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This approach to finding the reaction order is called the Method of Initial Rates, since it relies on fixing the concentration at specific initial values and measuring the initial rate associated with each concentration.
So far we have considered only reactions that have a single reactant. Consider a second example of the Method of Initial Rates involving the reaction of hydrogen gas and iodine gas:
H 2 (g) + I 2 (g) → 2HI (g)
In this case, we expect to find that the rate of the reaction depends on the concentrations for both reactants. As such, we need more initial rate observations to determine the rate law. In Table 3, observations are reported for the initial rate for three sets of initial concentrations of H 2 and I 2 .
| Experiment | [H 2 ] 0 (M) | [I 2 ] 0 (M) | Rate (M/sec) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 3.00×10 -4 |
| 2 | 0.20 | 0.10 | 6.00×10 -4 |
| 3 | 0.20 | 0.20 | 1.19×10 -3 |
Following the same process we used in the N 2 O 5 example, we write the general rate law for the reaction as
Rate = k[H 2 ] n [I 2 ] m
By comparing experiment 1 to experiment 2, we can write
This simplifies to (0.50) = (0.50) m (1.00) n from which it is clear that m = 1. Similarly, we can find that n = 1. The reaction is therefore first order with respect to each reactant and is second order overall.
Rate = k[H 2 ][I 2 ]
Once we know the rate law, we can use any of the data from Table 3 to determine the rate constant, simply by plugging in concentrations and rate into Equation (9). We find that k = 3.00×10 -2 /M∙sec.
This procedure can be applied to any number of reactions. The challenge is preparing the initial conditions and precisely measuring the initial change in concentration versus time. Table 4 provides an overview of the rate laws for several reactions. A variety of reaction orders are observed, and they cannot be easily correlated with the stoichiometry of the reaction.
| Reaction | Rate Law |
|---|---|
| 2NO(g) + O 2 → 2NO 2 (g) | Rate = k[NO] 2 [O 2 ] |
| 2NO(g) + 2H 2 → 2N 2 (g) + 2H 2 O(g) | Rate = k[NO] 2 [H 2 ] |
| 2ICl(g) + H 2 → 2HCl(g) + I 2 O(g) | Rate = k[ICl][H 2 ] |
| 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) | Rate = k[N 2 O 5 ] |
| 2NO 2 (g) + F 2 (g) → 2NO 2 F(g) | Rate = k[NO 2 ][F 2 ] |
| 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g) | Rate = k[H 2 O 2 ] |
| H 2 (g) + Br 2 (g) → 2HBr(g) | Rate = k[H 2 ][Br 2 ] 1/2 |
| O 3 (g) + Cl(g) → O 2 (g) + ClO(g) | Rate = k[O 3 ][Cl] |
Once we know the rate law for a reaction, we should be able to predict how fast a reaction will proceed. From this, we should also be able to predict how much reactant remains or how much product has been produced at any given time in the reaction. We will focus on the reactions with a single reactant to illustrate these ideas.
Let’s consider a first order reaction illustrated by A → products , where A is a typical reactant. The rate law must be
From Calculus, it is possible to integrate Equation (10) to find the function [A](t), which tells us the concentration [A]as a function of time. The result is
[A] = [A] 0 e -kt
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