Answer :
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.
