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

Would you like further details on any of these calculations?

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.