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Chemistry LibreTexts

8: Kinetics II (Worksheet)

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    Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

    This worksheet reviews some of the concepts we have seen concerning rates of reactions. Specifically, it covers once again the idea of how we define the Rate on the basis of the reaction’s stoichiometry, and it reviews the process of determining the form of the differential rate law from kinetic data. The differential rate law shows how the rate of the reaction changes with concentration of reactants (and sometime products). The integrated rate law, which is derived by means of calculus from the differential rate law, shows how the concentration of reactant changes with time. We will look at the form of the integrated rate law for a first order reaction, and in the next worksheet we will extend this to discuss the idea of half life.

    Learning Objectives

    • Review definition of Rate based on the stoichiometric relationships in a balanced equation
    • Review determining the form of the differential rate law from experimental data
    • Know and understand the meaning of first- and second-order integrated rate law expressions

    Success Criteria

    • Be able to write the defining expression for the rate of any reaction
    • Be able to determine the form of the differential rate law from experimental data
    • Be able to sketch a first-order integrated rate law plot and understand the information that can be obtained from it


    Consider the reaction:

    \[\ce{N2(g) + 3 H2(g) \rightarrow 2 NH3(g)} \nonumber\]

    At a certain point in the course of an experiment the rate of disappearance of \(N_2(g)\) is 3.56 x 10–3 mol L–1 s–1. How would you define the Rate on the basis of the disappearance of \(H_2(g)\) such that you would obtain the same numerical value at this point in the reaction? How would you define it on the basis of the appearance of \(NH_3(g)\)?


    For each of the following experimentally determined rate law expressions, give the units for \(k\), assuming that all species concentrations are in units of mol L–1 and all rates are in units of mol L–1 s–1.

    \[\ce{2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g) }\nonumber\]

    with \(rate = k[\ce{N2O5}]\)

    \[\ce{2 NO_2(g) + F_2(g) \rightarrow 2 NO_2F(g)}\nonumber\]

    with \(rate = k[\ce{NO2}][\ce{F2}]\)

    \[\ce{2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)} \nonumber\]

    with \(rate = k[\ce{NO}]^2[\ce{O2}]\)

    \[\ce{CH_3CHO(g) \rightarrow CH_4(g) + CO(g)}\nonumber\]

    with \(rate = k[\ce{CH3CHO}]^{3/2}\)


    Consider the reaction:

    \[\ce{2 NO(g) + 2 H_2(g) \rightarrow N_2(g) + 2 H_2O(g)} \nonumber\]

    1. The rate law for this reaction is second order in \(\ce{NO}\) and first order in \(\ce{H_2}\). Write the rate law expression.
    2. If the rate constant for this reaction at 1000 K is \(6.0 \times 10^4 M^{–2}s^{–1}\), what is the reaction rate when \([\ce{NO}] = 0.035 \,M\) and \([\ce{H_2}] = 0.015\, M\)?
    3. What is the reaction rate at 1000 K when \([\ce{NO}] = 0.070 \,M\) and \([H_2] = 0.015\, M\)?
    4. What is the reaction rate at 1000 K when \([\ce{NO}] = 0.035 \,M\) and \([H_2] = 0.030 \,M\)?
    5. What is the reaction rate at 1000 K when \([\ce{NO}] = 0.070 \,M\) and \([H_2] = 0.030 \,M\)?


    Consider the following kinetic data for the reaction

    \[\ce{2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)} \nonumber\]

    1. What is the rate law expression for the reaction?
    2. What is the value of the rate constant with the proper units?
    Exp. [NO] (M) [O2] (M) Initial Rate (M/s)
    1 0.0126 0.0125 \(1.41 \times 10^{–2}\)
    2 0.0252 0.0250 \(1.13 \times 10^{–1}\)
    3 0.0252 0.0125 \(5.64 \times 10^{–2}\)

    Integrated Rate Laws

    The differential rate laws we have been looking at show the dependence of rate on concentration. By using calculus, such rate laws can be converted into equations that show the dependence of concentration on time. These expressions are called the integrated rate laws. The form of the integrated rate law equation varies with the order of the differential rate law from which it is derived. In this worksheet we will only consider the form of the integrated rate law for a first-order reaction.

    For a reaction, \(\ce{A -> products}\), which is first-order in \(\ce{A}\), we can write

    \[ rate = \dfrac{-d[\ce{A}]}{dt} - k[\ce{A}]\]

    Suppose we consider the change in concentration of A from its initial value \([\ce{A}]_o\) to its value \([\ce{A}]\) at some later time \(t\). It can be shown by integrating the differential rate law (via calculus) that the dependence on time will be given by

    \[ - \ln \left( \dfrac{[\ce{A}]}{[\ce{A}]_o} \right)= kt \label{Ilaw1}\]


    \[ \ln \left( \dfrac{[\ce{A}]}{[\ce{A}]_o} \right) = - kt \label{Ilaw2}\]


    \[ \ln \left( \dfrac{[\ce{A}]_o}{[\ce{A}]} \right)= kt \label{Ilaw3}\]

    Equation \ref{Ilaw3} (and also Equations \ref{Ilaw1}-\ref{Ilaw2}) is the first-order integrated rate law equation. We can more easily see how concentration of \(\ce{A}\) varies with time by rearranging to give the following form:

    \[ \ln[\ce{A}] = –kt + \ln[\ce{A}]_o \nonumber\]

    This is a straight-line equation of the form

    \[y = mx + b \nonumber\]

    where \(\ln[\ce{A}]\) is the \(y\) variable, \(t\) is the \(x\) variable, \(–k\) is the slope, and \(\ln[\ce{A}]_o\) is the \(y\) intercept.

    If we plot the natural logarithm of \([\ce{A}]\) versus \(t\) we should get a straight line whose slope is \(-k\) and whose intercept is \(\ln[\ce{A}]_o\). Indeed, only a first-order reaction would give a linear plot. If the rate law were anything other than first order, such a plot would yield a curve.


    Consider a first order reaction, \(A \rightarrow products\). Sketch a plot of \(\ln [A]\) vs. \(t\) for such a reaction. Indicate on your sketch where the value of the logarithm of initial concentration of A, \(\ln [A]_o\), is located. Is the slope positive or negative? How is the slope determined and how is it related to \(k\)?


    We have been calculating the values of the rate constant, \(k\), from tables of kinetic data using the value of the rate for a specific set of concentrations in any one experiment. If the reaction were first order, why would it be better to use a plot of the logarithm of reactant concentration versus time to obtain the value of \(k\)?

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