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.
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 NO_{2} gas at elevated temperatures, which occurs according to the following reaction:
2NO_{2 }(g)Δ → 2NO (g) + O_{2 }(g)
Experimental data for this reaction at 330°C are listed in the table below; they are provided as [NO_{2}], ln[NO_{2}], and 1/[NO_{2}] versus time to correspond to the integrated rate laws for zeroth, first, and secondorder reactions, respectively. The actual concentrations of NO_{2} are plotted versus time in part (a) in Figure 7.4.1.
Time (s)  [NO_{2}] (M)  ln[NO_{2}]  1/[NO_{2}] (M1) 
0  1.00 × 10^{−2}  4.605  100 
60  6.83 × 10^{−3}  4.986  146 
120  5.18 × 10^{−3}  5.263  193 
180  4.18 × 10^{−3}  5.477  239 
240  3.50 × 10^{−3}  5.655  286 
300  3.01 × 10^{−3}  5.806  332 
360  2.64 × 10^{−3}  5.937  379 
Figure 7.4.1. These plots show the decomposition of a sample of NO_{2} at 330°C as (a) the concentration of NO_{2} versus t, (b) the natural logarithm of [NO_{2}] versus t, and (c) 1/[NO_{2}] versus t.
Because the plot of [NO_{2}] versus t is not a straight line, we know the reaction is not zeroth order in NO_{2}. A plot of ln[NO_{2}] versus t (part (b) in Figure 7.4.1) shows us that the reaction is not first order in NO_{2} because a firstorder reaction would give a straight line. Having eliminated zerothorder and firstorder behavior, we construct a plot of 1/[NO_{2}] versus t (part (c) in Figure 7.4.1). This plot is a straight line, indicating that the reaction is second order in NO_{2}.
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 NO_{2} concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.
Zeroth Order First Order Second Order 

Differential Rate = Δ[A]Δt=k
Rate = Δ[A]Δt=k [A] Rate = Δ[A]Δt=k[A]^{2} rate law 

Concentration vs. time 

Integrated [A] = [A]_{0} – kt [A] = [A]_{0 }e^{kt1[A]} = 1[A]0 + kt
rate law or ln[A] = ln[A]_{0} – kt 

Straightline plot to determine rate constant 

Relative rate vs. concentration


Halflife
t_{1/2} = [A]_{0}^{2}k t_{1/2} = 0.693k t_{1/2} = 1k[A]_{0} 

Units of k, M/s 1/s M^{1} s^{1}
rate constant 

Figure 7.4.2. Properties of reactions that obey zero, first, and secondorder rate laws.
Example 7.4.1 – Integrated Example – Describing the Reaction Kinetics for a Reaction
Dinitrogen pentoxide (N_{2}O_{5}) decomposes to NO_{2} and O_{2} at relatively low temperatures in the following reaction:
2N_{2}O_{5 }(soln) → 4NO_{2 }(soln) + O_{2 }(g)
This reaction is carried out in a CCl_{4} solution at 45°C. The concentrations of N_{2}O_{5} as a function of time are listed in the following table, together with the natural logarithms and reciprocal N_{2}O_{5} 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.
Time (s) 
[N_{2}O_{5}] (M) 
ln[N_{2}O_{5}] 
1/[N_{2}O_{5}] (M^{1}) 
0 
0.0365 
3.310 
27.4 
600 
0.0274 
3.597 
36.5 
1200 
0.0206 
3.882 
48.5 
1800 
0.0157 
4.154 
63.7 
2400 
0.0117 
4.448 
85.5 
3000 
0.00860 
4.756 
116 
3600 
0.00640 
5.051 
156 
Solution
Here are plots of [N_{2}O_{5}] versus t, ln[N_{2}O_{5}] versus t, and 1/[N_{2}O_{5}] versus t:
The plot of ln[N_{2}O_{5}] versus t gives a straight line, whereas the plots of [N_{2}O_{5}] versus t and 1/[N_{2}O_{5}] versus t do not. This means that the decomposition of N_{2}O_{5} is first order in [N_{2}O_{5}].
The rate law for the reaction is therefore
rate = k[N_{2}O_{5}]
Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a firstorder reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N_{2}O_{5}] 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
1,3Butadiene (CH_{2}=CH—CH=CH_{2}; C_{4}H_{6}) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C_{4}H_{6} as a function of time at 326°C are listed in the following table along with ln[C_{4}H_{6}] 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 C_{4}H_{6}, the rate law, and the rate constant for the reaction.
Time (s) 
[C_{4}H_{6}] (M) 
ln[C_{4}H_{6}] 
1/[C_{4}H_{6}] (M^{1}) 
0 
1.72 × 10^{−2} 
4.063 
58.1 
900 
1.43 × 10^{−2} 
4.247 
69.9 
1800 
1.23 × 10^{−2} 
4.398 
81.3 
3600 
9.52 × 10^{−3} 
4.654 
105 
6000 
7.30 × 10^{−3} 
4.920 
137 
Answer
second order in C_{4}H_{6}; rate = k[C_{4}H_{6}]^{2}; k = 1.3 × 10^{−2} M^{−1}·s^{−1}