00:00:00
right so this video is going to cover um
00:00:02
the A2 kinetics topic from uh unit four
00:00:06
um kinetics is really it's just sort of
00:00:08
a fancy term to describe rates of
00:00:10
reaction which is a topic that comes up
00:00:13
kind of really heavily within within
00:00:14
chemistry gtsc as where you looked at
00:00:17
things like Max Maxwell boltzman
00:00:19
distribution curves and then
00:00:22
A2 where it gets a little bit more
00:00:25
tricky um but actually once you've got
00:00:28
the kind of General Concepts um the the
00:00:32
ideas behind it are quite repetitive in
00:00:34
the way the exam questions work and so
00:00:36
actually you can quite quickly get to
00:00:38
grasp it and and get those extra marks
00:00:40
in the exam
00:00:42
so um
00:00:48
kinetics uh as I said we're essentially
00:00:51
looking at
00:00:52
rates of reaction and that's going to
00:00:55
play a huge huge role in uh in this
00:00:59
entire video and in all the calculations
00:01:01
that you do or when you have to do the
00:01:02
calculation part of it it's not all
00:01:04
based around
00:01:05
calculations uh but it does have quite a
00:01:08
heavy the calculation do have quite
00:01:10
heavy influence um on on actually the
00:01:13
way this sort of topic
00:01:16
works the the key thing here uh and
00:01:18
there's some sort of the idea what I'm
00:01:21
going to come on to particular when to
00:01:22
start talking about things like orders
00:01:24
and that I'll make sense hopefully in a
00:01:25
couple of minutes is the idea that and
00:01:28
back to GC read and I guess also I is
00:01:30
the idea that when you change
00:01:31
concentration you can or we've normally
00:01:34
said that we we would change the
00:01:36
reaction rate but we can change the
00:01:37
reaction rate um if we change the
00:01:40
concentration particularly when we're
00:01:41
looking at the the reactants involved
00:01:44
obviously generally sort of speaking
00:01:46
it's not necessarily going to be the
00:01:49
products that we're going to be looking
00:01:49
at although I guess depends on the kind
00:01:51
of question but looking at those
00:01:53
reactants so we could think of a generic
00:01:54
sort of reaction of a plus
00:01:57
b which reacts to ruce C so a plus b
00:02:02
react to produce C now if I just
00:02:04
concentrate on a initially within that
00:02:07
we could say
00:02:09
well I
00:02:11
guess changing the concentration of a
00:02:14
could have nothing to do with the rate
00:02:16
of the reaction that is occurring it's
00:02:18
the speed that this react changing the
00:02:20
concentration of a has no effect at all
00:02:24
um so that's our first
00:02:27
option no effect
00:02:30
this is changing the concentration of a
00:02:32
I should make that very clear
00:02:35
changing con of
00:02:38
a so it could have absolutely no effect
00:02:40
at all
00:02:42
second we could change the rate we could
00:02:45
change the concentration of a and it
00:02:46
could have an effect um so an effect and
00:02:51
that effect can really come in sort of
00:02:52
two ways we could say that the rate is
00:02:58
directly proportional to the
00:02:59
concentration of a um IE or I should say
00:03:04
sorry EG I
00:03:09
guess
00:03:11
double concentration of
00:03:15
a double the rate that would be a direct
00:03:19
proportionality between the two whatever
00:03:21
you do to a the concentration of a is
00:03:23
going to happen to the rate so you have
00:03:25
concentration of a you will have the
00:03:27
rate of the reaction Etc
00:03:30
um square brackets U if you are new to
00:03:33
those ideas depends if you've done any
00:03:35
of the other sort of topics square
00:03:36
brackets just uh a shorthand way really
00:03:38
of of saying concentration and that's
00:03:41
concentration in moles per decimet
00:03:45
cubed so we've
00:03:48
got this idea then that we rate could be
00:03:50
proportional to a we could also say that
00:03:53
the
00:03:54
rate is
00:03:57
proportional to a terrible draw on there
00:04:00
a squ uh and what we mean by that I know
00:04:03
where this arrow is going either what I
00:04:05
mean by that is that again if we say
00:04:08
that double
00:04:12
A in this case we
00:04:14
would
00:04:16
quadruple the
00:04:18
rate and these are really the three sort
00:04:22
of kinds of effect you're going to find
00:04:25
within the A2 so potentially no effect
00:04:29
at all or an effect and that effect is
00:04:31
going to come in two FS if you like one
00:04:34
where the rate is directly proportional
00:04:35
one where it's where it's proportional
00:04:37
to this a concentration
00:04:39
squared now if we say if we ignore the
00:04:42
no effect from minut and we'll go into
00:04:43
this one now whenever we have this
00:04:45
proportionality here we can actually
00:04:46
change that we can stick in a constant
00:04:48
so this is the same as really writing
00:04:50
here change the color we could say that
00:04:53
actually in this
00:04:56
case rate equals k
00:05:00
a and in this case we could say the rate
00:05:04
equals
00:05:06
K A squ and that's just taken
00:05:10
this um this this equation this this
00:05:14
whatever you would call this that's just
00:05:15
changed it basically got rid of the
00:05:16
proportionality sign stuck a K in and
00:05:18
then an equals rest Remains the Same
00:05:20
rate is equal now to k multip a squ now
00:05:23
I've ignored B obviously and I don't
00:05:24
want to start bringing B in at this
00:05:25
point um I'll do that in in a minute or
00:05:28
two's time um it's easier to do with one
00:05:30
and sort of think of these
00:05:31
ideas now to add onto that slightly
00:05:35
more the idea of orders of a reaction so
00:05:39
if we were talking about the order of
00:05:40
the reaction with respect to a no effect
00:05:45
is class as a zero
00:05:48
order an effect of this sort doubling
00:05:51
and doubling Haring and Haring this is a
00:05:54
first that be respect um order with
00:05:57
respect to a would be first order this
00:06:01
one would be a second order so
00:06:05
first
00:06:07
second and
00:06:09
really that links
00:06:12
into the indices that are that are next
00:06:15
to it if you imagine and I'll try and
00:06:17
explain this really clearly now you
00:06:20
could say that if a had no effect we
00:06:23
could say rate equals K of a
00:06:27
concentration of a to this zero and this
00:06:29
Z there is this
00:06:32
order well a to the 0 equal 1 therefore
00:06:38
there is no effect at all concentration
00:06:40
a is not going to do anything it's
00:06:41
always going to end up being one it's
00:06:43
going to affect the re re in no way at
00:06:44
all if we go here we could rewrite this
00:06:47
and we could say rate equals
00:06:50
k a to the one and that's come from
00:06:53
there the order is being shown
00:06:56
there well in this case yes
00:07:00
a to the^ of one is a so the rate is
00:07:03
equal to K multipli by a over here final
00:07:08
one
00:07:10
k a s and again that second order being
00:07:14
shown as the squaring there and this
00:07:16
again we can see now rate equals K
00:07:19
multipli by a squ concentration of a s
00:07:22
so these are classes the order of the
00:07:24
reaction if you're feeling a bit unsure
00:07:26
about this now as as the video goes on
00:07:28
it hopefully will make a little bit more
00:07:29
sense sense um with relation to sort of
00:07:33
what's actually kind of happening
00:07:35
here so as I said I've ignored a well
00:07:39
let's let's include sorry I've ignored B
00:07:41
I've definitely included a um let's take
00:07:44
this a little bit further let's say well
00:07:45
let's include B and let's just make a
00:07:48
really generic rate equation and what I
00:07:49
mean by that this is this is an equation
00:07:52
that um a rate equation that governs
00:07:55
this particular reaction that we're
00:07:56
talking about so just make this up and
00:07:57
I'll say the rate equation for the
00:07:59
reaction above um so this is the rate
00:08:06
equation for this reaction here I'm just
00:08:10
making this up okay this isn't no idea
00:08:14
you know really if this is true or not
00:08:15
obiously it's just a and b they don't
00:08:17
really
00:08:18
exist so the rate is equal to k multip a
00:08:21
s multiplied by B oh square bracket
00:08:24
there don't forget those bad what this
00:08:27
tells me then is that if I to double the
00:08:30
concentration of
00:08:34
a square brackets again I'd
00:08:37
quadruple the rate and if I were to
00:08:42
double concentration of B I would double
00:08:47
the rate because in this case the
00:08:50
reaction is first order with respect to
00:08:54
B so first
00:08:58
order with with
00:09:01
respect
00:09:03
to B just call it B if you like whatever
00:09:06
uh and this one
00:09:09
is second order again with respect but
00:09:12
in this case to a because we're obvious
00:09:14
talking about
00:09:16
a one fun little bit before we move on
00:09:19
actually look at some sort of numbers
00:09:20
which I think clarifies this slightly
00:09:22
you could give an
00:09:25
overall
00:09:28
order now in this case and in every case
00:09:30
actually the overall order is achieved
00:09:34
by adding together the individual orders
00:09:37
of the of the reactants or the the
00:09:41
components of the rate equation now in
00:09:43
this case we've got a second order so
00:09:44
we've got a two we've got a first order
00:09:47
we' got a one therefore overall order
00:09:49
here is equal to three that occasionally
00:09:53
has come up in
00:09:54
questions not a huge sort of amount
00:09:57
really but it's a fairly straightforward
00:09:59
if you do get it it's worth a Mark um
00:10:01
you've just got to add the numbers
00:10:02
together there so I mean another example
00:10:04
we could say um just to check on that
00:10:07
one rate is equal to c and d
00:10:14
and just both would be in first order of
00:10:17
respect to C and and also of D in this
00:10:19
case the overall order would be one + 1
00:10:24
therefore two so fairly simple stuff
00:10:27
adding these numbers together hopefully
00:10:29
nothing particularly difficult um there
00:10:32
it's a strange thing this idea of sort
00:10:34
of orders and the rest but it is just
00:10:35
hopefully I've kind of explained that in
00:10:37
in enough detail um as I've gone
00:10:41
up there is a a kind of a trick I guess
00:10:45
you could see when it comes to sort of
00:10:47
these rate equations um and that falls
00:10:49
into the following that imagine the
00:10:52
equation um let's A + B + C goes to D +
00:11:00
e what if this equation be for this what
00:11:04
I could say is and again I'm just making
00:11:05
up a rate equation I could set the rate
00:11:07
equation for this reaction is equal to
00:11:12
rate um
00:11:20
K and that would be a completely
00:11:22
legitimate um rate equation there the
00:11:25
what you we notice is that actually B
00:11:26
has not been included and the reason
00:11:28
there is is that
00:11:30
b does
00:11:33
not
00:11:35
affect rate or rather I I guess the
00:11:39
concentration of B does not affect the
00:11:42
rate so in that
00:11:45
case uh we could say
00:11:48
zero order there in the case of a first
00:11:52
order in the case of C it would be
00:11:55
second
00:11:57
order you're not always going to be get
00:11:59
get this sort of nice idea of having the
00:12:01
rate equation sometimes you will have to
00:12:03
make the rate equation which obviously
00:12:05
means you need to know what's involved
00:12:06
in the reaction what's Taken part and
00:12:07
all the rest and that's up to you to
00:12:08
deduce that uh which is hopefully what
00:12:11
I'm going to cover now um which I'm
00:12:16
hoping will make some sense so let's
00:12:18
have a look at this table here so this
00:12:21
table then shows
00:12:23
you four experiments I've taken this
00:12:26
from uh chem revised
00:12:31
.org rather than just make my own which
00:12:33
is a great website for um for notes
00:12:36
really if you are looking for actual
00:12:38
sort of hard copies of notes just print
00:12:39
them off and there they're all PDFs I
00:12:41
think but they're really good
00:12:43
website so four
00:12:46
experiments each these experiments we
00:12:49
have a rate here given and the rate is
00:12:52
mole per decimet cubed per second and
00:12:55
often this rate is denoting the loss of
00:12:58
something
00:12:59
um so the rate of decline so the the the
00:13:02
rate the concentration of something fall
00:13:04
that something Falls for example now
00:13:06
don't worry too much about what that is
00:13:07
but it's essentially just the rate and
00:13:09
so the higher the rate the the faster
00:13:12
the reaction which hopefully makes sense
00:13:15
we've got four experiments and you'll
00:13:16
notice that we've got three things we've
00:13:19
got a reaction that
00:13:20
involves
00:13:22
a b and c which is going to something
00:13:26
doesn't matter what it's going to it
00:13:28
doesn't affect the at all and you'll
00:13:30
notice that with each one of these
00:13:31
columns this is the concentration of a
00:13:32
in experiment 1 2 3 4 etc etc across to
00:13:36
including conservation of B and of C as
00:13:39
well so you'll notice these numbers are
00:13:43
some of them are the same some of them
00:13:44
are changing and this is very important
00:13:46
because the purpose of this is to use
00:13:49
this information or rather using this
00:13:52
information you could deduce your own
00:13:54
rate equation from what is here and they
00:13:57
do like to do this in exams they love to
00:14:00
give you this to do and actually once
00:14:02
you actually kind of think about what
00:14:03
could I do and some of you might be
00:14:05
watching this think and actually oh yeah
00:14:06
I understand exactly what I'm going to
00:14:07
do there um some may not the idea is
00:14:11
you're looking
00:14:12
for try and word this well you looking
00:14:15
to if you can keep two of them the same
00:14:18
we hold on let me start again you're
00:14:20
looking to compare two
00:14:22
experiments uh and you're looking to
00:14:24
start with one reactant at a time so
00:14:26
we'll start with a now we want the other
00:14:29
two to remain the same across those two
00:14:31
experiments uh whereas we want the one
00:14:33
that we're investigating to change so
00:14:36
for
00:14:37
example it works in this case that one
00:14:39
and two are are good experiments to
00:14:42
choose for a because these remain the
00:14:45
same 0.5 0.5 and this also Remains the
00:14:49
Same there 0.25 0.25 whereas a doubles
00:14:54
in experiment one we had a concentration
00:14:56
of 0.1 experiment two we had a
00:14:57
concentration of 0.2
00:15:00
now this comes back to thinking about
00:15:01
what we did earlier on it's the idea
00:15:02
that when I double the concentration of
00:15:05
a I then need to look I ignore these two
00:15:07
I'm looking directly now at the rate of
00:15:09
the reaction and I'm looking at how does
00:15:10
that affect the rate of reaction so I
00:15:12
I've doubled from 0.1 to 0.2 the
00:15:14
concentration of a oh well look the rate
00:15:17
has doubled from 0.1 to 0.2 it could go
00:15:20
from 0.2 to 0.4 the same thing is true
00:15:23
it has doubled therefore we could say
00:15:25
that again the a of course is proportion
00:15:29
therefore to the rate or vice versa rate
00:15:32
is proportional to a uh which is
00:15:35
actually a much better way to write it
00:15:36
this that's currently it's terrible run
00:15:38
it that way around so the rate is
00:15:41
proportional to a but of
00:15:44
course we could say there the rate
00:15:46
equals k a so we know in this case here
00:15:51
that a is first order so we could say
00:15:55
well let's write start writing a rate
00:15:57
equation we know that K is going to be
00:15:59
involved and we know that a is going to
00:16:01
be in there and it's not going to have
00:16:02
any IND to C there at all we look at B
00:16:05
next and then we'll look at
00:16:06
C so
00:16:10
B again same concept we need to keep
00:16:14
ideally keep the other two the same
00:16:16
across two experiments any two
00:16:19
experiments and look for B to have a
00:16:21
change now I can see that b will change
00:16:23
between here and here it doubles oh
00:16:26
brilliant this Remains the Same
00:16:29
0.1 0.1 0.25 again
00:16:35
0.25 when I
00:16:37
double the initial the well the
00:16:39
concentration of B I go from 0.5 to one
00:16:43
I quadruple the rate therefore B is
00:16:47
second order so I stick in my B here
00:16:51
stick up a little two there because it
00:16:53
is second
00:16:54
order finally we'll look at C try and
00:16:57
find another color running out now
00:16:59
difference would be between this one
00:17:01
then and this one uh where here we have
00:17:04
0.5 0.5 remaining constant there so we
00:17:08
double from 0.25 to 5 the concentration
00:17:12
C oh but look the rate Remains the Same
00:17:15
therefore this is zeroth I'm going to
00:17:17
call it zeroth no that's I think that's
00:17:19
right it's zeroth order with respect to
00:17:21
C therefore C is not included in the
00:17:24
rate equation so for this data given
00:17:28
there the rate equation would
00:17:30
be rate equals K multip the
00:17:34
concentration of a multip the
00:17:35
concentration of B squared and that is
00:17:38
the rate equation
00:17:41
done now through this you may have been
00:17:44
thinking yourself or you may not have
00:17:45
done it's not really any problem if not
00:17:49
what happens if you can't find an
00:17:51
experiment
00:17:53
where the other two that you are not
00:17:56
investigating or the other one the other
00:17:58
two and this this case the other one or
00:18:00
whatever remains constant when you're
00:18:02
changing the concentration of your
00:18:04
other um and exper an example would look
00:18:08
something like this so in this example
00:18:10
what we've got
00:18:11
then is a more tricky one to sort of get
00:18:15
your head round I guess in
00:18:18
that well initially perhaps wouldn't
00:18:20
know it's only once you start looking at
00:18:21
the data I mean we've got more weird
00:18:22
numbers here not really an issue doesn't
00:18:24
matter times 10 to Theus 6 don't care
00:18:26
ignore those we'll just treat not .5 0.3
00:18:29
and 2.4 three experiments this time
00:18:32
we've only got X and Y so we've got a
00:18:33
reaction that involves x + y going to
00:18:37
something we don't know what doesn't
00:18:38
really matter no one
00:18:39
cares okay so let's look at X I want to
00:18:43
investigate the
00:18:44
order with respect to X well brilliant
00:18:48
the first two here experiment one and
00:18:49
two y remains constant X doubles so it
00:18:52
doubles from 0.05 to 0.1 oh look the
00:18:55
rate doubles so we know that this rate
00:18:59
not written very well that really if I'm
00:19:02
honest we know the rate of the of the
00:19:05
reaction is going to have with with
00:19:07
respect to or the order I should say of
00:19:10
X is is first order so we could put a
00:19:13
one there if you like but we obviously
00:19:14
don't need
00:19:16
to so it's first order with respect to X
00:19:19
now we got to look at Y and we got to
00:19:20
say right let's look at y so the only
00:19:21
two we can use is this one and this one
00:19:25
there's a problem though this
00:19:29
is
00:19:31
quadrupling in the space this is
00:19:34
doubling now that's an
00:19:37
issue because that's not
00:19:40
constant the best thing to do about this
00:19:43
is if you know this is first order
00:19:45
because we've already decided that then
00:19:47
if this was to be quadrupled the rate
00:19:49
would be quadrupled so ignoring why for
00:19:51
now the initial rate here
00:19:54
0.15 when we quadruple it here we go to
00:19:57
2.4
00:19:59
now
00:20:01
if this alone were affecting the rate we
00:20:05
would expect
00:20:07
0.15 * 4 to go to
00:20:12
0.6 but we don't see
00:20:15
that that would only be true if this
00:20:18
reaction
00:20:19
were zero zeroth order with respect to Y
00:20:24
if y had no effect at all we would
00:20:25
expect to see this go from 0.15 0 six
00:20:29
that's clearly not not the case so we
00:20:30
know for a fact now why can only it can
00:20:34
be 0 one or two it cannot be zero now
00:20:38
can it be first
00:20:40
order we would expect this to go to 0.6
00:20:42
well if this was first order then
00:20:44
doubling this would double the rate so
00:20:46
we would expect this
00:20:48
0.6 to go to 1.2 well hold on it doesn't
00:20:51
it goes to 2.4 so therefore it can't be
00:20:54
a first order it must therefore be the
00:20:57
second order and when you look at this
00:20:58
you can kind of get you can kind of
00:21:00
understand it treat each one
00:21:02
individually to an extent do the
00:21:04
calculations one and then the other this
00:21:06
would be times four taken to 0.6 this
00:21:08
one though when it's
00:21:10
doubled this then goes up by four so we
00:21:14
go from 0.6 up to
00:21:16
2.4 that must mean this is second order
00:21:19
so if we were to finish the equation
00:21:20
then it's going to look like that K
00:21:25
multip by x^ of one or we can ignore
00:21:27
that doesn't matter multip by y^ of 2
00:21:31
y^2 so it's a more difficult example
00:21:33
this one it is more tricky and again the
00:21:35
key thing here work out one you will
00:21:38
always be able to work out one of them
00:21:40
easily and in this case it was working
00:21:42
out X first so we knew that X was first
00:21:45
order looking at y knowing that X is
00:21:48
first order we know what we would
00:21:50
expect if this was only the only thing
00:21:52
having the effect but of course it's not
00:21:55
because the way the the reaction changed
00:21:56
it's not 0.6 we would expect it's not
00:21:59
one 1.2 which would be true if this was
00:22:02
a first order it therefore must be
00:22:04
second order because it quadruples the
00:22:06
.6 that we would expect sorry it
00:22:09
quadruples the 0.6 from X up to 2.4 not
00:22:13
sure I've explained that brilliantly
00:22:15
well there towards the end um hopefully
00:22:17
that makes some sense and that actually
00:22:19
isn't too too much sort of trouble
00:22:24
um the final part really is calculating
00:22:27
is calculating various things you could
00:22:29
be asked to calculate K you could be
00:22:31
asked to calculate the and all the rest
00:22:33
and actually to do this I'm going to
00:22:34
show you a p
00:22:36
paper okay so this is a paper from
00:22:38
January 2013 and it's unit four paper
00:22:43
um now what you've got here you can see
00:22:46
that in this case our try where the pen
00:22:50
is oh it's there there we
00:22:52
go in this case we have a a reaction of
00:22:57
it looks like probably B and C there um
00:23:00
so B and C react and we're given a table
00:23:03
similar to what I showed you before
00:23:04
where we have an experiment here where
00:23:06
we have one and two we have an initial
00:23:08
concentration of B and of C between
00:23:10
different differing between the two uh
00:23:12
we have an initial rate there and we
00:23:13
need to calculate this one here so what
00:23:17
do we do well we need this data here
00:23:20
calculation wise what are we going to do
00:23:26
well initially
00:23:28
we're looking to calculate k for
00:23:31
experiment one and I've lost the pen
00:23:33
again oh there it is so looking to
00:23:35
calculate k for experiment one so we
00:23:37
given this equation
00:23:41
here this needs to be rearranged so
00:23:43
that's going to be your first step so
00:23:45
rearrange this to make K the subject so
00:23:47
K is going to be equal
00:23:51
to B
00:23:53
squared no it's not ignore that entirely
00:23:57
terrible K is going to be equal to
00:24:00
rate
00:24:04
over B ^2
00:24:06
multiplied by
00:24:09
c um so K so I've divided this side by
00:24:11
b^ S multip by C obviously therefore
00:24:13
goes under there putting in actual
00:24:15
numbers into this we're dealing with
00:24:17
experiment one as it says in the
00:24:18
question so rate well that's given to us
00:24:22
as 8.4 * 10- 5 okay b^ 2 well B is given
00:24:29
to us as 4.2 * 10- 2 C we are told is
00:24:37
2.6 *
00:24:40
10us 2 just bracket that to make it look
00:24:42
a little bit clearer or you go W into a
00:24:45
calculator and you're going to get an
00:24:47
answer that's going to be your value 4K
00:24:49
and that answer in this case comes out
00:24:50
to be
00:24:54
1.83 now if we scroll down here
00:24:56
obviously you'd get more of this in
00:24:58
terms of the paper you seal see a bit
00:25:00
more at once so asking us for the units
00:25:03
as well so we need to calculate units
00:25:05
for this I've lost the pen again oh
00:25:06
there it is okay so K units so let's
00:25:10
have a look we
00:25:21
have over
00:25:25
mole and I would do it like this I think
00:25:27
it make makes it a little
00:25:32
bit clearer oh I would say
00:25:37
anyway so we've got mole decimeter cubed
00:25:40
per second oh dodgy line oh oh there we
00:25:44
go mle decim Cub per second over mle per
00:25:47
decim cubed multip by mole per decim Cub
00:25:49
multip by mole per decimeter cubed and
00:25:51
I've lost the pen again there it is Tiny
00:25:54
um so what can we do well we can cancel
00:25:56
some stuff for a start so we know that
00:25:59
this will cancel with one of these
00:26:02
leaving us with basically this bit here
00:26:05
so you've essentially got second to the
00:26:08
minus one over these two that multiplied
00:26:12
by that with any sort of Indy kind of
00:26:15
calculation isn't you you've just got to
00:26:17
sort of know how to do these really um
00:26:20
multiply the moles by each other or as
00:26:22
you would expect it's going to give you
00:26:23
mole squared
00:26:26
DM multipli by DM dmus 3 m DM to the
00:26:31
minus 3 um is going to
00:26:35
be um DM to the minus 6 when you
00:26:38
multiply indices if you're not sure of
00:26:40
this multiply indices you add them
00:26:42
together so it's minus three add Min -3
00:26:45
which is going to give us minus 6 um the
00:26:47
mole it's if you imagine it's mole to
00:26:50
the one multip by mole to the one is
00:26:51
going to give you mole to the two
00:26:53
because this is over that um to give it
00:26:56
K into sort of to remove I guess the um
00:27:02
the element of of the fraction here I've
00:27:05
lost 10 again oh there it is um we're
00:27:10
basically going
00:27:12
to stick minuses in front of these so
00:27:15
what we're going to end up with
00:27:19
is mole to minus
00:27:22
2 dm6 basically minus and a minus can
00:27:25
end up with a plus second to Theus - 1
00:27:28
and they are going to be your units in
00:27:30
this case it's 1.83 mo- 2 dm6 secondus
00:27:36
one okay next bit calculate the value
00:27:39
for the initial rate in experiment two
00:27:41
if you can find where the pen is I I
00:27:44
need to work out how this is going to
00:27:46
work honestly made it any smaller um so
00:27:51
calculate value the initial rate in
00:27:53
experiment two so this time nice easy
00:27:55
one rate equals k um and it was try and
00:28:00
remember this one it
00:28:02
was e
00:28:04
s multip by C obviously on your paper
00:28:07
you would be able to look up a little
00:28:08
bit clearer and see that
00:28:10
there this time we're put numbers in and
00:28:12
we're going to calculate the rate sorry
00:28:15
I wasn't supposed to be
00:28:17
there K we can use from the previous
00:28:20
question because the the K is is is is a
00:28:24
constant so it's going to remain
00:28:25
constant in this it changes with things
00:28:26
like um temperature uh amongst other
00:28:30
things but for this
00:28:32
initial because we're talking
00:28:35
about or you can assume same conditions
00:28:37
betweener exper American experiment one
00:28:40
and experiment two the the constant
00:28:42
there for this particular reaction is
00:28:43
going to remain the same so in that case
00:28:46
you can use uh the constant here so
00:28:50
we've got
00:28:53
1.83 we've got to actually go up and get
00:28:55
our values so B for experiment two
00:28:58
remember not experiment 1 be 6.3 * 10 -
00:29:03
2 6.3 * 10-
00:29:07
2 squared multiplied by 7.8 * 10- 2 7.8
00:29:14
* 10 - 2 and that's going to give us an
00:29:18
answer of
00:29:21
5.67 * 10 to
00:29:24
Theus 4 uh in terms of rate we know
00:29:28
our units here mole DM cubed oh that's a
00:29:32
terrible three seconds minus
00:29:35
one okay so a bit of calculation that's
00:29:38
just sort of standard thing you expect
00:29:40
to be able to do and actually really
00:29:41
it's just putting numbers into into into
00:29:44
the equations given or equations that
00:29:45
you
00:29:47
found going down so this is one of those
00:29:50
ones that I showed you before so we've
00:29:51
got three more experiments here um this
00:29:54
taste we're looking at uh DNE so DNA at
00:29:58
at a constant temperature notice that's
00:29:59
a key thing it tells us the rate's going
00:30:01
to remain so not the rate soor the
00:30:02
constants going to remain um true or
00:30:05
remain equal for uh for all of
00:30:07
them so experiment we're looking for the
00:30:10
order of reaction with respect to D so
00:30:13
we're looking ready to keep these the
00:30:14
same excellent straight away I can see
00:30:16
that we kept them the same that's
00:30:18
perfect now whilst these remain the same
00:30:21
between those this goes up by three now
00:30:26
our initial reate initial rate our
00:30:29
initial rate if it was Zero order it
00:30:32
would not change so we know that it is
00:30:34
not zero order if it were first order
00:30:38
this 0.26 would be multiplied by three
00:30:40
and that would give us our value now I'm
00:30:42
pretty sure that is's not multiply by
00:30:45
three because I believe that multiply
00:30:46
three you would get
00:30:50
0.78 that's not true either so it's not
00:30:53
first order so it must therefore be
00:30:55
second order um in terms of second order
00:30:57
just to check that you
00:31:01
essentially this multip by three
00:31:04
remember because of the squared as this
00:31:07
If This Were say doubled the the rate
00:31:09
would be quadrupled this is tripled
00:31:11
therefore the rate is nine duped um
00:31:15
which isn't a word but essentially the
00:31:17
rate goes up by nine that's what we
00:31:19
would expect this was second order and
00:31:21
looking at these numbers I'm pretty
00:31:24
happy that that is going to be the case
00:31:25
yeah so 0.26 basically
00:31:29
time 10 minus 3 multip by 9 you should
00:31:33
get absolutely perfectly 2.34 * 10 to
00:31:37
Theus 3 a fairly easy um granted only
00:31:42
one Mark there but you know a reasonably
00:31:44
easy sort of Mark to get to be honest um
00:31:48
now we're looking at it with respect to
00:31:49
e now notice straight away this is one
00:31:52
of those dodgy ones this is one of those
00:31:54
that we looked at and we're like oh bit
00:31:57
weird we're looking at respect to e
00:32:00
again if I can find this pen I need to
00:32:03
work this out right there we go
00:32:06
oh right respect to e now in this
00:32:11
case look for E changing so here we go
00:32:15
between three and five and E is doubling
00:32:18
in that time so we're now looking
00:32:21
between three and five this one is
00:32:22
between three and five now this goes up
00:32:24
quite a lot this I would guess is is
00:32:28
going up what uh six times okay this is
00:32:31
going up six
00:32:35
times now we would expect if this is
00:32:38
increasing six times then we've got to
00:32:40
initially look at this before we even
00:32:42
think about looking at e if this alone
00:32:44
was to be responsible then we would
00:32:47
multiply the
00:32:50
0.26
00:32:51
by
00:32:55
six 36 he says realizing this error
00:32:59
again because of the fact that it's
00:33:00
squared so if this was doubled we would
00:33:03
quadruple the rate if this was tripled
00:33:05
we would nine duple the rate it's not
00:33:07
it's six it's six so we 36 the rate now
00:33:12
in terms of looking that that on a
00:33:14
calculator uh if we work that out we
00:33:16
actually get 2
00:33:18
9.36 * 10 to the minus 3 that means that
00:33:24
this must be zeroth order because this
00:33:26
alone is C cusing that reaction or the
00:33:29
rate change this goes up by six rate is
00:33:32
increased by by a factor of 36 therefore
00:33:35
e is just zeroth order so order of
00:33:39
reaction with respect to e zero there we
00:33:44
go okay question one continues on the
00:33:46
next page more question
00:33:48
one this is the other type of question
00:33:51
that you can get when it comes to um
00:33:53
particular of the kinetics aspect and
00:33:56
this is one rather than going through it
00:33:59
um on the on the bit before I thought it
00:34:01
best to Really Leave it to
00:34:02
this this is a question which talks
00:34:06
about sort of um processes and and I
00:34:08
guess what's called the rate determining
00:34:10
step so this phrase that you've
00:34:13
got um here so the rate determining step
00:34:18
ultimately the rate determining step is
00:34:20
the step that determines the rate so
00:34:22
it's not particularly sort of inventive
00:34:25
um as to what the um
00:34:28
the way they've worded it and it works
00:34:30
in Reactions where we have uh more than
00:34:33
one step so one that occurs in stages
00:34:35
and a lot of reactions do occur in
00:34:37
stages although they can be written as
00:34:38
an overall equation they do occur as as
00:34:41
multiple points and you can see this one
00:34:42
here the overall reaction is this one
00:34:47
here um but it can be broken down or it
00:34:50
suggested that there's there's there is
00:34:51
a two-step process that that could be
00:34:53
being followed where we have this
00:34:55
reaction uh breaks down to form the
00:34:57
carbocation bromo etc etc now it says
00:35:00
deduce the rate determining step in this
00:35:02
two-step process okay so what it's
00:35:04
saying is the reaction was found to be
00:35:07
first order with respect to
00:35:09
ch33 CBR so first order
00:35:13
here but zero order or zero order with
00:35:17
respect to the
00:35:19
hydroxide ions now one of the important
00:35:23
things here is the rate determining step
00:35:25
is the one that the rate equation uh
00:35:29
will adhere
00:35:30
to okay so rate determining step
00:35:36
uh I don't want to say equals rate
00:35:38
reaction but this is the so the rate
00:35:40
equation will
00:35:41
apply to this step and not to the other
00:35:45
steps so in terms of the rate equation
00:35:47
here we know that our rate equation is
00:35:50
basically involving this first order
00:35:53
there zero order there so our rate
00:35:55
equation would be something like POS
00:35:57
possibly could be k
00:36:02
um
00:36:05
ch3 3 C
00:36:08
BR um Clos brackets there that could be
00:36:11
our our rate equation in terms of the
00:36:13
steps well first step here step
00:36:17
one we'll start with step two because
00:36:20
the answer is step one step two just
00:36:23
can't be that because we know that the
00:36:25
hydroxide ions have zero order so this
00:36:28
step could not be determining the rate
00:36:30
of the reaction because actually
00:36:31
changing the concentration of the
00:36:32
hydroxide ions does nothing to the rate
00:36:35
therefore nothing can to happen there uh
00:36:37
and this being this sort of the chyoa IR
00:36:39
again isn't part of we're not told
00:36:41
anything about that um so that's not
00:36:43
going to be involved Diva this one
00:36:44
though step one the ch33
00:36:48
CBR actually we can see that is part of
00:36:51
our we know that is first order so if we
00:36:53
changed if we
00:36:55
hared the concentration of this and we
00:36:58
you know and we drop this down to you
00:37:03
know I don't know to a very very low
00:37:05
amount actually we find
00:37:08
that this would have an effect changing
00:37:10
the concentration of this guy would have
00:37:14
an effect whereas this wouldn't again
00:37:15
from the information given to us zero
00:37:17
order to respect the hydroxide ions
00:37:19
therefore this here so step one must in
00:37:23
fact be the rate determining step
00:37:24
because it is the one that the rate
00:37:26
equation which we can fix figure out
00:37:28
from the information given that's what
00:37:29
the rate equation aderes to has been a
00:37:32
bit of Whistle Stop tour there of
00:37:33
kinetics right through trying to explain
00:37:35
a bit about what sort of the um the idea
00:37:37
of the orders of the reactions are uh
00:37:39
then going through
00:37:41
and bit about how you can work out
00:37:44
orders bit of calculation there as well
00:37:45
finally with um this little guy here so
00:37:49
hopefully that's been of some help um
00:37:51
any problems do leave comments or get in
00:37:53
touch
