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Solutions 8B: More Chemical Kinetics

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  • S8B.1

    1. \(\mathrm{reaction\: rate = -\dfrac{1}{3}\dfrac{d[A]}{dt} = -\dfrac{1}{2}\dfrac{d[B]}{dt} = \dfrac{d[C]}{dt} = \dfrac{d[D]}{dt}}\)
      \(\mathrm{reaction\: rate = -\dfrac{1}{3}(-8.2 \times 10^{-4}\: M\, s^{-1}) = 2.73 \times 10^{-4}\, M\, s^{-1}}\).
    2. rate of disappearance of \(\mathrm{B = \dfrac{d[B]}{dt}}\)
      \(\mathrm{-\dfrac{1}{2}\dfrac{d[B]}{dt} = 2.73 \times 10^{-4}\, M\, s^{-1}}\)
      \(\mathrm{\dfrac{d[B]}{dt} = -2(2.73 \times 10^{-4}\, M\, s^{-1})}\)
      \(\mathrm{\dfrac{d[B]}{dt} = -5.46 \times 10^{-4}\, M\, s^{-1}}\)
    3. rate of formation of \(\mathrm{D = \dfrac{d[D]}{dt} = reaction\: rate = 2.73 \times 10^{-4}\, M \,s^{-1}}\)

    To review reaction rate laws, please visit the page "The Rate Law".


    \(\mathrm{average\: rate= \dfrac{\Delta[B]}{\Delta(t)}= \dfrac{0.567\,M-0.356\,M}{50.3\,s-31.6\,s}=0.011\,M/s}\)

    To review reaction rate laws, please visit the pages "The Rate Law" and "Rate of Reaction"


    Video Solution

    1. using the equation \(\mathrm{\textrm{rate of reaction}=k[A]^x[B]^y}\) we can solve for \(\ce{x}\) and \(\ce{y}\)
      \(\mathrm{R_2=k[A]^x[B]^y=k(0.150)^x(0.246)^y =1.408\times10^{-2}}\)
      \(\mathrm{\dfrac{R_2}{R_1} = \dfrac{k(0.150)^x(0.246)^y}{k(0.150)^x(0.123)^y} = \dfrac{1.408\times10^{-2}}{3.52\times10^{-3}} =}\)
      \(\mathrm{4=2y \Rightarrow y=2 \quad [B]}\) second order
      \(\mathrm{\dfrac{R_3}{R_1}=\dfrac{[k(0.300)^y(0.123)^y]}{[k(0.150)^x(0.123)^y]} = \dfrac{7.04\times10^{-3}}{3.52\times10^{-3}} =}\)

    \(\mathrm{2x=2 \Rightarrow x=1 \quad [A]}\)first order

    1. \(\mathrm{x+y=1+2= 3}\) third order overall reaction
    2. simply plug the values into \(\mathrm{\textrm{rate of reaction}=k[A]^x[B]^y}\) and solve for \(\ce{k}\)
      \(\mathrm{3.52\times10^{-3}=k(0.150)(0.123)^2 }\)
      \(\mathrm{k= \dfrac{3.52\times10^{-3}}{(0.150)(0.123)^2} =1.55\, M^{-2}s^{-1}}\)

    To review reaction rate laws, please visit the page "The Rate Law".








    \(\mathrm{\dfrac{2}{\left (\dfrac{4.55\times10^{-5}}{9.08\times10^{-5}} \right )}=4}\)





    To review reaction rate laws, please visit the page "The Rate Law".


    Step 1: Find k

    if 60% decomposes 40% remains

    \(\mathrm{\ln\dfrac{A_t}{A_o}=-kt \quad \ln(0.4)=-k(125\:min)}\)

    \(\mathrm{k=0.00733\: min^{-1}}\)

    Step 2: Solve for time using first order t1/2=ln2/k

    \(\mathrm{t_{1/2}=\dfrac{\ln 2}{0.00733}=94.56\: mins}\)

    To review first-order reactions, visit the page "First-Order Reactions".


    For first order reactions \(\mathrm{t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}}\)

    If the half life is \(\mathrm{t_{1/2} = 180 \, s}\), then the value of k for the reaction is \(\mathrm{k= \dfrac{t_{1/2}}{\ln 2} = \dfrac{0.693}{720\,s} = 9.625\times10^{-4}\,s^{-1}}\)

    1. since \(\mathrm{\ln[A_t] = -kt + \ln[A]_o}\), then \(\mathrm{\ln\dfrac{[A]_t}{[A]_t} = -kt}\), you can say \(\mathrm{\dfrac{[A]_t}{[A]_o} = e^{-kt}}\)
      this is convenient since the percent remaining is the “amount at time t” / “initial amount” which is equal to \(\mathrm{\dfrac{[A]_t}{[A]_o}}\)
      therefore \(\mathrm{\%\: remaining = e^{-kt} = e^{\large{(-(9.625\times10^{-4}s^{-1})(720\,s))}} = 0.50007 (100\%) = 50\%}\)
    2. given \(\mathrm{[A] = 0.25\,M}\), and we know that \(\mathrm{\textrm{rate of reaction} = -\dfrac{\Delta[A]}{\Delta t} = k[A]}\)\(\mathrm{k[A] = 9.625\times10^{-4}\,s^{-1}(0.25\,M) = 2.406\times10^{-4}\,M/s}\)

    To review the rate laws, visit the page "The Rate Law".


    1. \(\mathrm{\dfrac{12.24\,g}{2.04} = 8}\) Thus \(\ce{A}\) decomposes to a eighth of its original mass in 24 minutes.
      \(\mathrm{\dfrac{1}{8} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}}\) For \(\ce{A}\) to become a eight of its original mass, three half lives (\(\mathrm{t_{1/2}}\)) must have passed. Thus, \(\ce{A}\)’s half-life must be equal to \(\mathrm{\dfrac{24\: minutes}{3} = 8\: minutes}\).

    2. Step 1: Find k.
      For a first order reaction: half life \(\mathrm{(t_{1/2}) = \dfrac{\ln 2}{k}}\). In this reaction, \(\mathrm{8\: minutes = \dfrac{\ln 2}{k}}\).\(\mathrm{k = \dfrac{\ln 2}{8\:minutes} = 0.0866\: minutes^{-1}}\)
      Step 2: Use the following equation to find the amount of A at time = 1hour: ln(At/Ao) = -kt
      \(\mathrm{\ln(\dfrac{A_t}{12.24\,g}) = -(0.0866\: minutes^{-1}\times60\: minutes)}\)
      \(\mathrm{\dfrac{A_t}{12.24\,g} = e^{\large{-(0.0866\: minutes^{-1}\times60\: minutes)}}}\)
      \(\mathrm{A_t = 12.24\,g \times e^{\large{-(0.0866\: minutes^{-1}\times60\: minutes)}}}\)
      \(\mathrm{A_t = 0.06779\,g\textrm{ remaining after 1.00}\, hr}\)

    To review half life and first-order reactions , please visit the pages "Half Life" and "First-Order Reactions".


    \(\mathrm{\ln\dfrac{A_t}{A_o}=-kt}\) If 80 % decomposed, that means 20 % remains


    \(\mathrm{k=0.00749\: minutes^{-1}}\)

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

    \(\mathrm{t_{1/2}=\dfrac{\ln 2}{0.00749\: minutes^{-1}}= 92.6\: minutes}\)

    To review half life and first-order reactions , please visit the page "Half Life"


    1. The amount that it decreases by per second (slope) is a constant value so its zero order.
    2. The amount that it decreases by per second is a set percentage.
    3. The slope \(\mathrm{\dfrac{\left (\dfrac{1}{A^1}-\dfrac{1}{A^2} \right )}{(t^1-t^2)}}\) is the same value at all points.

    To review rate laws, visit the page "The Rate Law".


    We need to solve for \(\mathrm{t_{1/2} = \dfrac{0.693}{k}}\). So solving for \(\ce{k}\) in the integrated rate law for first order reactions...

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

    \(\mathrm{k = \dfrac{\ln\dfrac{[A]_t}{[A]_o}}{-t}}\)

    now we use any point of time and its corresponding concentration, like \(\mathrm{t = 100\,s}\) and \(\mathrm{[A] = 0.4\,M}\)

    \(\mathrm{k = \dfrac{\ln\left (\dfrac{0.4}{1} \right )}{-100} = 9.16\times10^{-3}\,s^{-1}}\)

    then use this in the half-life equation

    \(\mathrm{t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{9.16\times10^{-3}\,s^{-1}} = 75.66\,s}\)

    To review first-order reactions visit the page "First-Order Reactions".


    \(\mathrm{rate = k[A]^2[B]}\) (given in problem)
    \(\mathrm{rate = (0.0205\, M^{-1} min^{-1})(0.005\,M)^2(3.02\,M)}\)
    \(\mathrm{rate = 1.548 \times 10^{-6}\, M\, s^{-1}}\)

    To review second-order reactions, please visit the pages "Second-Order Reactions".


    Average rate of reaction: \(\mathrm{\dfrac{\Delta[H_2O_2]}{\Delta(t)}=\dfrac{(2\,M-1.26\,M)}{(500\,s)}=0.00148\,M/s}\)
    reaction order: zeroth order because plot of the concentration vs. time is linear

    To review zero-order reactions, please visit the page "Zero-Order Reactions"


    Ch24. 37 graph.PNG

    When we graphed it \(\ce{[A]}\)(M) vs Time(s) the graph was linear so we can determine that the reaction is a zero order reaction.

    The half life of this reaction can be found using \(\mathrm{t_{1/2}=\dfrac{[A]_0}{2k}}\)

    \(\mathrm{k=}\) the slope of the \(\ce{[A]}\)(M) vs Time(s) graph \(\mathrm{[A]_t=-kt+[A]_0}\)

    \(\mathrm{-k=\dfrac{[A]_t-[A]_0}{t} =\dfrac{0.085-0.715}{126-0}= -0.005}\)

    \(\mathrm{k=0.005\, Ms^{-1}}\)

    Using our \(\ce{k}\) solve for \(\mathrm{t_{1/2}}\) using \(\mathrm{t_{1/2}=\dfrac{[A]_0}{2k}}\)

    \(\mathrm{t_{1/2}=\dfrac{0.715}{2(0.005)}=71.5\: mins}\)

    To review zero-order reaction, visit the page "Zero-Order Reactions".


    The half life of zero-order reactions increase with the initial concentration, because the concentration is decreasing with the constant slope \(\ce{-k}\), so a larger value will be reduced by a smaller percent. For instance \(\mathrm{10-1=9/10\times100=90\%}\) of the original, while \(\mathrm{5-1=4/5\times100\%=80\%}\) of the original.

    The half life of second-order reactions decrease with the initial concentration, because when the concentration is higher, there is a higher probability that the molecules will interact. This causes the concentration to decrease faster. This relationship is shown by the equation \(\mathrm{t_{1/2}=\dfrac{1}{(k\times A_0)}}\).

    To review zero-order reactions, visit the page "Zero-Order Reactions".


    1. The collision frequency only determines the frequency of molecules to collide. If each collision of molecules resulted in a reaction, then the reaction’s rate would be extremely fast and would be determined solely by the collision frequency. But each collision does not result in a reaction, and therefore the reaction rate is only a fraction of the calculated collision frequency.
    2. A raise in temperature directly affects the gas molecules speed. This rise in speed makes a higher chance for molecules to collide and therefore raises the collision frequency of a reaction.
      The rise in temperature dramatically rises the reaction’s rate, however. This is because the reaction’s rate is already a very small fraction of the reactions collision frequency, and is directly affected by both the collision frequency and the fraction of molecules that become “activated”. Adding heat helps overcome the activation energy, so it not only raises the collision frequency but also the amount of activated molecules. This causes a much more dramatic rise to reaction rate with respect to collision frequency.
    3. A catalysts main purpose is to lower the activation energy in a reactions pathway. The lower the activation energy, the larger the fraction of energetic collisions and the faster the reaction. This speeds up reaction rate, regardless of how fast the molecules are sped up by an increase in temperature. They are independent of one another.

    To review this topic, visit the page "The Rate Law".


    1. \(\mathrm{activation\: energy_{reverse}= 66\, kJ/mol - 34\, kJ/mol = 32\, kJ/mol}\)

    24-47 - Copy.jpg

    To review the potential energy profile of reactions, please visit the page "Potential Energy Profile".


    1. 3 intermediates
    2. 4 transition states
    3. Endothermic
    4. Endothermic
    5. E
    6. C

    To review the potential energy profile of reactions, please visit the page "Potential Energy Profile".


    Using the modified Arrhenius Equation \(\mathrm{\ln\dfrac{k_2}{K_1}=-\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)}\)
    \(\mathrm{E_a= 50000\, j/mol=50\, kj/mol}\)

    To review activation energy, visit the page "Arrhenius Equation".


    Enzymes are much more specific with what reaction they can be catalysts for, while metals like rhodium can be catalysts for a wide variety of reactions. Enzymes also normally only operate at a very specific temperature range. With enzymes, the reactant attaches to the enzyme, the reaction occurs, and the products are released. With catalysts like rhodium (heterogeneous catalysis), tje catalysis absorbs the reactants, diffues them across its occurs, and then is desorbed.

    To review enzymes, visit the page "Catalytic Efficiency of Enzymes".


    The reaction must have a substrate that reacts with the enzyme at an active site to form the enzyme-substrate complex. This must happen for the complex to decompose and form products and regenerate the enzyme, as well as the enzyme must be operating at the correct temperature.

    To review enzymes, visit the page "Catalytic Efficiency of Enzymes".


    \(\textrm{Fast: }\ce{2ICl \rightarrow I2 + 2Cl}\)

    \(\underline{\textrm{Slow: }\ce{H2 + 2Cl \rightarrow 2HCl}}\)
    \(\textrm{Overall: }\ce{H2 + 2ICl \rightarrow I2 + 2HCl}\)

    The reaction rate is only as fast as its slowest step (aka, the rate limiting step). Thus, \(\mathrm{reaction\: rate = slow\: rate = k_2[H_2][Cl]^2}\)

    Since \(\ce{Cl}\) is not in the overall equation, we need to find out how to put it in terms of \(\ce{[ICl]}\) since \(\ce{[ICl]}\) is in the overall equation. The fast rate is nearly in equilibrium since it produces \(\ce{[I2]}\) and \(\ce{2Cl}\) faster than the slow reaction can use up the \(\ce{Cl}\). Thus, the forward rate of the fast reaction equals its own reverse rate.

    \(\mathrm{Fast\: rate = k_1[ICl_2]^2 = k_{-1}[I_2][Cl]^2}\)

    \(\mathrm{[Cl]^2= \dfrac{k_1[ICl_2]^2}{k_{-1}[I_2]}}\)

    Now, substitute this value in for \(\mathrm{[Cl]^2}\) in the slow rate equation.

    \(\mathrm{slow\: rate = k_2[H_2]\times\dfrac{k_1[ICl_2]^2}{k_{-1}[I_2]}}\)

    Since all the \(\ce{k}\)’s are constants, they can be combined into the overall reaction constant, \(\ce{k}\). Also according to the problem, the \(\ce{[I2]}\) is assumed not to affect the reaction rate.Thus, the reaction rate can be described as:

    \(\mathrm{Reaction\: rate = k[H_2][ICl_2]^2}\)

    The reaction rate confirms the experimentally determined reaction order.

    To review rate determining steps, visit the page "Rate Determining Step".


    We have observed that the units of \(\ce{k}\) depend on the overall reaction order. Derive a general expression for the units of \(\ce{k}\) using the units of the order of the reaction (o) units of concentration (M) and time (s)

    The units of \(\ce{k}\) change on the overall reaction

    \(\mathrm{reaction\: rate=k[A]^o}\)

    The units of reaction rate \(\mathrm{= Ms^{-1}}\) units of \(\mathrm{[A]^o=M^o}\)

    So the units of \(\mathrm{k= \dfrac{reaction\: rate}{[A]^o}= \dfrac{Ms^{-1}}{M^o}}\)

    which can be simplified to \(\mathrm{k=\dfrac{1}{M^{o-1}\,s}}\) where o is the overall reaction order

    To review reaction rates, visit the page "Measuring Reaction Rates".


    \(\mathrm{reaction\: rate=k_2[N_2O_2][O_2]}\)

    disappearance of \(\mathrm{N_2O_2 = k_{-1}[N_2O_2]+k_2[N_2O_2][O_2]}\)

    formation of \(\mathrm{N_2O_2 = k_1[NO]^2}\)

    We can assume that the formation of \(\ce{N2O2}\)is equal to its disappearance, so

    \(\mathrm{k_1[NO]^2 = k_{-1}[N_2O_2]+k_2[N_2O_2][O_2]}\)


    substituting for the reaction rate we get,

    assuming molarities are one,

    \(\mathrm{rate=\dfrac{k_1[NO]^2}{(k_{-1}+k_2[O_2])}k_2*[O_2] = 172}\)

    Since \(\mathrm{k_1 < k_2}\), \(\ce{NO + NO \leftrightarrow N2O2}\)is the slower reaction so,


    \(\mathrm{k= 172}\)

    \(\mathrm{rate = 172[NO]^2}\)

    To review reaction rates, visit the page "Measuring Reaction Rates".