Collision Theory Arrhenius Equation Activation Energy Chemical Kinetics

in this tutorial let's talk about the collision theory model so what's the basic idea behind the collision theory the basic idea is that for molecules to react they have to collide if they don't collide there's not going to be any chemical reaction if there's no physical contact between these molecules nothing's going to happen so molecules must collide in order to react the second thing is they have to have the correct molecular orientation in order for reaction to proceed so to illustrate this let's say if we have hydroxide and hydroxide is going to react with methyl bromide ch3br now bromine is more electronegative than carbon so it bears partial negative charge and carbon is less electronegative than bromine so it's partially positive the element that's more electronegative is going to pull the electrons toward itself and that's why bromine has the partial negative charge so hydroxide since it has negative charge it's attracted to the carbon in methyl bromide so it wants to attack this carbon thus expelling the bromide ion and so this will create methanol so if it approaches from the back of the methyl bromide if it attacks from the carbon side it has the correct molecular orientation so the reaction is going to work now let's say if it approaches it from the other direction so let's say this is methyl bromide and hydroxide is coming in this direction well the reaction is not going to happen hydroxide is repelled by the partial negative charge of the bromine atom and so when two negative charges come together they're going to fill force that's going to repel them and so that's going to prevent the reaction from occurring whereas when hydroxide attacks from the carbon side the negative charge on the hydroxide is attracted to the partially positive charge on the carbon atom and that electrostatic force of attraction accelerates them together which favors the collision of those two molecules and so if the two molecules do not have the correct molecular orientation the reaction is not going to happen but if they collide with the appropriate orientation then the reaction is going to happen so in order for reaction to occur molecules they have to collide and two they have to collide with the right orientation now the third thing is that they need enough energy because if the temperature is low even though these two may be attracted to each other if the hydroxyl ion doesn't move with enough momentum it's not going to be enough to knock off this bromide ion so it has to have enough energy to attack the carbon atom and expel the bromide ion out of that reaction to create methanol so there's three things that is required for reaction to take place the molecules must collide and they have to collide with the right orientation and they have to have sufficient energy to break some bonds and form some new bonds now this leads us into energy diagrams so on the y-axis is the energy of the system on the x-axis is going to be the reaction coordinate here we have the reactants and here we have the products the top is the transition state also known as the activated complex in order for this reaction to occur the system has to have enough energy the reactants need to gain enough energy there's something called activation energy which is the difference between the energy of the transition state and the energy of the reactants if the reactants do not have enough energy to overcome the activation energy the reaction will not happen so if they can't get above this hill there's no reaction that's going to take place so they have to have enough energy to get over that hill and so that minimum energy to get the reaction started is the activation energy so even if molecule collides with the right orientation if they don't have enough energy to make it to the activated complex then the reaction is just not going to happen so one way you can get up there is by increasing the temperature whenever you increase the temperature the rate of the reaction increases the fraction of molecules that can reach the activated complex will increase now one thing want to specify is this is known as the forward activation energy there's something called the reverse activation energy which is the energy difference between the products and the transition state so in this example it's easier to go this way than to go that way now the next thing you need to know is that notice that the difference between the energy of the products is greater than the energy of the reactants whenever the energy of the products is greater than the reactants what you have is an endothermic reaction in order for this reaction to occur energy has to be put into the system an exothermic reaction looks like this the energy of the reactants is going to be greater than the energy of the products so this is an exothermic reaction now initially you got to put in energy to get to the activated complex to get the reaction started and once you overcome that barrier then lot of energy is going to be released so you got to put in small amount of energy to get it started but you're going to get large amount of energy once the process is started and so that's an exothermic reaction now let's talk about the arrhenius equation is equal to zp raised to the negative over rt now is known as the collision frequency if you can increase the collision frequency of reaction the rate of the entire reaction will go up is the steric factor and this has to do with the fraction of collisions that have the right orientation so this is number between zero and one and it depends on the nature of the reactants now this part which is the inverse of the natural log function raised to the negative over rt that represents the fraction of molecules that have sufficient energy to get the reaction going so notice that it depends on temperature and activation energy so if you want reaction to occur there's two things you can do to help it to make it work you can increase the temperature and that can give the molecules the energy they need to overcome that barrier as you increase the temperature the rate of the reaction will increase the second thing you could do is add catalyst whenever you add catalyst catalyst provides an alternative pathway for the reaction and by doing so it lowers the activation energy which increases the rate of the reaction now the last thing we need to talk about is the product zp zp is equal to which is also known as the frequency factor is the product of the collision frequency and steric factor so now let's put this all together now let's consider first order reaction where we have rate is equal to the rate constant times the concentration of and we know that the rate constant is equal to the frequency factor times raised to the negative ea over rt and the frequency factor can be replaced with the collision frequency times the steric factor and then times this so the collision frequency represents the total number of collisions per second that's occurring in reaction the steric factor which we said was between zero and one that's the fraction of molecules with the proper orientation or more specifically that's the fraction of collisions that are occurring with the proper orientation so you can think of as the probability that two molecules will collide with the right orientation this portion here which we describe as the fraction of collisions with sufficient energy you could also think of it as the probability that two molecules that are colliding will have enough energy to overcome the activation barrier so you can think of these two values as probabilities or fractions so if is the collision frequency and is the probability that two collider molecules will have the right orientation represents the total number of collisions per second that's occurring in reaction with the right orientation now if we multiply the frequency factor by the fraction of molecules having sufficient energy we can thus come up with an understanding of the rate constant so we can say that the rate constant represents the total number of collisions per second having the proper molecular orientation as well as sufficient energy needed to overcome the activation energy barrier to get the reaction going to convert the reactants into products so the rate constant takes all of those factors into account now let's talk about what happens when we increase the concentration of the reactant as we increase the concentration of the reactant there's going to be more collisions the number of collisions will increase and therefore the rate of the reaction will increase now what about if we increase the temperature if we increase the temperature the average kinetic energy of the molecules will increase so molecules are going to be moving faster the molecular speed of the molecules will increase as well and therefore there's going to be more collisions and this too will increase the rate of the reaction now you can think about it another way too in this formula the temperature affects the rate constant now the temperature is in the bottom of the fraction and typically when you increase the denominator of fraction the value of the fraction goes down however notice that we have negative exponent and so because of that when you increase the temperature it has the effect of increase in it's like double negative you could think of it that way when you increase the temperature the rate constant goes up the fraction of molecules having sufficient energy to overcome the activation barrier increases and when goes up the rate constant goes up as well and let's show this with distribution curve so we're going to have energy on the x-axis and we're going to have the number of molecules on the y-axis it's not moles but molecules and we're going to have two curves one at lower temperature which we'll say look something like this and another curve which i'll draw in blue at higher temperature so this one will be let's say at temperature of 300 kelvin and the other one will be at higher temperature we'll say at 500 kelvin and somewhere along the x-axis we're going to have ea the activation energy so anything above ea represents the fraction of molecules have insufficient energy sufficient collision energy so to speak enough so that they can overcome the energy barrier to make the reaction going and let's put that in green so in green that represents the fraction of molecules at 300 kelvin having enough energy to make the reaction go but in red which encompasses the area in green as well that represents the fractional molecules have enough energy to overcome the energy barrier to also make the reaction work but notice that at 500 kelvin the fraction of molecules that have enough energy is greater than the fraction of molecules at 300 kelvin it's significantly greater and so that's what happens when you increase the temperature when you increase the temperature the average kinetic energy of the molecules go up so therefore the fraction of molecules having enough energy to make the reaction go and that fraction increases and so if more molecules have more energy to get the reaction going the overall rate of the reaction will increase now let's talk about what's going to happen if we were to add catalyst to the reaction what do you think is going to happen if we introduce catalyst catalyst has the effect of lowering the activation energy now here's visual description of that so this is potential energy diagram we have potential energy on the y-axis and the progress of the reaction on the x-axis so let's say we have something that looks like this where this represents the energy of the reactants the products and this is the transition state for catalyzed reaction it would look something like this so this is the uncatalyzed reaction and this is the catalyzed reaction the activation energy is the difference between the energy of the reactants and the transition state but notice that when you add catalyst the activation energy decreases catalyst achieves this by providing reaction with an alternative pathway it actually changes the reaction mechanism it allows the reactants to become products using by taking different route so to speak so that's how the catalyst can lower the activation energy it helps the reaction find an easier way to go from reactants to products now notice that activation energy is in the numerator of that exponential fraction whereas temperature is in the denominator now we know that an increase in temperature leads to an increase in the rate constant so what happens if we increase the activation energy well it should have the opposite effect it should decrease is in the denominator is in the numerator so if increase in leads to an increase in increase in ea will decrease likewise the reverse is true if we decrease the activation energy the rate constant should go up and that's what happens when you add catalyst the activation energy goes down the rate constant go the rate constant goes up and so the overall rate of the reaction increases so those are some ways in which you can increase the rate of chemical reaction so to summarize if we increase the concentration of the reactant the rate is going to go up if we increase the temperature the rate of the chemical reaction will go up and if we add catalyst the activation energy will go down and the rate of the reaction will go up so those are some different ways in which we can increase the rate of chemical reaction you can increase concentration raise the temperature or introduce catalyst to speed up the reaction now let's write down some equations that you'll need to know in order to solve problems so you're already familiar with this one the rate constant is equal to the frequency factor times raised to the negative over rt here's another one that you should be familiar with the natural log of the rate constant that's equal to negative over times one over plus the natural log of now in this form this is written in slope intercept form so corresponds to and the slope corresponds to negative over corresponds to one over and the intercept is lna so if you were to graph versus where is and so we'll put that on the axis mean the axis and one over is so we'll put one over on the axis you're going to get graph that looks like this it's going to be straight line plot but notice that the slope is negative so the slope of this line is going to equal negative over and the y-intercept which this is where the graph starts from that's going to be ln so for problems associated with this equation where if you have plot of ln versus 1 over just know that you can find the activation energy from the slope of the graph and the y-intercept is equal to the natural log of so you could find the frequency factor from the y-intercept now is 8.3145 joules per mole per kelvin so because has unit joules and moles the activation energy which is typically reported in kilojoules per mole when you use it in these formulas it needs to be in joules per mole so you need to convert it so when reporting the activation energy you want your answer to be in kilojoules per mole but whenever you plug it into one of these formulas you need to convert it to joules per mole so make sure to be aware of that now the slope is equal to negative over and the slope is y2 minus y1 divided by x2 minus x1 ln is on the y-axis 1 over is on the x-axis so if is equal to lnk we could say that this is ln k2 minus ln k1 all divided by 1 over t2 minus 1 over t1 so that's going to equal negative ea over now let's say if we have ln minus and using property of natural logs or logs we can convert this into single log expression by writing it like this ln over so we can write this as single long expression as well we can say that that's ln mean that's ln k2 over k1 and this is 1 over t2 minus 1 over t1 and that equals negative ea over now if we want to get activation energy by itself we can multiply both sides by negative if we do that will cancel on the right side as well as the negative sign and we'll get an equation for activation energy so the activation energy is going to be negative ln k2 over k1 divided by 1 over t2 minus 1 over t1 so you could use this equation to calculate activation energy now you might see variant of this equation sometimes this negative sign will be distributed to 1 over t2 minus 1 over t1 and that will basically flip the order of t1 and t2 so this is also equal to positive ln k2 over k1 and if you distribute the negative sign negative times negative one over becomes positive one over one and then here this will be negative so minus one over two so you might see it in that format too so just be aware of that but if you have the negative sign it's going to be one over two minus one one over one instead of the reverse now let's go back to this equation make sure to write this equation by the way because you're going to need it to solve problems now instead of multiplying both sides by negative let's multiply both sides by this term if we do that we'll get different equation we'll get this one ln k2 divided by k1 is equal to negative ea over and we're going to move this to the other side times one over two minus one over one so that's another form of the arrhenius equation that you might see it in now what we're going to do at this point is we're going to put both sides of this equation on the exponent of so i'm going to put all of this on the exponent of and all of this on the exponent of because is the same and these two are equal to each other both sides of this equation will be equivalent now the base of natural log is so on the left side these two will cancel and so we're just going to get k2 over k1 is equal to everything that we see here now if we multiply both sides by k1 we can get an expression for k2 so you want to write down this formula because sometimes you need to calculate new rate constant at new temperature so that new rate constant k2 is equal to k1 times raised to the negative over times one over two minus one over one so this is the second equation that you wanna write down so you have an equation that will help you to calculate the activation energy and this equation will help you to calculate the new rate constant given new temperature now let's go back to the equation that we started with on this screen alright that's as far as my software will let me go back now sometimes you need to calculate the second temperature so what we're going to do is we're going to isolate t2 to do that we're going to multiply both sides by negative over so those will cancel and we're going to have negative ln k2 divided by k1 over ea that's going to equal 1 over t2 minus 1 over t1 so we're going to move this to the other side where it will become positive so we have 1 over t1 minus ln k2 over k1 over ea that's equal to one over two now we're going to raise both sides to minus one one over two raised to the negative one power is just two so we get this equation two is equal to one over t1 minus ln k2 over k1 over ea and this is raised to the negative 1 power so this will help you to calculate the second temperature if you know the second rate constant and if you know the activation energy so those are some formulas that you want to use when solving problems associated with the uranius equation activation energy the rate constant and things like that now let's go ahead and put these equations to good use so we're going to work on few practice problems so take this one for example certain reaction has an activation energy of 50 kilojoules per mole the rate constant at 25 degrees celsius is .0039 minutes to the minus one what is the value of the rate constant at 75 degrees celsius so first let's make list of what we know so we have the activation energy it's 50 kilojoules per mole the rate constant which we'll call k1 is .0039 and that's at temperature of 25 degrees celsius which we need to convert to kelvin so add 273 to 25 that gives you 298 kelvin our goal is to calculate k2 and the second temperature is 75 plus 273 so that's 348 kelvin so what is the formula that will help us to calculate k2 it's k2 is equal to k1 times raised to the negative over multiplied by one over t2 minus one over t1 so anytime you need to find rate constant at different temperature that's the formula that you want to commit to memory now k1 we know it's 0.0039 the activation energy has to be in joules so we've got to multiply 50 by thousand so it's going to be fifty thousand and then it's one over t2 which is 348 minus one over t1 which is 298 and then divide that by is 8.3145 joules per mole per kelvin you don't want to use the gas constant for our 0.0826 it's not going to work for this equation so if you type this entire thing the way you see it or you can do it one step at time you should get this answer k2 is .0708 so notice that k2 is greater than k1 0.07 is greater than 0.003 now this answer makes sense because the second temperature is higher than the first temperature and as you increase the temperature of reaction the rate constant will increase and when the rate constant goes up the rate of the chemical reaction will go up as well and so that's it for this problem you