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(C) From the graph,at $t = 5 \ s$,the concentrations of $A$,$C$,and $B$ become constant,indicating that equilibrium has been reached.Initial concentra

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At $t = 5 \ s$ equilibrium has been reached and $	ext{percentage dissociation}$ of $A$ is $30\%$

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(C) From the graph,at $t = 5 \ s$,the concentrations of $A$,$C$,and $B$ become constant,indicating that equilibrium has been reached.Initial concentration of $A = 10 \ mol/L$.Equilibrium concentration of $A = 7 \ mol/L$.Amount of $A$ dissociated $= 10 - 7 = 3 \ mol/L$.$	ext{Percentage dissociation of } A = \frac{	ext{Amount dissociated}}{	ext{Initial amount}} 	imes 100 = \frac{3}{10} 	imes 100 = 30\%$.Now,let's calculate $K_c$ for the reaction $A_{(g)} ightleftharpoons 3C_{(g)} + 2B_{(g)}$.Equilibrium concentrations: $[A] = 7 \ M$,$[C] = 9 \ M$,$[B] = 6 \ M$.$K_c = \frac{[C]^3 [B]^2}{[A]} = \frac{(9)^3 (6)^2}{7} = \frac{729 	imes 36}{7} = \frac{26244}{7} \approx 3749.14 \ (mol/L)^4$.Comparing this with the options,option $C$ is correct.

Attainment of the equilibrium $A_{(g)} \rightleftharpoons 3C_{(g)} + 2B_{(g)}$ gave the following graph. Find the correct option. $(\text{Percentage dissociation} = \text{fraction dissociated} \times 100)$

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