in this video we're going to focus on reaction mechanisms how to write the rate law for the overall reaction and things like that so reaction mechanism is basically a step-by-step pathway by which a particular reaction occurs and each individual step in a reaction mechanism are known as elementary reactions so let me give an example so let's say this is the first step of a certain reaction a plus b becomes c plus d so i'm gonna put a one here to indicate that's step one now step two this is gonna be d plus e and that
turns into f plus b so the series of these two steps is known as a reaction mechanism each individual step is an elementary reaction and we can write the rate log expression for each elementary reaction based on the coefficients that we see for the reactants of that reaction so let's say if we want to write the rate law expression for the first step of this reaction mechanism notice the coefficients of a and b it's one to one and we're going to call the rate constant associated with the first step k1 the rate constant for the
second step we'll call it k2 so this is going to be rate is equal to k1 times the concentration of a times the concentration of b and this is going to be to the first power so that's the rate law expression for the first step of this reaction mechanism now for the second step or the second elementary reaction the rate law expression is going to be k2 times d times e so that's how you can write the rate law expression for a certain step in a reaction mechanism by the way in this reaction can you
identify the catalysts and the intermediate which species represents the catalyst and which one is the intermediate if we were to add these two elementary reactions the catalyst and the intermediate will be cancelled as we try to get the overall reaction to distinguish the catalyst from the intermediate you need to know that the catalyst is consumed first in the reaction and then it is produced later the reverse is true for the intermediate the intermediate is produced first in the reaction and then it is consumed later so notice that we can cancel species b and we can
cancel d one of them is the intermediate and the other is the catalyst so is b the catalyst or the intermediate what would you say notice that b appears first on the left side of the reaction so it is consumed first then it reappears on the right side of another reaction later so it's consumed first and then it's produced later that is the catalyst the catalyst is present in the beginning of the reaction and at the end of the reaction the intermediate is only present in the middle it is produced first and then it is
consumed later so if you look at d d is on the right side of step one it's on the right side of that reaction so it is produced first then it shows up later on the left side of the second reaction so it's consumed later it is produced first consumed later so it is the intermediate so remember the catalyst is at the beginning and at the end of the reaction it is consumed first and then produced later but the intermediate is in the middle it is produced first and then consumed later so that's a quick
and simple way that you could use to identify the intermediate and the catalyst in a reaction mechanism now remember a catalyst is a substance that can speed up a chemical reaction and the way that it accomplishes this is by providing the reaction with an alternative pathway for the reactants to become products and another benefit of a catalyst is that it lowers the activation energy and by those means it can speed up a chemical reaction so now let's write the overall reaction for this reaction mechanism b and d has been cancelled on the left we have
a and e so this is going to be a plus e and on the right we have c and f so this is the overall equation now sometimes you need to be able to determine the molecularity of each individual step in the reaction mechanism so the first time you need to be familiar with is a uni molecular reaction in a unimolecular reaction if you think of the word uni uni means one so we have one molecule that is going to be the reactant so a good example of a unimolecular elementary reaction is a going into
b the rate law expression for that is rate is equal to k times a raised to the first order the overall order of a unimolecular reaction is first order now the next thing we need to talk about is a bimolecular elementary reaction so remember an elementary reaction is simply one step in the overall reaction mechanism a good example of a bimolecular reaction and remember the prefix by means two so we're dealing with two reactants one example is a plus a turns into b another example is a plus b turns into c so both of these
elementary reactions are bi-molecular reactions in each case we have two reactants reacting with each other so the rate law expression for the first one is going to be k times a times a or we can just write a squared and for the second one it's going to be rate is equal to k times a to the first power times b to the first power in each case the overall order of the reaction in this case for each reaction it's second order so that's going to be the overall order for a bimolecular reaction now the next
type of reaction we need to consider is a term molecular reaction when i think of the prefix term for some reason i associate it with try and try means three now term molecular reactions are quite rare because it's statistically speaking it's difficult for three molecules to collide at the same time with the right amount of energy and with the right orientation for the reaction to occur so these reactions are very rare and they tend to be quite slow too but here's an example of a term molecular reaction a plus b plus c become in d
another example is 2a we can write as a plus a or just 2a plus b turning into c or we could say 3a becomes b all of these examples are term molecular reactions the rate law expression for the first one is going to be rate is equal to k times a times b times c for the second one we have rate is equal to k times a squared because we have 2a and then times b for the last one its rate is equal to k times a to the third power so the order of each
reactant is simply the coefficient of the reactant for an elementary reaction you can only do that for elementary actions and not the overall reaction so as we can see the overall order for each of these elementary reactions is three so a term molecular reaction the overall order for that type of reaction is third order it's third order overall now consider this reaction where we have o3 reacting with no2 to become no3 plus o2 so that's the first step and for the second step we have no3 plus no2 turn it into n2 o5 now let's say
the first step is slow but the second step is fast with this information go ahead and write the overall reaction and determine the rate law for the overall reaction and identify any intermediate or catalyst if applicable so first let's add up this reaction notice that no3 cancels is this a catalyst or intermediate so no3 doesn't appear in the beginning or at the end it's only in the middle of the reaction it is produced first and then it is consumed later so because it only shows up in the middle of the reaction this is going to
be an intermediate now let's write down what's left over so on the left side we have 103 molecule and we have two no2 molecules on the right side we have one n2o5 molecule and one o2 molecule so that's the overall reaction for this reaction mechanism now how can we write the rate law expression for the overall reaction the rate of the overall reaction is going to be dependent on the rate determinant step which step is the rate determining step is it the first step or is it the second step the rate determining step is going
to be the first elementary reaction because that's a slow step the rate will always depend on the slow step so the rate for the overall reaction will be equal to the rate of the slow step which is the rate of the slow step is k1 times o3 times no2 so that's a bimolecular reaction so they were to ask you what is the molecularity of the first step you would say it's bi-molecular because the overall order for that step is two we have two molecules reacting with each other so this is the rate law expression for
the slow step to write the rate law for the overall reaction instead of writing k1 we can simply write k so we can say it's k times o3 times nl2 so k would be just a generic term uh for a constant for the overall reaction but this is the rate log expression that we're looking for so it's first order with respect to o3 first order with respect to no2 and it's second order overall go ahead and try this example problem so write the overall reaction for the mechanism shown below identify any catalysts and intermediates that
may be interaction and determine the molecularity for each elementary reaction and then write the rate law for the overall reaction as well so let's begin let's identify any catalysts or intermediates that may be present first notice that we can cancel iodide and we can cancel io minus so io minus is the hypo iodine ion which one of these is the catalyst and which one is the intermediate what would you say so looking at i minus it is present at the beginning of the reaction and at the end it is consumed first and then it is
produced later so this this is the catalyst io minus is in the middle of the reaction it is produced first and then it is consumed later so this is going to be the intermediate so that's how you can distinguish from the the two from each other and that's how you can also identify them as well so now let's go ahead and write the overall reaction on the left side we have two peroxide molecules reacting with each other so that's hydrogen peroxide on the right it's going to produce two water molecules and we're going to get
an oxygen molecule so the decomposition of hydrogen peroxide into water and oxygen gas is accelerated by the presence of an iodide catalyst now what is the molecularity for the first elementary reaction and what about for the second step for the first step we have two reactant molecules so it's by molecular for the first step and it's also by molecular for the second step so that's the molecularity for each step in this reaction mechanism now to write the rate law expression i need to tell you which one is slow and which one is fast let's say
the first step is the slow step and the second step is the fast step so with that information what is the rate law expression for the overall reaction feel free to pause the video and work on that the rate law expression for the overall reaction is going to be equal to the rate of the slow step the rate determinist step so let's write the rate law expression for the slow step so this is k1 and this is k2 so the rate for the slow step is going to be k1 times the concentration of hydrogen peroxide
the coefficient for that is a one and then times the concentration of iodide which is also to the first power so that's the rate law expression for the first step but now to write the rate law expression for the overall reaction which is this reaction here notice that iodide is not part of the overall reaction the overall reaction contains only one reactant h2o2 so therefore the rate law expression should only contain that reactant so instead of writing k1 we can simply write k for the rate constant for the overall reaction even though this k looks
big this should be lower case k now this part is going to stay the same the rate law for the overall reaction is going to equal the concentration is proportional to the concentration of h2o2 to the first power now since iodide is not in the overall rate log expression we can get rid of it so this will be the rate law expression for the overall reaction and this is the rate law expression for the first step with iodide included so that's it for this problem you
