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

To solve this problem, we need to determine the equilibrium concentration of \(B_2(g)\) given the equilibrium constant \(K_c = 99.0\), the initial moles of \(B_2 = 2\), and the reaction: \[A_2(g) \rightleftharpoons B_2(g)\]

Assume initially, the concentration of \(A_2\) is 0, and the concentration of \(B_2\) is 2 mol/L since the container is 1 L. Given the reaction's stoichiometry, we set up an ICE (Initial, Change, Equilibrium) table:

	\([A_2(g)]\)	\([B_2(g)]\)
Initial	0	2
Change	+x	-x
Equilibrium	x	2-x

From the equilibrium expressions, we have: \[K_c = \frac{[B_2]}{[A_2]}\Rightarrow K_c = \frac{2-x}{x} = 99.0\] Solving for \(x\): \[99x = 2-x\] \[100x = 2\] \[x = \frac{2}{100} = 0.02\] Substitute \(x\) back to find \([B_2]_{eq}\): \[[B_2]_{eq} = 2-x = 2-0.02 = 1.98\text{ mol/L}\] Therefore, the equilibrium concentration of \(B_2\) is \(1.98\text{ mol L}^{-1}\). 

Answer 2

The given reaction is: 

A₂(g) ⇌ B₂(g)

The equilibrium constant for the reaction is:

K_c = 99.0 at temperature T (K)

Initial moles of B₂ = 2 moles in a 1 L flask
→ Initial concentration of B₂ = 2 mol / 1 L = 2 M

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

• [A₂] = x
• [B₂] = 2 − x

Equilibrium expression:

K_c = [B₂]² / [A₂]

99.0 = (2 − x)² / x

Solving this equation gives the equilibrium value of x ≈ 0.0204.

Therefore, the concentration of B₂ at equilibrium is:

[B₂] = 2 − x = 2 − 0.0204 ≈ 1.98 M

Final Answer: The concentration of B₂ at equilibrium is 1.98 M.