Chemical Equilibrium

To analyze the equilibrium concentrations for the reaction $A_2(g) \rightleftharpoons B_2(g)$, where the equilibrium constant $K_c$ at temperature $T(K)$ is 99.0, follow these steps:Given:$K_c = 99.0$Initial moles of $A_2: 2$ molesVolume of the flask: 1 LCalculation:Initial Conditions and Definition:Since the reaction starts with 2 moles of $A_2(s)$ which sublimes to $A_2(g)$, the initial concentration of $A_2$ is $2\ \text{mol/L}$ (as the flask is 1 L in volume).Setting up the Equilibrium Expression:For the equilibrium constant expression:$K_c = \frac{[B_2]}{[A_2]}$Let the concentration of $B_2$ at equilibrium be $x\ \text{mol/L}$. Therefore, the concentration of $A_2$ becomes $(2-x)\ \text{mol/L}$ at equilibrium.Solving the Equilibrium Equation:Substitute the equilibrium values into the $K_c$ expression:$99 = \frac{x}{2-x}$Solving for $x$ :$\;99(2-x)=x$$\;198-99x=x$$\;198=100x$$\;x=1.98$Thus, the concentration of $B_2$ is $[B_2] = 1.98\ \text{mol/L}$.Finding the Concentration of $A_2$ :Calculate $[A_2]$ using the expression $2-x$ :$[A_2] = 2-1.98=0.02\ \text{mol/L}$Conclusion:The concentrations at equilibrium are:$\; [A_2] = 0.02\ \text{mol/L}$$\; [B_2] = 1.98\ \text{mol/L}$