The reaction N2O4(g) ↔ 2NO2(g) is an equilibrium reaction at some temperature with an equilibrium?

Question

constant K = 4.00 x 10−7.
If the intial condition is pure N2O4(g) at a concentration of 1.59 mol/L what is the equilibrium concentration of NO2 in mol/L ?




HINT : Since K is small you can use the approximation for Small Equilibrium constants, as in your text


HINT : in the equilibrium expression the product is written as 2x
Since the numerator in the expression for the equilibrium constant
is the product concentration raised to the stochiometric coefficient,
it becomes (2x)2

Answer 1

Initial concentration is 1.59
N2O4->2NO2
I=1.59 ........................ 0
C= -x ........................ +2x
E= 1.59-x ......................2x

Kc=[NO2]^2/[N2O4]=4.00x10^ -7
(2x)^2/(1.59-x)=4.00x10^ -7
4x^2=4.00x10^ -7(1.59-x)
4x^2=6.36x10^ -7 -4.00x10^-7 x
4x^2 + 4.00*10^ -7 x -6.36*10^-7=0
by quaratic formula
x=.000399
therefore concentration of NO2= 2*.000399=.000798 M
that is answer i believe good luck
note I=initial, C=change E=equilibrium