A study of the rate of dimerization of \({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}}\) gave the data shown in:

\({\bf{2}}{{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}} \to {{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}\)

  1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
  2. Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus (\({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}}\)). What are the units of this rate?

(c) Determine the average rate of formation of \({{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}\) at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).

Short Answer

Expert verified
  1. The average rate of dimerization between 0s and 1600s was found to be\({\bf{3}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M/s and rate of dimerization between 1600s and 3200s was found to be\({\bf{1}}{\bf{.04 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M/s
  2. The instantaneous rate of dimerization at 3200s is\({\bf{ - 7}}{\bf{.83 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}\)M/s and units of this rate is M/s
  3. The average rate of formation of \({{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}\) at 3200s is \({\bf{1}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\) M/s

Step by step solution

01

Average rate of dimerization

Rate of dimerization= -\(\frac{{{\bf{\Delta (C}}{}_{\bf{4}}{\bf{H}}{}_{\bf{6}}{\bf{)}}}}{{{\bf{Vt}}}}\)

Rate of dimerization between 0s and 1600s

\(\begin{aligned}{}{\bf{ = }} - \frac{{{\bf{5}}{\bf{.04 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}} - {\bf{1}}{\bf{.00 \times 10}}{}^{{\bf{ - 2}}}{\bf{M}}}}{{{\bf{1600s}} - {\bf{0s}}}}\\{\bf{ = 3}}{\bf{.10 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M/s}}\end{aligned}\)

Rate of dimerization between 1600s and 3200s

\(\begin{aligned}{}{\bf{ = }} - \frac{{{\bf{3}}{\bf{.37 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}} - {\bf{5}}{\bf{.04 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}}}}{{{\bf{3200s}} - {\bf{1600s}}}}\\{\bf{ = 1}}{\bf{.04 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M/s}}\end{aligned}\)

02

Instantaneous rate of dimerization      

Graph of time versus (\({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}}\))

Slope between 1600s ­­­­­­­­and 3200s

Use points (1600s,\({\bf{5}}{\bf{.04 \times 10}}{}^{{\bf{ - 3}}}\)M) and (3200s,\({\bf{3}}{\bf{.37 \times 10}}{}^{{\bf{ - 3}}}\)M)

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{3}}{\bf{.37 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{M - 5}}{\bf{.04 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}}}}{{{\bf{3200s - 1600s}}}}\\{\bf{ = - }}\frac{{{\bf{0}}{\bf{.00167M}}}}{{{\bf{1600s}}}}\\{\bf{ = - 1}}{\bf{.04 \times 10}}{}^{{\bf{ - 6}}}{\bf{M/s}}\end{aligned}\)

Slope between 3200s ­­­­­­­­and 4800s

Use points (3200s,\({\bf{3}}{\bf{.37 \times 10}}{}^{{\bf{ - 3}}}\)M) and (4800s, \({\bf{2}}{\bf{.53 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}}\))

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{2}}{\bf{.53 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{M - 3}}{\bf{.37 \times 10}}{}^{{\bf{ - 3}}}{\bf{M}}}}{{{\bf{4800s - 3200s}}}}\\{\bf{ = - }}\frac{{{\bf{0}}{\bf{.00084M}}}}{{{\bf{1600s}}}}{\bf{ = - 5}}{\bf{.25 \times 10}}{}^{{\bf{ - 7}}}{\bf{M/s}}\end{aligned}\)

Take two slope and find average of instantaneous rate of dimerization

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{ - 1}}{\bf{.04 \times 10}}{}^{{\bf{ - 6}}}{\bf{ + }}\,\,{\bf{5}}{\bf{.25 \times 10}}{}^{{\bf{ - 7}}}}}{{\bf{2}}}\\{\bf{ = - 7}}{\bf{.83 \times 10}}{}^{{\bf{ - 7}}}{\bf{M/s}}\end{aligned}\)

03

Average rate of formation of \({{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}\) at 1600 s and the instantaneous rate of formation at 3200 s

Linked equation between average rate of dimerization and average rate of formation.

Equation:

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{ - 1}}}}{{\bf{2}}}\frac{{{\bf{\Delta (}}{{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}}{\bf{)}}}}{{{\bf{\Delta t}}}}\\{\bf{ = }}\frac{{{\bf{\Delta (}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{\Delta t}}}}\end{aligned}\)

Average rate of formation for \({{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}\) at 1600s and use rate of dimerization between 0s and 1600s

Equation:

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{ - 1}}}}{{\bf{2}}}{\bf{ \times }}\frac{{{\bf{3}}{\bf{.10 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M/s}}}}{{}}\\{\bf{ = }}\frac{{{\bf{\Delta (}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{\Delta t}}}}\\{\bf{ = }}\frac{{{\bf{\Delta (}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{\Delta t}}}}\\{\bf{ = 1}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M/s}}\end{aligned}\)

Linked equation between instantaneous rate of formation and instantaneous rate of dimerization.

Equation:

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{ - 1}}}}{{\bf{2}}}\frac{{{\bf{d(}}{{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{6}}}{\bf{)}}}}{{{\bf{dt}}}}\\{\bf{ = }}\frac{{{\bf{d(}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{dt}}}}\end{aligned}\)

Instantaneous rate of formation for 3200s and use instantaneous rate of dimerization at 3200s into equation

\(\begin{aligned}{}{\bf{ = }}\frac{{{\bf{ - 1}}}}{{\bf{2}}}{\bf{ \times }}\frac{{{\bf{7}}{\bf{.83 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M/s}}}}{{\bf{1}}}\\{\bf{ = }}\frac{{{\bf{d(}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{dt}}}}\\{\bf{ = }}\frac{{{\bf{d(}}{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{12}}}}{\bf{)}}}}{{{\bf{dt}}}}\\{\bf{ = - 3}}{\bf{.92 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M/s}}\end{aligned}\)

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Most popular questions from this chapter

Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as \({\bf{CC}}{{\bf{l}}_{\bf{2}}}{{\bf{F}}_{\bf{2}}}\), catalyse the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{Cl}}\to {{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{ClO}}\\{\bf{ClO}}{\rm{ }} + {\rm{ }}{\bf{O}}\to {\bf{Cl}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:

\({\bf{2}}{{\bf{O}}_{\bf{3}}}\mathop {}\limits^{}\to {\bf{3}}{{\bf{O}}_{\bf{2}}}\)

(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism: Is NO a catalyst for the decomposition? Explain your answer.

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{\rm{ }}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{NO}}\rightarrow {\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\\{\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\rightarrow {\bf{NO}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

For each of the following pairs of reaction diagrams, identify which of the pair iscatalyzed:

Thefollowingdatahave been determined for the reaction: \({{\bf{I}}^{\bf{ - }}}{\bf{ + OC}}{{\bf{l}}^{\bf{ - }}} \to {\bf{I}}{{\bf{O}}^{\bf{ - }}} + {\bf{C}}{{\bf{l}}^{\bf{ - }}}\)

1

2

3

\({{\bf{(}}{{\bf{I}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.10

0.20

0.30

\({{\bf{(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.050

0.050

0.010

Rate(mol/l/s)

\({\bf{3}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{6}}.{\bf{2*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{1}}.{\bf{83*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

Determine the rate equation and the rate constant for this reaction.

Account for the increase in reaction rate brought about by a catalyst.

The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t1/2?

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