Question

For the following reaction of N2O4, the equilibrium constant is 0.593 at a particular temperature.

Answer 1

Answer:

"0.583" is the appropriate answer.

Explanation:

Let,

The initial constant of $N_2O_4$ be "C".

Amount of $N_2O_4$ dissociated into $NO_2$ be "x".

now,

                                     $N_2O_4 \rightleftharpoons 2NO_2$

Initial constant               C            -

Equilibrium constant     C          2x

The Kc is given as:

⇒ $K_c = \frac{[NO_2]^2}{[N_2O_4]}$

         $=\frac{(2x)^2}{C-x}$

$0.593=\frac{4x^2}{0.88-x}$

  $4x^2=0.593(0.88-x)$

  $4x^2=0.512-0.593\ x$

     $x=0.291$

hence,

The constant of $NO_2$ will be:

= $2x$

= $0.583$