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ChemWiki: The Dynamic Chemistry E-textbook > Wikitexts > UC Davis > UCD Chem 2C > UCD Chem 2C: Larsen > Homework Problems > Extra Problems (not official) > Unit III: Chemical Kinetics > Kinetics > Kinetics 4

Kinetics 4

Table of Contents
  1. 1. Q3.
  2. 2. S3.
  3. 3. Q11.
  4. 4. S11.
  5. 5. Q13.
  6. 6. S13.
  7. 7. Q17.
  8. 8. S17.
  9. 9. Q19.
  10. 10. S19.
  11. 11. Q21.
  12. 12. S21.
  13. 13. Q23.
  14. 14. S23.
  15. 15. Q27.
  16. 16.  S27.
  17. 17. Q29.
  18. 18. S29.
  19. 19. Q33.
  20. 20. S33.
  21. 21. Q45.
  22. 22. S45.
  23. 23. Q47.
  24. 24. S47.
  25. 25. Q35.
  26. 26. S35.
  27. 27. Q37.
  28. 28. S37.
  29. 29. Q51.
  30. 30. S51.
  31. 31. Q59.
  32. 32. S59.
  33. 33. Q61.
  34. 34. S59.
  35. 35. Q63.  
  36. 36.  S63.
  37. 37. Q67. 
  38. 38. S67.
  39. 39. Q78.
  40. 40. S78.

Q3.

In the reaction \(\mathrm{A \rightarrow B}\), \(\mathrm{[A]}\) is found to be 0.585 M at 81.5 s, and 0.574 M at 83.4 s. What is the average rate of the reaction during this time interval?

S3.

Average rate of reaction for reactants is:

\(\mathrm{- \dfrac{\Delta[A]}{\Delta t} =  - \dfrac{0.574\, M - 0.585\, M}{83.4\, s - 81.5\, s} = 0.006\, M/s}\)

The Rate of a Chemical Reaction

Q11.

The initial rate of the reaction \(\mathrm{A + B \rightarrow C + D}\) is determined for different initial conditions, with the results listed in the table.

Expt

\(\mathrm{[A]}\), M

\(\mathrm{[B]}\), M

Initial Rate, Ms-1

1

0.0617

0.0443

1.1167 x 10-4

2

0.0617

0.0886

4.5 x 10-4

3

0.1234

0.0443

2.25 x 10-4

4

0.1234

0.0886

9.00 x 10-4

 

  1. What is the order of reaction with respect to \(\mathrm{A}\) and to \(\mathrm{B}\)?
  2. What is the overall reaction order?
  3. What is the value of the rate constant, \(\mathrm{k}\)?

S11.

  1. First, write out the respective rate laws, where \(\ce{m}\) and \(\ce{n}\) are the orders with respect to \(\mathrm{A}\) and \(\mathrm{B}\):

\(\mathrm{R_1 = 1.1167 \times 10^{-4} = [0.0617]^m [0.0443]^n}\)

\(\mathrm{R_2 = 4.5 \times 10^{-4} = [0.0617]^m [0.0443\times 2]^n}\)

\(\mathrm{R_3 = 2.25 \times 10^{-4} = [0.0617\times2]^m [0.0443]^n}\)

\(\mathrm{R_4 = 9.00 \times 10^{-4} = [0.0617\times2]^m [0.0443\times2]^n}\)

Now, we can take put rate laws together and solve –

\(\mathrm{\dfrac{R_2}{R_1} = \dfrac{4.5 \times 10^{-4}}{1.1167 \times 10^{-4}}}\)

\(\mathrm{= \dfrac{[0.0617]^m [0.0443\times 2]^n}{[0.0617]^m [0.0443]^n}}\)

\(\mathrm{\dfrac{R_2}{R_1} = 4 = 2^n;\, n = 2}\).

Therefore, the reaction is second order with respect to \(\mathrm{B}\). Now, take \(\mathrm{\dfrac{R_3}{R_1}}\):

\(\mathrm{\dfrac{R_3}{R_1} =  \dfrac{2.25 \times 10^{-4}}{1.1167 \times 10^{-4}} }\)

\(\mathrm{= \dfrac{[0.0617\times2]^m [0.0443]^n}{[0.0617]^m [0.0443]^n}}\)

\(\mathrm{\dfrac{R_3}{R_1} = 2 = 2^m;\, m = 1}\)

Therefore, the reaction is first order with respect to \(\mathrm{A}\).

  1. Now that we have determined the orders with respect to \(\mathrm{A}\) and \(\mathrm{B}\), we can find the overall rate law:

\(\mathrm{rate\: law = k [A][B]^2}\)

The overall reaction order is the sum of the exponents – 2 for \(\mathrm{B}\) and 1 for \(\mathrm{A}\), \(\mathrm{2+1 = 3}\). Therefore the overall reaction is a third order reaction.

  1. Now that we have determined the overall rate law and orders with respect to A and B, we can sub in any of the values in the table to solve for rate constant \(\mathrm{k}\).

\(\mathrm{rate = k [A][B]^2}\)

rearrange for \(\mathrm{k}\) –

\(\mathrm{k = \dfrac{rate}{[A][B]^2}}\)

Let’s use the data from experiment 1:

\(\mathrm{k = \dfrac{1.1167\times10^{-4}}{[0.0617][0.0443]^2}}\)

\(\mathrm{k = \dfrac{0.92224}{M^2s}}\)

The Rate of a Chemical Reaction

Q13.

The following rates of reaction were obtained in three experiments with the reaction

\(\ce{2NO(g) + Cl2(g) \rightarrow 2NOCl (g)}\)

What is the rate law for this reaction?

Expt

\(\ce{[NO]}\),M

\(\ce{[Cl2]}\),M

Initial Rate, Ms-1

1

0.025

0.051

4.54x10-5

2

0.025

0.102

9.1x10-5

3

0.05

0.051

1.816x10-4

S13.

Let’s write out the respective rate laws for each experiment:

\(\mathrm{R_1 = 4.54\times10^{-5} = [0.025]^m[0.051]^n}\)

\(\mathrm{R_2 = 9.1\times10^{-5} = [0.025]^m[0.051\times2]^n}\)

\(\mathrm{R_3 = 1.816\times10^{-4} = [0.025\times2]^m[0.051]^n}\)

Now let’s solve for \(\ce{m}\) and \(\ce{n}\), the orders of the reaction with respect to \(\mathrm{[NO]}\) and \(\mathrm{[Cl2]}\)

\(\mathrm{\dfrac{R_2}{R_1} = \dfrac{9.1\times10^{-5}}{4.54\times10^{-5}} = 2}\)

\(\mathrm{= \dfrac{[0.025]^m[0.051\times2]^n}{[0.025]^m[0.051]^n}}\)

so, \(\mathrm{2 = 2^n}\) ; \(\mathrm{n = 1}\).

\(\mathrm{\dfrac{R_3}{R_1} = \dfrac{1.816\times10^{-4}}{4.54\times10^{-5}} = 4}\)

\(\mathrm{= \dfrac{[0.025\times2]^m[0.051]^n}{[0.025]^m[0.051]^n}}\)

so, \(\mathrm{4 = 2^m}\) ; \(\mathrm{m = 2}\)

Now we know that the overall rate law \(\mathrm{= k [NO]^2[Cl_2]}\) and we can solve for \(\mathrm{k}\), the rate constant with any set of experimental data.

Let’s use experiment 1 –

\(\mathrm{4.54\times10^{-5} = k [0.025]^2[0.051]}\)

rearrange and solve for \(\mathrm{k}\) –

\(\begin{align}
\mathrm k &= \dfrac{4.54\times10^{-5}}{[0.025]^2[0.051]} \\
&= 1.424\, \mathrm{M^{-2}s^{-1}}
\end{align}\)

Now we know that the \(\mathrm{rate = 1.424\,M^{-2}s^{-1} [NO]^2[Cl_2]}\)

The Rate Law

Q17.

The first order reaction \(\mathrm{C \rightarrow products}\) has \(\mathrm{t_{1/2} = 150\: s}\).

  1. What percent of a sample of \(\mathrm{C}\) remains unreacted 900 s after a reaction has been started?
  2. What is the rate of reaction when \(\mathrm{[C] = 0.40\: M}\)?

S17.

  1.  

\(\mathrm{t_{1/2}\: (first\: order) = \dfrac{\ln 2}{k}}\)

rearrange:

\(\mathrm{k = \dfrac{\ln 2}{t_{1/2}} = \dfrac{\ln 2}{150\, s} = 0.0046/s}\)

% unreacted is \(\mathrm{\dfrac{[A]_t}{[A]_0}}\)

using the rate law:

\(\mathrm{ln \dfrac{[A]_t}{[A]_0} = - kt}\)

rearrange:

\(\begin{align}
\mathrm{\dfrac{[A]_t}{[A]_0}} &= \mathrm{e^{-kt}}\\
&= \mathrm{e^{\large{(-0.0046/s \,^*\, 900s)}}}\\
&= \mathrm{0.0159 \times 100 = 1.59\%\textrm{ remains unreacted}}
\end{align}\)

  1. For a first order reaction,

\(\begin{align}
\mathrm{rate} &= \mathrm{k[C]} \\
&= \mathrm{0.0046/s * 0.40\, M} \\
&= \mathrm{0.00184\, M/s}
\end{align}\)

The Rate Law

Q19.

The reaction \(\mathrm{C \rightarrow products}\) is first order in \(\mathrm{C}\).

  1. If 2.4 g \(\mathrm{C}\) is allowed to decompose for 40 minutes, the mass of \(\mathrm{C}\) remaining undecomposed is found to be 0.3 g. What is the half life, t1/2, of this reaction?
  2. Starting with 2.40 g \(\mathrm{C}\), what is the mass of \(\mathrm{C}\) remaining undecomposed after 1.50 hours?

S19.

  1.  

\(\mathrm{\dfrac{2.4\, g\, C}{8} = 0.3\, g}\) (original 2.4 g of \(\mathrm{C}\) has decreased 1/8 of original value)

Notice that \(\mathrm{\dfrac{1}{8} = \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2}}\) (this means that 3 half lives have elapsed)

\(\mathrm{3 * t_{1/2} = 40\, minutes}\), \(\mathrm{t_{1/2} = 120\, minutes}\).

  1.  

\(\mathrm{t_{1/2} = \dfrac{\ln 2}{k} (first\: order\: reaction)}\)

rearrange – \(\mathrm{k = \dfrac{\ln 2}{t_{1/2}}}\)

        \(\mathrm{= \dfrac{\ln 2}{120\: minutes} = \dfrac{0.005776}{minutes}}\)

We know that \(\mathrm{\ln \dfrac{[A]_t}{[A]_0} = -kt}\)

         \(\begin{align}
&= \mathrm{- \dfrac{0.005776}{minutes} * (1.50\: hr) * \dfrac{60\: minutes}{1\: hour}}\\
&= -0.52
\end{align}\)

Now cancel out ln –

\(\begin{align}
\mathrm{\dfrac{[A]_t}{[A]_0}}&= \mathrm{e^{-kt}}\\
&= \mathrm e^{-0.52}\\
&= 0.5945\\
\mathrm{[A]_t}&= [A]_0 * \mathrm{e^{-kt}}\\
&= 2.4 * 0.5945 \\
&=1.43\textrm{ g of C remaining undecomposed}
\end{align}\)

The Rate Law

Q21.

In the first order reaction \(\mathrm{C \rightarrow products}\), it is found that 98% of the original amount of reactant \(\mathrm{C}\) decomposes in 138 minutes. What is the half-life, t1/2, of this decomposition reaction?

S21.

Remember that \(\mathrm{\dfrac{[A]_t}{[A]_0}}\)is the remaining of reactant \(\mathrm{C}\) (what’s left over/ unreacted).

If 98% of the original amount of reactant \(\mathrm{C}\) has decomposed, that means there is 2% left over (0.02)

So,

\(\mathrm{0.02 = \dfrac{[A]_t}{[A]_0} = e^{-kt}}\)

\(\mathrm{\ln (0.02) = \ln e^{-kt}}\)

\(\mathrm{\ln (0.02) = - kt}\)

\(\mathrm{\ln (0.02) = - k (138\: minutes)}\)

\(\mathrm{\dfrac{0.028}{minutes} = k}\)

Now, using the t1/2 equation for a first order reaction – \(\mathrm{t_{1/2} = \dfrac{\ln2}{k}}\)

        \(\begin{align}
&= \mathrm{\dfrac{\ln 2}{(0.028/minutes)}}\\
&= \mathrm{2.475\: minutes}
\end{align}\)

Q23.

Acetoacetic acid, \(\ce{CH3COCH2COOH}\), a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\ce{CO2 (g)}\).

\(\ce{CH3COCH2COOH (aq) \rightarrow CH3COCH3 (aq) + CO2 (aq)}\)

The first order decomposition has a half life of 145 minutes.

How long will it take for a sample of acetoacetic acid to be 70% decomposed?

S23.

\(\mathrm{rate = k[A]}\) (where \(\mathrm{A}\) is the acetoacetic acid)

First find the rate constant \(\mathrm{k}\) using the t1/2 equation – \(\mathrm{t_{1/2}  = \dfrac{\ln 2}{k}}\)

\(\begin{align}
\mathrm{k} &= \mathrm{\dfrac{\ln 2}{t_{1/2}}}\\
\mathrm{k} &= \mathrm{\dfrac{\ln 2}{145\: minutes}}\\
\mathrm{k} &= \mathrm{\dfrac{0.00478}{minutes}}
\end{align}\)

\(\mathrm{\dfrac{[A]_t}{[A]_0} = 0.30}\)(there is 30% remaining: 70% decomposed)

\(\mathrm{\ln\dfrac{[A]_t}{[A]_0} = \ln (0.30) = - kt}\)

\(\mathrm{\dfrac{\ln(0.30)}{k} = t}\)

\(\mathrm{\dfrac{\ln (0.30)}{0.00478/ minutes} = t}\)

\(\mathrm{t = 252\: minutes}\)

Q27.

Which of these sets of date corresponds to a (a) zero-order reaction, (b) first-order reaction, (c) second-order reaction? 

I

 

II

 

III

 

Time (s)

\(\mathrm{[A]}\) (M)

Time (s)

\(\mathrm{[A]}\) (M)

Time (s)

\(\mathrm{[A]}\) (M)

0

1.00

0

2.00

0

0.50

25

0.287

25

1.75

25

0.44

50

0.223

50

1.50

50

0.40

75

0.174

75

1.25

75

0.36

100

0.135

100

1.00

100

0.33

 S27.

  1. II
  2. I
  3. III

Explanation:

There are two ways to solve this problem:

  1. Graphically
  2. Analytically

Case I – First Order

Time (s)

\(\mathrm{\ln [A]}\) (M)

0

-1

25

-1.25

50

-1.5

75

-1.75

100

-2

  1. Plot the following using the data for each case and see which yields the straight line.

 

  1. \(\mathrm{[A]}\) vs. t \(\rightarrow\) zero order; \(\mathrm{m = -k}\)
  2. \(\mathrm{\ln [A]}\) vs. t \(\rightarrow\) first order; \(\mathrm{m = -k}\)
  3. \(\mathrm{\dfrac{1}{[A]}}\) vs. t \(\rightarrow\) second order; \(\mathrm{m = k}\)
  4. See graphs below


kinetics 5 case 1.jpg

 

Case II – Zero Order

Time (s)

\(\mathrm{[A]}\) (M)

0

2

25

1.75

50

1.5

75

1.25

100

1

kinetics 5 case 2.jpg

Case III – Second Order

Time (s)

\(\mathrm{\dfrac{1}{[A]}}\) (M)

0

2

25

2.25

50

2.5

75

2.75

100

3

kinetics 5 case 3.jpg

  1. Look at the increments between the concentrations in the data and see if they increase linearly. If the increments are even for each step then the data points must form a line. By definition of a line, data points are correlated by even steps called the slope (m) (rise over run).

              i.      Case I

  1. \(\mathrm{[A]}\) increases step wise by 0.25 when \(\mathrm{\ln[A]}\) is compared to t
  2. ln(1); 1n(0.287); ln(0.223); ln(0.174); ln(0.135) corresponds to
  3. 0, -1.25, -1.5, -1.75, -2 when you round the numbers slightly i.e. -1.24~-1.25
  4. Therefore, it must be first order.

               ii.      Case II

  1. \(\mathrm{[A]}\) increases step wise by 0.25 when \(\mathrm{[A]}\) is compared to t. It must be zero order.

               iii.      Case III

  1. \(\mathrm{[A]}\) increases step wise by 0.25 when \(\mathrm{\dfrac{1}{[A]}}\) is compared to t.
  2. \(\dfrac{1}{0.5}\); \(\dfrac{1}{0.44}\); \(\dfrac{1}{0.4}\); \(\dfrac{1}{0.36}\); \(\dfrac{1}{0.33}\) corresponds to
  3. 2, 2.25, 2.50, 2.75, 3
  4. Therefore, it must be second order

Q29.

What is the half-life of the first order reaction? \(\mathrm{Rate = K[A]}\) where the \(\mathrm{k = 2.3E\textrm{-3}\,s^{-1}}\) 

S29.

\(\mathrm{t_{1/2} =301\,s}\)

Explanation:

 i.      \(\mathrm{T_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}}\) for a first order reaction

\(\mathrm{t_{1/2} = \dfrac{\ln2}{2.3E\textrm{-3}\, s^{-1}}}\)

Q33.

The half-life of a first order reaction does NOT depend on the concentration. However, the half-lives of both the zero-order and the second order reactions do depend on initial concentration. In one instance, the half-life gets longer as the initial concentration increases and in the other instance, the half-life gets shorter as the initial concentration increases. Which is which?

S33.

  1. Zero Order – as concentration increases, the initial half-life increases
  2. Second order – as the concentration increase, the initial half-life decreases

Explanation:

  1. Zero Order

\(\mathrm{t_{1/2} = \dfrac{[A]_0}{2K}}\)

 \(\mathrm{t_{1/2}\propto[A]}\) where “∝” means “proportional to”

Since half-life is directly proportional to concentration, as concentration increases to infinity, the half-life increases

  1.  Second Order

\(\mathrm{t_{1/2} = \dfrac{1}{k[A]_0}}\)

 \(\mathrm{t_{1/2} \propto \dfrac{1}{[A]_0}}\)

Since half-life is inversely proportional to concentration, as concentration increases to infinity, the half-life decreases to 0 as \(\mathrm{\dfrac{1}{\infty} = 0}\)

Q45.

Explain why:

  1. One cannot calculate the reaction rate from collision frequency alone.
  2. When the temperature increases, the rate of a chemical reaction may increase drastically, but the collision frequency increase more slowly.
  3. When the concentration increase, the collision frequency increases quickly but the rate of a chemical reaction may increase more slowly.
  4. The addition of a catalyst dramatically affects the rate of a reaction even if temperature is held constant.

S45.

  1. The reaction rate cannot be calculated from the collision frequency alone. In order to have a reaction, the energy of the collision must be higher than the activation energy and the molecules must be aligned correctly.  
  2. When the temperature is increased, the molecules do not really hit each other more often. However, when the molecules do collide, they collide with a greater energy because they are moving faster due to the higher temperature. Thus, collision frequency does not increase but the rate of the chemical reaction does increase.
  3. When the concentration is increased, the number of collisions is increased due to the greater number of particles. However, the number of reactions (the reaction rate) increase more slowly since the energy of the molecules is still the same.
  4. A catalyst speeds up a reaction by lowering the activation energy of the reaction.

Q47.

In the reversible reaction, \(\mathrm{A + B \leftrightarrow  A + B}\), the enthalpy of the forward reaction is 25 kJ/mol and the activation energy of the forward reaction is 100 kJ/mol.

  1. What is the activation energy of the reverse reaction?
  2. Sketch the reaction profile for the above reaction
  3. What if the enthalpy of the forward reaction is -25kJ/mol?

S47.

  1. 75 kJ/mol
  2. See sketch
  3. 125kJ/mol

Explanation: See sketch  

The graph looks like a parabola with the products higher than the reactants by 25 KJ/mol. The top of the parabola is 100 kJ/mol taller than the reactants and 75 KJ/mol taller than the products. If the enthalphy were negative then the products would be lower than the reactants by 25 KJ/mol. 

Q35.

The decomposition of \(\mathrm{HI (g)}\) at 700 K is followed for 400 s, yielding the following data:

time (s)

\(\mathrm{[HI]}\), M

\(\mathrm{\ln [HI]}\)

\(\mathrm{\dfrac{1}{[HI]}}\), M-1

0

1

0

1

100

2/3

-0.405

1.5

200

0.5

-0.301

2.0

300

0.4

-0.916

2.5

400

1/3

-1.1

3.0

S35.

Looking at this set of data, we can see that plotting time vs. \(\mathrm{\dfrac{1}{[HI]}}\) gives you a straight line with positive slope, hence the decomposition of \(\mathrm{HI (g)}\) is a second order reaction.

Kinetics 35 graph.jpg

We can also calculate the slope of this equation numerically, treating change in x (time) and change in y (1/molarity):

\(\mathrm{\dfrac{1.5 - 1}{100-0}= 0.005}\),

thus proving the equation of the line valid.

Q37.

For the reaction \(\mathrm{C \rightarrow products}\), the following data were obtained:

Time (Seconds)

\(\mathrm{[C]}\), M

\(\mathrm{\dfrac{1}{[C]}}\)

\(\mathrm{\ln [C]}\)

0

0.8150

1.23

-0.205

20

0.695

1.4388

-0.364

40

0.551

1.815

-0.596

60

0.455

2.1978

-0.787

S37.

Here, you can see that the concentration in column 2 is decreasing consistently by 0.1 increments. The graph that relates time with concentration that gives a negative slope is a zero order reaction.

37 kinetics graph.jpg

We can also numerically calculate the slope to be -0.0061, by taking two points and doing change in y / change in x:

\(\mathrm{\dfrac{0.695 - 0.8150}{20-0} = -0.006}\) (negative sign indicates negative slope)

Q51.

The rate constant for the reaction \(\ce{H2 (g) + I2 (g) \rightarrow 2HI (g)}\) has been determined at the following temperatures: 600 K, \(\mathrm{k = 5.5 \times 10^{-4}\, M^{-1}s^{-1}}\); 685 K, \(\mathrm{k = 2.9 \times 10^{-2}\,M^{-1}s^{-1}}\). Calculate the activation energy for the reaction.

S51.

Use the Arrhenius equation: \(\mathrm{\ln \dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left (\dfrac{1}{T_1} - \dfrac{1}{T_2} \right )}\)

\(\mathrm{\ln \dfrac{5.5 \times 10^{-4}\, M^{-1}s^{-1}}{2.9 \times 10^{-2}\, M^{-1}s^{-1}} = \dfrac{E_a}{8.3145\, J\, mol^{-1}K^{-1}}\left (\dfrac{1}{685} - \dfrac{1}{600}\right )}\)

Rearrange to solve for \(\mathrm{E_a}\) –

\(\mathrm{E_a = \ln \dfrac{5.5 \times 10^{-4}\, M^{-1}s^{-1}}{2.9 \times 10^{-2}\, M^{-1}s^{-1}}*\dfrac{8.3145\, J\, mol^{-1}K^{-1}}{\dfrac{1}{685K} - \dfrac{1}{600K}}}\)

\(\mathrm{E_a = 1.594 \times 10^5\, J / mol}\)

Q59.

Here are two statements about catalysis, are there any modifications you would make in them?

  1. A catalyst is defined as a substance that speeds up a chemical reaction but it is not known to take part in the reaction.
  2. The role of a catalyst is to lower the activation energy of a chemical reaction.

S59.

  1. A catalyst does take part in the reaction, but it does not always speed up the chemical reaction. Negative catalysts that slow down the speed of a chemical reaction are known as inhibitors.
  2. The function of a catalyst is to change the mechanism of a reaction. One of the mechanisms that a catalyst can change is changing the activation energy of the original reaction.

Q61.

Discuss the similarities and differences between the catalytic activity of platinum metal and of an enzyme.

S59.

One similarity between the catalytic activity of platinum metal and an enzyme is that they both have metal centers that act as the active site. One difference is when they are placed in solution – the platinum metal forms a heterogeneous solution (does not dissolve) and the enzyme forms a homogeneous solution (soluble). A significant difference is that the platinum metal is nonspecific, and it can catalyze many different reactions. An enzyme is very specific and usually only works for one reaction.

Q63.  

The following graph depicts the effect of enzyme concentration on the rate of an enzyme-meditated reaction. What are the reaction conditions necessary to account for this graph?

kinetics 63.jpg

 S63.

There must be an unlimited amount of substrate such that the enzyme is always saturated and is working at its maximum rate. If there was a finite amount of substrate, then the reaction rate will drop off as substrate is used up. 

Q67. 

Consider the following hypothetical reaction: \(\mathrm{2A + 2B \rightarrow C + 2D}\). This reaction is second order in \(\mathrm{[A]}\) and first order in \(\mathrm{[B]}\). A three-step mechanism has been proposed and the elementary steps are as follows:

                                                            1) \(\mathrm{2A \leftrightarrow  I_1 \quad (fast)}\)

                                                            2) \(\mathrm{???\quad(slow)}\)

                                                            3) \(\mathrm{I_2+ B \rightarrow C + D \quad(fast)}\)

Find the entire three-step mechanism and show that it conforms to the experimentally determined reaction order.

Note: That \(\mathrm{I_n}\)are intermediates and that the forward reaction of step 1 is in equilibrium with the reverse reaction of step 1. In other words, \(\mathrm{k_1=k_{-1}}\).

S67.

  1. Find the experimentally determined reaction order:

\(\mathrm{Rate = k[A]^x[B]^y}\)

Since the reaction is second order in \(\mathrm{[A]}\), \(\mathrm{x = 2}\).

Since the reaction is first order in \(\mathrm{[B]}\), \(\mathrm{y = 1}\)

Therefore, the experimentally determined reaction order is as follows:

\(\mathrm{Rate = k[A]^2[B]^1}\)

  1. Find step 2.

We know that the three elementary steps must sum up to the overall reaction. In addition, the intermediates cannot show up in the final reaction. Solve by inspection.

  1. \(\mathrm{2A \leftrightarrow  I_1 \quad (fast)}\)
  2. \(\mathrm{I_1+ B \rightarrow D + I_2 \quad (slow)}\)
  3. \(\mathrm{I_2+ B \rightarrow C + D \quad(fast)}\)

When you add up the reactions, elements will cancel out and you will be left with the overall reaction.

  1. \(\mathrm{2A \leftrightarrow  \not{I_1}\quad(fast)}\)
  2. \(\mathrm{\not{I_1}+ B \rightarrow D + \not{I_2}\quad(slow)}\)
  3. \(\mathrm{\not{I_2}+ B \rightarrow C + D \quad(fast)}\)

The overall reaction when you cancel out the intermediates is:

\(\mathrm{2A + 2B \rightarrow C + 2D}\)

  1. Prove that the rate of the three-step mechanism conforms to the experimentally determined reaction order.

We know that the rate of the three-step mechanism is the reaction order of the slowest (rate-determining) step. Note that in an elementary reaction, the coefficients are also the exponents of the reaction.

For step 2, the rate is as follows:

\(\mathrm{Rate = k[I_2][B]}\)

However, we cannot have intermediates in the rate law. So we have to solve for \(\mathrm{[I_2]}\) using step 1.

From step 1, the rate is as follows:

\(\mathrm{Rate = k_1[A]^2}\)

However, since this step is at equilibrium, the rate can also be expressed as follows:

\(\mathrm{Rate = k_{-1}[I_1]}\)

Therefore,

\(\mathrm{k_1[A]^2 = k_{-1}[I_1]}\)

Solving for \(\mathrm{[I_1]}\) we get

\(\mathrm{[I_1] = k_1k_{-1}[A]^2}\)

We can combine \(\mathrm{k_1}\) and \(\mathrm{k_{-1}}\)since they are both constants into \(\mathrm{K}\).

 Thus, we get

\(\mathrm{[I_1] = K [A]^2}\)

Substituting this back into the rate in step 2 and combining \(\mathrm{k}\) and \(\mathrm{K}\) into \(\mathrm{K}\) we get

\(\mathrm{Rate = K [A]^2[B]}\)

This is the same as the experimentally determined reaction order! Success!!! 

Q78.

The unit of \(\mathrm{k}\) (rate of formation constant) depends on the overall order of a reaction. Here is a derivation of a general expression for the units of \(\mathrm{k}\) for a reaction of any overall order, based on the order of reaction (o), and the units of concentration (M), and time (s).

S78.

We know that rate has units of M/s always. Concentration is always in terms of molarity (M). We can look at several different cases and derive a pattern:

Case 1: The general rate equation of a first order; \(\mathrm{rate = k [A]}\).

If we just look at the units of this equation, you get: \(\mathrm{M/s = \{\textrm{units of k}\} * M}\)

rearrange for\(\mathrm{\{\textrm{units of k}\} = \dfrac{M/s}{M}= s^{-1}}\)

Case 2: The general rate equation of a second order rate; \(\mathrm{= k [A]^2}\).

In terms of units: \(\mathrm{M/s = \{\textrm{units of k}\} * M^2}\)

Rearrange: \(\mathrm{\{\textrm{units of k}\} = \dfrac{M/s}{M^2}  = s^{-1}M^{-1}}\)

Case 3: The general rate law for a zero order reaction is \(\mathrm{rate = k}\)

In terms of units: \(\mathrm{M/s = \{\textrm{units of k}\}}\)

Now, by looking at all three cases, you can see that the general expression for the units of \(\mathrm{k}\) is:

\(\mathrm{\{\textrm{units of k}\} = \dfrac{1}{M^{o-1} s}}\)

where \(\mathrm{o}\) stands for order of a reaction, M stands for concentration (molarity), and s is time (seconds)

For more information on these topics check out this link:  The Rate Law

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18:34, 28 Mar 2014

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