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Chapter 12: Q12.9CYL (page 673)

Does the following data fit a second-order rate law?

Trial

Time(s)

(A) (M)

1

5

0.952

2

10

0.625

3

15

0.465

4

20

0.370

5

25

0.308

6

35

0.230

Short Answer

Expert verified

Yes. The graph between 1 / (A) vs. t is linear.

Step by step solution

01

Rate of a Reaction

The rate of reaction may be defined as the speed of the reactant reacting to obtain a product in a particular reaction at a particular time. The concentration of the reactant and product are represented into mole/L.

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}^{}\)

02

Second-Order Reaction

In this reaction, the second-order reaction depend upon only one concentration of the reactant:

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{2}}}\)

Second-order reactions having the integrated rate law:

\({\bf{1 / }}\left( {\bf{A}} \right){\bf{ = kt + 1 / }}{\left( {\bf{A}} \right)_{\bf{0}}}\)

03

Graph

Trial

Time(s)

(A) (M)

1 / (A)

1

5

0.952

1.050

2

10

0.625

1.60

3

15

0.465

2.15

4

20

0.370

2.70

5

25

0.308

3.25

6

35

0.230

4.34

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Most popular questions from this chapter

Use the data provided in a graphical method to determine the order and rate constant of the following reaction:\({\bf{2P}} \to {\bf{Q}} + {\bf{W}}\)

Time (s)

9.0

13.0

18.0

22.0

25.0

(P) (M)

1.077 × 10−3

1.068 × 10−3

1.055 × 10−3

1.046 × 10−3

1.039 × 10−3

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation: \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)(g) ⟶\({\bf{C}}{{\bf{H}}_{\bf{4}}}\)(g) +\({\bf{CO}}\)(g)

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial

(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)

\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls−1)

1.

1.75 × 10−3

2.06 × 10−11

2.

3.50 × 10−3

8.24 × 10−11

3.

7.00 × 10−3

3.30 × 10−10

How will each of the following affect the rate of the reaction:

\({\bf{CO}}\left( {\bf{g}} \right){\bf{ + \;N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {{\bf{O}}_{\bf{2}}}{\bf{\;}}\left( {\bf{g}} \right){\bf{ + NO}}\left( {\bf{g}} \right)\) if the rate law for the reaction is rate = \({\bf{k(NO}}{}_{\bf{2}}{\bf{)(CO)}}\)?

  1. Increasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.1 atm to 0.3 atm
  2. Increasing the concentration of CO from 0.02 M to 0.06 M.

What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is \({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\)L/mol/s.

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \({{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}{\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{ + }}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\) follows a first-order rate equation for the disappearance of sucrose: \({\bf{Rate = k}}\left( {{{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}} \right)\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

  1. In neutral solution, \({\bf{k = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 27 °C and \({\bf{8}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
  2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is\({\bf{1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\). How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
  3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
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