Question

At temperature \( T \) (K), the equilibrium constant \( K_c \) for the reaction: \[ A_2(g) \rightleftharpoons B_2(g) \] is 99.0. Two moles of \( A_2(g) \) were heated in a 1L closed flask to reach equilibrium. What are the equilibrium concentrations (in \( mol \ L^{-1} \)) of \( A_2(g) \) and \( B_2(g) \)?

• A. 1.86, 0.0187
• B. 1.98, 0.02
• C. 0.0187, 1.86
• D. 0.02, 1.98

Hint:

For equilibrium problems, always set up an ICE table and use the given \( K_c \) to solve for unknown concentrations.

Answer 1

The Correct Option is D

Step 1: Define the ICE Table
Let the initial concentration of \( A_2 \) be \( 2 \ mol/L \) and let \( x \) be the amount dissociated:
\[ A_2(g) \rightleftharpoons B_2(g) \]
 

Step 2: Apply the Equilibrium Expression
\[ K_c = \frac{[B_2]}{[A_2]} \]
\[ 99 = \frac{x}{2-x} \]

 

Species	Initial (M)	Equilibrium (M)
\( A_2 \)	2	\( 2 - x \)
\( B_2 \)	0	\( x \)

Step 3: Solve for \( x \)
\[ 99(2-x) = x \]
\[ 198 - 99x = x \]
\[ 198 = 100x \]
\[ x = 1.98 \]

Step 4: Find Equilibrium Concentrations
\[ [A_2] = 2 - 1.98 = 0.02 \ mol/L \]
\[ [B_2] = 1.98 \ mol/L \]