Respuesta :
The table with the data is in the picture attached.
Answer:
- [tex]k=0.0033M^{-2}.s^{-1}[/tex]
Explanation:
The reaction equation suggests that the law could have this form:
- [tex]rate=k[A]^a[B]^b[C]^c[/tex]
Then, the work is to find the values of the exponents that satisfy the initial rate data.
A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.
Then, so far you can say:
- [tex]rate=k[A]^a[C]^c[/tex]
When you use trials 1 and 2, you get:
[tex]\frac{r_2}{r_1}=\frac{27M/s}{9M/s}=\frac{(0.3M)^a(0.3M)^b(0.9M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=3^c\\\\ 3=3^c\\ \\ 1=c[/tex]
Now, you can use trials 1 and 3 to determine the other exponent:
[tex]\frac{r_3}{r_1}=\frac{36M/s}{9M/s}=\frac{(0.6M)^a(0.3M)^b(0.3M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=2^a\\\\ 4=2^a\\ \\ 2^2=2^a\\ \\ 2=a[/tex]
Thus, you have the rate law:
- [tex]r=k[A]^2[C][/tex]
Now, you just use any trial to obtain k. Using trail 1:
- [tex]k(0.3M)^2(0.3M)=9.10^{-5}M/s[/tex]
Which yields:
- [tex]k=0.0033M^{-2}.s^{-1}[/tex]
The rate constant for the given reaction has been 0.0033 [tex]\rm \bold{M^-^1.s^-^1}[/tex].
The given reaction has been:
A + B +C [tex]\rightarrow[/tex] D + E
The rate of reaction can be given by:
Rate = [tex]\rm k\;[A]^a\;[B]^b\;[C]^c[/tex]
The value of the coefficients can be calculated from the given trials, When the concentration for two reactants is same, the coefficient for the third reactant can be calculated from the ratio of the two reactions.
- For calculating a:
Ratio of trial 1 to trial 3
[tex]\rm \dfrac{Rate\;3}{Rate\;1}[/tex] = [tex]\rm \dfrac{[A]^a\;[B]^b\;[C]^c}{[A]^a\;[B]^b\;[C]^c}[/tex]
[tex]\rm \dfrac{3.6\;\times\;10^-^4}{9.0\;\times\;10^-^5}[/tex] = [tex]\rm \dfrac{(0.60)^a\;(0.30)^b\;(0.30)^c}{(0.30)^a\;(0.30)^b\;(0.30)^c}[/tex]
4 = [tex]\rm (2)^a[/tex]
[tex]\rm (2)^2\;=\;(2)^a[/tex]
2 = a
- For calculating b:
By changing the concentrations of B in 3rd and 4th trial, the rate has not been changed. Thus the rate has been independent of the concentration of B.
- For calculating c:
Ratio of trials 2 and 1.
[tex]\rm \dfrac{Rate\;2}{Rate\;1}[/tex] = [tex]\rm \dfrac{[A]^a\;[B]^b\;[C]^c}{[A]^a\;[B]^b\;[C]^c}[/tex]
[tex]\rm \dfrac{2.7\;\times\;10^-^4}{9.0\;\times\;10^-^5}[/tex] = [tex]\rm \dfrac{(0.30)^a\;(0.30)^b\;(0.90)^c}{(0.30)^a\;(0.30)^b\;(0.30)^c}[/tex]
3 = [tex]\rm (3)^c[/tex]
c = 1.
The rate of reaction can be given by:
Rate = [tex]\rm k\;[A]^a\;[C]^c[/tex]
From the trial 1:
[tex]\rm 9.0\;\times\;10^-^5\;=\;k\;[0.30]^2\;[0.30]^1[/tex]
0.00009 = k 0.09 [tex]\times[/tex] 0.30
k = 0.0033 [tex]\rm \bold{M^-^1.s^-^1}[/tex].
The rate constant for the given reaction has been 0.0033 [tex]\rm \bold{M^-^1.s^-^1}[/tex].
For more information about rate constant, refer to the link:
https://brainly.com/question/23184115
