Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of [tex]N_2O_4[/tex] = 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration [tex]N_2O_4[/tex].
[tex]\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M[/tex]
The given equilibrium reaction is,
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
Initially c 0
At equilibrium [tex](c-c\alpha)[/tex] [tex]2c\alpha[/tex]
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
[tex]K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}[/tex]
where,
[tex]\alpha[/tex] = degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
[tex]K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}[/tex]
[tex]K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}[/tex]
[tex]K_c=1.066\aprrox 1.1[/tex]
Therefore, the value of equilibrium constant for this reaction is, 1.1
