NEED HELP ASAP!! medal/fan any helpers!!
Consider the following equilibrium and the table of experimental data:
N2O4(g) Ý 2NO2(g)
Which of the following represents the Keq value?
A)0.18
B)0.0067
C)0.131
D)0.0046

Question

NEED HELP ASAP!! medal/fan any helpers!!
Consider the following equilibrium and the table of experimental data:
N2O4(g) Ý 2NO2(g)
Which of the following represents the Keq value?
A)0.18
B)0.0067
C)0.131
D)0.0046

anonymous · Answer

Initial                             Equilibrium
________________________________________
[N2O4]      [NO2]              [N2O4]     [No2]
0.0200      0.0600            0.0429    0.0141

joannablackwelder · Answer

Keq= [NO2]^2/[N2O4] at equilibrium. Do you see how I got that formula?

photon336 · Answer

$$\frac{ [NO2]^{2} }{ [N2O4] } = K _{eq}$$

photon336 · Answer

$$K _{eq} = Products/Reactants  = [C]^{a}[D]^{b}/[A]^{c}[B]^{d} $$