As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Chapter 11-3, 18-1 and 18-2, and/or Chapter 9.6-9.12 in the Additional Reading Materials section.

D19.1 Integrated Rate Laws

Another way to determine the rate law and rate constant from experimental data for a reaction is the integrated rate law method. An integrated rate law relates the concentration of reactant or product  in a reaction mixture to the elapsed time of the reaction. Thus, an integrated rate law can be used to determine the concentration of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For a given rate equation, calculus can be used to derive an appropriate integrated rate law. If you are not familiar with calculus and integration, don’t worry—you can still use the integrated rate law for a reaction without deriving it yourself.

For a generic reaction:

The rate of the reaction can be expressed as:

[latex]\displaystyle{\rm{rate }} = {\rm{ }} - \frac{{\Delta {\rm{[A]}}}}{{\Delta t}}[/latex]

In calculus, the definition of a derivative is

[latex]\displaystyle\frac{{d{\rm{[A]}}}}{{dt}}{\rm{ }} = {\rm{ }}\mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta {\rm{[A]}}}}{{\Delta t}}[/latex]

So a rate law for the reaction can be written as

Using the rules of calculus, both sides of this rate equation can be integrated to

or, assuming that the reaction starts at t = 0,

where [A]₀ is the initial reactant concentration and [A]_t is the reactant concentration at time t. The solution to the left side of this equation has a different mathematical form depending on the order of the reaction with respect to A (value of m). We will explore each value of m in the following sections.

D19.2 First-Order Reaction

When m = 1,

which can be alternatively expressed as:

Raising e (the base of the natural logarithm system) to the power of each side of this equation gives,

The integrated rate law for a first order reaction can also be rearranged to a standard linear equation format:

Hence, if a reaction is first order, a plot of ln[A]_t vs. t must be a straight line. The slope of the plot is −k and the y-intercept is ln[A]₀. If the plot is not a straight line, the reaction is not first order in A.

D19.3 Second-Order Reaction

When m = 2,

which can be alternatively expressed as:

The integrated rate law for second-order reaction also has a standard linear equation format:

A plot of [latex]\dfrac{1}{[A]_t}[/latex] vs t is a straight line for a second-order reaction; the slope equals k and the y-intercept is [latex]\dfrac{1}{[A]_0}[/latex]. If the plot is not a straight line, then the reaction is not second order with respect to A.

Example 4

Determination of Reaction Order via Graphical Method

Test the data given to show whether the dimerization of C₄H₆ is a first- or a second-order reaction.

Trial	Time (s)	[C4H6] (M)
1	0	1.00 × 10−2
2	1600	5.04 × 10−3
3	3200	3.37 × 10−3
4	4800	2.53 × 10−3
5	6200	2.08 × 10−3

Solution

In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C₄H₆] versus t and compare it with a plot of [latex]\frac{1}{[\text{C}_4\text{H}_6]}[/latex] versus t. The values needed for these plots follow.

Time (s)	[latex]\frac{1}{[\text{C}_4\text{H}_6]}(M^{-1})[/latex]	ln[C4H6]
0	100	−4.605
1600	198	−5.289
3200	296	−5.692
4800	395	−5.978
6200	481	−6.175

The plots are shown below. The left plot of ln[C₄H₆] versus t is not linear; therefore the reaction is not first order. The right plot of [latex]\frac{1}{[\text{C}_4\text{H}_6]}[/latex] versus t is linear, indicating that the reaction follows second-order kinetics.

Check Your Learning

Do these data fit a second-order rate law?

Trial	Time (s)	[A] (M)
1	5	0.952
2	10	0.625
3	15	0.465
4	20	0.370
5	25	0.308
6	35	0.230

Answer:

Yes. The plot of [latex]\frac{1}{[A]}[/latex] vs. t is linear:

D19.4 Zeroth-Order Reaction

When m = 0,

A zeroth-order reaction exhibits a constant reaction rate, regardless of the concentration of reactant A:

The integrated rate law for a zeroth-order reaction also has has a standard linear equation format:

A plot of [A]_t vs t for a zeroth-order reaction is a straight line with a slope of −k and an intercept of [A]₀.

Figure 1 shows a plot of [NH₃] vs t for the decomposition of ammonia on a hot tungsten (W) wire and on hot quartz (SiO₂). The decomposition on hot tungsten is zeroth-order; the plot is a straight line. From the slope, we can determine the rate constant:

The decomposition on hot quartz is not zeroth order (analysis of the data shows that it is first order).

D19.5 Flooding: Pseudo-Order Methods

The integrated rate laws are quite useful when determining the reaction order and rate constants using all the data from a single experimental trial run. However, thus far we’ve only considered example reactions involving one reactant, whereas many reactions involve two or more reactants. How can we use integrated rate laws to find the reaction order and rate constants for those reactions?

Let’s consider the reaction

where

Flooding refers to running a reaction that involves two or more reactants with a large excess of all but one reactant. For example, run the reaction above with a large excess of reactant B so that A is the limiting reactant by a significant amount. If [B]₀ >> [A]₀, reactant B concentration effectively remains constant during the reaction. (If [B]₀ = 0.100 M and [A]₀ = 0.00100, then at a time t when all the A has reacted, [B]_t = (0.100-0.001) M = 0.099 M, which is almost no change in the B concentration.) Under this condition of [B]_t ≈ [B]₀ = constant, the rate law becomes

where k_obs, the rate constant we observe during the flooded experiment, is k_obs = k[B]₀ⁿ. This rate equation allows us to use the integrated rate laws we’ve just discussed to determine the reaction order with respect to reactant A and to determine k_obs.

Here is an example: Flood the reaction mixture with a large excess of reactant B ([latex][B]_{0_1}[/latex]) and measure [A]_t as the reaction progresses. Then plot ln[A]_t vs t. If the graph is a straight line, then the reaction is first order with respect to A. (In this case, we call the flooded reaction a pseudo-first-order reaction.) The negative of the slope of the graph is [latex]k_{obs_1}[/latex].

Then run another trial with a different, large excess concentration of B, [latex][B]_{0_2}[/latex]. The plot of ln[A]_t vs t has a different slope corresponding to [latex]-k_{obs_2}[/latex]. The ratio of the two k_obs

allows us to determine n, the reaction order with respect to reactant B, because all other variables in the equation are known or have been experimentally determined. Finally, the actual rate constant for the reaction, k, can be determined from the relationship k_obs = k[B]₀ⁿ.

D19.6 Radioactive Decay

Determining the rate law and rate constant has many real world applications. An important one of these is the rate of radioactive decay, the spontaneous change of an unstable nuclide into a different nuclide. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide (Figure 2). The daughter nuclide may be stable, or it may decay itself.

The atomic representations used when discussing radioactive decay reactions have two numbers written to the left of the atomic symbol, for example, [latex]^{238}_{\;92}\text{U}[/latex] and its isotope [latex]^{235}_{\;92}\text{U}[/latex]. The subscript denotes the atomic number of the element (number of protons), and the superscript denotes the mass number of the isotope (number of protons + number of neutrons).  Both subscript and superscript are necessary for balancing nuclear equations.

D19.7 Types of Radioactive Decay

Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure 3) helped him determine that one type of radiation consisted of positively charged and relatively massive α particles, which are high-energy helium nuclei ([latex]_2^4\text{He}[/latex], or [latex]_2^4{\alpha}[/latex]); a second type consisted of negatively charged and much lighter β particles, which are high-energy electrons ([latex]_{-1}^{\;\;0}{\beta}[/latex], or [latex]_{-1}^{\;\;0}\text{e}[/latex]); and a third was uncharged electromagnetic waves, γ rays, which are are photons of very high energy. Note that for beta particles, which do not contain any protons, the subscript denotes the charge of the particle.

Alpha (α) decay is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay:

Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). The loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide.  Note that the sum of protons (subscripts) on the product side is equal to that on the reactant side. Theame is true for the mass number (superscripts).  This nuclear reaction is balanced.

Beta (β) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay:

Beta decay involves the conversion of a neutron into a proton and a β particle. The β particle emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Emission of a β particle does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons.

Gamma (γ) decay is the emission of a γ-ray photon from a nucleus. The energies of γ rays are quite high, in the range of hundreds of thousands to tens of millions of kJ/mol, so γ rays can easily break chemical bonds if they are absorbed by matter. (In comparison, x-rays have energies roughly 10 times smaller than γ rays.) Cobalt-60 is an example nuclide that decays via β emission as well as γ emission:

The asterisk (*) denotes that Nickel-60 is in an excited state. It decays to its ground state with the emission of another γ photon:

There is no change in mass number or atomic number during γ emission unless it is accompanied by one of the other modes of decay.

Positron (β⁺) decay is the emission of a positron, or antielectron, from a nucleus. In the process, a proton is converted into a neutron. Oxygen-15 is an example of a nuclide that undergoes positron emission:

Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus and transforms a proton into a neutron. For example, potassium-40 undergoes electron capture:

Electron capture has the same effect on the nucleus as does positron emission. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors.

Figure 4 summarizes these types of decay.