Suppose a 500mL flask is filled with 1.0 mol of N2 and 1.5 mol of O2. The following reaction becomes possible:
N2 + O2 -><- 2NO
The equilibrium constant K for this reaction is 3.24 at the temperature of the flask. 
Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

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.

Suppose a 500mL flask is filled with 1.0 mol of N2 and 1.5 mol of O2. The following reaction becomes possible: N2 + O2 -><- 2NO The equilibrium constant K for this reaction is 3.24 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

Calculate the equilibrium molarity of NO in a reaction between N2 and O2 in a 500mL flask. Using chemical equilibrium principles, stoichiometry, and quadratic equations, this problem provides a step-by-step solution. Ideal for college-level chemistry students.

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