Question

For the reaction: \[ A_2(g) \rightleftharpoons B_2(g) \] The equilibrium constant \( K_c \) is given as 99.0. In a 1 L closed flask, two moles of \( B_2(g) \) is heated to T(K). What is the concentration of \( B_2(g) \) (in mol L\(^{-1}\)) at equilibrium?

• A. \( 0.02 \)
• B. \( 1.98 \)
• C. \( 0.198 \)
• D. \( 1.5 \)

Hint:

To solve equilibrium problems, use an ICE (Initial-Change-Equilibrium) table and apply the equilibrium constant expression. Ensure the correct interpretation of equilibrium concentrations.

Answer 1

The Correct Option is B

The given reaction is:

A₂(g) ⇌ B₂(g)

The equilibrium constant for the reaction is given by:

K_c = 99.0 at temperature T (K).

Initial moles of B₂(g) are 2 moles in a 1 L flask. Therefore, initial concentration of B₂ is:

Initial concentration of B₂ = 2 mol / 1 L = 2 M.

Let 'x' be the moles of A₂ that dissociate at equilibrium. At equilibrium:

Concentration of A₂ = x M, and concentration of B₂ = 2 - x M (since one mole of B₂ is produced for each mole of A₂ dissociated).

The equilibrium constant expression is:

K_c = [B₂]² / [A₂]

Substitute the values:

99.0 = (2 - x)² / x²

Solving for 'x' gives the concentration of B₂(g) at equilibrium as:

[B₂] = 1.98 M