[Borek]

Way of expressing numbers in the so called scientific notation.

3.2e-5 = 3.2*10^-5

Every calculator displays numbers this way.

[Donaldson Tan]

2 NCl₃ <-> N₂ +3 Cl₂
Kc=3.3E-12.
Kc = [[ N₂ ]] [[ Cl₂ ]] ³ / [[ NCl₃ ]] ²

	NCl3		N2	Cl2
Initial	X0		0	0
Change	-2x		+x	+x
Equilibrium	(X0 - 2x)		(4.6E-4)	x

Using the ICE table above,
x = 4.6E-4

Remembering that Kc is expressed in terms of concentration, you must convert the quantity of each substance into its corresponding concentration, before plugging into the Kc expression.

Upon algebraic manipulation, you will be able to find X₀, ie. the initial amount of NCl₃ initially present in the flask.