To determine the order of the reaction with respect to NO, we will follow these steps:
1. Formulate the Rate Law:
For a reaction of the form [tex]\( aA + bB \to \text{products} \)[/tex], the rate law is generally expressed as:
[tex]\[ \text{Rate} = k [A]^x [B]^y \][/tex]
For our specific reaction, it can be written as:
[tex]\[ \text{Rate} = k [NO]^m [Cl_2]^n \][/tex]
where [tex]\( m \)[/tex] is the order of the reaction with respect to NO, and [tex]\( n \)[/tex] is the order of the reaction with respect to [tex]\( Cl_2 \)[/tex].
2. Choose Data Sets to Compare:
To find the order of the reaction with respect to NO ([tex]\( m \)[/tex]), we need to choose experiments where the concentration of [tex]\( [Cl_2] \)[/tex] is constant. We can then look at how changes in [tex]\( [NO] \)[/tex] affect the rate.
From the data:
\begin{tabular}{|c|c|c|}
\hline Initial [NO], & Initial [tex]$\left[ Cl_2 \right]$[/tex], & Initial Reaction Rate \\
\hline 0.2 M & 0.3 M & [tex]\( 2.2 \times 10^{-5} \)[/tex] M/s \\
\hline 0.4 M & 0.3 M & [tex]\( 8.8 \times 10^{-5} \)[/tex] M/s \\
\hline
\end{tabular}
We can use these two sets of data because [tex]\( [Cl_2] \)[/tex] is the same (0.3 M).
3. Set Up the Rate Equations and Divide:
Let's denote the rates from the experiments as [tex]\(\text{Rate}_1\)[/tex] and [tex]\(\text{Rate}_2\)[/tex]:
For the first experiment:
[tex]\[ \text{Rate}_1 = k [NO_1]^m [Cl_2]^n \][/tex]
For the second experiment:
[tex]\[ \text{Rate}_2 = k [NO_2]^m [Cl_2]^n \][/tex]
Now divide [tex]\(\text{Rate}_2\)[/tex] by [tex]\(\text{Rate}_1\)[/tex]:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [NO_2]^m [Cl_2]^n}{k [NO_1]^m [Cl_2]^n} \][/tex]
Simplifying, we get:
[tex]\[ \frac{8.8 \times 10^{-5}}{2.2 \times 10^{-5}} = \left( \frac{[NO]_2}{[NO]_1} \right)^m \][/tex]
[tex]\[ \frac{8.8 \times 10^{-5}}{2.2 \times 10^{-5}} = \left( \frac{0.4}{0.2} \right)^m \][/tex]
[tex]\[ 4 = 2^m \][/tex]
4. Solve for [tex]\( m \)[/tex]:
To isolate [tex]\( m \)[/tex], take the logarithm on both sides:
[tex]\[ \log(4) = m \log(2) \][/tex]
[tex]\[ m = \frac{\log(4)}{\log(2)} \][/tex]
Since [tex]\(\log(4) = 2 \log(2)\)[/tex]:
[tex]\[ m = 2 \][/tex]
Thus, the order of the reaction with respect to NO is 2.
