3- \( \mathrm{SO}_{2} \) and \( \mathrm{SO}_{3} \) and \( \mathrm{O}_{2} \) are taken in a 5 liters flask. The equilibrium constant is found to be 50 for this reaction \( 2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftarrows 2 \mathrm{SO}_{3(\mathrm{~g})} \), If \( \left[\mathrm{SO}_{2}\right] \) and \( \left[\mathrm{SO}_{3}\right] \) are equal at equilibrium, calculate the number of moles of \( \mathrm{O}_{2} \) at equilibrium \[ \begin{array}{cccc}\na-0.01 & b-0.02 & c-0.2 & d-0.1 \end{array} \]

To solve this equilibrium problem, let's analyze the given reaction: \[ 2 \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftarrows 2 \mathrm{SO}_{3(g)} \] **Given:** - Equilibrium constant, \( K = 50 \) - Volume of the flask, \( V = 5 \) liters - At equilibrium, \([ \mathrm{SO}_2 ] = [ \mathrm{SO}_3 ]\) **Steps:** 1. **Define Concentrations:** Let’s denote the equilibrium concentration of both \(\mathrm{SO}_2\) and \(\mathrm{SO}_3\) as \( x \) mol/L. Let the equilibrium concentration of \(\mathrm{O}_2\) be \( y \) mol/L. 2. **Express the Equilibrium Constant:** The equilibrium expression for the reaction is: \[ K = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 [\mathrm{O}_2]} = \frac{x^2}{x^2 y} = \frac{1}{y} \] Given that \( K = 50 \): \[ \frac{1}{y} = 50 \implies y = \frac{1}{50} = 0.02 \text{ mol/L} \] 3. **Calculate Moles of \( \mathrm{O}_2 \) at Equilibrium:** \[ \text{Moles of } \mathrm{O}_2 = [\mathrm{O}_2] \times V = 0.02 \text{ mol/L} \times 5 \text{ L} = 0.1 \text{ mol} \] **Answer:** d) 0.1 mol