At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×10^3. If the initial concentration of NO is 0.175 M, what is the equilibrium concentration of NO?

To solve for the equilibrium concentration of NO in the reaction \(2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)\), we can use an ICE (Initial, Change, Equilibrium) table along with the equilibrium constant expression.

### Step 1: Write the balanced equation and the equilibrium expression

\[
2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)
\]

The equilibrium constant expression for this reaction is:

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

Given:
- \(K_c = 2.4 \times 10^3\)
- Initial concentration of NO, \([\text{NO}]_0 = 0.175 \, M\)
- Initial concentrations of \( \text{N}_2 \) and \( \text{O}_2 \) are 0 M.

### Step 2: Set up the ICE table

| Species        | Initial (M)       | Change (M)        | Equilibrium (M) |
|----------------|-------------------|-------------------|-----------------|
| \(\text{NO}\)  | 0.175             | \(-2x\)            | \(0.175 - 2x\)  |
| \(\text{N}_2\) | 0                 | \(+x\)             | \(x\)           |
| \(\text{O}_2\) | 0                 | \(+x\)             | \(x\)           |

### Step 3: Substitute into the equilibrium expression

\[
K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} = \frac{x \cdot x}{(0.175 - 2x)^2} = 2.4 \times 10^3
\]

This simplifies to:

\[
K_c = \frac{x^2}{(0.175 - 2x)^2} = 2.4 \times 10^3
\]

### Step 4: Solve for \(x\)

Taking the square root on both sides:

\[
\sqrt{K_c} = \frac{x}{0.175 - 2x}
\]

\[
\sqrt{2.4 \times 10^3} = \frac{x}{0.175 - 2x}
\]

\[
49 = \frac{x}{0.175 - 2x}
\]

Solving for \(x\):

\[
49(0.175 - 2x) = x
\]

\[
8.575 - 98x = x
\]

\[
8.575 = 99x
\]

\[
x = \frac{8.575}{99} \approx 0.0866 \, M
\]

### Step 5: Calculate the equilibrium concentration of NO

\[
[\text{NO}]_{eq} = 0.175 - 2x = 0.175 - 2(0.0866) \approx 0.175 - 0.1732 = 0.0018 \, M
\]

So, the equilibrium concentration of NO is approximately \(0.0018 \, M\).

Would you like more details on this problem or have any questions?

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Here are some related questions to consider:

1. How does temperature affect the equilibrium constant \(K_c\) in this reaction?
2. What happens to the equilibrium concentration of NO if the initial concentration is doubled?
3. How would adding an inert gas at constant volume affect the equilibrium?
4. What is the relationship between \(K_p\) and \(K_c\) for this reaction?
5. How does Le Chatelier's principle apply to this reaction if the concentration of O₂ is increased?
6. What is the significance of a large equilibrium constant, like \(K_c = 2.4 \times 10^3\)?
7. How would the equilibrium shift if the pressure is increased by decreasing the volume?
8. Can you derive the expression for \(K_c\) from the standard Gibbs free energy change?

**Tip:** In equilibrium calculations, approximations like neglecting \(x\) in the denominator can simplify calculations when \(x\) is very small compared to the initial concentration.

Learn how to calculate the equilibrium concentration of NO in the chemical reaction 2NO(g) ⇌ N2(g) + O2(g) at 2000 °C with a detailed step-by-step solution. This problem covers equilibrium chemistry, ICE tables, and equilibrium constant expressions. Suitable for undergraduate students and chemistry enthusiasts.

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