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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 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 .

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 . 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. 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.

The half-life of a second order reaction differs from the half-life of a first order reaction. From , if we take [ A ] 1 2 [ A ] 0 , 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 howmuch material we start with. From , the more concentrated the reactant is, the shorter the half-life.

Observation 3: temperature dependence of reaction rates

It is a common observation that reactions tend to proceed more rapidly with increasing temperature. Similarly,cooling reactants can have the effect of slowing a reaction to a near halt. How isthis change in rate reflected in the rate law equation, e.g. ? One possibility is that there is a slight dependence on temperature ofthe concentrations, since volumes do vary with temperature. However, this is insufficient to account for the dramatic changesin rate typically observed. Therefore, the temperature dependence of reaction rate is primarily found in the rate constant, k .

Consider for example the reaction of hydrogen gas with iodine gas at high temperatures, as given in . The rate constant of this reaction at each temperature can be found using the method of initial rates,as discussed above, and we find in that the rate constant increases dramatically as the temperature increases.

Rate constant for hydrogen gas and iodine gas
T (K) k ( 1 M s )
667 6.80 -3
675 9.87 -3
700 3.00 -2
725 8.43 -2
750 2.21 -1
775 5.46 -1
800 1.27

As shown in , the rate constant appears to increase exponentially with temperature. After a littleexperimentation with the data, we find in that there is a simple linear relationship between k and 1 T .

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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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