Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

In a study to determine the rate of the following reaction:
[tex]\[ 2 NO(g) + O_2(g) \longrightarrow 2 NO_2(g) \][/tex]
the concentration of NO was [tex]\( 0.0300 \, M \)[/tex] at [tex]\( t = 5.0 \, s \)[/tex] and [tex]\( 0.0225 \, M \)[/tex] at [tex]\( t = 650.0 \, s \)[/tex]. What is the average rate of the reaction during this time period?

Express your answer in [tex]\( M/s \)[/tex].

Sagot :

To determine the average rate of the reaction:
[tex]\[ 2 \text{NO} (g) + \text{O}_2 (g) \longrightarrow 2 \text{NO}_2 (g) \][/tex]

given the concentration data, we proceed as follows:

1. Identify the initial and final concentrations and times:

- Initial concentration of NO: [tex]\([NO]_{\text{initial}} = 0.0300 \, M\)[/tex] at [tex]\( t_{\text{initial}} = 5.0 \, \text{s}\)[/tex]
- Final concentration of NO: [tex]\([NO]_{\text{final}} = 0.0225 \, M\)[/tex] at [tex]\( t_{\text{final}} = 650.0 \, \text{s}\)[/tex]

2. Calculate the change in concentration of NO ([tex]\(\Delta [NO]\)[/tex]):

[tex]\[ \Delta [NO] = [NO]_{\text{initial}} - [NO]_{\text{final}} \][/tex]
[tex]\[ \Delta [NO] = 0.0300 \, M - 0.0225 \, M \][/tex]
[tex]\[ \Delta [NO] = 0.0075 \, M \][/tex]

3. Calculate the change in time ([tex]\(\Delta t\)[/tex]):

[tex]\[ \Delta t = t_{\text{final}} - t_{\text{initial}} \][/tex]
[tex]\[ \Delta t = 650.0 \, \text{s} - 5.0 \, \text{s} \][/tex]
[tex]\[ \Delta t = 645.0 \, \text{s} \][/tex]

4. Determine the average rate of reaction:

The average rate of the reaction is given by the change in concentration per unit time. For the reaction involving NO, we use the rate of decrease of [tex]\([NO]\)[/tex].

[tex]\[ \text{Average rate} = \frac{\Delta [NO]}{\Delta t} \][/tex]
[tex]\[ \text{Average rate} = \frac{0.0075 \, M}{645.0 \, s} \][/tex]
[tex]\[ \text{Average rate} = 1.1627906976744185 \times 10^{-5} \, \text{M/s} \][/tex]

Therefore, the average rate of the reaction during this time period is [tex]\(1.1627906976744185 \times 10^{-5} \, \text{M/s}\)[/tex].