7.4 – Reaction Kinetics: Summary

Collision Frequency does not solely determine the rate of the reaction because in order to overcome the activation energy to react, temperature and concentrations will also dictate the rate of the reaction.

A rise in temperature results in higher kinetic energies in the molecules thus increasing the % chance of colliding AND reacting rather than colliding and staying inert. Having a steady collision frequency with higher kinetic energy will result in spiked rates of reactions.

Catalysts induce reactants to react with one another by lowering the activation energy of the chemical reaction. This can be done without any sort of flux in the enthalpy of the system.

1) Whether or not the collisions occurring have enough energy to get over the activation energy and become products.

2) Whether or not the molecules involved in the collisions are situated properly for the reaction to occur.

Collision frequency does not increase dramatically with temperature; however, the percentage of molecules with enough energy to get over the activation energy and become products does.

The net effect of the addition of a catalyst is the decreasing of the energy barrier to products. The catalyst does so by enabling an alternative mechanism with a lower activation energy.

You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.

We will illustrate the use of these graphs by considering the thermal decomposition of NO₂ gas at elevated temperatures, which occurs according to the following reaction:

2NO₂ (g)Δ  →  2NO (g) + O₂ (g)

Experimental data for this reaction at 330°C are listed in the table below; they are provided as [NO₂], ln[NO₂], and 1/[NO₂] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO₂ are plotted versus time in part (a) in Figure 7.4.1.

Figure 7.4.1. These plots show the decomposition of a sample of NO₂ at 330°C as (a) the concentration of NO₂ versus t, (b) the natural logarithm of [NO₂] versus t, and (c) 1/[NO₂] versus t.

Because the plot of [NO₂] versus t is not a straight line, we know the reaction is not zeroth order in NO₂. A plot of ln[NO₂] versus t (part (b) in Figure 7.4.1) shows us that the reaction is not first order in NO₂ because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO₂] versus t (part (c) in Figure 7.4.1). This plot is a straight line, indicating that the reaction is second order in NO₂.

We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure 7.4.2, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO₂ concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.

Figure 7.4.2. Properties of reactions that obey zero-, first-, and second-order rate laws.

Example 7.4.1 – Integrated Example – Describing the Reaction Kinetics for a Reaction

Dinitrogen pentoxide (N₂O₅) decomposes to NO₂ and O₂ at relatively low temperatures in the following reaction:

2N₂O₅ (soln) → 4NO₂ (soln) + O₂ (g)

This reaction is carried out in a CCl₄ solution at 45°C. The concentrations of N₂O₅ as a function of time are listed in the following table, together with the natural logarithms and reciprocal N₂O₅ concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.

Solution

Here are plots of [N₂O₅] versus t, ln[N₂O₅] versus t, and 1/[N₂O₅] versus t:

The plot of ln[N₂O₅] versus t gives a straight line, whereas the plots of [N₂O₅] versus t and 1/[N₂O₅] versus t do not. This means that the decomposition of N₂O₅ is first order in [N₂O₅].

The rate law for the reaction is therefore

rate = k[N₂O₅]

Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N₂O₅] versus t. Using the points for t = 0 and 3000 s,

Thus k = 4.820 × 10⁻⁴ s⁻¹.

Check Your Learning 7.4.1 – Integrated Example – Describing the Reaction Kinetics for a Reaction

1,3-Butadiene (CH₂=CH—CH=CH₂; C₄H₆) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C₄H₆ as a function of time at 326°C are listed in the following table along with ln[C₄H₆] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C₄H₆, the rate law, and the rate constant for the reaction.

Answer

second order in C₄H₆; rate = k[C₄H₆]²; k = 1.3 × 10⁻² M⁻¹·s⁻¹