Math Snap

STEP 1

1. The reaction is at equilibrium with a given equilibrium constant $K_c = 4.8$.
2. The initial concentrations of $\mathrm{SO}_2$ and $\mathrm{NO}_2$ are both $0.36 \, \mathrm{mol/L}$.
3. The volume of the container is $5.0 \, \mathrm{L}$.
4. We need to find the equilibrium concentration of $\mathrm{SO}_3$.

STEP 2

1. Set up the equilibrium expression for the reaction.
2. Use an ICE (Initial, Change, Equilibrium) table to express concentrations at equilibrium.
3. Solve for the equilibrium concentration of $\mathrm{SO}_3$.
4. Calculate the amount of $\mathrm{SO}_3$ in moles.

STEP 3

The equilibrium expression for the reaction is given by:
$$K_c = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]}$$

STEP 4

Set up an ICE table:
- Initial concentrations: $[\mathrm{SO}_2] = 0.36 \, \mathrm{mol/L}$, $[\mathrm{NO}_2] = 0.36 \, \mathrm{mol/L}$, $[\mathrm{NO}] = 0 \, \mathrm{mol/L}$, $[\mathrm{SO}_3] = 0 \, \mathrm{mol/L}$.
- Change in concentrations: $-x$ for $[\mathrm{SO}_2]$ and $[\mathrm{NO}_2]$, $+x$ for $[\mathrm{NO}]$ and $[\mathrm{SO}_3]$.
- Equilibrium concentrations: $[\mathrm{SO}_2] = 0.36 - x$, $[\mathrm{NO}_2] = 0.36 - x$, $[\mathrm{NO}] = x$, $[\mathrm{SO}_3] = x$.

STEP 5

Substitute the equilibrium concentrations into the equilibrium expression:
$$K_c = \frac{x \cdot x}{(0.36 - x)(0.36 - x)} = 4.8$$ Simplify and solve for $x$:
$$\frac{x^2}{(0.36 - x)^2} = 4.8$$ Take the square root of both sides:
$$\frac{x}{0.36 - x} = \sqrt{4.8}$$ Solve for $x$:
$$x = 0.36 \cdot \frac{\sqrt{4.8}}{1 + \sqrt{4.8}}$$

Calculate the amount of $\mathrm{SO}_3$ in moles:
$$x = [\mathrm{SO}_3] = 0.36 \cdot \frac{\sqrt{4.8}}{1 + \sqrt{4.8}}$$ $$\text{Amount of } \mathrm{SO}_3 = x \times 5.0 \, \mathrm{L}$$ Calculate the numerical value to find the amount in moles.
The amount of $\mathrm{SO}_3$ in the container at equilibrium is approximately:
$$\boxed{0.72 \, \text{moles}}$$