Question

An equilibrium mixture taken in 2 litre vessel of the reaction: \[ \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \] has 4 moles of $\mathrm{SO_2}$, 3 moles of $\mathrm{O_2}$ and 6 moles of $\mathrm{SO_3}$ then the value of equilibrium constant $(K_c)$ will be:

• A. $0.75\ \mathrm{L\ mol^{-1}}$
• B. $0.15\ \mathrm{mol\ L^{-1}}$
• C. $1.5\ \mathrm{L\ mol^{-1}}$
• D. $15\ \mathrm{mol\ L^{-1}}$

Hint:

Before applying the equilibrium constant formula, always convert moles into molar concentration by dividing by the volume of the container.

Answer 1

The Correct Option is C

Concept: The equilibrium constant in terms of concentration is: \[ K_c = \frac{\text{Product concentrations raised to stoichiometric coefficients}}{\text{Reactant concentrations raised to stoichiometric coefficients}} \] For the reaction: \[ \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \] the equilibrium constant expression is: \[ K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \]

Step 1: Calculate Equilibrium Concentrations Volume of vessel: \[ V = 2\ \mathrm{L} \] Given moles: \[ \mathrm{SO_2} = 4\ mol \] \[ \mathrm{O_2} = 3\ mol \] \[ \mathrm{SO_3} = 6\ mol \] Concentration is: \[ \mathrm{Concentration = \frac{Moles}{Volume}} \] Therefore: \[ [SO_2] = \frac{4}{2} = 2\ \mathrm{M} \] \[ [O_2] = \frac{3}{2} = 1.5\ \mathrm{M} \] \[ [SO_3] = \frac{6}{2} = 3\ \mathrm{M} \]

Step 2: Substitute in Equilibrium Expression \[ K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \] Substituting values: \[ K_c = \frac{(3)^2}{(2)^2(1.5)} \] \[ K_c = \frac{9}{4 \times 1.5} \] \[ K_c = \frac{9}{6} \] \[ K_c = 1.5\ \mathrm{L\ mol^{-1}} \] Hence, the correct answer is: \[ \boxed{(C)\ 1.5\ \mathrm{L\ mol^{-1}}} \]