Unit 2 Kinetics
2.4 Exercises
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Section 2.4 Exercises
- Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
- What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
- Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
- The rate of a certain reaction doubles for every 10 °C rise in temperature.
(a) How much faster does the reaction proceed at 45 °C than at 25 °C?
(b) How much faster does the reaction proceed at 95 °C than at 25 °C? - The rate constant at 325 °C for the decomposition reaction [latex]\text{C}_4\text{H}_8\;{\longrightarrow}\;2\text{C}_2\text{H}_4[/latex] is 6.1 × 10−8 s−1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
- An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
- Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
Temperature (K) k (M−1 s−1) 555 6.23 × 10−7 575 2.42 × 10−6 645 1.44 × 10−4 700 2.01 × 10−3 What is the value of the activation energy (in kJ/mol) for this reaction?
- The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
[latex]\text{C}_{12}\text{H}_{22}\text{O}_{11}\;+\;\text{H}_2\text{O}\;{\longrightarrow}\;\text{C}_6\text{H}_{12}\text{O}_6\;+\;\text{C}_6\text{H}_{12}\text{O}_6[/latex]follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
(a) In neutral solution, k = 2.1 × 10−11 s−1 at 27 °C and 8.5 × 10−11 s−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).
(b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10−7 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.
(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)? - Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first [latex]A\;+\;BC\;{\longrightarrow}\;AB\;+\;C[/latex] reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
Solutions
- The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.
- The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.
- After finding k at several different temperatures, a plot of ln k versus [latex]\frac{1}{T}[/latex], gives a straight line with the slope [latex]\frac{-E_{\text{a}}}{R}[/latex] from which Ea may be determined.
- (a) 4-times faster
(b) 128-times faster - [latex]3.8\;\times\;10^{15}\;\text{s}^{-1}[/latex]
- 43.0 kJ/mol
- 180 kJ/mol
- (a) Ea = 108 kJ
A = 1.3 × 108 s−1
k = 3.2 × 10−10 s−1
(b) 6.52 × 1011 s or 1.81 × 108 h or 7.6 × 106 day.
(c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state. - The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.