Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270k to decompose.

(i) The order of decomposition reaction of cyclobutane is first order.

(ii) It requires 53.15 milliseconds for 99% decomposition of cyclobutane.

(d) Rate law = k [\rm Cl_2Cl

2

, as chlorine is the reactant.

(e) Thus student claim is false as Chlorine gas is the reactant, not a catalyst.

(i) The order of reaction can be determined with the help of a graph given below. The graph depicts the reduction in the concentration of cyclobutane with time depicting decomposition.

The graph depicts the reaction to be of first-order kinetics. The general concentration expression can be:

\rm C_(_t_)\;=\;C_(_0_)\;e^-^k^t

Where, \rm C_(_t_) = concentration at time t

\rm C_(_0_) = initial concentration 1.60 mol/liter

k = rate constant

for first order reaction,

k = In \rm \frac{2}{Half\;time}

Halftime

2

From the graph, half time  = 0.008 sec i.e. time when the concentration reduced to 50%.

k = In \rm \frac{2}{0.008}

0.008

2

k = 86.64 / sec.

(ii) The time required for 99% decomposition can be calculated, by calculating the \rm C_(_t_) for left 1 % solution.

0.01 \rm C_(_0_)  =  \rm e^-^k^t

-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 [\rm Cl_2Cl

2

(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.

uh since this is clearly a troll i'll just answer for points.

your answer is oof.

answer:

sans lol

explanation: