4.10: Rate Laws - Differential
- Page ID
- 35142
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- Describe the variation of concentration vs. time for 1st order reactions.
- Describe the variation of concentration vs. time for 2nd order reactions.
- Figure out order of reaction from concentration vs. time plots.
Rate and Order of Reactions
The rate of a chemical reaction is the amount of substance reacted or produced per unit time. The rate law is an expression indicating how the rate depends on the concentrations of the reactants and catalysts. The power of the concentration in the rate law expression is called the order with respect to the reactant or catalyst.
This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics.
Reaction Rates and Stoichiometry
In acidic solutions, hydrogen peroxide and iodide ion react according to the equation:
\(\ce{H2O2 + 2H+ + 3I- \rightarrow 2 H2O + I3-}\).
In this reaction, the reaction Rate can be expressed as
- decreasing Rate of \(\ce{H2O2}\), \(\ce{- \dfrac{d[H2O2]}{dt}}\)
- decreasing Rate of \(\ce{H+}\), \(\ce{- \dfrac{d[H+]}{dt}}\)
- decreasing Rate of \(\ce{I-}\), \(\ce{- \dfrac{d[I- ]}{dt}}\)
- increasing Rate of \(\ce{H2O}\), \(\ce{+ \dfrac{d[H2O]}{dt}}\)
- increasing Rate of \(\ce{I3-}\), \(\ce{\dfrac{d[I3- ]}{dt}}\)
However, from the stoichiometry, you can easily see the following relationship:
\(\ce{-\dfrac{d[H2O2]}{dt}}=\ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}}=\ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}\)
In reality, if the concentration of \(\ce{H2O2}\) is low, the changes in concentrations of \(\ce{H+}\) and \(\ce{H2O}\) are very difficult to detect because their quantities are so large in the solution. The merit in this equation is to show you that the rates of decreasing of reactant concentrations are governed by the stoichiometry. So are the rates of increasing of product concentrations.
There are many ways to express the reaction Rates, for example
\(\ce{-\dfrac{d[H2O2]}{dt}}= \ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}} = \ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}= Rate\)
or
\(\ce{-\dfrac{d[I- ]}{dt}}= Rate\,'\)
Obviously,
\(Rate = 3 Rate\,'\)
But both Rate and Rate' are reasonable expressions.
To generalize it, let the chemical reaction be represented by
\(a\, \ce A + b\, \ce B \rightarrow c\, \ce C + d\, \ce D\)
then the rate is represented by any one of the following
\(rate = -\dfrac{1}{a}\ce{\dfrac{d[A]}{dt}}= -\dfrac{1}{b}\ce{\dfrac{d[B]}{dt}}=\dfrac{1}{c}\ce{\dfrac{d[C]}{dt}}=\dfrac{1}{d}\ce{\dfrac{d[D]}{dt}}\)
Unless the rate expression is specified.
Differential Rate Laws and Integrated Rate Laws
For simplicity, let us consider the reactions:
\(\mathrm{A + other\: reactants} \xrightarrow{\:\:\large{k}\:} \mathrm{products}\)
where \(\ce{A}\) is one of the reactants, and k is the rate constant.
For most experiments in chemical kinetics, the concentration of one reactant or product is monitored. If the concentrations of other reactants are high, they are not greatly changed. Thus, we have a pseudo decomposition reaction.
For the decomposition of \(\ce{A}\) with a rate constant k,
\(\mathrm{A \rightarrow products}\)
The concentration \(\ce{[A]}\) can be monitored. Let the order of the reaction be n, then the expression
\(\mathrm{-\dfrac{d[A]}{dt}= \mathit k [A]^n}\)
is called the differential rate law.
The differential expressions can be integrated to give an explicit relation of \(\ce{[A]}\) with respect to time t. These explicit relations are called integrated rate laws.
Depending on the value of n, the integrated equations are different. If the reaction is first order with respect to \(\ce{[A]}\), integration with respect to time, t, gives:
\(\mathrm{[A] = [A]_o}\, e^{\large{-k\, t}} \tag{1}\)
where \(\mathrm{[A]_o}\) is the concentration of \(\ce{A}\) at t = 0, and \(\ce{[A]}\) is the concentration at time t.
For a second order reaction, the integrated rate law is:
\(\mathrm{\dfrac{1}{[A]}=\dfrac{1}{[A]_o}}+ k t \tag{2}\)
Derive the above equations yourself.
Determination of Rate Constants Using the Integrated Rate Laws
The usual approach to calculate the rate constant k makes use of the differential rate law. A series of experiments are performed with various initial concentrations, and their rates measured. The rate constants are calculated from the initial concentration and time of measurement. Often, you should construct a graph for their evaluation. However, results so obtained contain errors due to the approximation whereas values for k calculated using the integrated rate laws (1) and (2) are more accurate. Today, calculations can easily be performed with the aid of calculators and computers.
Thus, we emphasize that you use the integrated rate laws whenever possible. Whether you use the differential rate laws or the integrated laws, you have to evaluate the order first.
Equation (1) may be rewritten as
\(k =\mathrm{\dfrac{\ln [A]_o - \ln [A]}{t}}\tag{1'}\)
In a real experiment, you should plot \(\ce{- \ln [A]}\) vs. t and find a line to best fit your data. The slope of the line is k.
Similarly, equation (2) may be rewritten as:
\(k =\mathrm{\dfrac{\dfrac{1}{[A]}-\dfrac{1}{[A]_o}}{t}}\tag{2'}\)
Thus, plot of \(\ce{\dfrac{1}{[A]}}\) vs. t should yield a straight line, and the slope is k.
Half Lives of First and Second Order Reactions
The half-life (\(t_{\frac{1}{2}}\)) of a reaction is the time period required to reduce the reactant to half of its original value. The half life of a first order reaction is a constant, independent of the initial concentration. The rate constant and half-life have the relationship:
\(t_{\frac{1}{2}} = \dfrac{\ln(2)}{k}\)
\(t_{\frac{1}{2}} \times k = \ln(2)\)
For 2nd order reactions, the half life depends on the initial concentration, \(\mathrm{[A]_o}\), and the rate constant k:
\(t_{\frac{1}{2}} = \dfrac{1}{k \mathrm{[A]_o}}\)
\(t_{\frac{1}{2}} \times k \mathrm{[A]_o} = 1\)
Since the concentration is reduced to half of its original value at the end of its first half-life, the second half-life is twice as long as the first half life. Thus, a plot of \(\ce{[A]}\) vs. t easily reveals the order of the reaction by tracking its half life.
Either plotting \(\ce{[A]}\) vs. t or plotting the appropriate linear relationship will reveal the order of the reaction.
A summary of the relationships for first and second order reactions is given below:
| Differential rate law |
Integrated rate law |
Linear plot | Half life | |
|---|---|---|---|---|
| first order |
\(\mathrm{-\dfrac{d[A]}{dt} = \mathit k [A]}\) | \(\mathrm{[A] = [A]_o} e^{-\large{k\, t}}\) |
\(\mathrm{\ln [A]\: vs}\: t\) \(\textrm{slope} = - k\) |
\(t_{\frac{1}{2}}=\dfrac{\ln(2)}{k}\) |
| second order |
\(\mathrm{- \dfrac{d[A]}{dt} = \mathit k [A]^2}\) |
\(\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t\) |
\(\mathrm{\dfrac{1}{[A]}\: vs}\: t\) \(\textrm{slope} = k\) |
\(t_{\frac{1}{2}}=\dfrac{1}{k\mathrm{[A]_o}}\) |
Further applications of rate laws will be discussed in Integrated Rate Laws. Students must analyze the problems carefully, and appropriately apply the differential or integrated rate laws to derive the desirable results.
Confidence Building Questions
- If the reaction is zeroth (0th) order with respect to \(\ce{[X]}\), which of the following quantities when plotted vs. t should be a straight line: \(\ce{\ln [X]}\), \(\ce{\dfrac{1}{[X]}}\), \(\ce{[X]}\), or \(\ce{[X]^2}\)?
Discussion -
Since: \(\mathrm{- \dfrac{d[X]}{dt} = k}\); => \(\mathrm{[X] - [X]_o = - \mathit k t}\); The plot of \(\ce{[X]}\) and t has a linear relationship. - The half-lives for all first order reactions are independent of the initial concentration, true or false?
Discussion -
All radioactive decays follow first order kinetics. - For a first order reaction, the product of rate constant and half-life is a constant. What is the constant? (give value)
Discussion -
\(k \times t_{\frac{1}{2}} = \ln(2) = 0.693\) - Tritium (T) decays to helium (\(\ce{He}\) of mass 3) with a half-life of 12.5 years. How long will it take 1.0 g of T to reduce to 0.25 g of T (and 0.75 g of \(\ce{He}\))?
Discussion -
The first 12.5 y reduced 1.0 g of T to 1/2 g. Another 12.5 y reduced 0.5 g of T to 1/4 g.
Calculate the time for 1.0 g T to reduce to 0.9 g. Calculation of half-life from experiment is also a common exercise. - \(\ce{2 HI \rightarrow H2 + I_{2\large{(s)}}}\)
In an experiment starting at 8:00 hr, the partial pressure of \(\ce{HI}\) decreases from 30.0 Pa to 15.0 Pa at 8:45 hr (at a constant temperature). What shall the time be for the partial pressure of \(\ce{HI}\) to reduce to 3.75 Pa?
Discussion -
Amount left: 30, 15, 7.5, 3.75 Time: 8:00 8:45 10:15 13:15 t'increment: 45 90 180 (min)Calculate the time when 10 percent of the \(\ce{HI}\) has decomposed. - The dust explosion is a
- homogeneous reaction
- heterogeneous reaction
- catalytic reaction
- chain reaction
Discussion -
Chain reactions may lead to explosion, but dust explosion is not a chain reaction. Chemists explained the causes of a grain elevator explosion. - \(\ce{2 NO + O2 \rightarrow 2 NO2}\),
the following results were obtained:
run [NO] [O2] Rate 1 0.12 0.05 0.12 2 0.12 0.10 0.24 3 0.24 0.05 0.48
What is the order of the reaction with respect to \(\ce{[NO]}\)?Hint: 2
Discussion -
What is the order with respect to \(\ce{[O2]}\)?