Question

At temperature $ T(K) $, the equilibrium constant $ K_c $ for the reaction: $$ \mathrm{AO_2(g) + BO_2(g) \rightleftharpoons AO_3(g) + BO(g)} $$ is 16. In a 1 L flask, one mole each of $\mathrm{AO_2}$, $\mathrm{BO_2}$, $\mathrm{AO_3}$, and $\mathrm{BO}$ are taken and heated to $ T(K) $. Identify the correct statements:

• Total number of moles at equilibrium is 4
• At equilibrium, the ratio of moles of $\mathrm{AO_2}$ and $\mathrm{AO_3}$ is 1:4
• Total number of moles of $\mathrm{AO_2}$ and $\mathrm{BO_2}$ at equilibrium is 0.8

• A. I only
• B. I, III only
• C. II, III only
• D. I, II, III

Hint:

For equilibrium problems, always start with the ICE (Initial–Change–Equilibrium) method, apply the equilibrium constant expression, and solve algebraically.

Answer 1

The Correct Option is D

Step 1: Initial moles (all species 1 mole each): \[ [\mathrm{AO_2}] = [\mathrm{BO_2}] = [\mathrm{AO_3}] = [\mathrm{BO}] = 1\ \text{mol} \] Let the change in moles at equilibrium be \( x \): \[ \mathrm{AO_2 + BO_2 \rightleftharpoons AO_3 + BO} \] \[ \text{Equilibrium: } [\mathrm{AO_2}] = 1 - x,\ [\mathrm{BO_2}] = 1 - x,\ [\mathrm{AO_3}] = 1 + x,\ [\mathrm{BO}] = 1 + x \]
Step 2: Apply \( K_c = 16 \) \[ K_c = \frac{(1 + x)^2}{(1 - x)^2} = 16 \Rightarrow \frac{1 + x}{1 - x} = 4 \Rightarrow 1 + x = 4(1 - x) \Rightarrow 1 + x = 4 - 4x \Rightarrow 5x = 3 \Rightarrow x = 0.6 \]
Step 3: Final concentrations \[ [\mathrm{AO_2}] = 0.4,\quad [\mathrm{BO_2}] = 0.4,\quad [\mathrm{AO_3}] = 1.6,\quad [\mathrm{BO}] = 1.6 \] Check statements: I. Total moles = \( 0.4 + 0.4 + 1.6 + 1.6 = 4 \) II. Ratio \(\mathrm{AO_2 : AO_3} = 0.4 : 1.6 = 1 : 4 \) III. \(\mathrm{AO_2 + BO_2} = 0.4 + 0.4 = 0.8 \)