Respuesta :
Answer:
(c)(i) The order of the reaction based on the graph provided is first order.
(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.
d) Rate = K [Cl₂]
K = rate constant
The justification is presented in the Explanation provided below.
e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.
Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.
So, this shows why the student's claim is false.
Explanation:
To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.
So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.
The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.
C(t) = C₀ e⁻ᵏᵗ
where C(t) = concentration of the reactant at any time
C₀ = Initial concentration of cyclobutane = 1.60 mol/L
k = rate constant
The rate constant for a first order reaction is given
k = (In 2)/T
where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.
From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s
k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s
(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose
C(t) = C₀ e⁻ᵏᵗ
when 99% of the cyclobutane has decomposed, there's only 1% left
C(t) = 0.01C₀
k = 86.64 /s
t = ?
0.01C₀ = C₀ e⁻ᵏᵗ
e⁻ᵏᵗ = 0.01
In e⁻ᵏᵗ = In 0.01 = -4.605
-kt = -4.605
t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.
d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as
Cl₂ → 2Cl (slow)
Cl + C₅H₁₀ → HCl + C₅H₉ (fast)
C₅H₉ + Cl → C₅H₉Cl (fast)
The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.
For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is
Rate = K [Cl₂]
K = rate constant
Since, no intermediates appear in this rate law, no further simplification is necessary.
The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.
e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.
Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.
So, this shows why the student's claim is false.
Hope this Helps!!!
(i) The order of reaction is first order.
(ii) It requires 53.15 miliseconds for 99% decomposition.
(d) Rate law = k [[tex]\rm Cl_2[/tex]].
(e) Chlorine gas is the reactant. Thus student claim is false.
(i) The order of reaction can be determined with the help of a graph. The graph depicts the exponential behavior of decomposition of cyclobutane with respect to time.
The graph depicts the reaction to be of first-order kinetics. The general concentration expression can be:
[tex]\rm C_(_t_)\;=\;C_(_0_)\;e^-^k^t[/tex]
Where, [tex]\rm C_(_t_)[/tex] = concentration at time t
[tex]\rm C_(_0_)[/tex] = initial concentration 1.60 mol/liter
k = rate constant
for first order reaction,
k = In [tex]\rm \frac{2}{Half\;time}[/tex]
From the graph, half time = 0.008 sec
k = In [tex]\rm \frac{2}{0.008}[/tex]
k = 86.64 / sec.
(ii) The time required for 99% decomposition can be calculated, by calculating the [tex]\rm C_(_t_)[/tex] for left 1 % solution.
[tex]\rm C_(_t_)\;=0.01\;C_(_0_)[/tex]
0.01 [tex]\rm C_(_0_)[/tex] = [tex]\rm C_(_0_)[/tex] [tex]\rm e^-^k^t[/tex]
-kt = -4.605
t = 53.15 miliseconds
(d) The rate law is obtained against the reaction.
In rate law, the intermediates that appear in the reaction are removed, and the equation has been formed.
The slow step is the formation of Chlorine. Thus the rate law will be:
Rate = k [[tex]\rm Cl_2[/tex]]
(e) The catalyst function in improvising the rate of the reaction. Since, the reaction has chlorine as the part of the reactant, it is consumed in the reaction and thus does not act as a catalyst. Thus the claim of the student is false.
For more information, refer the link:
https://brainly.com/question/1368306?referrer=searchResults
