Question

At T(K), the value of \( K_c \) for the reaction 
\[ AO_2(g) + BO_2(g) \leftrightarrow{} AO_3(g) + BO(g) \] is 16. In a one-litre closed flask, 1 mole each of \( AO_2(g) \), \( BO_2(g) \), \( AO(g) \), and \( BO(g) \) were taken and heated to T(K). 
What are the equilibrium concentrations (in mol L\(^{-1}\)) of \( BO_2(g) \) and \( BO(g) \) respectively? 
 

• A. 1.6, 0.4
• B. 0.467, 1.533
• C. 1.533, 0.467
• D. 0.4, 1.6

Hint:

Equilibrium calculations often involve setting up an ICE table (Initial, Change, Equilibrium) and solving for the unknown change in concentration.

Answer 1

The Correct Option is D

Step 1: Write the expression for \( K_c \) \[ K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = 16 \] Step 2: Set up the initial concentrations and changes Initial concentrations:
\([AO_2] = 1 \, {mol/L}\)
\([BO_2] = 1 \, {mol/L}\)
\([AO] = 1 \, {mol/L}\)
\([BO] = 1 \, {mol/L}\)
Let \( x \) be the change in concentration of \( AO_2 \) and \( BO_2 \) that react to form \( AO_3 \) and \( BO \). 
At equilibrium: - \([AO_2] = 1 - x\) - \([BO_2] = 1 - x\) - \([AO_3] = 1 + x\) - \([BO] = 1 + x\) 
Step 3: Substitute into the equilibrium expression \[ 16 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = \frac{(1 + x)^2}{(1 - x)^2} \] Taking the square root of both sides: \[ 4 = \frac{1 + x}{1 - x} \] Solving for \( x \): \[ 4(1 - x) = 1 + x \] \[ 4 - 4x = 1 + x \] \[ 3 = 5x \] \[ x = \frac{3}{5} = 0.6 \] 
Step 4: Calculate equilibrium concentrations \([BO_2] = 1 - x = 1 - 0.6 = 0.4 \, {mol/L}\)
\([BO] = 1 + x = 1 + 0.6 = 1.6 \, {mol/L}\)