Let's break down the question into three parts and solve each one step by step:
### Part 1: Determine the rate law for this reaction
The rate law expresses the rate [tex]\( R \)[/tex] of the reaction in terms of the concentration of the reactants raised to a power, typically represented as:
[tex]\[ R = k[\text{NO}_2][\text{CO}] \][/tex]
In this problem, the rate law for the reaction is given as:
[tex]\[ \text{Rate} = k[\text{NO}_2][\text{CO}] \][/tex]
### Part 2: Calculate the value of the rate constant [tex]\( k \)[/tex] at [tex]\( 225^{\circ} C \)[/tex]
The rate constant [tex]\( k \)[/tex] is provided in this problem. The value of [tex]\( k \)[/tex] at [tex]\( 225^{\circ} C \)[/tex] is:
[tex]\[ k = 0.00005 \, M^{-1} s^{-1} \][/tex]
### Part 3: Calculate the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] when [tex]\([ \text{NO}_2 ] = [ \text{CO} ] = 0.510 \, M \)[/tex]
We use the rate law determined in Part 1 to find the rate of the reaction:
[tex]\[ \text{Rate} = k[\text{NO}_2][\text{CO}] \][/tex]
Given:
- [tex]\( k = 0.00005 \, M^{-1} s^{-1} \)[/tex]
- [tex]\([ \text{NO}_2 ] = 0.510 \, M \)[/tex]
- [tex]\([ \text{CO} ] = 0.510 \, M \)[/tex]
We substitute these values into the rate equation:
[tex]\[ \text{Rate} = 0.00005 \times 0.510 \times 0.510 \][/tex]
The rate of appearance of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ \text{Rate} \approx 1.3005 \times 10^{-5} \, M/s \][/tex]
Therefore, the detailed steps to arrive at the rate of appearance of [tex]\( \text{CO}_2 \)[/tex], are as follows:
1. Substitute [tex]\([ \text{NO}_2 ] = 0.510 \, M \)[/tex] and [tex]\([ \text{CO} ] = 0.510 \, M \)[/tex] into the rate law.
2. Multiply the rate constant [tex]\( k \)[/tex] by the concentrations of [tex]\([ \text{NO}_2 ]\)[/tex] and [tex]\([ \text{CO} ] \)[/tex].
3. Calculate the product to obtain the rate of appearance of [tex]\( \text{CO}_2 \)[/tex].
Thus, the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] when both [tex]\([ \text{NO}_2 ]\)[/tex] and [tex]\([ \text{CO} ]\)[/tex] are at [tex]\( 0.510 \, M \)[/tex] is [tex]\( 1.3005 \times 10^{-5} \, M/s \)[/tex].
