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Chemical Kinetics: Rate Laws, Rate Constants, and Reaction Mechanisms, Exams of Physical Chemistry

A comprehensive overview of chemical kinetics, covering key concepts such as rate laws, rate constants, reaction mechanisms, and the arrhenius equation. It includes numerous examples and exercises to illustrate the principles and applications of chemical kinetics. Particularly useful for students studying chemistry at the university level.

Typology: Exams

2023/2024

Uploaded on 03/11/2025

K_Dirt87
K_Dirt87 🇺🇸

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1) 3 s 1. State the rates of
formation and consumption of A, B, and D. (b) The rate law for this reaction was reported as v = kr[A][B][C]
with the molar concentrations in mol dm 3 and the time in seconds. What are the units of kr?
(a) The reaction stoichiometry is used to determine the rates of consumption and formation of the other
participants in the reaction from the known rate of formation of C.
(b) Rearrange the rate law, , to give:
The reaction rate has dimensions of concentration/time and in this case its dimensions are mol dm 3 s1.
The units of the rate constant are therefore:
2) The initial rate of a reaction depended on the concentration of a substance J as follows:
[J]
0
/(mmol dm
3
)
5.0
10.2
17
30
v
0
/(10
7
mol dm
3
s
1
)
3.6
9.6
41
130
Find the order of the reaction with respect to J and the rate constant.
Assume that the initial rate law has the form: . In this form, we can take the natural logarithm of
both sides of the equation to yield: . j is the reaction order. We can then use this
construct the following table:
[J]
0
/(mmol dm
3
)
5.0
10.2
17
30
l
n([J]
0
/(mol dm
3
))
-
5.29
-
4.58
4.07
-
3.5
v
0
/(10
7
mol dm
3
s
1
)
3.6
9.6
41
130
l
n(v
0
/(mol dm
3
s
1
))
-
14.83
-
13.85
12.40
-
11.25
The plot of ln v0 against ln [J]0 is shown below.
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  1. 3 s 1. State the rates of formation and consumption of A, B, and D. (b) The rate law for this reaction was reported as v = kr[A][B][C] with the molar concentrations in mol dm 3 and the time in seconds. What are the units of kr? (a) The reaction stoichiometry is used to determine the rates of consumption and formation of the other participants in the reaction from the known rate of formation of C. (b) Rearrange the rate law, , to give: The reaction rate has dimensions of concentration/time and in this case its dimensions are mol dm 3 s 1. The units of the rate constant are therefore:
  2. The initial rate of a reaction depended on the concentration of a substance J as follows: [J] 0 /(mmol dm 3 ) 5.0 10.2 17 30 v 0 /(10 7 mol dm 3 s 1 ) 3.6 9.6 41 130 Find the order of the reaction with respect to J and the rate constant. Assume that the initial rate law has the form:. In this form, we can take the natural logarithm of both sides of the equation to yield:. j is the reaction order. We can then use this construct the following table: [J] 0 /(mmol dm 3 ) 5.0 10.2 17 30 ln([J] 0 /(mol dm 3 )) - 5.29 - 4.58 - 4.07 - 3. v 0 /(10 7 mol dm 3 s 1 ) 3.6 9.6 41 130 ln(v 0 /(mol dm 3 s 1 )) - 14.83 - 13.85 - 12.40 - 11. The plot of ln v 0 against ln [J] 0 is shown below.

, j = 2. From the intercept, ln kr = -4.125, so, with units restored, kr = 1.6 × 10 2 dm^3 mol 1 s 1.

  1. The following measurements were recorded for a sample of the radioisotope 57 Co: t/day 0 4 8 12 16 20 Activity/(disintegrations/min) 10000 9899 9797 9702 9603 9501 Determine the rate constant for decay and the half-life for 57 Co. Radioactive decay is a first-order process, and as the activity is proportional to the mass of 57 Co: where kd is the rate constant for decay. We can construct the following table: t/day 0 4 8 12 16 20 Activity/(disintegrations/min) 10000 9899 9797 9702 9603 9501 ln(Activity/(disintegrations/min)) 9.2103 9.2002 9.1898 9.1801 9.1698 9. The plot of ln Activity against t is shown below:
  1. -order rate law with kr = 1.24 cm^3 mol 1 s 1. Calculate the time required for the concentration of A to change from 0.260 mol dm 3 to 0.026 mol dm 3. -order kinetics, then: where the factor of 2 arises from the stoichiometric coefficient of A in the chemical equation. With this in mind, we can use our integrated rate law, slightly modified: We need to rearrange this equation to calculate the time required for the change in concentration:
  2. Establish the integrated form of a third-order rate law of the form v = kr[A]^3. What would it be appropriate to plot to confirm that a reaction is third-order? -order kinetics, then: Rearrange and integrate from t = 0, when the concentration of A is [A] 0 , to a time t when the concentration of A is [A] to give: Use the standard integral form: , where Applying this to our problem: Hence, 2 A reaction is third order if a plot of 1/[A]^2 against t is a straight line.
  1. The oxidation of ethanol to ethanal (acetaldehyde) by NAD+ in the liver in the presence of the enzyme liver alcohol dehydrogenase: CH 3 CH 2 OH(aq) + NAD+(aq) + H 2 3 CHO(aq) + NADH(aq) + H 3 O+(aq) is effectively zeroth-order overall as the ethanol is in excess and the concentration of the NAD+^ is maintained at a constant level by normal metabolic processes. Calculate the rate constant for the conversion of ethanol to ethanal in the liver if the concentration of ethanol in body fluid drops by 50 per cent from 1.5 g dm 3 , a level that results in lack of coordination and slurring of speech, in 49 min at body temperature. Express your answer in units of g dm 3 h 1. For a zeroth-order reaction: for Rearrange: And, since after 49 min:
  2. The reaction mechanism (fast) (slow) involves an intermediate A. Deduce the rate law for the formation of P. The two-step mechanism, taking the first step to be a pre-equilibrium, is: with with rate constant From K, we can solve for [A]: It then follows that: If we lump together our constant parameters: ,where
  1. A rate constant is 1.78 × 10 4 dm^3 mol 1 s 1 at 19 °C and 1.38 × 10 3 dm^3 mol 1 s 1 at 37 °C. Evaluate the Arrhenius parameters of the reaction. From the equation for the temperature dependence of the rate constant: Rearrange and solve for the activation energy, Ea: Rearrange to obtain the frequency factor A:
  2. Which reaction responds more strongly to changes of temperature, one with an activation energy of 52 kJ mol 1 or one with an activation energy of 25 kJ mol 1? The greater Ea, the greater the response of kr to an increase in T. Hence the reaction with the activation energy of 52 kJ mol 1 responds more strongly to changes in temperature (Based pm the Arrhenius Equation).
  1. Make an appropriate Arrhenius plot of the following data for the binding of an inhibitor to the enzyme carbonic anhydrase and calculate the activation energy for the reaction. T/K 289.0 293.5 298.1 303.2 308.0 313. kr/(10^6 dm^3 mol 1 s 1 ) 1.04 1.34 1.53 1.89 2.29 2. From the Arrhenius equation: Draw up the following table and plot ln kr against 1/T: T/K (10^3 K)/T kr/(10^6 dm^3 mol 1 s 1 ) ln(kr/(10^6 dm^3 mol 1 s 1 )) 289.0 3.460 1.01 0. 293.5 3.407 1.34 0. 298.1 3.355 1.53 0. 303.2 3.298 1.89 0. 308.0 3.247 2.29 0. 313.5 3.190 2.84 1. The plot of ln kr against 1/T is shown below: