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The Rate of a Chemical Reaction: Kinetics

Chemical kinetics is essential for the understanding of air pollution.

Very often in chemistry - as in physics - we want to measure the rate of something. Chemists usually want to measure the rate of a chemical reaction. This empirical rate law often is a measure of how the rate of a chemical reaciton depends on the amounts (# of particles) of reactants combined at a given temperature. This empirical relationship - which must be determined experimentally - is termed the differential rate law or experimental rate law. As you will learn (and appreciate) when you take chemistry this year (I hope!), knowledge of the empirical differential rate law is often the first step in elucidating the sequence of molecular events that the reactants in a chemical reaction go through to form products. This sequence of hypothesized molecular events is termed the (proposed) mechanism of a reaction. Each of the molecular events that constitute the mechanism is termed an elementary reaction or elementary step.

So, it is perhaps easy to understand why we want to determine the differential rate law of a chemical reaction. But how do we do this? We try to set-up a correspondence between the number of reactant particles that react per unit time and the number of molecules of reactant remaining (or of product molecules formed). Usually we express particle numbers in units of moles or of something proportional to moles. For many reactions (solution reactions, gas-phase reactions, etc.), we can express particle numbers in units of moles per volume (L in SI units). This unit of concentration (i.e. moles per Liter of solution) is termed Molarity and is abbreviated M. For an ideal gas, its concentration is the same as its number of moles (n) divided by the volume of the container (an ideal gas has the entire container volume available to it). So - for a substance "X":

For a substance (solute) dissolved in a solvent to form a solution, the concentration in Molarity units is given by:

In either case, we symbolize the molarity of a substance "A" in moles per Liter as [A], i.e. "A" with square brackets around it.

So, for a general chemical reaction (among gases, say - although the results are general) we can write:

,

where "A", "B", "C", and "D" are substances (atoms or molecules) and a, b, c, and d are their stoichiometric mole coefficients in the balanced (atoms conserved) chemical reaction. Now, when we measure the rate of a reaction - as mentioned above - we are usually measuring the change in concentration (i.e. per change in time (i.e. . As mentioned in your physics and mathematics lectures, these quantities are finite differences, i.e. "average" changes. We will see that the change in concentration is not usually constant over time - but rather depends (in a detailed way) on how far the reaction has progressed, etc. Thus, when we make a measurement of the rate of change of chemical "C" above, we write ("f" and "i" denote "final" and "initial", respectively):

Note that since C is being formed (as the reaction above proceeds from reactants to products) and the above rate should be positive. Convince yourself that is also positive whereas and would both be negative. Ideally, we would like to determine the instantaneous rate of concentration change - which can be approached as we measure the average rate (above) over smaller and smaller time and concentration intervals. Thus, we would - for mathematical modeling purposes - like to consider the instantaneous rates, i.e.

As hinted at previously, the first derivatives (early in the reaction) should be negative and the second two (early in the reaction) should be positive. Also, the magnitude (numerical value without sign) will not be the same - due to the stoichiometry of the chemical reaction. In other words, for every a moles of A that reacts, b moles of B react and c moles of C and d moles of D form. The changes in concentration are thus proportional to the stoichiometric coefficients of the balanced chemical reaction! So what does a chemist mean when she measures the "rate of a chemical reaction"? From the above discussion it is obvious that it depends on which chemical you are talking about (and - as we'll see later - on where you are in the progress of the reaction). To make the rate independent of the stoichiometry of the reaction (i.e. of which chemical we monitor) we "normalize" the rates by dividing by the stoichiometric coefficient and we add the appropriate sign (plus or minus) to ensure that each rate is a positive quantity. In other words, we can write:

For the instanteneous rates, we can similarly write:

We've stoichiometrically corrected everything. But - as discussed in the physics and mathematics sessions - these rates change with time - the "slopes are not constant! Before we address this issue, let's talk about the right-hand side (r.h.s.) of these differential equations, i.e. what are these rates (derivatives) equal to? From "intuition" it seems reasonable that the time rate of change of concentration of a given species should depend on how much of everything is present. This is true. In general, we should be able to write:

As we've expressed above, the rate (derivative) should be a function of the amount of reactants, and the amount of products present at a given instant (t). For completeness, I also mentioned that the rate will depend on the temperature. For this discussion, we will presume that everything is carried out at constant temperature. But what does this function look like? For reasons that will become more apparent as you study the law of mass action in chemistry, the functional dependence on concentrations is that of a product of concentrations. However, in order for this simple functional form to agree with nature, a few "adjustable parameters" must be included. It turns out that for each reactant or product that enters the r.h.s. of the above differential equation, we must include an emperically-determined exponent - termed the order of the reaction with respect to that component. Also there is a proportionality constant on the r.h.s. of the differential equation - termed the rate constant - "k". Before we write down the form of the r.h.s. of this differential rate law, we need to simplify things a bit. First of all, writing down and solving the differential equation that governs the entire course of the reaction is a daunting task! Usually, chemists are just interested in modelling the differential rate law "early in the reaction", i.e. before too much product forms. This rate, the so-called initial rate is (to a first approximation) a function of the amounts of reactants only! Thus, we can simplify our above equation to:

Putting everything together, we can write for our "general" reaction above:

Recall, we can write down the rates (left-hand side, l.h.s. of the above equation) in the following ways:

The circumstances of the particular situation will usually make it clear whether we are talking about the instantaneous rate or the average rate.

Now, even after all of this simplification, we have to determine the values of k, x, and y above. From the onset it is important that the following be made clear:

One way of determining the rate constant (k) and the orders x and y is by the isolation method. An example may serve to aid our class discussion. See below. Note: for this method to work, volumes and temperatures must be kept constant from trial to trial! Why??

In general, for two trials, m and n, the RATIO of the Rates (using the expression for the experimental rate law) is given by:

We take the RATIO of two (2) trials where ONLY ONE of the species is CHANGING CONCENTRATION.
Taking the ratio of Trials #1 and #2 (i.e. m = 2 and n = 1 in Equation 1) - to ISOLATE Br_2:

or: 1.414 = 2^y. Thus, using log (base 10) or ln to solve for y, we have: ln(1.414) = y * ln(2) (or: log_{10}(1.414) = y * log_{10}(2)).

Solving gives : y = 0.5 or 1/2. This is the ORDER with respect to BR_2

Similarly, taking the ratio of Trials #1 and #3 (i.e. m=3 and n=1 in Equation 1) - to ISOLATE H_2:

or: 2^1 = 2^x.. By "inspection" it is easy to see that x = 1. This is the ORDER with respect to H_2. Thus, the experimental RATE LAW (for any trial) is

        R = Rate = k_{exp}[H_2]^1[Br_2]^{1/2}

We say (as mentioned above) that the ORDER with respect to a given species is just the value of its exponent (x or y, etc.) in the (experimental) RATE LAW. We define the OVERALL ORDER as the sum of the exponents in the (experimental) RATE LAW. Here, we have

Order with respect to H_2 = x = 1 and Order with respect to Br_2 = y = 1/2 or 0.5 . Overall Order = x + y = 1 + 1/2 = 3/2 or 1.5.

Now, using any trial, we can "plug in" the Rate (R) and the [H_2] and [Br_2] - and solve for k_{exp}. IT SHOULD BE THE SAME FOR EVERY TRIAL - AS LONG AS WE MAINTAIN CONSTANT TEMPERATURE. In a "real" experiment, we may determine k_{exp} for EACH trial then calculate the AVERAGE k_{exp}.

Using Trial #1, for example, we have

    2.000 x 10^{-5} M/s = k_{exp}(0.10 M)^1(0.10 M)^{1/2}  Solving for k_{exp} gives -
    k_{exp} = 6.324 x 10^{-4} M^{-1/2}s^{-1}.              (Determine k_{exp} for the other
                                                           trials for yourself and compare.)

• The solutions of the differential rate laws - the integrated rate laws i.e. Concentration as an explicit function of time.

• Case I: For A --» Products
  Found experimentally: Rate = -d[A]/dt = k[A] ==» First-order (k in s^{-1})

  Initial condition: [A] at t = 0 is [A]_0.

  Solution: ln[A] = ln[A]_0 - kt or [A] = [A]_0 exp(-kt)

where slope = (change in y)/(change in x) and the y-intercept is the point where x (time t in this case) is 0.

Half-life: when [A] = [A]_0/2 ==» t = t_{1/2}. So,

t_{1/2} = ln(2)/k = 0.693/k .

• Case II: For A --» Products
  Found experimentally: Rate = -d[A]/dt = k[A]^2 ==» Second-order (k in M^{-1}s^{-1})
  Initial condition: [A] at t = 0 is [A]_0.

  Solution: 1/[A] = 1/[A]_0 + kt

where: slope = (change in y)/(change in x) and the y-intercept is the point where x (t in this case) is 0.

Half-life: when [A] = [A]_0/2 ==» t = t_{1/2} so t_{1/2} = 1/(k[A]_0).

Note: t_{1/2} depends inversely on [A]_0.

• Case III: For A --» Products
  Found experimentally: Rate = -d[A]/dt = k[A]^0 = k ==» Zero-order (k in M/s)
  Initial condition: [A] at t = 0 is [A]_0.

  Solution: [A] = [A]_0 - kt

where: slope = (change in y)/(change in x) and the y-intercept is the point where x (t in this case) is 0.

Half-life: when [A] = [A]_0/2 ==» t = t_{1/2}, so t_{1/2} = [A]_0/(2k).
Note: t_{1/2} depends directly on [A]_0.

An example may clarify things a bit:

Example: Consider the reaction: A --» Products; for which the numerical value of k (in units of M and s) is 1.72 x 10^{-5} and [A]_0 = 0.10 M. Determine k (in units of M and s) and t_{1/2} (in s), if the rate law of the above reaction,
Rate = k[A]^x, is:
(1) First order in A; (2)Second order in A; (3) Zero order in A.

Solutions:

1. For a first order rate law, x = 1 above and Rate = k[A]. Thus, k has units of s¯ ¹ .
   t_{1/2} is given by: t_{1/2} = ln(2)/k = 0.693/(1.72 x 10^{-5} s¯ ¹ ) = 4.03 x 10^4 s.
2. For a second order rate law, x = 2 above and Rate = k[A]² . So, k has units M¯ ¹ s¯ ¹ .
   t_{1/2} is given by: t_{1/2} = 1/(k[A]_0) = 1/[1.72 x 10^{-5} M¯ ¹ s¯ ¹ )(0.10 M)] = 5.81 x 10^5 s - depends on [A]_0.
3. For a zero order rate law, x = 0 above and Rate = k. So, k has units of M s¯ ¹ or (M/s).
   t_{1/2} is given by t_{1/2} = [A]_0/(2k) = (0.10 M)/[(2)(1.72 x 10^{-5} M/s)] = 2.91 x 10³ s - depends on [A]_0.

• Case IV: For A --» Products General Order

  Rate = -d[A]/dt = k[A]^n ==» nth-order, n not equal to 1.

  Initial condition: [A] at t = 0 is [A]_0. Show, using MAPLE, that the solution is:
  or
  

  Figure out for yourself what would have to be plotted (what function of [A] versus time) in order to arrive at a linear plot. When you have done this, determine the formulas for

  
    . Slope        . Y-intercept         . Half-life (t_{1/2})
    . Finally, determine the units of the rate constant k.
  

  A general procedure of how to figure out "n" experimentally is termed the van't Hoff method. This method, using Maple, is presented here.

• Case V: For A + B --» C Second order overall, First order in A and in B
  We will call our products something specific here, i.e. "C". You'll see why in a bit.

  Rate = -d[A]/dt = -d[B]/dt = +d[C]/dt = k[A][B]

  Notice that we have two (2) dependent variables in our rate law, [A] and [B].

  A          +           B      ---»     C
[A]_0                 [B]_0              0   initial
-x                     -x                +x  progress or change
[A]_0 - x          [B]_0 - x             +x  final (time, t)

Remember the initial conditions from above and: x(t) = x(0) = 0 at t = 0.

This equation can be solved - in terms of x, as a function of [A]_0, [B]_0, and t. After some cumbersome algebra - remembering the definition of x, the solution can be written as follows:

Stay tuned for some other interesting situations!