Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Sure! Let's go through each part of the problem step by step:
### Part (a): Determining the Rate Law
The rate law for a reaction involving [tex]\( I^- \)[/tex] and [tex]\( OCl^- \)[/tex] can generally be expressed as:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
Where:
- [tex]\( k \)[/tex] is the rate constant.
- [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are the orders of the reaction with respect to [tex]\( I^- \)[/tex] and [tex]\( OCl^- \)[/tex], respectively.
To find the orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we compare the rates from different experiments where only one concentration changes at a time.
#### Finding [tex]\( m \)[/tex]:
Compare experiments 1 and 2 where [tex]\([OCl^-]\)[/tex] is constant:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \left(\frac{[I^-]_1}{[I^-]_2}\right)^m \][/tex]
[tex]\[ \frac{7.91 \times 10^{-2}}{3.95 \times 10^{-2}} = \left(\frac{0.12}{0.060}\right)^m \][/tex]
[tex]\[ 2 = 2^m \][/tex]
[tex]\[ m = 1.0018250414083267 \approx 1 \][/tex]
#### Finding [tex]\( n \)[/tex]:
Compare experiments 1 and 3 where [tex]\([I^-]\)[/tex] is constant:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \left(\frac{[OCl^-]_1}{[OCl^-]_3}\right)^n \][/tex]
[tex]\[ \frac{7.91 \times 10^{-2}}{9.88 \times 10^{-3}} = \left(\frac{0.18}{0.090}\right)^n \][/tex]
[tex]\[ 8 = 2^n \][/tex]
[tex]\[ n = 3.001094747775477 \approx 3 \][/tex]
Thus, the rate law is:
[tex]\[ \text{Rate} = k [I^-] [OCl^-]^3 \][/tex]
### Part (b): Calculating the Rate Constant [tex]\( k \)[/tex]
Using the rate law derived and the data from any experiment, we can calculate [tex]\( k \)[/tex]. We'll take the average [tex]\( k \)[/tex] from all four experiments for better accuracy.
Using the rate law:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
[tex]\[ k = \frac{\text{Rate}}{[I^-]^m [OCl^-]^n} \][/tex]
Calculate [tex]\( k \)[/tex] for each experiment:
[tex]\[ k_1 = \frac{7.91 \times 10^{-2}}{(0.12)^1 (0.18)^3} \][/tex]
[tex]\[ k_2 = \frac{3.95 \times 10^{-2}}{(0.060)^1 (0.18)^3} \][/tex]
[tex]\[ k_3 = \frac{9.88 \times 10^{-3}}{(0.030)^1 (0.090)^3} \][/tex]
[tex]\[ k_4 = \frac{7.91 \times 10^{-2}}{(0.24)^1 (0.090)^3} \][/tex]
Taking the average [tex]\( k \)[/tex] from these values, we get:
[tex]\[ k = 284.42355843931466 \ \text{L}^3 \ \text{mol}^{-3} \ \text{s}^{-1} \][/tex]
### Part (c): Calculating the Initial Rate for Given Concentrations
Given [tex]\( [I^-] = 0.15 \ \text{mol/L} \)[/tex] and [tex]\( [OCl^-] = 0.15 \ \text{mol/L} \)[/tex]:
Using the rate law:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
[tex]\[ \text{Rate} = 284.42355843931466 \times (0.15)^1 \times (0.15)^3 \][/tex]
Calculate the rate:
[tex]\[ \text{Rate} = 284.42355843931466 \times 0.15 \times 0.15^3 \][/tex]
[tex]\[ \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \][/tex]
So, the initial rate for the given concentrations is:
[tex]\[ \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \][/tex]
To summarize:
- Rate Law: [tex]\( \text{Rate} = 284.42355843931466 [I^-] [OCl^-]^3 \)[/tex]
- Rate Constant: [tex]\( k = 284.42355843931466 \ \text{L}^3 \ \text{mol}^{-3} \ \text{s}^{-1} \)[/tex]
- Initial Rate: [tex]\( \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \)[/tex] when both [tex]\( [I^-] \)[/tex] and [tex]\( [OCl^-] \)[/tex] are initially 0.15 mol/L.
### Part (a): Determining the Rate Law
The rate law for a reaction involving [tex]\( I^- \)[/tex] and [tex]\( OCl^- \)[/tex] can generally be expressed as:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
Where:
- [tex]\( k \)[/tex] is the rate constant.
- [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are the orders of the reaction with respect to [tex]\( I^- \)[/tex] and [tex]\( OCl^- \)[/tex], respectively.
To find the orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we compare the rates from different experiments where only one concentration changes at a time.
#### Finding [tex]\( m \)[/tex]:
Compare experiments 1 and 2 where [tex]\([OCl^-]\)[/tex] is constant:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \left(\frac{[I^-]_1}{[I^-]_2}\right)^m \][/tex]
[tex]\[ \frac{7.91 \times 10^{-2}}{3.95 \times 10^{-2}} = \left(\frac{0.12}{0.060}\right)^m \][/tex]
[tex]\[ 2 = 2^m \][/tex]
[tex]\[ m = 1.0018250414083267 \approx 1 \][/tex]
#### Finding [tex]\( n \)[/tex]:
Compare experiments 1 and 3 where [tex]\([I^-]\)[/tex] is constant:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \left(\frac{[OCl^-]_1}{[OCl^-]_3}\right)^n \][/tex]
[tex]\[ \frac{7.91 \times 10^{-2}}{9.88 \times 10^{-3}} = \left(\frac{0.18}{0.090}\right)^n \][/tex]
[tex]\[ 8 = 2^n \][/tex]
[tex]\[ n = 3.001094747775477 \approx 3 \][/tex]
Thus, the rate law is:
[tex]\[ \text{Rate} = k [I^-] [OCl^-]^3 \][/tex]
### Part (b): Calculating the Rate Constant [tex]\( k \)[/tex]
Using the rate law derived and the data from any experiment, we can calculate [tex]\( k \)[/tex]. We'll take the average [tex]\( k \)[/tex] from all four experiments for better accuracy.
Using the rate law:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
[tex]\[ k = \frac{\text{Rate}}{[I^-]^m [OCl^-]^n} \][/tex]
Calculate [tex]\( k \)[/tex] for each experiment:
[tex]\[ k_1 = \frac{7.91 \times 10^{-2}}{(0.12)^1 (0.18)^3} \][/tex]
[tex]\[ k_2 = \frac{3.95 \times 10^{-2}}{(0.060)^1 (0.18)^3} \][/tex]
[tex]\[ k_3 = \frac{9.88 \times 10^{-3}}{(0.030)^1 (0.090)^3} \][/tex]
[tex]\[ k_4 = \frac{7.91 \times 10^{-2}}{(0.24)^1 (0.090)^3} \][/tex]
Taking the average [tex]\( k \)[/tex] from these values, we get:
[tex]\[ k = 284.42355843931466 \ \text{L}^3 \ \text{mol}^{-3} \ \text{s}^{-1} \][/tex]
### Part (c): Calculating the Initial Rate for Given Concentrations
Given [tex]\( [I^-] = 0.15 \ \text{mol/L} \)[/tex] and [tex]\( [OCl^-] = 0.15 \ \text{mol/L} \)[/tex]:
Using the rate law:
[tex]\[ \text{Rate} = k [I^-]^m [OCl^-]^n \][/tex]
[tex]\[ \text{Rate} = 284.42355843931466 \times (0.15)^1 \times (0.15)^3 \][/tex]
Calculate the rate:
[tex]\[ \text{Rate} = 284.42355843931466 \times 0.15 \times 0.15^3 \][/tex]
[tex]\[ \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \][/tex]
So, the initial rate for the given concentrations is:
[tex]\[ \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \][/tex]
To summarize:
- Rate Law: [tex]\( \text{Rate} = 284.42355843931466 [I^-] [OCl^-]^3 \)[/tex]
- Rate Constant: [tex]\( k = 284.42355843931466 \ \text{L}^3 \ \text{mol}^{-3} \ \text{s}^{-1} \)[/tex]
- Initial Rate: [tex]\( \text{Rate} = 0.1431940465235448 \ \text{mol} \ \text{L}^{-1} \ \text{s}^{-1} \)[/tex] when both [tex]\( [I^-] \)[/tex] and [tex]\( [OCl^-] \)[/tex] are initially 0.15 mol/L.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
