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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. [link] 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 [link] 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
[link] 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.

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 1 2 . At that time, [ A ] 1 2 [ A ] 0 . From [link] and using the properties of logarithms, we find that, for a first orderreaction

t 1 2 2 k
This equation tells us that the half-life of a first order reaction does not depend on how much material we startwith. It takes exactly the same amount of time for the reaction to proceed from all of the starting material to half of the startingmaterial as it does to proceed from half of the starting material to one-fourth of the starting material. In each case, we halve theremaining material in a time equal to the constant half-life in [link] .

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

Rate t [ A ] k [ A ] 2
Again, we can use Calculus to find the function [ A ] t from [link] . The result is most easily written as
1 [ A ] 1 [ A ] 0 k t
Note that, as t increases, 1 [ A ] increases, so [ A ] decreases. [link] 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 . Again, if we know the rate constant and the initial concentration, we canfind the concentration [ A ] at any time of interest during the reaction.

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Source:  OpenStax, Concept development studies in chemistry 2012. OpenStax CNX. Aug 16, 2012 Download for free at http://legacy.cnx.org/content/col11444/1.4
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