This problem involves solving for the equilibrium concentrations of the gases in a reaction where the equilibrium constant $K = 3.37$. The reaction is:

$\text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g)$

Given:

• Initial concentrations of all gases are $[\text{SO}_2] = [\text{NO}_2] = [\text{SO}_3] = [\text{NO}] = 0.800 \, \text{M}$.
• Equilibrium constant: $K = 3.37$.

The equilibrium expression is:

[ K = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]} ]

Let the change in concentration for the forward reaction be $x$. The concentration changes are as follows:

• $[\text{SO}_2] = [\text{NO}_2] = 0.800 - x$
• $[\text{SO}_3] = [\text{NO}] = 0.800 + x$

At equilibrium:

$K = \frac{(0.800 + x)(0.800 + x)}{(0.800 - x)(0.800 - x)}$

This expands to:

$3.37 = \frac{(0.800 + x)^2}{(0.800 - x)^2}$

Taking the square root of both sides:

$\sqrt{3.37} = \frac{0.800 + x}{0.800 - x}$

Solving:

1. Calculate $\sqrt{3.37}$: $\sqrt{3.37} \approx 1.836$

2. Write the equation: $1.836 = \frac{0.800 + x}{0.800 - x}$

3. Cross-multiply: $1.836 (0.800 - x) = 0.800 + x$

4. Distribute: $1.4688 - 1.836x = 0.800 + x$

5. Combine like terms: $1.4688 - 0.800 = x + 1.836x$ $0.6688 = 2.836x$

6. Solve for $x$: $x = \frac{0.6688}{2.836} \approx 0.236$

Equilibrium Concentrations:

• $[\text{SO}_2] = 0.800 - x = 0.800 - 0.236 = 0.564 \, \text{M}$
• $[\text{NO}_2] = 0.800 - x = 0.800 - 0.236 = 0.564 \, \text{M}$
• $[\text{SO}_3] = 0.800 + x = 0.800 + 0.236 = 1.036 \, \text{M}$
• $[\text{NO}] = 0.800 + x = 0.800 + 0.236 = 1.036 \, \text{M}$

Final Answer:

• $[\text{SO}_2] = 0.564 \, \text{M}$
• $[\text{NO}_2] = 0.564 \, \text{M}$
• $[\text{SO}_3] = 1.036 \, \text{M}$
• $[\text{NO}] = 1.036 \, \text{M}$

Would you like a detailed explanation of any step?

Related Questions:

1. How does changing the initial concentrations affect equilibrium concentrations?
2. How is the equilibrium constant $K$ affected by temperature changes?
3. Why does taking the square root simplify this particular equilibrium problem?
4. What happens to the equilibrium if the reaction vessel volume is doubled?
5. How would the calculation change if the initial concentrations of gases were not all the same?

Tip:

When solving equilibrium problems, always start by clearly defining the changes in concentration ($x$) and using the balanced equation to write the equilibrium concentrations systematically.