Question

At T(K), \( K_c \) value for \( \text{AO}_2(g) + \text{BO}_2(g) \rightleftharpoons \text{AO}_3(g) + \text{BO}(g) \) is 16. In a closed 1 L flask, one mole each of \( \text{AO}_2 \), \( \text{BO}_2 \), \( \text{AO}_3 \), and \( \text{BO} \) are taken and heated to T(K). What is the concentration (in mol L\(^{-1}\)) of \( \text{AO}_3 \) at equilibrium?  

• A. 0.4
• B. 0.6
• C. 1.6
• D. 1.4

Hint:

When equilibrium concentrations are unknown, use the ICE table method and apply the equilibrium constant expression to solve for changes.

Answer 1

The Correct Option is C

Step 1: Write the equilibrium expression.
\[ K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = 16 \] Step 2: Initial moles of all species = 1 mole in 1 L volume. Let equilibrium shift by \(x\) moles towards products. \[ [AO_2] = 1 - x, \quad [BO_2] = 1 - x, \quad [AO_3] = 1 + x, \quad [BO] = 1 + x \] Step 3: Apply values in \(K_c\) expression. \[ \frac{(1+x)^2}{(1-x)^2} = 16 \Rightarrow \frac{1+x}{1-x} = 4 \Rightarrow 1+x = 4(1-x) \Rightarrow 1+x = 4 - 4x \Rightarrow 5x = 3 \Rightarrow x = 0.6 \] \[ [AO_3] = 1 + x = 1.6 \, \text{mol L}^{-1} \]