Answer:
"0.583" is the appropriate answer.
Explanation:
Let,
The initial constant of [tex]N_2O_4[/tex] be "C".
Amount of [tex]N_2O_4[/tex] dissociated into [tex]NO_2[/tex] be "x".
now,
[tex]N_2O_4 \rightleftharpoons 2NO_2[/tex]
Initial constant C -
Equilibrium constant C 2x
The Kc is given as:
⇒ [tex]K_c = \frac{[NO_2]^2}{[N_2O_4]}[/tex]
[tex]=\frac{(2x)^2}{C-x}[/tex]
[tex]0.593=\frac{4x^2}{0.88-x}[/tex]
[tex]4x^2=0.593(0.88-x)[/tex]
[tex]4x^2=0.512-0.593\ x[/tex]
[tex]x=0.291[/tex]
hence,
The constant of [tex]NO_2[/tex] will be:
= [tex]2x[/tex]
= [tex]0.583[/tex]
