Question

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 \[ \begin{array}{cccc}\na-0.01 & b-0.02 & c-0.2 & d-0.1 \end{array} \]

Ask by Gross Ford. in Iraq
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

At equilibrium, there are 0.1 moles of \( \mathrm{O}_2 \).

Solution

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

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Let’s denote the concentration of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) at equilibrium as \(x\). According to the balanced equation, the relationship between the concentrations at equilibrium can be expressed as follows: \[ K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 \cdot [\mathrm{O}_2]} \] Given that at equilibrium \( [\mathrm{SO}_{2}] = [\mathrm{SO}_{3}] = x \) and substituting \(K_c = 50\): \[ 50 = \frac{x^2}{x^2 \cdot [\mathrm{O}_2]} \] This simplifies to: \[ 50 = \frac{1}{[\mathrm{O}_2]} \] Thus: \[ [\mathrm{O}_2] = \frac{1}{50} = 0.02 \, \text{M} \] Now, to find the number of moles of \(\mathrm{O}_{2}\) in the 5-liter flask: \[ \text{Moles of } \mathrm{O}_{2} = [\mathrm{O}_{2}] \times \text{Volume} = 0.02 \times 5 = 0.1 \, \text{mol} \] So, the answer is \(d-0.1\). If you need any more help or explanations regarding concepts or calculations, feel free to ask!

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