At a certain concentration of N2 and O3, the initial rate of reaction is 0.830 M / s. What would the initial rate of the reaction be if the concentration of N2 were doubled? Round your answer to 3 significant digits.

To determine the new initial rate of the reaction when the concentration of N2 is doubled, we need to understand the rate law for the reaction. The rate law is generally expressed as:

$\text{Rate} = k [\text{N}_2]^m [\text{O}_3]^n$

where:

• $k$ is the rate constant,
• $[\text{N}_2]$ is the concentration of N2,
• $[\text{O}_3]$ is the concentration of O3,
• $m$ and $n$ are the reaction orders with respect to N2 and O3, respectively.

Given that the initial rate of the reaction is 0.830 M/s, we can write:

$0.830 = k [\text{N}_2]^m [\text{O}_3]^n$

When the concentration of N2 is doubled, the new concentration of N2 becomes $2[\text{N}_2]$. The new rate of the reaction can be expressed as:

$\text{New Rate} = k (2[\text{N}_2])^m [\text{O}_3]^n$

This simplifies to:

$\text{New Rate} = k 2^m [\text{N}_2]^m [\text{O}_3]^n$

Since $k [\text{N}_2]^m [\text{O}_3]^n = 0.830$, we can substitute this into the equation:

$\text{New Rate} = 2^m \times 0.830$

To find the new rate, we need to know the value of $m$. However, since the value of $m$ is not provided, we will assume that the reaction is first-order with respect to N2 (i.e., $m = 1$) for the sake of this calculation. If $m = 1$:

$\text{New Rate} = 2^1 \times 0.830$ $\text{New Rate} = 2 \times 0.830$ $\text{New Rate} = 1.66$

Therefore, the new initial rate of the reaction, when the concentration of N2 is doubled, is approximately 1.66 M/s, rounded to three significant digits.