Respuesta :
Answer:
[tex]0.00442 s^{-1}[/tex] is the value of the rate constant.
Explanation:
[tex]2 H_2O_2\rightarrow 2 H_2O + O_2[/tex]
Let the order of the reaction be x.
The rate law of the reaction can be written as:
[tex]R=k[H_2O_2]^x[/tex]
1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.
[tex]R=-\frac{0.697 M-0.882 M}{60 s-0 s}=0.00308 M/s[/tex]
[tex]0.00308 M/s=k[0.697 M]^x[/tex]..[1]
2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.
[tex]R=-\frac{0.236M-0.372M}{120s-60 s}=0.00227 M/s[/tex]
[tex]0.00218 M/s=k[0.236 M]^x[/tex]..[2]
[1] ÷ [2]
[tex]\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}[/tex]
Solving fro x:
x = 0.92 ≈ 1
[tex]R=k[H_2O_2]^1[/tex]
[tex]0.00308 M/s=k[0.697 M]^1[/tex]
[tex]k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}[/tex]
[tex]0.00442 s^{-1}[/tex] is the value of the rate constant.
The value of the rate constant is 0.00442 [tex]s^{-1}[/tex]
Balanced chemical reaction:
[tex]H_2O_2 --- > 2 H_2O+O_2[/tex]
Calculation for rate constant:
Let the order of the reaction be x.
The rate law of the reaction can be written as:
Rate = [tex]k[H_2O_2]^x[/tex]
1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.
[tex]R=-\frac{0697M-0.882 M}{60-0} \\\\R=0.00308M/s\\[/tex]
0.00308 M/s = [tex]k[0.697]^x[/tex]................(i)
2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.
[tex]R=-\frac{0.236M-0.372 M}{120-60} \\\\R=0.00227M/s\\[/tex]
0.00277 M/s = [tex]k[0.239]^x[/tex]................(ii)
On dividing ii by i
x = 0.92 ≈ 1
[tex]\text{Rate}=k[H_2O_2]^1\\\\0.00308=k[0.697]\\\\k=\frac{0.00308}{0.697} \\\\k=0.00442s^{-1}[/tex]
Thus, The value of the rate constant is 0.00442 [tex]s^{-1}[/tex]
Find more information about Rate constant here:
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