Rate Law Notes – link to pdf
Rate Law is essentially a topic that investigates the RATE (speed) at which a reaction will occur with a reaction that is ALREADY SPONTANEOUS. There are many factors that affect the Rate of a Chemical reaction and they all deal with changing the frequency of collisions with chemicals THAT ALEADY HAVE A PATHWAY to proceed or react.
In order for a reaction to occur there must an effective collision between the reaction particles.
An effective collision is a physical collision between reacting elements, compounds, or ions that results in the breaking of old bonds and the formation of new ones.
An effective collision requires 2 conditions:
1. High enough kinetic energy to overcome the repulsive forces of the
cloud of electrons all chemical species have on their outside.
– if the reactants are not free atoms then some energy is needed to be consumed to
break the bonds.
2. Correct orientation of the reacting chemicals. Larger molecules especially
have areas where chemical reactions can and cannot occur. Thus the
collisions of the reacting particles can occur at certain positions allowed
by the second law of thermodynamic.
Kinetic Energy factor:
Orientation Factor:
When Effective collision are increased then the rate at which the chemical reactions occur also increases.
Now in general or statistically, if we increase the frequency of collisions, we will increase the number of effective collisions. The factors that will increase the frequency of collisions include:
1. Temperature – particle motion (velocity) is proportional to temperature:
Average Speed (velocity) = urms = √3RT/M
The faster reactants move the higher the frequency of collisions.
Light Sticks Demo: https://youtu.be/oCXzWF1EOfc
2. Pressure – If a reactant is a gas then increasing its partial pressure or decreasing its volume of the reactant container will increase the frequency of the collisions.
Lazy Susan Demo: https://youtu.be/6VnZLa2EZQM
3. Catalyst – Catalysts will provide a new reaction mechanism that will have a Net lower Ea (activation
energy) or will help with the orientation like biological enzymes or metal catalysts.
Catalysts lowers the energy requirement to start reactions and thus makes the frequency of effective collisions increases with increasing the the overall frequency of collisions. It increases the percentage of effectiveness of total collisions.
Elephants toothpaste demo: https://youtu.be/8ZnNSTzl1lc
4. Surface area – powdered forms of reactants will increase the frequency of collisions more than reactants
in solid lumps because more reactants are exposed to collisions.
Elevator Explosion demo: https://youtu.be/u3CixGBlBN0
dragons breath demo: https://youtu.be/WLcEFWTXWpk
5. Concentrations – The amount of available reacting particle. will affect the frequency of collisions.
This is the factor that Rate Law investigates.
Iodine Clock Demo: https://youtu.be/CRYGyDyXiSo
Liquid oxygen: https://youtu.be/JQXGwExP-lM
No matter how effective a collision is the reaction MUST BE SPONTANEOUS for it to occur.
It must have a favorable Thermodynamic pathway to occur before we can investigate how the concentration (Molarity) can affect the rate of a reaction.
The rate of a reaction depends on a Path Function and thus depends on the path a reaction takes to reach the end of reaction (either by running out of a limiting reagent or by reaching equilibrium). Rate is Path function unlike the thermodynamic STATE FUNCTIONS of ∆H (kJ/mol), ∆S (J/mol), and ∆G (kJ/mol). Notice these proceeding quantities are all measured by ENERGY per mol NOT ENERGY per time!
This means that if a reaction occurs quickly (high rate) the amount of heat released if exothermic would be very high Because WE USED A LOT OF MOLES of reactant in a short period of time!
The Energy per mole stays constant because it is a STATE FUNCTION!
Conversely if the same reaction occurs slowly (slow rate) the amount of heat released would be lower because used less moles of the reactant in the same time period. ∆H (kJ/mol), ∆S (J/mol), and ∆G (kJ/mol) all remain constant and thus THE RATE OF A REACTION HAS NOTHING TO DO with how spontaneous the reaction is!
∆G is measure of the Spontaneity of reaction and Rate of reaction is independent of its value!
As long as a reaction has a thermodynamic pathway to proceed (-∆G) the rate of reaction can have infinite possible speeds based on changing the infinite possible pathways that chemicals can collides with each other.
Example: Methane (natural gas, CH4 ) reacts very spontaneously with oxygen, ∆G = -801 kJ/mol
Because the ∆G = -801 kJ/mol is a very large negative number this reaction like most combustion reactions has equilibrium constant that favors products so much we would call this reaction a completion reaction. This reaction cannot reach equilibrium because of the extreme forward position of the Keq of this reaction.
Now Methane sitting in reaction with oxygen is said to be very spontaneous even is it not reacting. Remember that ∆G measures the ability to move in a preferred direction NOT how fast it moves in that direction.
CH4 (g) + 2 O2 (g) —–> CO2 (g) + 2 H2O (g) + *802 kj/mol
*this is the ∆Hrxn which is not the same as ∆G
The rate of a reaction is thus based on the particle pathways to collide
and not on the ability to react (∆G).
Even though that ∆G is a state function and the Rate of a Reaction is a path function they are connected only by how the equilibrium is established in the reversible process unlike the combustion reaction above.
Okay what is Rate Law? Rate Law is a mathematical equation that chemists use to determine how to increase the speed of a reaction BASED on initial concentrations of reactants.
What good is a spontaneous reaction if the rate of reaction is low? The point of industrial chemists is to produce products from reactants. That is essentially what chemistry is!
Rate Law helps chemists
1) calculate / predicts rates of reactions
2) Study reaction mechanisms – how the reaction actually occurs.
Rate Law is based on experimental evidence BUT the coefficient in the reaction does play a role!
The mathematical equation for the rate of a reaction based on initial concentrations of a reactants
RATE LAW = R = k [ X ]y [ Y ]x
Speed of reaction = ( Rate Law Constant) [M of reactants] orders of the reactant
The orders of each reactant MUST be found experimentally!! If we determine FROM experimental evidence that the reactant is in a elemental reaction the orders are the stoichiometric coefficients!!
Lets look at the three most common elementary reactions:
Elementary reaction- SINGLE STEP REACTIONS
UNIMOLECULAR – XY —–> X + Y
If this is the only step in the reaction: If we double the [XY] then we will create twice the amount of products in the same time period. Doubling the [XY] will double the Rate! If the Rate changes as much as the [reactants] then we say the order (exponent) of the reactant in the Rate Law is 1.
RATE LAW = R = k [ XY ]1
This reaction is 1st order! Changing the initial concentration of XY will change the Rate the same amount!
For example if we have the following experimental evidence:
[XY] Rate of formation of Y
_______mol/L________ M/sec__________
Experiment 1 .03 .20 M/sec
Experiment 2 .09 .60 M/sec
The [XY] tripled (3 x .o3 =) .09 and the Rate tripled (3 x .20= ) .60
So the Rate changed by the power of 1 of the change of the reactant (XY)
The power of 1 = the value below the exponent ——> [2]1 = 2 , [17]1 = 17
Rate (increased 3x) = [ XY increased 3 x ]1
Experimental evidence suggests that the Rate Law is: R = k [XY]1
The coefficient will only become the exponent in Elementary Reactions (single step) but we need evidence to prove the possibility of the elementary reaction.
BIMOLECULAR – X + Y —–> XY
If this is the only step in the reaction: If we double the [X] and keep [Y] constant then we will create twice the amount of products in the same time period. Doubling the [X] will double the Rate! If the Rate changes as much as the [reactants] then we say the order (exponent) of the reactant X in the Rate Law is 1 WITH RESPECT TO [X].
If this is the only step in the reaction: If we double the [Y] and keep [X] constant then we will create twice the amount of products in the same time period. Doubling the [Y] will double the Rate! If the Rate changes as much as the [reactants] then we say the order (exponent) of the reactant Y in the Rate Law is 1 WITH RESPECT TO [Y].
Of course this comes from experimental evidence:
[X] [Y] Rate of formation of Y
_______mol/L_ mol/L_______ M/sec__________
Experiment 1 .03 .02 .15 M/sec
Experiment 2 .06 .02 .30 M/sec
Experiment 3 .03 .04 .30 M/sec
Notice in Experiment 1 and Experiment 2 above:
The [X] doubled (2 x .03 =) .06 and the Rate doubled ( 2 x .15 = ) .30 while [Y] stays constant.
So the Rate changed by the power of 1 of the change of the reactant (X)
R = k [X]1 The reaction order with respect to [X] = 1
Notice in Experiment 1 and Experiment 3 above:
The [Y] doubled (2 x .02 =) .04 and the Rate doubled (2 x .15 = ) .40 while [X] stays constant.
So the Rate changed by the power of 1 of the change of the reactant (Y)
R = k [Y]1 The reaction order with respect to [Y] = 1
The rate law of the entire reaction: R = k [X]1 [Y]1
The experimental evidence supports the elementary reaction where the coefficients = exponents.
This reaction is considered to be a second order reaction because the overall order is 2 (add exponents).
TERMOLECULAR – 2X + Y ——> X2Y
Termolecular reactions are probably NOT elementary reactions as it is very hard for 3 particle to collide at one time. This means that this reaction probably occurs in multiple steps which we call a reaction mechanism.
Experimental evidence can help us understand what these steps or reaction could be?
So given the reaction above lets run an experiment to identify the reactions orders. They will probably not be equal to the coefficients as we saw in the last reactions THAT WERE ELEMENTARY reactions. Experimental evidence is needed!
(Rate Law Test 3)
[X] [Y] Rate of formation of X2Y
_______mol/L_ mol/L_______ M/min__________
Experiment 1 .25 .16 3.0 M/min
Experiment 2 .25 .32 3.0 M/min
Experiment 3 .50 .16 12.0 M/min
Skill Number 1 – Identify the orders of the reaction from the experimental evidence
Okay so the easiest thing here is to use the EYEBALL method! Keep one reactant constant and see what how the other reactant changes AS compared to the changes in the Rate.
If we look at Experiment 1 and Experiment 3 above (rate law test 3 above):
[X] increases from .25 M to .50 M and thus is doubling while Rate of reaction of Experiment 1 and Experiment 3 increases from 3.0 to 12.0 which means its is quadrupling!
The [X] doubled (2 x .25 =) .50 and the Rate quadrupled (4 x 3.0 = ) 12.0 while [Y] stays constant.
So the Rate changed by the power of 2 (square) of the change of the reactant (X)
(doubling initial concentrations)2 = quadruple the RATE!
[X]2 The order with respect to [X] is 2.
If we look at Experiment 1 and Experiment 2 above: (rate law test 3 above):
[Y] increases from .16 M to .32 M and thus is doubling while Rate of reaction of Experiment 1 and Experiment 2 remained the same (at 3.0) while [X] was held constant.
So the Rate changed by the power of 0 of the change of the reactant (Y)
[Y]o The order with respect to [X] is 0.
This means that changing the concentration of Y does not effect the rate of the reaction. It also means that reactant Y is not in the Rate Determining Step of a Mechanism. I will get to this below.
So the Rate Law for the reaction based on the experimental evidence:
R = k [X]2 [Y]0
Since anything to a zero power is 1 we can also write
the Rate Law equation like this: R = k [X]2
The reaction that we tested was : 2X + Y ——> X2Y
Notice that the coefficients do not become exponents or reaction orders! This is very telling! It tells us that this reaction has multiple steps and it NOT an single step reaction.
So lets look at 2 proposed reaction mechanisms for this reaction above. A reaction mechanism represents a series of elementary steps that occur that result in a final overall chemical reaction.
Most chemical reactions do not occur in a single elementary step AND the Rate Law formula data from the experiment of the termolecular reaction supports this.
The rate law of 2X + Y ——> X2Y is equal or R = k [X]2 [Y]0 and because [Y]0 is
THE COEFFICIENTS DO NOT EQUAL THE EXPONENTS (ORDERS)
Thus this reaction cannot occur in one elementary step. How can the order or the exponent of [Y] = 0?
Remember that we attained that order from the Rate Law Test 3 above which revealed when we kept [X] constant and doubled [Y] the Rate did not change (when comparing experiment 1 and 2). This is what gave us a reaction order of 0 for [Y] .
Skill Number 2 – Determining catalysts and intermediates in mechanisms
Lets look at 2 proposed mechanisms for this reaction to see which one is best supported from the Rate Law that was determined from experimental evidence.
Mechanism 1: X + Z —-> XZ
XZ + X —-> X2 + Z
X2 + Y —-> X2Y
__________________________
This mechanism is an example of a Catalyst driven mechanism. A Catalyst is a chemical that starts a whole series of elementary reactions that normally would occur. Notice in step 1 Z reacts with X to create the intermediate XZ which reacts with Y to start the second elementary step. The second elementary step reproduces Z and another intermediate XY that starts the step 3.
Catalysts do not get consumed in chemical reactions THUS THEY are always reproduced in a later step to be re-used over and over (if enough of X is still present). Therefore Z is a catalyst!
Intermediates are produced and Start future steps. They LINK one step to another in a mechanism.
Chemical XZ is the intermediate that link step 1 to step 2 and chemical XY links step 2 to step 3.
Skill Number 3 – Determining Rate Law from mechanisms or evaluating the best mechanism
There are 2 ways prove or check that a proposed mechanism can support a reaction and
its Rate Law formula:
1: Does the Mechanism actually cancel out to equal the OVERALL REACTION?
2. Does the Mechanism support the Rate Law determined?
Lets take mechanism 1 and cancel out the Catalysts and Intermediates to see if we actually get the overall reaction.
Mechanism 1: X + Z —-> XZ
XZ + X —-> X2 + Z
X2 + Y —-> X2Y
_________________
2X + Y —-> X2Y
Okay so the overall reaction does appear when you cancel so step one check out!
Because of the cancelling only the reactants are expressed in the RATE LAW!
Now lets test step 2:
Looking back at cancelled mechanism we need to understand that intermediate are usually VERY REACTIVE
and in Step 2 XZ reacts very quickly BUT X in step 2 has to wait till XZ in step 1 . Since step 1 does not use a high reactive intermediate it reacts slower and all of the other steps have to wait for the production of the first intermediate XZ . Waiting for step one makes step one the RATE DETERMINING STEP (RDS). This is the slowest step of the 3 step mechanism that the whole process IS LIMITED BY! The overall reaction can only go as fast as the slowest step!
X + Z —-> XZ SLOWEST STEP! RDS
XZ + X —-> X2 + Z Fast
X2 + Y —-> X2Y Fast
___________________________
2X + Y —-> X2Y
SO we find the slowest step in in are mechanism that we determine from the production of the intermediates or sometimes questions will point out the slowest step.
Since the slowest step DETERMINES THE OVERALL RATE then the RATE LAW comes from the single elementary step that is the RATE DETERMINING STEP (RDS)!
STEP 1 is the RDS : X + Z —-> XZ
Since the Rate LAW is based on this single elementary step
THE COEFFICIENTS Equal the Exponents (orders of reactions)
So we get a RATE LAW from this mechanism: R = k [X]1
This does not match the Rate Law that we got from the experimental evidence (rate law test 3) and thus the second test FAILS prove that the proposed mechanism is possible.
The Rate Law based on the experimental evidence: = k [X]2 [Y]0
The Rate Law based on the proposed mechanism: = k [X]1
So lets see if another proposed mechanism supports the evidence in (rate law test 3).
Mechanism 2: X + X —-> X2
X2 + Y —-> X2Y
_________________
The first test is to see if the proposed mechanism ( Mechanism 2: ) will cancel out to the overall overall reaction.
Mechanism 2: X + X —-> X2 SLOWEST STEP! RDS
X2 + Y —-> X2Y FAST
_________________
2X + Y —-> X2Y
So yes the entire individual elementary steps do cancel out to the overall reaction tested.
Now does the mechanism support the experimental Rate Law formula?
Well we have analyze the mechanism and when we do we find that the first step CREATES an intermediate X2 that quickly will reacts with the Y to produce the final product. Because intermediates are highly reactive, the second step is the faster step that MUST WAIT FOR THE SLOWER 1st STEP!
Mechanism 2: X + X —-> X2 SLOWEST STEP! RDS
X2 + Y —-> X2Y FAST
_________________
2X + Y —-> X2Y
So since the slowest step must be step 1: X + X —-> X2
And that each step in a mechanism is an elementary step.
Then the Rate Law formula is determined from the coefficients in this single elementary step!
We get from step 1 of this mechanism: R = k [X]2 [Y]0 or R = k [X]2
The Rate Law based on the experimental evidence: = k [X]2 [Y]0
This Rate Law matches what we get from the data in (Rate Law Test 3) and thus this mechanism does support the experimental evidence.
Skill Number 4 – Determine the rate law constant with correct Units
So lets use the last reaction and its experimentally determined rate law to calculate its rate law constant:
2X + Y —-> X2Y R = k [X]2 [Y]0
The rate law was determined from the Rate law Test 3:
(Rate Law Test 3) [X] [Y] Rate of formation of X2Y
______mol/L_ mol/L_______ M/min__________
Experiment 1 .25 .16 3.0 M/min
Experiment 2 .25 .32 3.0 M/min
Experiment 3 .50 .16 12.0 M/min
Once you determine the Rate Law orders from the experiment (Skill 1) then you pick any one of experiments (rows) and plug in the value of the [X] and [Y] to solve for the Rate Law Constant.
Given the Rate Law for the reaction: R = k [X]2 [Y]0 and using experiment 1 values:
3.0 M/min = k [ .25 M ]2 [ .16 M ]0
[.0625 M2 ] [ 1 ] anything to the exponent of zero = 1
3.0 M/min = k = rate law constant (not Keq!!!)
.0625 M2
Careful with your units as most AP questions will evaluate the VALUE and UNITS in this type of question.
The units will be different for different reactions with different orders! You must
48 1 = k = 48 M-1 min-1
M min
Notice that 48 1 is equivalent to 48 M-1 min-1 (exponent -1 = inverse or reciprocal)
M min
Skill Number 5 – Calculate the Rate for the initial concentrations that are not in the experiment.
Staying with rate law test 3 for the following reaction and Rate Law we will now calculate a value for the Rate given ANY INITIAL CONCENTRATION of the reactants, now that we determined the rate law constant
in Skill 4 above.
2X + Y —-> X2Y R = k [X]2 [Y]0 k = 48 M-1 min-1
This skill is really what we use rate law test primarily for, which is to calculate the Rate of a reaction with ANY initial concentrations. Chemical factories want the rate of reaction to stay high (time is money) and thus will use the formula for their reactions to maintain a high rate.
So now this skill will ask you calculate the rate from any arbitrary values of X and Y.
Calculate the Rate of the reaction if the initial concentration of X is .47 M and Y is .28 M. We have determined from the experiment that the rate order of Y is 0 and so the concentration of Y is not a factor in this calculation BUT Y is still required. I would be a mistake to think that because Y has a reaction order of 0 that the reaction does not need Y to react! In truth you need a significant amount of Y to react.
It’s that changing [Y] that is already present in a significant amount will not change the Rate!
Lets Solve: R = k [X]2 [Y]0
R = 48 M-1 min-1 [.47]2 [.28]0
R = 48 1 [.2209 M2 ] [ 1 ]
M min
R = 10.6 M/min
Notice that 1 M cancels in the equation to give correct units for Rate. Also notice that the Rate is consistent with the values in Rate Law Test 3.
(Rate Law Test 3) [X] [Y] Rate of formation of X2Y
______mol/L_ mol/L_______ M/min__________
Experiment 1 .25 .16 3.0 M/min
Experiment 2 .25 .32 3.0 M/min
.47 .28 10.6 M/min
Experiment 3 .50 .16 12.0 M/min
Skill Number 6 – Determine the Stoichiometric loss of reactants to gain of products
Working with same chemical reaction: 2X + Y —-> X2Y
You could be asked to determine the rate of decrease of X from one of the experiments in the Rate law test.
This is a stoichiometric relationship:
Lets solve for the rate of decrease of X from experiment 2 below:
(Rate Law Test 3) [X] [Y] Rate of formation of X2Y
______ mol/L_ mol/L_______ M/min__________
Experiment 1 .25 .16 3.0 M/min
Experiment 2 .25 .32 3.0 M/min
Experiment 3 .50 .16 12.0 M/min
Because of the coefficients : 2X + Y —-> 1 X2Y
lose 2 gain 1
Twice as much of X is lost to the gain of one X2Y
The rate of formation of X2Y in experiment 2 = 3.0 M/min and thus
the rate of decrease X is Twice the formation of X2Y:
2 x 3.0 M/min = – 6.0 M/sec = should be negative because of the loss.
Another way to express the stoichiometric relationships on how the decrease of reactants relate to the increase of the products based the above stoichiometric ratios.
– 1 ∆[X] = ∆1[X2Y]
2 t t
Some people thought I had this wrong in my lecture but when you think about this if I gave the rate of decrease of X and you were to solve for the Rate of increase of X2Y this equation works! The rate of increase of X2Y is half of the rate of decrease of X if : 2X + Y —-> 1 X2Y.
Now back to solving the original problem:
– 1 ∆[X] = ∆[3.0]
2 t t
When you solve: ∆[X] = – ∆ 2[3.0] t
∆ t
∆[X] = – 6.0 M/min
What if we were to solve for the rate of increase of X2Y from the rate of decrease X in experiment 2.
Remember that these types of problems is based on the ratios: 2X + Y —-> 1 X2Y.
Rate of decrease X: 6.0 M/min ——-> Rate of increase of X2Y: 1/2 (6.0) = 3.0 M/min
Skill Number 7 – Solve for the Rate Law orders without EYEBALL method – NON-EYEBALL METHOD
Okay this is the same skill as skill # 1 but the eyeball test will not always work if the experimental numbers do not give clear observable reaction orders.
I have changed the values of Rate Law Test 3 to reflect the same equation and Rate Law that we have been using:
2X + Y —-> X2Y R = k [X]2 [Y]0
These new data will help demonstrate how we solve for the orders if we cannot identify the orders by the eyeball method.
(Rate Law Test 3) [X] [Y] Rate of formation of X2Y
_______mol/L_ mol/L_______ M/min__________
Experiment 1 .056 .16 .18 M/min
Experiment 2 .056 .32 .18 M/min
Experiment 3 1.12 x 10-1 .16 .72 M/min
Lets solve for the rate order of X by comparing experiment 3 to experiment 1.
experiment 3 : R = k [X]m [Y]n
_________________
experiment 1 : R = k [X]m [Y]n
experiment 3 : .72 . = k [ 1.12 x 10-1 ]m [.16]n
_____________________________
experiment 1 : .18 = k [.056]m [.16]n
cancel :
experiment 3 : .72 = k [ 1.12 x 10-1 ]m [.16]n
____________________________
experiment 1 : .18 = k [.056]m [.16]n
Even if we do not know the order of [Y] it will cancel out as it is the same for all experiments!
Also if the [Y] values are not the same value we can still solve for [X]m algebraically.
experiment 3 : .72 = k [ 1.12 x 10-1 ]m [.16]n
______ ______________________
experiment 1 : .18 = k [.056]m [.16]n
↓ ↓
simplify: 4 = [ 2 ]m
2 to what power = 4? m = 2 ( 2 squared = 4! )
The order of the reaction with respect to X is 2 : [X]2
Skill Number 8 – Solve for the Rate Law orders with Graphs
Remember that our rate law formula that we have been using ONLY MEASURES THE INITIAL RATE of the reaction from INITIAL CONCENTRATIONS. Although it is powerful to know the instantaneous rate of a reaction, that point in time is very small because as soon as we get past that instantaneous point, the RATE HAS DECREASED and our Rate Law Equation is no longer helpful.
For instance, lets look at a graph of the second order reaction:
2X + Y —-> X2Y R = k [X]2 [Y]0 (second order reaction = 2 + 0 = 2)
We could measure the [X] and some time period AFTER time zero and calculate the Rate of the reaction from that point. This would give us any point on the graph at time zero BUT it would not give us the time it takes to get to that lowered [X]!!
If want to answer questions regarding the time interval that will give us changes in the [X] based on a certain rate law equation then we really have a more powerful tool!
When we create a differential equation that links ∆ [X] / ∆ t to the Rate law equation we add 2 powerful dimensions to the RATE Law equation:
[X]0 = Initial value , [X]t = value of X at some time (final value), and t (time).
These new equation allow us to NOW answer more powerful questions:
How long does it take for the reaction to be completed?
How long does it take for the concentration of my reactant to reach a certain value?
Given the time, how much of the reactant remains?
What is the shelf life of the chemical? (how long does it take to decompose?)
*Notice that each of these questions have a time linked to changes in the concentration of the reactant!
These integrated formulas are solved below:
Here they are in our reference tables:
 |
It is usually very helpful to utilize these equations when you arrange them into a linear equation. 1st Order: X –> Y
R = K [X]1
ln[A]t – ln[A]0 = -kt
ln[A]t = -kt + ln[A]0
y = mx + b
(negative slope)
2nd Order: X + X –> X2
1 – 1 = kt
[A]t [A]0
1 = kt + 1
[A]t [A]0
y = mx + B
(positive slope)
|
Notice that the derivation of the 2 posted integrated formulas are only for a
1st order reaction with a specific reaction: X —-> Y
And a second order reaction with a specific reaction: X + X —-> X2
*Notice that they do not provide a zero integrated rate law formula.
This means that the AP will not ask you to solve for zero order reaction but you should understand it.
Okay now that we have these new equations that link time with changes of reactant concentrations we can look at the plot of these NEW graphs to Identify the RATE ORDER of the reaction.
 |
 |
In order to solve whether a reaction has a reaction order of 2 or 1 all we have to do is plot the
graph of either the ln [X] vs time or 1/[X] vs time. The graph that is linear with the correct slope
will determine the order.
Skills 1, 7 Determined the reaction orders by Experimental Rate Law DATA
Skill 3 Determined the reaction orders by using reaction mechanisms
Skill 8 Determines the reaction order by graphing (using integrated y axix values)
For Example:
What is the order with respect to X given the following data for a new Reaction: X + Y —> XY
Rate Law Test 4:
actually measured |
The data actually collected in this Rate Law experiment was the [X] with time. The second and third columns are just values that were calculated from the [X] value to plot integrated graphs with y-axis values to determine help determine the order of the reaction graphically. We will do this in our Rate Law Lab with Crystal Violet and hydroxide. |
If we plot the graphs of ln[X] vs time AND 1/[X] vs time we can determine the order of the reaction:
 |
 |
ln [X]t = – k t + ln [X]0 1/[X]t = k t + 1/[X]o
y = – mx + b y = mx + b
Will be a linear line with a negative Will be a linear line with a positive
slope IF THE REACTION is 1st order slope IF THE REACTION is 2nd order
with respect to X! with respect to X!
This graph proves the reaction that
was tested is 1st order!
Based on the above data and graphs: R = k [X]1
This data would not support the reaction I have been using throughout these note.
Skill Number 9 – Solve for time, concentration, or k using integrated Rate Law formulas
Now that we have determined the order of the reactant by either Skill 1, 6, 3, or 7, we can use the integrated rate law formulas to solve for:
[X]0 = Initial concentration
[X]t = Concentration of reactant X at some period of time
t = Time for reactant to change from [X]0 to [X]t
k = Rate law constant (watch your units here!)
Lets solve for the the rate law constant using the appropriate integrated rate law formula for Rate Law Test 4:
Graphically we determined the reaction was 1st order so we use:
ln [X]t – ln [X]0 = – k t given in your reference tables
rewrite equivalent: ln [X]t = – k t
________
ln [X]0
I chose the 0 minute to 20 minute interval in
Rate Law Test 4 table above:
ln[.00427 M]20 = – k (20 min)
______________
ln[.00633 M]0
Cancel units. Be careful. Correct units for
Rate Law Constant are usually a separate
point earned.
ln [.6745] = – k (20 min)
– [-.3937 ] = k (20 min)
[ .3937 ] = k (20 min)
[ .3937 ] = k (20 min)
.01968 min-1 = k
Skill Number 10: Solve for the reaction orders by using the graph of [X] vs time
by observing half-life.
Half life = the time interval that a reactant decreases it concentration in half
We can determine whether a reaction is 0th , first, or second order
based on whether the half lives
Decrease = [X]0
Stay the same = [X]1
Increase = [X]2
This is the final way we can determine the reaction orders:
0 order half life:
1st order half life:
2nd order half life:
©MrGrodskiChemisry 2021