Chemical Kinetics (Part II): Rate Law & Order of Reaction

In our previous post, we looked at the overview of the rates of chemical reactions, where we studied the concepts and factors that affect the rates of reactions. In this post, we will be studying the rate law and the different orders of reaction, and how to determine them.
Rate Law
Consider the reaction:
            mA + nB ----> Products
The rate law states that the rate of a reaction is directly proportional to the active masses of the reactants. This implies that the concentration of the reacting species will determine how fast and how far a reaction can go. Using the above equation, the rate law can be expressed as:
            rate & [A]^m[B]^n ..........(i)
            rate = k[A]^m[B]^n ..........(ii)
where,
[A] = concentration of reactant A
[B] = concentration of reactant B
   & = sign of proportionality
   k = rate constant
The rate law is also known as the law of mass action.
Orders of Reaction
In chemical kinetics, an order is the index or power of the concentration of a reactant in a rate law. The overall order of a reaction is the sum of the orders of the respective reactants in the rate law. The three common types of orders of reaction are:
a) zeroth-order
b) first-order
c) second-order
From equation (ii) above, m is the order with respect to reactant A, while n is the order with respect to reactant B, and the overall order of the reaction is given as (m+n), and can either be 0, 1, 2 etc. However, it is important to note that the order of reaction can only be determined experimentally and NOT from the stoichiometric coefficients of the reactants in a given reaction.
For instance, according to the stoichiometry for the decomposition of nitrogen (I) oxide [N2O], to form nitrogen and oxygen in the presence of platinum,
             2N2O(g) ----> N2(g) + O2(g)
             rate = k[N2O]^2 .....(iii)
the order of the reaction is 2, i.e, second order with respect to the reactant, N2O. Meanwhile, the reaction is a zero order reaction when determined experimentally, i.e,
            rate = k[N2O]^0 ......(iv)
Orders of Reactions
Zeroth-Order Reaction
A reaction is said to be zeroth-order, if the rate of the reaction is independent of the concentration of the reactants, i.e,
            rate = k[reactants]^0
From our example on the decomposition of N2O, the rate of reaction is not dependent on the concentration of the reactant, but on the surface area of the platinum catalyst. This explains why the reaction will still proceed at the same rate, even when the concentration of N2O is very low.
So, how can we know that a given reaction is zeroth-order?
Let us consider the hypothetical reaction:
              A ----> Products
For a zeroth-order,
              rate = k[A]^0..............(v)
              rate = k (since [A]^0 = 1) ………(vb)
Recall,
             rate = -d[A]/dt ............(vi)
Therefore,
        -d[A]/dt = k
          d[A]/dt = -k
               d[A] = -k dt..................(vii)
Integrating equation (vii), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
             $d[A] = -k $dt
       [A] - [Ao] = -k (t - 0)
       [A] - [Ao] = -kt
                 [A] = [Ao] - kt ..................(viii)
where [Ao] = initial concentration of A at time, t = 0
              [A] = concentration of A at time, t = t
Equation (vii) is the differential rate law, while equation (viii) is the integrated rate law of a zeroth-order reaction.
Therefore, a plot of the reactant concentration, [A] against time, t, will give a straight-line graph with a negative slope, -k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equivalent to the initial concentration of the reactant, [Ao].
Similarly, from equation (v), a plot of the reaction rate against concentration of reactant will give a straight-line graph, of uniform rate of reaction. From equation (vb), the rate constant of a zeroth-order reaction is equivalent to its rate of reaction. Hence, its unit is mol dm^-3 s^-1.
Thus, any reaction, whose graphs possess the abovementioned features can be said to be a zeroth-order reaction.
First-Order Reaction
A first-order reaction is one, whose rate is directly proportional to the concentration of only one reactant.
Let us consider the following reactions:
            A + B ----> Products .................(ix)
                  A ----> Products ........................(x)
From the above equations, if the rate of reaction is dependent on the concentration of A only, then, the rate law of equation (ix) will reduce to that of equation (x), as shown:
               rate = k[A]^1[B]^0
               rate = k[A] ..................(xi)
Substituting equation (vi) into (xi) gives
        -d[A]/dt = k[A]
          d[A]/dt = -k[A] .................(xii)
Rearranging equation (xii) gives us
          d[A]/[A] = -k dt .................(xiii)
Integrating equation (xiii), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
   $(1/[A])d[A] = -k $dt
ln [A] - ln [Ao] = -k(t - 0)
ln [A] - ln [Ao] = -kt ...................(xiv)
    ln ([A]/[Ao]) = -kt
           [A]/[Ao] = e^-kt
                   [A] = [Ao]e^-kt ..........(xv)
From equation (xiv), equation (xv) can be written as:
               ln [A] = ln [Ao] - kt .......(xvi)
where [A] and [Ao] retain their usual meanings.
Equation (xii) is the differential rate law, while equations (xv) and (xvi) are the integrated rate laws of a first-order reaction. While the former is the exponential form, the latter is the logarithmic form.
Therefore, a plot of the natural logarithm of the concentration of the reactant, ln [A] against time, t, will give a logarithmic curve with a negative slope, -k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equivalent to the natural logarithm of the initial concentration of the reactant, ln [Ao].
Similarly, from equation (xii), a plot of the rate of reaction against the concentration of reactant will give a straight-line graph, passing through the origin, with a negative slope equal to the rate constant of the reaction. The unit of the rate constant of a first-order reaction is s^-1 (i.e, mol dm^-3 s^-1/mol dm^-3)
Hence, any reaction, whose graphs show these features can be classified as a first-order reaction.
Second-Order Reaction
A reaction is said to be second-order when its rate is directly proportional to the square of the concentration of only one reactants, or the product of the concentration of two reactants.
Consider the hypothetical reaction below:
            2A ----> Products .................(xvii)
           A + B ----> Products ……………(xviib)
From the above, the reaction rate can depend on the concentration of A only or those of A and B. Hence, the rate law can be stated as:
               rate = k[A]^2 ..................(xviii), or
               rate = k[A]^1[B]^1............(xix)          
For our scope, we will focus on equation (xviii), where the rate of reaction is dependent on the concentration of only one reactant.
So, substituting equation (vi) into (xviii) gives
        -d[A]/dt = k[A]^2
          d[A]/dt = -k[A]^2 .................(xx)
Rearranging equation (xx) gives us
          d[A]/[A]^2 = -k dt .................(xxi)
Integrating equation (xxi), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
   $(1/[A]^2)d[A] = -k $dt
-([A]^-1) - (-[Ao]^-1) = -k(t - 0)
-1/[A] - (-1/[Ao]) = -kt ...................(xxii)
    -1/[A] + 1/[Ao] = -kt
Rearranging the above equation gives us
                    1/[A] = 1/[Ao] + kt .............(xxiii)
where [A] and [Ao] retain their meanings.
Equation (xx) gives the differential rate law, while equation (xxiii) is the integrated rate law of a second-order reaction.
Thus, a plot of the reciprocal of the concentration of the reactant, 1/[A] against time, t, will give a straight-line graph with a positive slope, k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equal to the reciprocal of the initial concentration of the reactant, 1/[Ao].
Similarly, from equation (xx), a plot of the rate of reaction against the square of the concentration of reactant will give a parabolic curve, with a tangent equal to the rate constant, k. The unit of a second-order reaction is dm^3 mol^-1 s^-1 (i.e, mol dm^-3 s^-1/mol^2 dm^-6).
Therefore, any reaction, whose graphs possess these features can be said to be a second-order reaction.
Summary
Reaction Order      Differential Law       Integrated Law
Zeroth                     d[A]/dt =  -k                   [A] = [Ao] - kt
First                         d[A]/dt = -k[A]           ln [A] = ln [Ao] - kt
Second                   d[A]/dt = -k[A]^2        1/[A] = 1/[Ao] + kt
How to Determine the Rate Law and Rate Constant from a Set of Experimental Data
Given a set of experimental data of a reaction, the following steps can be used to determine the rate law and the rate constant of the reaction.
1. Compare the reactant concentrations and reaction rates of two experiments to see the effect of concentration on the rate of reaction.
(a) If a change in reactant concentration has no effect on the reaction rate, then, it is a zeroth-order reaction.
(b) If a change in the reactant concentration leads to a corresponding change in the rate of reaction, then, it is first order with respect to the reactant.
(c) If the reaction rate is quadrupled when the reactant concentration is doubled, then, the reaction is second order with respect to the reactant.
Alternatively, you can use the concentration and rate data from two experiments to find the reaction order with regards to a given reactant in the rate law.
2. Write the appropriate rate law for the reaction.
3. Use the measured concentration and rate data from any of the experiments to find the rate constant using the rate law.
Example 1
At high temperatures, 1-chloroethane produces hydrogen chloride and ethene by the following reaction:
C2H5Cl(g) + heat ----> HCl(g) + C2H4(g)
Using the data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the 1-chloroethane, and determine the rate constant for the reaction.
Exp    [C2H5Cl]o (M)    Initial Rate (M s^-1)
1             0.0050                     0.8 × 10^-8
2             0.0075                     1.2 × 10^-8
3             0.0150                     2.4 × 10^-8
4             0.0200                     3.2 × 10^-8
Answer
From Exp 1 and 3, as the concentration is tripled, the reaction rate also triples; and from Exp 1 and 4, the rate of reaction quadruples as the concentration quadrupled. Therefore, the reaction is a first-order with respect to 1-chloroethane, because the rate is directly proportional to its concentration.
Hence, the rate law is:
            Rate = k[C2H5Cl]
To find k, substitute the data from Exp 1 into the rate law:
          0.8 × 10^-8 = k (0.005)
                            k = 8 × 10^-9/5 × 10^-3
                               = 1.6 × 10^-6 s^-1
Alternatively, we can use the data from Exp 1 and 3 to find the order of reaction with respect to 1-chloroethane using:
            Rate = k[C2H5Cl]^m
where, m = order of reaction with respect to C2H5Cl.
2.4 × 10^-8/0.8 × 10^-8 = k(0.015)^m/k(0.005)^m
                                       3 = (0.015/0.005)^m
                                   3^1 = 3^m
             Equate the powers of 3
                                      m = 1
Therefore, the reaction is first-order with respect to the reactant, and the rate law is:
            Rate = k[C2H5Cl]
Example 2
If a sample of the 1-chloroethane in Example 1 above, with an initial concentration of 0.0100 mol dm^-3 is heated at 650°C,
(a) what will be its concentration after 5 hours?
(b) How many hours at 650°C must elapse for the concentration to decrease to 0.0025 mol dm^-3.
Answer
Here, we will make use of the integrated law for a first-order reaction:
             ln [A] = ln [Ao] - kt
Recall that we have already calculated the rate constant, k = 1.6 × 10^-6 s^-1. So, substitute the given parameters into the equation and solve for the unknown.
(a) [Ao] = 0.0100 mol dm^-3, [A] =?
            t = 5 × 3600s = 18000s
ln [A] = ln 0.01 - (1.6 × 10^-6)(1.8 × 10^4)
                  ln [A] = -4.6052 - 0.0288
                  ln [A] = -4.6340
                      [A] = e^-4.6340
                            = 0.00972 mol dm^-3
(b) [Ao] = 0.0100 mol dm^-3, [A] = 0.0025 mol dm^-3, t =?
ln 0.0025 = ln 0.01 - (1.6 × 10^-6)t
               t = (ln 0.01 - ln 0.0025)/(1.6 × 10^-6)
                 = -4.6052 - (-5.9915)/(1.6 × 10^-6)
                 = 1.3863/(1.6 × 10^-6)
                 = 866437.5 s/3600 s hr^-1
                 = 240.68 hrs
Example 3
1-Bromopropane is a colourless liquid that reacts with S2O3-- according to the equation:
C3H7Br + S2O3-- ----> C3H7S2O3- + Br-
The reaction is first order in 1-bromopropane and first order in S2O3--, with a rate constant of 8.05 x 10^-4 mol dm^-3 s^-1.
(a) If the reaction began with 0.4 mol dm^-3 of C3H7Br and an equivalent concentration of S2O3--, what will be the initial reaction rate?
(b) If the concentration of each reactant was decreased to 0.2 mol dm^-3, what will be the initial reaction rate?
Answer
Since the reaction is first order to C3H7Br and S2O3--, then, the rate law of the reaction will be given as:
             Rate = k[C3H7Br][S2O3--], and the overall order = 1 + 1 = 2
(a) If [C3H7Br] = [S2O3--] = 0.4 mol dm^-3, k = 8.05 × 10^-4 dm^3 mol^-1 s^-1, then:
Initial rate = 8.05 × 10^-4 × 0.4 × 0.4
                   = 1.288 × 10^-4 mol dm^-3 s^-1
(b) If [C3H7Br] = [S2O3--] = 0.2 mol dm^-3, then:
Initial rate = 8.05 × 10^-4 × 0.2 × 0.2
                   = 3.22 × 10^-5 mol dm^-3 s^-1
From the above calculations, we can see that a reduction of the reactants concentrations by half decreases the rate of reaction by a quarter, which agrees with the behaviour of a second-order reaction.
Example 4
The peroxydisulfate ion (S2O8--) is a potent oxidizing agent that reacts rapidly with iodide ion in water as represented by the equation:
S2O8--(aq) + 3I-(aq) ----> 2SO4--(aq) + I3-(aq)
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant
Exp    [S2O8--]o (M)   [I-]o (M)     Initial Rate (M s^-1)
1            0.27                   0.38               2.05
2            0.40                   0.38               3.06  
3            0.40                   0.22               1.76
Answer
Here, we are going to use the alternative method by using the data from Exp 1 and 2, and Exp 2 and 3 to determine the orders of reaction with respect to the reactants and the rate law. So, using:
            Rate = k[S2O8--]^m[I-]^n
where, m = order of reaction with regards to S2O8--, and n = order of reaction with respect to I-
(a) From Exp 1 and 2;
3.06/2.05 = k(0.40)^m(0.38)^n/k(0.27)^m(0.38)^n
3.06/2.05 = (0.40/0.27)^m
         1.49 = 1.48^m
             m = 1.01, approximately 1
Similarly, from Exp 2 and 3
3.06/1.76 = k(0.40)^m(0.38)^n/k(0.40)^m(0.22)^n
3.06/1.76 = (0.38/0.22)^n
         1.74 = 1.73^n
              n = 1.01, approximately 1
Therefore, the rate law is:
            Rate = k[S2O8--][I-], and the overall order of reaction = 1 + 1 = 2
(b) To find k, substitute the data from Exp 1 into the rate law, i.e,
          Rate = k[S2O8--][I-]
where rate = 2.05; [S2O8--] = 0.27; [I-] = 0.38
                 2.05 = k(0.27)(0.38)
                       k = 2.05/0.1026
                          = 19.98 dm^3 mol^-1 s^-1
Do These:
Question 1
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 according to the reaction:
SO2Cl2(g) ----> SO2(g) + Cl2(g)
Data for the reaction at 320°C are given in the table. Calculate the reaction order with respect to SO2Cl2, and determine the rate constant for the reaction.
Exp    [SO2Cl2]o (M)     Initial Rate (M s^-1)
1             0.0100                    2.2 × 10^-7
2             0.0150                    3.3 × 10^-7
3             0.0200                    4.4 × 10^-7
4             0.0250                    5.5 × 10^-7
Question 2
Iodide reduces iron (III) according to the equation:
2Fe3+(aq) + 2I-(aq) ----> 2Fe2+(aq) + I2(s)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4.
(a) What is the reaction order with respect to each reactant?
(b) Determine the overall rate law.
(c) Determine the overall order of reaction.
Question 3
The reaction of nitrogen (II) oxide, (NO) and oxygen gas, (O2) is found to be second order with respect to NO and first order with respect to O2.
(a) What is the overall reaction order?
(b) What is the effect of doubling the concentration of each reactant on the reaction rate?
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