Let's solve the problem step-by-step.

Step 1: Write down the balanced chemical equation

$\text{N}_2 + \text{O}_2 \leftrightarrow 2\text{NO}$

Step 2: Define the initial conditions

• Volume of the flask, $V = 0.5$ L
• Moles of $\text{N}_2$ = 1.0 mol
• Moles of $\text{O}_2$ = 1.5 mol
• Equilibrium constant, $K_c = 3.24$

Step 3: Initial concentrations

Since the volume of the flask is 0.5 L, we calculate the initial concentrations of $\text{N}_2$ and $\text{O}_2$:

$[\text{N}_2]_0 = \frac{1.0 \, \text{mol}}{0.5 \, \text{L}} = 2.0 \, \text{M}$ $[\text{O}_2]_0 = \frac{1.5 \, \text{mol}}{0.5 \, \text{L}} = 3.0 \, \text{M}$

Step 4: Set up the equilibrium expression

Let $x$ be the change in molarity of $\text{N}_2$ and $\text{O}_2$ that reacts to form $2x$ molarity of $\text{NO}$. The equilibrium concentrations are:

$[\text{N}_2] = 2.0 - x$ $[\text{O}_2] = 3.0 - x$ $[\text{NO}] = 2x$

Step 5: Write the equilibrium expression using $K_c$

The equilibrium constant expression is:

[ K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} ]

Plugging in the equilibrium concentrations:

$3.24 = \frac{(2x)^2}{(2.0 - x)(3.0 - x)}$

Step 6: Solve for $x$

$3.24 = \frac{4x^2}{(2.0 - x)(3.0 - x)}$

Cross-multiply and solve the quadratic equation:

$3.24 \cdot (2.0 - x)(3.0 - x) = 4x^2$ $3.24 \cdot (6.0 - 5x + x^2) = 4x^2$ $19.44 - 16.2x + 3.24x^2 = 4x^2$

Combine like terms:

$19.44 - 16.2x - 0.76x^2 = 0$

Rearrange into standard quadratic form:

$0.76x^2 + 16.2x - 19.44 = 0$

Step 7: Solve the quadratic equation

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 0.76$, $b = -16.2$, and $c = -19.44$:

$x = \frac{-(-16.2) \pm \sqrt{(-16.2)^2 - 4 \cdot 0.76 \cdot (-19.44)}}{2 \cdot 0.76}$

Step 8: Calculate the values

Solving this gives:

$x \approx 1.11$

Step 9: Determine the equilibrium concentration of NO

$[\text{NO}] = 2x = 2 \times 1.11 = 2.22 \, \text{M}$

Therefore, the equilibrium molarity of $\text{NO}$ is 2.22 M.

Would you like more details or have any questions?

Related Questions

1. How would the equilibrium shift if the temperature were increased?
2. What would happen to the equilibrium concentration of $\text{NO}$ if more $\text{O}_2$ were added initially?
3. How does the value of the equilibrium constant $K_c$ change with temperature?
4. What are the units of $K_c$ in this reaction, and why?
5. How do you use the ICE (Initial, Change, Equilibrium) table to find equilibrium concentrations?

Tip

Always check if your quadratic equation solution makes sense by substituting the value back into the equilibrium expression to verify accuracy.