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

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UpStudy AI · Accepted Answer

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

3- and and are taken in a 5 liters flask. The equilibrium constant is found to be 50 for this reaction
, If and are equal at equilibrium, calculate the number of moles of at equilibrium

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3- and and are taken in a 5 liters flask. The equilibrium constant is found to be 50 for this reaction , If and are equal at equilibrium, calculate the number of moles of at equilibrium