Skip to main content
Chemistry LibreTexts

Solutions 5

  • Page ID
    47376
  • S1

    a. Ans:

    $$ A{\overset{k_1}{\rightarrow}}B {\overset{k_2}{\rightarrow}}C$$

    $$-\frac{d[A]}{dt}=k_{1}[A]$$

    $$ \dfrac{d[B]}{dt}=k_{1}[A]-k_{2}[B]$$

    $$\frac{d[C]}{dt}=k_{2}[B]$$

    b. Ans:

    $$ A \underset{k_{-1}} {\overset{k_1}{\rightleftharpoons}}B {\overset{k_2}{\rightarrow}}C$$

    $$\frac{d[A]}{dt}=-k_{1}[A]+k_{-1}[B]$$

    $$ \dfrac{d[B]}{dt}=k_{1}[A]-k_{-1}[B]-k_{2}[B]$$

    $$\frac{d[C]}{dt}=k_{2}[B]$$

    c. Ans:

    $$ A{\overset{k_1}{\rightarrow}} B   \underset{k_{-2}} {\overset{k_2}{\rightleftharpoons}}C$$

    $$-\frac{d[A]}{dt}=k_{1}[A]$$     

    $$ \dfrac{d[B]}{dt}=k_{1}[A]-k_{2}[B]+k_{-2}[C]$$

    $$\frac{d[C]}{dt}=k_{2}[B]-k_{-2}[C]$$

    d. Ans:

    $$A+B{\overset{k_1}{\rightarrow}}C {\overset{k_2}{\rightarrow}}D      and      A+D{\overset{k_3}{\rightarrow}}C     and      C{\overset{k_4}{\rightarrow}} A$$

    $$\frac{d[A]}{dt}=-k_{1}[A][B]-k_{3}[A][D]+k_{4}[C]$$

    $$\frac{d[B]}{dt}=-k_{1}[A][B]$$

    $$\frac{d[C]}{dt}=k_{1}[A][B]-k_{2}[C]+k_{3}[A][D]-k_{4}[C]$$

    $$\frac{d[D]}{dt}=k_{2}[C]-k_{3}[A][D]$$

    S2

    It is easy to write the differential rate laws (problem 5.1), but it is hard to write the integrated laws that describe the concentration of populations with time. This can be simplified if approximations are use: specifically "rate determine step" steady-state approximation" and "equilibrium" or specifically "pre-equilibrium". Each approximation has a range of rate constants that allow it to be used and work differently.

    a. Ans: First integrate the rate law equation for [A], substitute it into that of [B] and solve for [B]. In the end, solve for [C]. You can also assume steady state condition \(\dfrac{d[B]}{dt}=0\) and solve for [B] and [C]. 

    b. Ans: Assume steady state condition  \(\dfrac{d[B]}{dt}=0\) or equilibrium condition 

    c. Ans:  Assume steady state condition  \(\dfrac{d[B]}{dt}=0\) or equilibrium condition

    d. Ans: Assume steady state condition  \(\dfrac{d[C]}{dt}=0\)

    (Notes: There can be different approximations and assumptions, as long as you can explain and justify your reasoning, it will be acceptable.)

    S3

    According to Arrhenius equation, the rate constant of a reaction can be described in the following way:

    $$k=A e^{\dfrac{E_{a}}{RT}}$$

    Here A is the preexponential factor, \(E_{a}\) is the activation energy and RT is the average kinetic energy.

    The activation energy (\(E_{a}\)), is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa.

    Overcoming the energy barrier from thermal energy involves addressing the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law. According to the Maxwell-Boltzman distribution law, as the temperature increases, the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly. This is the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.

    S4

    a. Electrostatic repulsions:

    Activation energy is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion. Therefore, stronger electrostatic repulsion will lead to a higher activation energy.

    b. Bond formation in the activated complex:

    A minimum amount of energy is required to break chemical bonds so that new ones can form. Therefore, higher activation energy will lead to less bonds formed in the activated complex.

    c. The nature of the activated complex:

    Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complex or the transition state of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily. So it will be difficult for systems with high activation energy to form activated complex.

    S5

    The reaction rate will approximately double when the temperature increase from 20°C to 30°C, i.e., the reaction rate increases by about 21 = 2. When the temperature increases from 20°C to 70°C, the reaction rate increases by about 25 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2ΔT/10.

    S6

    No. Because as the data shows, when temperature increases from 720 K to 740 K, the reaction rate increases by 0.051/0.024=2.125 times. If acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate, then the reaction rate should become \(2^{2}=4\) times larger instead of 2.125 when temperature rises from 720 K to 740 K.

    S7

    \(E_{a}=100 kJ/mol\)

    To solve this question, first plot ln(k) vs \(\dfrac{1}{T}\), find the best fit line for the data points and calculate the slope. According to Arrhenius equation, 

    $$k=A e^{\dfrac{E_{a}}{RT}}$$

    Taking the natural log on the both sides:

    $$ln(k)=ln(A)-\dfrac{E_{a}}{RT}$$

    So the slope of the plot will be equal to \(-\dfrac{E_{a}}{R}\):

    $$slope=-\dfrac{E_{a}}{R}$$

    Solve for \(E_{a}\), we get:

    $$E_{a}=-R slope$$