0.18 Reaction rate laws  (Page 5/5)

or equivalently

ln[A] = ln[A} ₀ - kt

Equation (12) reveals that, if a reaction is first order, we can plot ln[A] versus time and get a straight line with slope equal to –k . This would be a way to test whether a reaction is first order or not. Moreover, if we know the rate constant and the initial concentration, we can predict the concentration at any time during the reaction.

An interesting point in the reaction is the time at which exactly half of the original concentration of A has been consumed. We call this time the “half-life” of the reaction and denote it as t ½ . At that time, . From Equation (12) and using the properties of logarithms, we find that, for a first order reaction

t _1/2 = \frac{\mathrm{ln\; 2}}{k}

This equation tells us that the half-life of a first order reaction does not depend on how much material we start with. It takes exactly the same amount of time for the reaction to proceed from all of the starting material to half of the starting material as it does to proceed from half of the starting material to one-fourth of the starting material. In each case, we halve the remaining material in a time equal to the constant half-life in Equation (13).

These conclusions are only valid for first order reactions. Consider then a second order reaction, such as the butadiene dimerization discussed above. The general second order reaction A → products has the rate law

Rate = - \frac{\mathrm{d[A]}}{dt} = k[A] ²

Again, we can use Calculus to find the function [A](t) from Equation (14). The result is most easily written as

1/[A] = 1/[A] ₀ + kt

Note that, as t increases, 1/[A] increases, so [A]decreases. Equation (15) reveals that, for a reaction which is second order in the reactant A, we can plot 1/[A] as a function of time to get a straight line with slope equal to k . This would be a way to test whether a reaction is second order. And, if we know the rate constant and the initial concentration, we can find the concentration [A] at any time of interest during the reaction.

The half-life of a second order reaction differs from the half-life of a first order reaction. From Equation (15), if we take [A] = [A] ₀ /2, we get

t _1/2 = 1/k[A] ₀

This shows that, unlike a first order reaction, the half-life for a second order reaction depends on how much material we start with. From Equation (16), the more concentrated the reactant is, the shorter the half-life.

These data tell us a great deal about how reaction rates depend on concentrations, but they do not tell us why they depend on concentrations, and in particular, they do not tell us why the exponents on the concentrations differ from one reaction to another and one reactant to another. In the next Concept Development Study, we will develop a model to account for this data.

Review and discussion questions

1. When C ₆₀ O ₃ in toluene solution decomposes, O ₂ is released leaving C ₆₀ O in solution.
   1. Based on the data in Figures 2 and 3, plot the concentration of C ₆₀ O as a function of time. How would you define the rate of the reaction in terms of the slope of the graph in part (a)?
   2. How is the rate of appearance of C ₆₀ O related to the rate of disappearance of C ₆₀ O ₃ ? Based on this, plot the rate of appearance of C ₆₀ O as a function of time.
2. The reaction 2N ₂ O ₅ (g) → 4NO ₂ (g) + O ₂ (g) was found in this study to have rate law given by k[N ₂ O ₅ ] with k =0.070 s ^-1 .
   1. How is the rate of appearance of NO ₂ related to the rate of disappearance of N ₂ O ₅ ?  Which rate is larger?
   2. Based on the rate law and rate constant, sketch a plot of [N ₂ O ₅ ], [NO ₂ ], and [O ₂ ] versus time all on the same graph.
3. Consider two decomposition reactions for two hypothetical materials, A and B. The decomposition of A is found to be first order, and the decomposition of B is found to be second order.
   1. Assuming that the two reactions have the same rate constant at the same temperature, sketch [A] and [B]versus time on the same graph for the same initial conditions, i.e. [A] ₀ =[B] ₀ .
   2. Compare the half-lives of the two reactions. Under what conditions will the half-life of B be less than the half-life of A? Under what conditions will the half-life of B be greater than the half-life of A?
4. We found that the rate law for the reaction H ₂ + I ₂ (g) → 2HI(g) is Rate = k[H ₂ ][I ₂ ] . Therefore, the reaction is second order overall but first order with respect to H ₂ . Imagine that we start with [H ₂ ] ₀ = [I ₂ ] ₀ and we measure [H ₂ ] versus time. Will a graph of ln([H ₂ ]) versus time be linear or will a graph of 1/[H ₂ ] versus time be linear? Explain your reasoning.