| title | layout |
|---|---|
Rate Laws |
page |
As described in the previous module, the rate of a reaction is affected by the concentrations of reactants. Rate laws{: data-type="term"} or rate equations{: data-type="term"} are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:
in which [A], [B], and [C] represent the molar concentrations of reactants, and k{: data-type="term"} is the rate constant{: data-type="term"}, which is specific for a particular reaction at a particular temperature. The exponents m, n, and p are usually positive integers (although it is possible for them to be fractions or negative numbers). The rate constant k and the exponents m, n, and p must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the concentration of A, B, or C, but it does vary with temperature and surface area.
The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order{: data-type="term"}. Consider a reaction for which the rate law is:
If the exponent m is 1, the reaction is first order with respect to A. If m is 2, the reaction is second order with respect to A. If n is 1, the reaction is first order in B. If n is 2, the reaction is second order in B. If m or n is zero, the reaction is zero order in A or B, respectively, and the rate of the reaction is not affected by the concentration of that reactant. The overall reaction order{: data-type="term"} is the sum of the orders with respect to each reactant. If m = 1 and n = 1, the overall order of the reaction is second order (m + n = 1 + 1 = 2).
The rate law:
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
describes a reaction that is second order in C4H6 and second order overall. The rate law:
describes a reaction that is first order in H+, first order in OH−, and second order overall.
Solution The reaction will have the form:
Check Your Learning The rate law for the reaction:
It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates{: data-type="term"}, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.
{: #CNX_Chem_12_03_OzoneHole}
| Trial | [NO] (mol/L) | [O3] (mol/L) | Δ[NO2]Δt(molL−1s−1)
| {: valign="middle"}|---------- | 1 | 1.00 ×
10−6 | 3.00 ×
10−6 | 6.60 ×
10−5 | {: valign="middle"}| 2 | 1.00 ×
10−6 | 6.00 ×
10−6 | 1.32 ×
10−4 | {: valign="middle"}| 3 | 1.00 ×
10−6 | 9.00 ×
10−6 | 1.98 ×
10−4 | {: valign="middle"}| 4 | 2.00 ×
10−6 | 9.00 ×
10−6 | 3.96 ×
10−4 | {: valign="middle"}| 5 | 3.00 ×
10−6 | 9.00 ×
10−6 | 5.94 ×
10−4 | {: valign="middle"}{: .medium .unnumbered summary="This table has four columns and six rows. The first row is a header row, and it labels each column: “Trial,” “[ N O ] ( mol / L),” “[ O subscript 3 ] ( mol / L ),” and “capital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).” Under the “Trial” column are the numbers: 1, 2, 3, 4, and 5. Under the “[ N O ] ( mol / L)” column are the numbers: 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 2.00 times ten to the negative six power; and 3.00 times ten to the negative six power. Under the “[ O subscript 3 ] ( mol / L )” column are the numbers: 3.00 times ten to the negative six; 6.00 times ten to the negative six; 9.00 times ten to the negative six; 9.00 times ten to the negative six; and 9.00 times ten to the negative six. Under the column “capital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )” are the numbers: 6.60 times ten to the negative five; 1.32 times ten to the negative four; 1.98 times ten to the negative four; 3.96 times ten to the negative four; and 5.94 times ten to the negative four." data-label=""}
Determine the rate law and the rate constant for the reaction at 25 °C.
Solution The rate law will have the form:
-
Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
-
Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:* * * {: data-type="newline"}
rate=k[NO]1[O3]1=k[NO][O3] -
Determine the value of k from one set of concentrations and the corresponding rate.* * * {: data-type="newline"}
k=rate[NO][O3]=6.60×10−5mol L−1s−1(1.00×10−6mol L−1)(3.00×10−6molL−1)=2.20×107Lmol−1s−1The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough. {: data-number-style="arabic" .stepwise}
Check Your Learning Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
| Trial | [CH3CHO] (mol/L) | −Δ[CH3CHO]Δt(molL−1s−1)
| {: valign="middle"}|---------- | 1 | 1.75 ×
10−3 | 2.06 ×
10−11 | {: valign="middle"}| 2 | 3.50 ×
10−3 | 8.24 ×
10−11 | {: valign="middle"}| 3 | 7.00 ×
10−3 | 3.30 ×
10−10 | {: valign="middle"}{: .medium .unnumbered summary="This table has three columns and four rows. The first row is a header row, and it labels each column, “Trial,” “[ C H subscript 3 C H O ] ( mol / L),” and “negative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).” Under the “Trial” column are the numbers: 1, 2, and 3. Under the “[ C H subscript 3 C H O ] ( mol / L)” are the numbers: 1.75 times ten to the negative three; 3.50 times ten to the negative three; and 7.00 times ten to the negative three. Under the column “negative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )” are the numbers: 2.06 times ten to the negative 11; 8.24 times ten to the negative 11; and 3.30 times ten to the negative ten." data-label=""}
with k = 6.73 ×
10−6 L/mol/s
| {: valign="middle"}|---------- | 1 | 0.10 | 0.10 | 0.00300 | {: valign="middle"}| 2 | 0.10 | 0.15 | 0.00450 | {: valign="middle"}| 3 | 0.15 | 0.10 | 0.00675 | {: valign="middle"}{: .medium .unnumbered summary="This table has four columns and four rows. The first row is a header row, and it labels the columns, “Trial,” “[ N O ] ( mol / L),” “[ C l subscript 2 ] ( mol / L ),” and “negative capital delta [ N O] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).” Under the column “Trial” are the numbers: 1, 2, and 3. Under the column “[ N O ] ( mol / L)” are the numbers: 0.10, 0.10, and 0.15. Under the column “[ C l subscript 2 ] ( mol / L )” are the numbers: 0.10, 0.15, and 0.10. Under the column “negative delta [ N O] divided by delta t ( mol L superscript negative 1 s superscript negative 1” are the numbers: 0.00300, 0.00450, 0.00675." data-label=""}
Solution The rate law for this reaction will have the form:
-
Determine the value of m from the data in which [NO] varies and [Cl2] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:* * * {: data-type="newline"}
ratexratey=k[NO]xm[Cl2]xnk[NO]ym[Cl2]ynUsing the third trial and the first trial, in which [Cl2] does not vary, gives:* * * {: data-type="newline"}
rate 3rate 1=0.006750.00300=k(0.15)m(0.10)nk(0.10)m(0.10)nAfter canceling equivalent terms in the numerator and denominator, we are left with:* * * {: data-type="newline"}
0.006750.00300=(0.15)m(0.10)mwhich simplifies to:* * * {: data-type="newline"}
2.25=(1.5)mWe can use natural logs to determine the value of the exponent m:* * * {: data-type="newline"}
ln(2.25)=mln(1.5)ln(2.25)ln(1.5)=m2=mWe can confirm the result easily, since:* * * {: data-type="newline"}
1.52=2.25 -
Determine the value of n from data in which [Cl2] varies and [NO] is constant.* * * {: data-type="newline"}
rate 2rate 1=0.004500.00300=k(0.10)m(0.15)nk(0.10)m(0.10)nCancelation gives:* * * {: data-type="newline"}
0.00450.0030=(0.15)n(0.10)nwhich simplifies to:
1.5=(1.5)nThus n must be 1, and the form of the rate law is:* * * {: data-type="newline"}
Rate=k[NO]m[Cl2]n=k[NO]2[Cl2] -
Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol−2 L2/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:* * * {: data-type="newline"}
0.00300molL−1s−1=k(0.10molL−1)2(0.10molL−1)1k=3.0mol−2L2s−1
{: data-number-style="arabic" .stepwise}
Check Your Learning Use the provided initial rate data to derive the rate law for the reaction whose equation is:
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
{: data-type="newline"}
2.00 = 2.00y* * * {: data-type="newline"}
y = 1* * * {: data-type="newline"}
rate 1rate 2=0.001840.00092=k(0.0040)x(0.0020)yk(0.0020)x(0.0040)y
{: data-type="newline"}
2.00=2x2y2.00=2x214.00=2xx=2
{: data-type="newline"}
Substituting the concentration data from trial 1 and solving for k yields:* * * {: data-type="newline"}
rate=k[OCl−]2[I−]10.00184=k(0.0040)2(0.0020)1k=5.75×104mol−2L2s−1
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.
Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
Reaction orders also play a role in determining the units for the rate constant k. In [link], a second-order reaction, we found the units for k to be Lmol−1s−1,
whereas in [link], a third order reaction, we found the units for k to be mol−2 L2/s. More generally speaking, the units for the rate constant for a reaction of order (m+n)
are mol1−(m+n)L(m+n)−1s−1.
[link] summarizes the rate constant units for common reaction orders.
| Rate Constants for Common Reaction Orders | |
|---|---|
| Reaction Order | Units of k |
| (m+n) | mol1−(m+n)L(m+n)−1s−1 |
| zero | mol/L/s |
| first | s−1 |
| second | L/mol/s |
| third | mol−2 L2 s−1 |
Note that the units in the table can also be expressed in terms of molarity (M) instead of mol/L. Also, units of time other than the second (such as minutes, hours, days) may be used, depending on the situation.
Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
(a) What is the order of the reaction with respect to that reactant?
(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
(a) What is the order of the reaction with respect to that reactant?
(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant?
if the rate law for the reaction is rate=k[NO2]2?
(a) Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm.
(b) Increasing the concentration of CO from 0.01 M to 0.03 M.
if the rate law for the reaction is rate=k[NO2][CO]
?
(a) Increasing the pressure of NO2 from 0.1 atm to 0.3 atm
(b) Increasing the concentration of CO from 0.02 M to 0.06 M.
is first order with respect to both NO and O3 with a rate constant of 2.20 ×
107 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 ×
10−6 M and [O3] = 5.9 ×
10−7 M?
10−5 mol/L/s
1532P⟶ 1632S+e−
{: data-type="newline"}
Rate = 4.85 ×
10−2 day−1[32P]
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?
10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles):* * * {: data-type="newline"}
614C⟶714N+e−
{: data-type="newline"}
rate=k[614C]
What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 ×
10−9 M?
10−13 mol/L/year
10−8 L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 ×
10−4 M?
| [C2H5OH] (M) | 4.4 ×
10−2 | 3.3 ×
10−2 | 2.2 ×
10−2 | {: valign="middle"}| Rate (mol/L/h) | 2.0 ×
10−2 | 2.0 ×
10−2 | 2.0 ×
10−2 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column serves as a header and labels each row, “[ C subscript 2 H subscript 5 O H ] ( M ),” and, “Rate ( mol / L / h ).” To the right of the “Initial Rate ( mol / L / s)” header are the numbers: 4.4 times ten to the negative 2; 3.3 times ten to the negative 2; 2.2 times ten to the negative 2. To the right of the “Rate ( mol / L / h )” header are the numbers: 2.0 times ten to the negative 2, 2.0 times ten to the negative 2, and 2.0 times ten to the negative 2." data-label=""}
Determine the rate equation, the rate constant, and the overall order for this reaction.
10−2 mol/L/h (about 0.9 g/L/h for the average male); The reaction is zero order.
| [NH3] (M) | 1.0 ×
10−3 | 2.0 ×
10−3 | 3.0 ×
10−3 | {: valign="middle"}| Rate (mol/L/h1) | 1.5 ×
10−6 | 1.5 ×
10−6 | 1.5 ×
10−6 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column labels each row, “[ N H subscript 3 ] ( M ),” and “Rate ( mol / L / h superscript 1 ).” The numbers to the right of the header, “[ N H subscript 3 ] ( M )” are: 1.0 times ten to the negative three; 2.0 times ten to the negative three; and 3.0 times ten to the negative three. The numbers to the right of the header, “Rate ( mol / L / h superscript 1” are: 1.5 times ten to the negative six; 1.50 times ten to the negative 6; and 1.5 times ten to the negative six." data-label=""}
Determine the rate equation, the rate constant, and the overall order for this reaction.
2NOCl(g)⟶2NO(g)+Cl2(g)
Determine the rate equation, the rate constant, and the overall order for this reaction from the following data:
| [NOCl] (M) | 0.10 | 0.20 | 0.30 | {: valign="middle"}| Rate (mol/L/h) | 8.0 ×
10−10 | 3.2 ×
10−9 | 7.2 ×
10−9 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column labels each row, “[ N O C l ] ( M ),” and, “Rate ( mol / L / h ).” To the right of the header, “[ N O C l ] ( M )” are the numbers: 0.10, 0.20, and 0.30. To the right of the header, “Rate ( mol / L / h )” are the numbers: 8.0 times ten to the negative ten; 3.2 times ten to the negative 9; and 7.2 times ten to the negative 9." data-label=""}
10−8 L/mol/h; second order
| [A] (M) | 1.33 ×
10−2 | 2.66 ×
10−2 | 3.99 ×
10−2 | {: valign="middle"}| Rate (mol/L/h) | 3.80 ×
10−7 | 1.52 ×
10−6 | 3.42 ×
10−6 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column labels each row, “[ A ] ( M ),” and “Rate ( mol / L / h).” To the right of the header, “[ A ] ( M ),” are the numbers: 1.33 times ten to the negative 2; 2.66 times ten to the negative 2; and 3.99 times ten to the negative 2.” To the right of the header, “Rate ( mol / L / h),” are the numbers 3.80 times ten to the negative 7; 1.52 times ten to the negative 6; and 3.42 times ten to the negative 6." data-label=""}
2NO(g)+Cl2(g)⟶2NOCl(g)
The following initial rates of reaction have been observed for certain reactant concentrations:
| [NO] (mol/L1) | [Cl2] (mol/L) | Rate (mol/L/h) | {: valign="middle"}|---------- | 0.50 | 0.50 | 1.14 | {: valign="middle"}| 1.00 | 0.50 | 4.56 | {: valign="middle"}| 1.00 | 1.00 | 9.12 | {: valign="middle"}{: .medium .unnumbered summary="This table has three columns and four rows. The first row serves as a header row and labels each column, “[ N O ] ( mol / L superscript 1 ),” “[ C l subscript 2 ] ( mol / L ),” and, “Rate ( mol / L / h ).” Under the column “[ N O ] ( mol / L superscript 1 )” are the numbers: 0.50, 1.00, and 1.00. Under the column “[ C l subscript 2 ] ( mol / L)” are the numbers: 0.50, 0.50, and 1.00. Under the column “Rate ( mol / L / h )” are the numbers: 1.14, 4.56, and 9.12." data-label=""}
What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?
Determine the rate equation, the rate constant, and the orders with respect to each reactant from the following data:
| [NO] (M) | 0.30 | 0.60 | 0.60 | {: valign="middle"}| [H2] (M) | 0.35 | 0.35 | 0.70 | {: valign="middle"}| Rate (mol/L/s) | 2.835 ×
10−3 | 1.134 ×
10−2 | 2.268 ×
10−2 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and three rows. The first column labels each row, “[ N O ] ( M ),” “[ H subscript 2 ] ( M ),” and “Rate ( mol / L / s ).” To the right of the header “[ N O ] ( M )” are the numbers: 0.30, 0.60, and 0.60. To the right of the header “[ H subscript 2 ] ( M )” are the numbers: 0.35, 0.35, and 0.70. To the right of the header, “Rate ( mol / L / s )” are the numbers 2.835 times ten to the negative 3; 1.134 times 10 to the negative two; and 2.268 times ten to the negative 2.”" data-label=""}
the following data were obtained at 30 °C:
| [A] (M) | 0.230 | 0.356 | 0.557 | {: valign="middle"}| Rate (mol/L/s) | 4.17 ×
10−4 | 9.99 ×
10−4 | 2.44 ×
10−3 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column labels each row, “[ A ] ( M ),” and “Rate ( mol / L / s ).” To the right of the header, “[ A ] ( M ),” are the numbers: 0.230, 0.356, and 0.557. To the right of the header, “Rate ( mol / L / s ),” are the numbers: 4.7 times ten to the negative 4; 9.99 times ten to the negative 4; and 2.44 times ten to the negative 3." data-label=""}
(a) What is the order of the reaction with respect to [A], and what is the rate equation?
(b) What is the rate constant?
10−3 L mol−1 s−1
the following data were obtained at 30 °C:
| [Q]initial (M) | 0.170 | 0.212 | 0.357 | {: valign="middle"}| Rate (mol/L/s) | 6.68 ×
10−3 | 1.04 ×
10−2 | 2.94 ×
10−2 | {: valign="middle"}{: .column-header .medium .unnumbered summary="This table has four columns and two rows. The first column labels each row, “[ Q ] subscript initial ( M ),” and “Rate ( mol / L / s ).” To the right of the header, “[ Q ] subscript initial ( M ),” are the numbers: 0.170, 0.212, and 0.357. To the right of the header, “Rate ( mol / L / s ),” are the numbers: 6.68 times ten to the negative 3; 1.04 times ten to the negative 2; and 2.94 times ten to the negative 2." data-label=""}
(a) What is the order of the reaction with respect to [Q], and what is the rate equation?
(b) What is the rate constant?
10−4 min−1.* * * {: data-type="newline"}
2N2O5⟶4NO2+O2
What is the rate of the reaction when [N2O5] = 0.40 M?
10−4 mol/L/min
(a) 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)
(b) 2NO(g)+O2(g)⟶2NO2(g)
(c) 3NO2(g)+H2O(l)⟶2HNO3(aq)+NO(g)
The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 ×
10−6 L2/mol2/s.
I−+OCl−⟶IO−+Cl−
| | 1 | 2 | 3 | {: valign="middle"}|---------- | [I−]initial
(M) | 0.10 | 0.20 | 0.30 | {: valign="middle"}| [OCl−]initial
(M) | 0.050 | 0.050 | 0.010 | {: valign="middle"}| Rate (mol/L/s) | 3.05 ×
10−4 | 6.20 ×
10−4 | 1.83 ×
10−4 | {: valign="middle"}{: .medium .unnumbered summary="This table has four columns and four rows. The first column labels each row. The first cell in the first column is blank, and then each row is labeled, “[ I superscript negative sign ] subscript initial ( M ),” “[ O C l superscript negative sign ] subscript initial ( M ),” and, “Rate ( mol / L / s ).” To the right of the blank cell are the numbers: 1, 2, and 3. To the right of the header, “[ I superscript negative sign ] subscript initial ( M ),” are the numbers: 0.10, 0.20, and 0.30. To the right of the header, “[ O C l superscript negative sign ] subscript initial ( M ),” are the numbers: 0.050, 0.050, and 0.010. To the right of the header, “Rate ( mol / L / s ),” are the numbers: 3.05 times ten to the negative 4; 6.20 times ten to the negative 4; and 1.83 times ten to the negative 4." data-label=""}
Determine the rate equation and the rate constant for this reaction.
10−2 L mol −1 s−1
method of initial rates : use of a more explicit algebraic method to determine the orders in a rate law ^
overall reaction order : sum of the reaction orders for each substance represented in the rate law ^
rate constant (k) : proportionality constant in the relationship between reaction rate and concentrations of reactants ^
rate law : (also, rate equation) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants ^
reaction order : value of an exponent in a rate law, expressed as an ordinal number (for example, zero order for 0, first order for 1, second order for 2, and so on)