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This approach to finding reaction order is called the method of initial rates, since it relies on fixing theconcentration at specific initial values and measuring the initial rate associated with each concentration.

So far we have considered only reactions which have a single reactant. Consider a second example of the method ofinitial rates involving the reaction of hydrogen gas and iodine gas:

H 2 ( g ) + I 2 ( g ) 2 H I ( 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 , observations are reported for the initial rate for three sets ofinitial concentrations of H 2 and I 2 .

Hydrogen gas and iodine gas initial rate data at 700k
Experiment [ H 2 ] 0 (M) [ I 2 ] 0 (M) Rate (M/sec)
1 0.10 0.10 3.00 -4
2 0.20 0.10 6.00 -4
3 0.20 0.20 1.19 -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
Rate 1 Rate 2 k [ H 2 ] 1 n [ I 2 ] 1 m k [ H 2 ] 2 n [ I 2 ] 2 m 3.00 -4 M s 6.00 -4 M s k 0.10 M m 0.10 M n k 0.20 M m 0.10 M n
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 in each reactant and issecond order overall.
Rate k [ H 2 ] [ I 2 ]
Once we know the rate law, we can use any of the data from to determine the rate constant, simply by plugging in concentrationsand rate into . We find that k 3.00 -2 1 M s .

This procedure can be applied to any number of reactions. The challenge is preparing the initial conditions andmeasuring the initial change in concentration precisely versus time. provides an overview of the rate laws for several reactions. A variety ofreaction orders are observed, and they cannot be easily correlated with the stoichiometry of the reaction.

Rate laws for various reactions
Reaction Rate Law
2 N O ( g ) + O 2 ( g ) 2 N O 2 ( g ) Rate k [ N O ] 2 [ O 2 ]
2 N O ( g ) + 2 H 2 ( g ) 2 N 2 ( g ) + 2 H 2 O ( g ) Rate k [ N O ] 2 [ H 2 ]
2 I Cl ( g ) + H 2 ( g ) 2 H Cl ( g ) + I 2 ( g ) Rate k [ I Cl ] [ H 2 ]
2 N 2 O 5 ( g ) 4 N O 2 ( g ) + O 2 ( g ) Rate k [ N 2 O 5 ]
2 N O 2 ( g ) + F 2 ( g ) 2 N O 2 F ( g ) Rate k [ N O 2 ] [ F 2 ]
2 H 2 O 2 ( aq ) 2 H 2 O ( l ) + O 2 ( g ) Rate k [ H 2 O 2 ]
H 2 ( g ) + Br 2 ( g ) 2 H Br ( g ) Rate k [ H 2 ] [ Br 2 ] 1 2
O 3 ( g ) + Cl ( g ) O 2 ( g ) + Cl O ( g ) Rate k [ O 3 ] [ Cl ]

Concentrations as a function of time and the reaction half-life

Once we know the rate law for a reaction, we should be able to predict how fast a reaction will proceed. Fromthis, we should also be able to predict how much reactant remains or how much product has been produced at any given time in thereaction. We will focus on the reactions with a single reactant to illustrate these ideas.

Consider a first order reaction like A products , for which the rate law must be

Rate t [ A ] k [ A ]
From Calculus, it is possible to use to find the function [ A ] t which tells us the concentration [ A ] as a function of time. The result is
[ A ] [ A ] 0 k t
or equivalently
[ A ] [ A ] 0 k t
reveals that, if a reaction is first order, we can plot [ A ] versus time and get a straight line with slope equal to k . Moreover, if we know the rate constant and the initialconcentration, we can predict the concentration at any time during the reaction.

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Source:  OpenStax, General chemistry ii. OpenStax CNX. Mar 25, 2005 Download for free at http://cnx.org/content/col10262/1.2
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