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To solve this, let's use the provided information about the reaction orders. The reaction is given as:

- Second-order in \( N_2 \)
- First-order in \( H_2 \)

The rate law for such a reaction can be written as:
\[
\text{Rate} = k [\text{N}_2]^2 [\text{H}_2]
\]
where \( k \) is the rate constant.

Given the first row in the table:

\[
[\text{N}_2] = 2.13 \, M, \quad [\text{H}_2] = 1.15 \, M, \quad \text{Rate} = 69.0 \, \text{M/s}
\]

we can solve for \( k \) as follows:

\[
69.0 = k \cdot (2.13)^2 \cdot (1.15)
\]
Solving for \( k \) will allow us to find the rate for the other two scenarios in the table.

### Step 1: Calculate \( k \)
1. Square the concentration of \( \text{N}_2 \): \( (2.13)^2 = 4.5369 \)
2. Multiply by the concentration of \( \text{H}_2 \): \( 4.5369 \times 1.15 = 5.217435 \)
3. Solve for \( k \): 
   \[
   k = \frac{69.0}{5.217435} \approx 13.23 \, \text{M}^{-2}\text{s}^{-1}
   \]

### Step 2: Use \( k \) to Find Rates for the Other Rows

1. **Second row:** \( [\text{N}_2] = 2.13 \, M \), \( [\text{H}_2] = 2.32 \, M \)
   \[
   \text{Rate} = 13.23 \times (2.13)^2 \times (2.32)
   \]
   Calculating:
   \[
   \text{Rate} \approx 13.23 \times 4.5369 \times 2.32 = 139.9 \, \text{M/s}
   \]

2. **Third row:** \( [\text{N}_2] = 3.48 \, M \), \( [\text{H}_2] = 0.703 \, M \)
   \[
   \text{Rate} = 13.23 \times (3.48)^2 \times (0.703)
   \]
   Calculating:
   \[
   \text{Rate} \approx 13.23 \times 12.1104 \times 0.703 = 112.5 \, \text{M/s}
   \]

### Final Table

| \([ \text{N}_2 ]\) | \([ \text{H}_2 ]\) | Initial Rate of Reaction (M/s) |
|-------------------|-------------------|-------------------------------|
| 2.13 M           | 1.15 M           | 69.0 M/s                      |
| 2.13 M           | 2.32 M           | 139.9 M/s                     |
| 3.48 M           | 0.703 M          | 112.5 M/s                     |

Would you like further details on any of these calculations?

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**Related Questions:**

1. What happens to the reaction rate if the concentration of \( \text{H}_2 \) doubles?
2. How would the rate change if the concentration of \( \text{N}_2 \) triples?
3. What units does the rate constant \( k \) have for a reaction of this order?
4. Can you determine the reaction mechanism based on this rate law?
5. How would the rate change if the temperature increased, assuming \( k \) depends on temperature?

**Tip:** Remember, each reaction order corresponds to the exponent in the rate law for each reactant, influencing the rate’s sensitivity to changes in concentration.

A certain reaction is second order in N2 and first order in H2. Use this information to complete the table below, ensuring each entry has the correct number of significant digits.

This guide explains how to calculate the initial rate of reaction for a chemical reaction that is second order in N2 and first order in H2, using the rate law. Learn how to determine the rate constant k and complete the reaction rate table with the correct significant figures.

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