Answer:
[tex]Kc=0.14[/tex]
Explanation:
Hello,
In this case, for the given reaction, the equilibrium expression is:
[tex]Kc=\frac{[NO]^2[O_2]}{[NO_2]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] is written as (initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L):
[tex]Kc=\frac{(2*x)^2(x)}{(0.24M-2*x)^2}[/tex]
Moreover, since at the equilibrium 0.14 moles of NO are present (a 0.14-M concentration), we can compute the reaction extent as shown below:
[tex][NO]=2*x=0.14M[/tex]
[tex]x=0.14M/2=0.07M[/tex]
In such a way, knowing [tex]x[/tex], we compute Kc as shown below:
[tex]Kc=\frac{(2*0.07)^2(0.07)}{(0.24M-2*0.07)^2}\\\\Kc=0.14[/tex]
Regards.
The kc is 0.14.
Since
[tex]Kc = \frac{[NO]^2[O_2]}{[NO_2]^2}\\\\[/tex]
Here the reaction extent x should be written like the initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L.
Now
[tex]Kc = \frac{(2\times\ x)^2}{(0.24M - 2\times x)^2}[/tex]
Since at the equilibrium 0.14 moles of NO are presented so the reaction should be
NO = 2*x = 0.14m
x = 0.07
Now kc should be
[tex]= \frac{(2\times 0.07)^2 (0.07)}{(0.24M - 2\times 0.07)^2}[/tex]
= 0.14
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