xt{K}) $ is 99.0, follow these steps:

Given:

$ K_c = 99.0 $

Initial moles of $ A_2 $: 2 moles

Volume of the flask: 1 L

Calculation:

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 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 $ mol/L. Therefore, the concentration of $ A_2 $ becomes $ (2 - x) $ 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 $ 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} $

Chemical Equilibrium

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