Explanation:
As the given reaction is as follows.
     [tex]2A(aq) + B(aq) \rightleftharpoons 2C(aq)[/tex]
Initial: Â Â 0.039 Â Â Â Â 0.039 Â Â Â Â Â Â Â Â 0
Equilbm: 0.01365 Â 0.02535 Â Â Â Â 0.02535
Now, expression for [tex]K_{c}[/tex] of the given reaction is as follows.
      [tex]K_{c} = \frac{[C]^{2}}{[A]^{2}[B]}[/tex]
Putting the given values into the above formula as follows.
       [tex]K_{c} = \frac{[C]^{2}}{[A]^{2}[B]}[/tex]
      [tex]K_{c} = \frac{(0.02535)^{2}}{(0.01365)^{2} \times (0.02535)}[/tex]
           = [tex]1.360 \times 10^{2}[/tex]
           = 136
Thus, we can conclude that the equilibrium constant for the given reaction is 136.
