Collision Theory - Arrhenius Equation & Activation Energy - Chemical Kinetics - Free Educational videos for Students in K-12

Collision Theory - Arrhenius Equation & Activation Energy - Chemical Kinetics - By The Organic Chemistry Tutor

Transcript


00:00	in this tutorial let's talk about the collision theory model
00:05	. So what's the basic idea behind the collision theory
00:09	? The basic idea is that from molecules to react
00:12	they have to collide if they don't collide , there's
00:16	not gonna be any chemical reaction . If there's no
00:19	physical contact between these molecules , nothing's going to happen
00:22	. So molecules must collide in order to react .
00:26	The second thing is they have to have the correct
00:29	molecular orientation in order for a reaction to proceed .
00:34	So to illustrate this , let's say if we have
00:38	hydroxide and hydroxide is going to react with methyl bromide
00:46	. Ch three br now bro ming is more electoral
00:51	negative than carbon . So it bears a partial negative
00:54	charge and carbon is less electric negative than grooming .
00:58	So it's partially positive . The element that's more electro
01:02	negative is gonna pull the electrons toward itself . And
01:05	that's why brahman has the partial negative charge . So
01:08	, hydroxide , since it has a negative charge ,
01:12	it's attracted to the carbon in methyl bromide . So
01:15	it wants to attack this carbon does expelling the bromide
01:19	ion and so this will create methanol . So if
01:23	it approaches from the back of the methyl bromide ,
01:27	if it attacks from the carbon side , it has
01:30	the correct molecular orientation . So the reaction is going
01:33	to work . Now let's say if it approaches it
01:36	from the other direction . So let's say this is
01:39	methyl bromide and hydroxide is coming in this direction .
01:44	Well , the reaction is not going to happen .
01:47	Hydroxide is repelled by the partial negative charge of the
01:51	bromine atom . And so when two negative charges come
01:54	together , they're going to fill a force that's going
01:57	to repel them . And so that's going to prevent
02:00	the reaction from occurring . Whereas when hydroxide attacks from
02:04	the carbon side , the negative charge on the hydroxide
02:08	is attracted to the partial positive charge on the carbon
02:11	atom and that electrostatic force of attraction accelerates them together
02:16	, which favors the collision of those two molecules .
02:20	And so if the two molecules do not have the
02:22	correct molecular orientation , the reaction is not going to
02:26	happen , but if they collide with the appropriate orientation
02:31	, then the reaction is going to happen . So
02:34	in order for a reaction to occur , molecules ,
02:36	they have to collide and two , they have to
02:39	collide with the right orientation . Now , the third
02:42	thing is that they need enough energy because if the
02:46	temperature is low , even though these two may be
02:49	attracted to each other , if the hydroxyurea and doesn't
02:52	move with enough momentum , it's not going to be
02:55	enough to knock off this bromide ion , so it
02:58	has to have enough energy to attack the carbon atom
03:02	and expel the bromine iron out of that reaction to
03:05	create methadone . So there's three things that is required
03:11	for a reaction to take place . The molecules must
03:13	collide and they have to collide with the right orientation
03:16	and they have to have sufficient energy . Two break
03:21	some bonds and form some new bonds . Now this
03:26	leads us into energy diagrams . So on the Y
03:30	axis is the energy of the system , on the
03:33	X axis is going to be the reaction coordinate .
03:39	Mhm . Here we have the reactant and here we
03:42	have the products . The top is the transition state
03:46	, also known as the activated complex . In order
03:49	for this reaction to occur , the system has to
03:53	have enough energy the reactions need to gain of energy
04:00	. There's something called activation energy , which is the
04:03	difference between the energy of the transition state and the
04:06	energy of the react . It's if the reactant do
04:10	not have enough energy to overcome the activation energy ,
04:13	the reaction will not happen . So if they can't
04:17	get above this hill , there's no reaction that's going
04:21	to take place . So they have to have enough
04:23	energy to get over that hill . And so that
04:26	minimum energy to get the reaction started as the activation
04:29	energy . So even if a molecule collides with the
04:32	right orientation , if they don't have enough energy to
04:36	make it to the activated complex , then the reaction
04:39	is just not going to happen . So one way
04:42	you can get up there is by increasing the temperature
04:44	. Whenever you increase the temperature , the rate of
04:46	the reaction increases the fraction of molecules that can reach
04:52	the activate complex will increase . Now one thing I
04:56	want to specify is this is known as the forward
04:59	activation energy . There's something called the reverse activation energy
05:04	, which is the energy difference between the products and
05:07	the transition state . So in this example , it's
05:10	easier to go this way then to go that way
05:18	. Now the next thing you need to know is
05:19	that notice that the difference between the energy of the
05:22	products is greater than the energy of the reactions .
05:25	Whenever the energy of the products is greater than the
05:27	reactant , what you have is an end a thermic
05:29	reaction . In order for this reaction to occur ,
05:33	energy has to be put into the system . An
05:36	extra thermic reaction looks like this . The energy of
05:39	the reactant is going to be greater than the energy
05:42	of the products . So this is an exotic reaction
05:49	. Now , initially you gotta put in energy to
05:52	get to the activated complex to get the reaction started
05:55	. Once you overcome that barrier , then a lot
05:58	of energy is gonna be released . So you got
06:00	to put in a small amount of energy to get
06:02	started but you're gonna get a large amount of energy
06:04	once the process is started . And so that's a
06:07	an extra thermic reaction . Now let's talk about the
06:11	Iranians equation K . Is equal to ZP E .
06:18	Raised to the negative E . A . Over artie
06:23	. Now see is known as the collision frequency .
06:28	If you can increase the collision frequency of reaction ,
06:31	the rate of the entire reaction will go up .
06:35	P is hysteric factor and this has to do with
06:45	The fraction of collisions that have the right orientation .
06:48	So this is a number between zero and 1 and
06:52	it depends on the nature of the reactant . Now
06:59	this part E , which is the inverse of the
07:03	natural log function , raised to the negative Ea over
07:05	RT . That represents the fraction of molecules that have
07:09	sufficient energy to get the reaction going . So notice
07:13	that it depends on temperature and activation energy . So
07:17	if you want a reaction to occur , there's two
07:19	things you can do to help it to make it
07:22	work . You can increase the temperature and that can
07:25	give the molecules the energy they need to overcome that
07:28	barrier . As you increase the temperature , the rate
07:31	of the reaction will increase . The second thing you
07:34	could do is add a catalyst . Whenever you add
07:36	a catalyst . A catalyst provides an alternative pathway for
07:40	the reaction and by doing so it lowers the activation
07:43	energy , which increases the rate of the reaction .
07:49	Now , the last thing when you talk about is
07:52	the product cp Z . P is equal to aim
07:56	, which is also known as the frequency factor ,
07:59	is the product of the collision frequency and hysteric factor
08:07	. So now let's put this all together . Now
08:11	let's consider a first order reaction . Where we have
08:15	rate is equal to the rate constant Kane times the
08:20	concentration of A . And we know that the rate
08:25	constant K . Is equal to the frequency factor times
08:31	E . Raised the negative E . A . Over
08:34	artie . And the frequency factor can be replaced with
08:39	the collision frequency times hysteric factor and then times this
08:47	. So Z . The collision frequency represents the total
08:52	number of collisions per second . That's a current in
08:55	the reaction . P The hysteric factor which we said
08:59	was between zero and 1 . That's the fraction of
09:03	molecules with the proper orientation . Or more specifically ,
09:08	that's the fraction of collisions that are occurring with the
09:12	proper orientation . So you can think of PS the
09:15	probability that two molecules will collide with the right orientation
09:22	. This portion here which we describe as the fraction
09:25	of collisions with sufficient energy . You could also think
09:29	of it as the probability that two molecules that are
09:32	colliding will have enough energy to overcome the activation barrier
09:38	. So you can think of these two values as
09:40	probabilities or fractions . So is the is the collision
09:46	frequency and P . Is the probability that tomb colliding
09:51	molecules will have the right orientation . A represents the
09:55	total number of collisions per second that's occurring . Interview
09:59	action with the right orientation . Now , if we
10:04	multiply the frequency factor by the fraction of molecules having
10:09	sufficient energy , we can dust come up with an
10:13	understanding of the very constant K . So we can
10:15	say that the rate constant K . Represents the total
10:19	number of collisions per second having the proper molecular orientation
10:24	, as well as sufficient energy needed to overcome the
10:29	activation energy barrier to get the reaction going to convert
10:33	the reactions into products . So the very constant K
10:37	. Takes all of those factors into account . Now
10:43	, let's talk about what happens when we increase the
10:47	concentration of the reaction . A as we increase the
10:51	concentration of the reactant , there's gonna be more collisions
10:55	. The number of collisions will increase and therefore the
11:00	rate of the reaction will increase . Now what about
11:04	if we increase the temperature , if we increase the
11:07	temperature , the average kinetic energy of the molecules will
11:12	increase . So molecules are going to be moving faster
11:16	, the molecular speed of the molecules will increase as
11:19	well . And therefore there's gonna be more collisions .
11:24	And this too will increase the rate of the reaction
11:29	. Now you can think about it another way to
11:31	and this formula , the temperature affects the re constantly
11:36	. Now the temperature is in the bottom of the
11:38	fraction and typically when you increase the denominated refraction ,
11:44	the value of the fraction goes down . However ,
11:48	notice that we have a negative exponents . And so
11:54	because of that , when you increase the temperature it
11:58	has the effect of increasing K . It's like a
12:01	double negative . You can think of it that way
12:03	. When you increase the temperature , the rate constant
12:06	K goes up , the fraction of molecules having sufficient
12:10	energy to overcome the activation barrier increases . And when
12:16	K goes up , the rate constant goes up as
12:20	well . And let's show this with a distribution proof
12:34	. So we're going to have energy on the X
12:37	axis and we're going to have the number of molecules
12:45	on the y axis . It's not moles but molecules
12:50	. And we're gonna have two curves , one at
12:53	a lower temperature which will say look something like this
12:57	, and another curve which I'll drawn blue at a
13:01	higher temperature . So this one will be , let's
13:09	say at a temperature of 300 Kelvin And the other
13:12	one will be at a higher temperature will say at
13:14	500 Calvin And somewhere along the X axis we're going
13:20	to have E . A . The activation energy .
13:27	So anything above E . A . Represents the fraction
13:31	of molecules have in sufficient energy , sufficient collision energy
13:36	, so to speak enough so that they can overcome
13:40	the energy barrier to make the reaction going . And
13:45	let's put that in green . So in green ,
13:52	that represents the fraction of molecules at 300 Calvin ,
13:55	having enough energy to make the reaction go . But
13:59	in red , which encompasses the area in green as
14:04	well , that represents the fraction of molecules have enough
14:07	energy to overcome the energy barrier to also make the
14:11	reaction work . But notice that at 500 Calvin the
14:16	fraction of molecules that have enough energy is greater .
14:20	Then the fracture of molecules that 300 Kelvin is significantly
14:23	greater . And so that's what happens when you increase
14:27	the temperature . When you increase the temperature , the
14:30	average kinetic energy of the molecules go up . So
14:32	therefore the fraction of molecules have enough energy to make
14:38	the reaction go , and that fraction increases . And
14:43	so if more molecules have more energy , two get
14:47	the reaction go on . The overall rate of the
14:49	reaction will increase . Now let's talk about what's going
14:55	to happen if we were to add a catalyst to
15:00	the reaction , what do you think is going to
15:02	happen if we introduce the catalysts ? A catalyst has
15:09	the effect of lowering the activation energy . Now ,
15:16	here's a visual description of that . So this is
15:21	a potential energy diagram . We have potential energy on
15:25	the Y axis and the progress of the reaction on
15:29	the X axis . So let's say we have something
15:35	that looks like this , where this represents the energy
15:38	of the reactant since the products . And this is
15:41	the transition state . For a catalyzed reaction , it
15:48	would look something like this . So this is the
15:55	un catalyzed reaction and this is the catalyzed reaction .
16:09	The activation energy is the difference between the energy of
16:11	the reactant and the transition state . But notice that
16:16	when you add a catalyst , the activation energy decreases
16:23	, a catalyst achieves this by providing a reaction with
16:28	an alternative pathway . It actually changes the reaction mechanism
16:34	. It allows the reactant to become products using by
16:39	taking a different route , so to speak . So
16:42	that's how the catalysts can lower the activation energy .
16:46	It helps the reaction find an easier way to go
16:49	from reactions to products . Now notice that activation energy
16:55	is in the numerator of that exponential fraction . West
16:59	temperature is in the denominator . Now we know that
17:04	an increase in temperature leads to an increase in the
17:09	very constant K . So what happens if we increase
17:15	the activation energy ? Well it should have the opposite
17:19	effect . It should decrease K . T . Is
17:25	in the denominator E . A . S . In
17:27	the numerator . So if increase in T . Leads
17:30	to an increase in K . Increase in E .
17:33	A . Will decrease Kane . Likewise the reverse is
17:36	true . If we decrease the activation energy the very
17:40	constant case should go up and that's what happens when
17:44	you add a catalyst , the activation energy goes down
17:47	, the rate cost and go the red costing K
17:50	. Goes up . And so the overall rate of
17:52	the reaction increases . So those are some ways in
17:57	which you can increase the rate of a chemical reaction
18:01	. So to summarize if we increase the concentration of
18:04	the reacted , the rate is gonna go up .
18:08	If we increase the temperature , the rate of the
18:11	chemical reaction will go up and if we add a
18:14	catalyst the activation energy will go down and the rate
18:19	of the reaction will go up . So those are
18:22	some different ways in which we can increase the rate
18:25	of a chemical reaction . You can increase concentration ,
18:28	raise the temperature or introduce the catalysts to speed up
18:32	the reaction . Now , let's write down some equations
18:36	that you need to know in order to solve problems
18:39	. So you're already familiar with this one . The
18:41	rate constant K . Is equal to the frequency factor
18:45	times E . Race to the negative E . A
18:49	. Over artie . Here's another one that you should
18:53	be familiar with . The natural log of the very
18:57	constant K . That's equal to negative E . A
19:04	. Over our times . One over T . Plus
19:10	the natural log of ain . Now , in this
19:17	form , this is written in slope intercept form .
19:22	So why corresponds to Ellen K . The slope M
19:26	corresponds to negative E . A . Of art X
19:30	corresponds to one of the T . And the Y
19:33	intercept B . Is Ellen A . So if you
19:38	were to graph why versus X . Where why is
19:43	Ellen K . So we'll put that on the X
19:45	axis , I mean the Y axis and one of
19:49	the T . S . X . So we'll put
19:50	one of the T . On the X axis .
19:56	You're gonna get a graph that looks like this .
19:58	It's going to be a straight line plot . But
20:00	notice that the slope is negative . So the slope
20:05	of this line is going to equal negative E .
20:09	A . Over our and the Y intercept which this
20:15	is where the graph starts from . That's going to
20:19	be Ln . A . So for problems associated with
20:25	this equation , if you have a plot of Ellen
20:27	K versus one of the T . Just know that
20:30	you can find the activation energy from the slope of
20:33	the graph . And the white intercept Is equal to
20:39	the natural log of eight . So you could find
20:41	a frequency factor from the Y intercept . Now are
20:47	is 8.3145 jules Permal per Calvin . So because our
20:54	has the jewels and moles . The activation energy which
20:59	is typically reported in colleges promote when you use it
21:02	in these formless , it needs to be in jules
21:05	promote , so you need to convert so when reporting
21:10	the activation energy you want your answer to the intelligence
21:13	promote . But whenever you plug it into one of
21:15	these formulas you need to convert it to jules promote
21:18	. So make sure to be aware of that .
21:21	Yeah , now the smoke em is equal to negative
21:25	E . A . Over our . And the slope
21:29	is why two minus Y . One Divided by X
21:33	2 - . Excellent . Ellen K . Is on
21:36	the Y axis . One of her T . Is
21:38	on the X axis . So if Y is equal
21:41	to Ellen K . We could say that this is
21:43	Ellen K two - Ellen K one all divided by
21:49	one over T two -1 over T . one .
21:54	So that's going to equal negative E . A .
21:57	Over our . Now let's see if we have L
22:00	N A minus L N B . Using the property
22:03	of natural logs or logs , we can convert this
22:06	into a single log expression by writing it like this
22:10	. Ellen A over B . So we can write
22:13	this as a single long expression as well . We
22:15	can say that that's LNA . I mean that's Ellen
22:18	K two over K one And this is one over
22:25	T 2 -1 over team one . Mhm . And
22:30	that equals negative E . A . Over our .
22:35	Now if we want to get activation energy by itself
22:39	we can multiply both sides by negative art . If
22:45	we do that our will cancel on the right side
22:48	as well as the negative sign . Mhm . And
22:51	we'll get an equation for activation energy . So the
22:57	activation energy is gonna be negative art . Ellen K
23:03	two over K . one Divided by one over T
23:09	2 -1 of the T . one . So you
23:12	could use this equation to calculate activation energy . Now
23:16	you might see a variant of this equation . Sometimes
23:19	this negative sign will be distributed to one of the
23:22	T . Too modest one of the T . One
23:24	and that will basically flip The order of T .
23:27	one and T . two . So this is also
23:28	equal to positive art . Ellen K two overcame run
23:34	. And if you distribute the negative sign negative times
23:37	negative one of the T becomes positive for one of
23:40	the T . One . And then here this will
23:42	be negative So -1 of the T . two .
23:46	So you might see in that format to so just
23:49	be aware of that . But if you have the
23:50	negative sign it's going to be one of the T
23:52	. Two minus one of the T . One instead
23:55	of the reverse . Now let's go back to this
23:59	equation . Mhm . Make sure to write this equation
24:03	by the way because you're gonna need it to solve
24:05	problems . Now instead of multiplying both sides by negative
24:26	art , let's multiply both sides by this term .
24:31	If we do that we'll get a different equation we'll
24:37	get this one Ln K two Divided by K .
24:43	one is equal to negative E . A . Over
24:50	our . And we're going to move this to the
24:52	other side , Times one over T two -1 over
24:57	T . one . So that's another form of the
25:02	Iranians equation that you might see it in . Now
25:08	what we're gonna do at this point is we're gonna
25:10	put both sides of this equation on the exponents of
25:15	E . So I'm gonna put all of this on
25:20	the exponents of E . And all of this on
25:24	the X . Point of E . Because he is
25:28	the same and these two are equal to each other
25:32	. Both sides of this equation will be equivalent now
25:35	the base of a natural log Izzy . So on
25:37	the left side These two will cancel . And so
25:42	we're just gonna get K . two over K1 is
25:45	equal to everything that we see here . Now if
25:50	we multiply both sides by K . one We can
25:54	get an expression for K . two . So you
25:57	want to write down this formula because sometimes you need
25:59	to calculate a new very constant at a new temperature
26:03	. So that new rate constant K two is equal
26:05	to K . One times E . Raised to the
26:09	negative E . A . Over our Times one over
26:12	T 2 -1 of the T . one . So
26:17	this is the second equation that you want to write
26:18	down . So you have an equation that will help
26:21	you to calculate the activation energy . And this equation
26:24	will help you to calculate the new very constant given
26:28	a new temperature . Now let's go back to the
26:36	equation that we started with on this screen . All
26:43	right . That's as far as uh my software will
26:46	let me go back . Now sometimes you need to
26:56	calculate the second temperature . So what we're gonna do
26:59	is we're gonna isolate T . two to do that
27:02	. We're going to multiply both sides by negative are
27:05	over E . A . So those will cancel we're
27:17	going to have negative art . Ellen K two Divided
27:24	by K . one over E . A . That's
27:27	going to equal one over T 2 -1 over T
27:30	one . So we're going to move this to the
27:32	other side where it will become positive . So we
27:36	have run over T1 minus R . Ellen K .
27:42	two over K . one Overeat eh That's equal to
27:46	one over T . two . That we're going to
27:49	raise both sides to -1 . One of the T
27:53	two . Raised to the negative one power is just
27:55	T . Two . So we get this equation T
27:57	two is equal to one over T . One minus
28:02	R . Ellen K two Overcame 1 . Overeat eh
28:09	And this is raised to the -1 power . So
28:13	this will help you to calculate the second temperature if
28:16	you know the secondary constant . And if you know
28:19	the activation energy . So those are some formulas that
28:22	you want to use when solving problems associated with the
28:27	Iranians equation , activation energy , very constant and things
28:30	like that . Now let's go ahead and put these
28:33	equations to good use . So we're gonna work on
28:36	a few practice problems . So take this one for
28:40	example , A certain reaction has an activation energy of
28:43	50 killer juice promotes The rate constant at 25°C is
28:48	.0039 Minutes to the -1 . What is the value
28:53	of the rate constant ? At 75°C. . So first
28:59	let's make a list of what we know . So
29:02	we have the activation energy . It's 50 kg ,
29:05	promote The rate constant which will call K . one
29:10	is .0039 . And that's why the temperature of 25°C,
29:16	, which we need to convert to Kelvin . So
29:19	add 73-25 . That gives you 2 98 Kelvin .
29:23	Mhm . Our goal is to calculate K tune And
29:27	the second temperature is 75 plus 2 73 . So
29:31	that's 348 Calvin . So what is the formula that
29:35	will help us to calculate K . 2 ? It's
29:40	K . two is equal to K . one times
29:43	E . Raised to the negative E . A .
29:46	Over our Multiplied by one over T two -1 over
29:51	T . one . So any time you need to
29:56	find a very constant at a different temperature . That's
30:00	the formula that you want to commit to memory .
30:03	Now came one we know 2.0039 . The activation energy
30:08	has to be in jewels . So we got to
30:11	multiply 50 by 1000 . So it's gonna be 50,000
30:15	and then it's one over T . Two which is
30:17	348 -1 over T . one which is 2 98
30:22	. And then divide that by our Our is 8.3145
30:27	jules promote per Calvin . You don't want to use
30:30	the gas constant for our .0826 . It's not going
30:34	to work for this equation . So if you type
30:46	this entire thing the way you see it , what
30:50	you doing one step at a time , you should
30:52	get this answer . K two is .0708 . So
30:59	notice that K two is greater than K one .07
31:04	is greater than .003 . Now this answer makes sense
31:08	because the second temperature is higher than the first temperature
31:13	. And as you increase the temperature of a reaction
31:18	, the rate constant will increase . When the rate
31:20	constant goes up , the rate of the chemical reaction
31:23	will go up as well . And so that's it
31:26	for this problem .

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