rate of change in concentration per unit time (M/s=mol/Ls)
rate of appearance of products equals negative the rate of disappearance of reactants
rates of appearance/disappearance in a reaction follow the mole ratio calculations
Example
In the reaction equation CX6HX12OX6+6OX26COX2+6HX2O, suppose the rate of disappearance of CX6HX12OX6 is xLsmolCX6HX12OX6
Then by mole ratio, we can calculate the rate of appearance of HX2O to be xLsmolCX6HX12OX6⋅molCX6HX12OX66molHX2O=6xLsmolCX6HX12OX6
Factors that Affect Reaction Rates
Concentration: if concentration is a part of the rate law, it will affect reaction rate
rate law written in the form Rate=k[A]x[B]y[C]z, where k is the rate constant
exponents must be found experimentally (with data)
Arrhenius equation: k=Ae−Ea/RT, where k is the rate constant, A is a proportionality constant, Ea is the activation energy, R is the gas constant, and T is the Kelvin temperature
the natural log of the above results in lnk=RT−Ea+lnA→ straight-line graph of lnk vs 1/T
activation energy is always positive
Ability to Meet: gas/solution/fine solids react mre easily
Catalyst Presence: increases reaction rate
Example
Given the table of data below, determine the rate law
Experiment Number
[A] (mol/L)
[B] (mol/L)
Initial Rate of Reaction (mol/Ls)
1
0.100
0.200
1.1⋅10−6
2
0.100
0.600
9.9⋅10−6
3
0.400
0.600
9.9⋅10−6
The general form is Rate=k[A]x[B]y
Divide the first two equations: Rate1Rate2=k(0.100)x(0.200)yk(0.100)x(0.600)y
This simplifies to 1.1⋅10−69.9⋅10−6=(0.200)y(0.600)y→9=3.00y→y=2
Next, divide the second two equations: Rate2Rate3=k(0.100)x(0.600)yk(0.400)x(0.600)y
This simplifies to 9.9⋅10−69.9⋅10−6=(0.100)x(0.400)x→1=4.00y→x=0
Finally, we use any one of the equations to find k. We use the first experiment: 1.1⋅10−6=k(0.100)0(0.200)2→k=2.8⋅10−5
So the final rate law is Rate=(2.8⋅10−5molsL)[B]2
Reaction Order
Zero-Order Reactions
Rate=k [A]=−kt+[A]0
does not depend on concentration
concentration vs time graph is a straight line
half-life: t1/2=[A]0/2k
First-Order Reactions
Rate=k[A] ln[A]=−kt+ln[A]0
ln concentration vs time graph is a straight line
half-life: t1/2=ln2/k
radioactive decay is a first-order process
Second-Order Reactions
Rate=k[A]2 or Rate=k[A][B] [A]1=kt+[A]01, when [A]=[B]
reciprocal concentration vs time graph is straight
half=life: t1/2=1/k[A]0
Kinetic Theories
Collision Theory: reaction rate equals frequency of effective collisions between reactants
when molecules collide, kinetic energy is converted into potential energy
if Δ(Potential Energy)>Ea and collide in the right direction, reaction occurs
by kinetic molecular theory, fraction of collisions reaching Ea increases as temperature increases
Rate=Nfefo, where N is collisions per second, fe is fraction of collisions with minimum energy, and fo is fraction of collisions in the right direction
Transition-State Theory: colliding molecules distort orbitals that weaken bonds, breaking and forming new ones
otherwise similar to collision theory
potential energy diagram's height difference is Ea
k is inversely proportional to Ea
heat of reaction, ΔH=PEproducts−PEreactants→ endothermic/exothermic
catalyst reduces Ea
Reaction Mechanisms
set of elementary steps that make up a reaction
elementary reactions: one molecule breaks apart or two molecules collide
coefficients are exponents in the rate law
rate-determining/rate-limiting step: slowest step that determines overall reaction rate
intermediate: substance that is part of a mechanism but not the overall reaction
the correct reaction mechanism:
adds up to the overall reaction
has a slow step that matches the observed reaction rate
to determine the rate law of an elementary step
use the same general formula, converting coefficients to exponents
if an intermediate appears, assume the forward and reverse reactions are equal and substitute
Example
What is the reaction rates of the following elementary steps for the reaction 2NO+OX22NOX2? NO+OX2NOX3 NOX3+NO2NOX2
The first reaction is simply Rate=k[NO][OX2]
The second is Rate=k[NOX3][OX2], but since NOX3 is not in the overall reaction, we must substitute.
From the first equation, kf[NO][OX2]=kr[NOX3]→[NOX3]=krkf[NO][OX2]
So, the second is Rate=k(krkf)[NO][OX2][NO]=k′[NO]2[OX2]