Question

The tabulated data were collected for this reaction:


CH3Cl(g) + 3Cl2(g) → CCl4(g) + 3HCl(g)



[CH3Cl] (M) [Cl2] (M) Initial Rate (M/s)


0.050 0.050 0.014


0.100 0.050 0.029


0.100 0.100 0.041


0.200 0.200 0.115



(a) Write an expression for the reaction rate law and calculate the value of the rate constant, k.


(b) What is the overall order of the reaction?

Answer 1

Answer:

ai) Rate law,  $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$, the rate of backward reaction is given by:  

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b$

$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$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

Answer 2

The rate law is for this reaction is    $r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}$

where the rate constant is k = $1.25 M^{-0.5} s^{-1}$ 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:

$r = k [CH_3Cl] [Cl_2]^{0.5}$

Let's use the values of the first experiment to find the value of the rate constant.

$k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}$

The final rate law is:

$r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}$

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    $r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}$

where the rate constant is k = $1.25 M^{-0.5} s^{-1}$ and the overall order of the reaction is 1.5.

Learn more about the rate law here: https://brainly.com/question/14945022