Respuesta :
Answer:
ai) Rate law, [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
aii) Rate constant, k = 1.25
b) Overall order of reaction = 1.5
Explanation:
Equation of Reaction:
[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]
If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:
[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]
k is constant for all the stages
Using the information provided in lines 1 and 2 of the table:
[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]
Using the information provided in lines 3 and 4 of the table and insering the value of a:
[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]
[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]
The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:
[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]
b) Overall order of reaction = a + b
Overall order of reaction = 1 + 0.5
Overall order of reaction = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Let's consider the following reaction.
CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)
What is the rate law?
The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.
The rate law for this reaction is:
r = k [CH₃Cl]ᵃ [Cl₂]ᵇ
where,
- r is the initial rate.
- k is the rate constant.
- a is the reaction order for CH₃Cl.
- b is the reaction order for Cl₂.
If we write the ratio r₂/r₁, we get:
r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ
r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ
(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ
a ≈ 1
If we write the ratio r₃/r₂, we get:
r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ
r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ
(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ
b ≈ 0.5
So far, the rate law is:
[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]
Let's use the values of the first experiment to find the value of the rate constant.
[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]
The final rate law is:
[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
What is the overall order of the reaction?
It is the sum of the individual orders of reaction.
a + b = 1 + 0.5 = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Learn more about the rate law here: https://brainly.com/question/14945022
