Solve this step by step: The reaction of nitrogen with oxygen in the combustion chamber of an automobile engine produces a variety of toxic oxides of nitrogen. One such reaction can be represented by the following equation:

2N2(g) + O2(g) 2N2O(g)

At a very high temperature, nitrogen at an initial concentration of 1.20mol/L is mixed with oxygen at an initial concentration of 1.00mol/L. This reaction mixture reaches equilibrium when the concentration of the produced dinitrogen oxide, commonly known as nitrous oxide, is 1.00mol/L.

6. For the reaction described above, calculate the equilibrium concentration of O2(g).

0.20 mol/L

1.20 mol/L

0.00 mol/L

0.50 mol/L

7. For the reaction described above, calculate the numerical value of the equilibrium constant.

100

50

75

10

### Step-by-Step Solution

#### Reaction:
\[ 2\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{N}_2\text{O}(g) \]

#### Initial Conditions:
- Initial concentration of \(\text{N}_2\) = 1.20 mol/L
- Initial concentration of \(\text{O}_2\) = 1.00 mol/L

#### At Equilibrium:
- Equilibrium concentration of \(\text{N}_2\text{O}\) = 1.00 mol/L

### Step 1: Write the ICE Table

| Species   | Initial (mol/L) | Change (mol/L)         | Equilibrium (mol/L)   |
|-----------|------------------|------------------------|-----------------------|
| \(\text{N}_2\)  | 1.20             | \(- x \)                | \( 1.20 - x \)        |
| \(\text{O}_2\)  | 1.00             | \(- \frac{x}{2} \)      | \( 1.00 - \frac{x}{2} \)  |
| \(\text{N}_2\text{O}\) | 0.00             | \(+ x \)                | \( x \)               |

Given that the equilibrium concentration of \(\text{N}_2\text{O}\) is 1.00 mol/L:

\[ x = 1.00 \]

### Step 2: Calculate Equilibrium Concentrations of \(\text{N}_2\) and \(\text{O}_2\)

For \(\text{N}_2\):

\[ \text{N}_2\)_{eq} = 1.20 - x \]
\[ \text{N}_2\)_{eq} = 1.20 - 1.00 \]
\[ \text{N}_2\)_{eq} = 0.20 \]

For \(\text{O}_2\):

\[ \text{O}_2\)_{eq} = 1.00 - \frac{x}{2} \]
\[ \text{O}_2\)_{eq} = 1.00 - \frac{1.00}{2} \]
\[ \text{O}_2\)_{eq} = 1.00 - 0.50 \]
\[ \text{O}_2\)_{eq} = 0.50 \]

### Step 3: Calculate the Equilibrium Constant (\(K_c\))

The equilibrium constant expression for the reaction is:

\[ K_c = \frac{[\text{N}_2\text{O}]^2}{[\text{N}_2]^2 [\text{O}_2]} \]

Substitute the equilibrium concentrations into the expression:

\[ K_c = \frac{(1.00)^2}{(0.20)^2 \cdot (0.50)} \]
\[ K_c = \frac{1.00}{0.04 \cdot 0.50} \]
\[ K_c = \frac{1.00}{0.02} \]
\[ K_c = 50 \]

### Summary of Answers:

6. The equilibrium concentration of \(\text{O}_2\) is **0.50 mol/L**.
7. The numerical value of the equilibrium constant is **50**.

### Additional Questions for Further Understanding:
1. What does the equilibrium constant tell us about the reaction?
2. How would changing the temperature affect the value of \(K_c\) for this reaction?
3. What are the units of the equilibrium constant for different reactions?
4. How can the reaction quotient (\(Q\)) be used to predict the direction of the reaction shift?
5. What role does pressure play in the equilibrium of reactions involving gases?

### Tip:
Always verify the consistency of your equilibrium calculations by cross-checking the calculated concentrations and ensuring they satisfy the initial conditions and the stoichiometry of the reaction.

Learn how to calculate the equilibrium concentration of O2 and the numerical value of the equilibrium constant for a chemical reaction involving nitrogen and oxygen. Detailed step-by-step solution provided. Suitable for college-level chemistry students.

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