Reaction Rate Laws - Concept
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
Reaction rate laws give an equation for finding the rate of a reaction using the concentration of the reactants and the stoichiometric coefficients. For the chemical equation aA +bB => C the reaction rate law is reaction rate = k[A]^a [B]^b where k is a constant.
Alright so there're a few things that affect how fast or slow a reaction takes place, one of them being concentration we know that concentration, the higher more concentrated the reactants are the faster they're going to collide, the more often they're going to collide and also the nature of the reactants how reactant a particular metal or substance is when it comes across something else. So taking those two things into account, we're going to actually make Mathematical rate laws and the rate laws are actually unique for each particular reaction. So let's take this reaction for example we're going to say a+b yields a product and what my products are and why do I care what they are. They do not affect how fast this reaction is, the reactants are the ones that are very important so we're going to deal with just those. Okay, so this is our backbone of our Mathematical expression for the rate law, rate equals k times the concentration of in this case a times the concentration of in this case b raised to a certain power. And let's dissect what this actually means, so the rate is going to be in molarity per unit of time in this case we're going to talk about seconds but they can be any time, minutes or hours or whatever it maybe depending on the reaction or how fast this law will take place.
In this particular instance it's going to be fast so I put seconds, if it's slow you put something like hours or days. This is going to be our rate constant, it's a k, it's a k value, it's not the same k you've seen in equilibrium or at the bases when you're dealing with those, it's a lower case k and it does have the units so you're going to have to express those units and it actually varies with temperature. So this is actually going to have to happen at a certain temperature the rate law, this particular rate law will change if the temperature is raised or lowered. This thing is dependent, this k is dependent on temperature. This is a concentration of that bracket means concentration of in molarity mols per liters so this is the concentration of this reactant and this is the brackets mean the concentration of this particular reactant. And these are raised to a certain power, this is what you call the reaction order, this will tell us how important these guys are in terms of how fast they go. So the higher are these numbers, the more important they are in terms of how fast the reaction will take place.
They notice, I want to make sure you do know this is really, really important that these superscripts exponents are not the coefficients in the reaction okay. So these are only going to take place in we're talking about reaction mechanisms which, there's another video on if you want to take a look at that. But talking about rate laws these numbers are not going to be the coefficients of a reaction. Alright, so let's about how we can actually find the rate law, so you're going to have to find empirical data meaning experimental data in order to find the rate law for each reaction. So in all the reaction that we originally had which I'll write again aa+bb yields products we have 2 reactants a and b. So I'm going to have 3 different trials, I'm going to have 3 different experiments, and the first experiment, in my first trial I'm going to put 0.1 mols per liter of a and this ambiguous 0.1 mols per liter of b and I know that to get to my product the rate is 2 times a negative third molarity per second that's how fast the products are being formed.
Then I'm like okay I want how dependent each one is like if I put more of a in is the reaction going to go faster, well I have to check it out. So I make sure b was the same, this is going to be my constant right here and then if I double the concentration of a and I notice the rate actually changed, it increased. Then I did my third trial this time I'm going to keep a the same and I'm going to increase the rate of b and see how much that affects the rate law. And so I'm going to use all this information to do the overall rate law for this reaction. So let's do this together, so remember the rate law is rate equals k time a to the concentration of m, times b to the order of n, so these are going to tell us how much each reactant is influencing the rate. So in order to find these guys I have to do this so you have to write the ratio of 2 trials. I'm going to pick trial 1 and 2 because b is the same here. So unaffected, so I'm just trying to find this guy, so I'm going to see how much it affects this. So if I doubled the concentration of a, so I went from 0.100 to 0.200 what's going to happen to my rate? My rate went from 2 times 10 to the negative third to 4 times 10 to the negative third.
Okay so if I reduce this, this is equal to one half to the m equals one half. In this case m is 1 okay, so right now I can erase this m and say now my m is 1. Okay great, then I'm going to see what n is, so I need to see how much b influences the rate of reaction. So I'm going to say the same thing for b so I'm going to go okay, so in this case a stays the same here but b changes so I'm going to use trial 2 and 3 to compare so this concentration of b was 0.1 to 0.2 to some certain power and the rate went from 4 times 10 to the negative third to 16 times 10 to the negative third. What is this, so this is one half to the n equals one fourth n is going to equal 2 okay. This then is going to be 2, these are what we call the orders of reactions okay. So b actually influences the rate of reaction more so if I put more, the higher the concentration of b faster the rate is going to go compared to a. So the overall order of this reaction is a third order reaction we're going to call this, so 1+2=3 so it's a third order reaction and this actually is not important in this particular simple case but when you get to Chemistry and you're talking about integrated rate laws, this sort of order reaction will tell us a lot of information.
But right now we're just going to note that it is a third order reaction because it's 1+2, the first order and second order equals third order okay. So right now we know this is our rate law, but we don't, we need find k because we said k is important and it's a constant for this particular rate law at this temperature. So we need to find that, so I'm actually going to erase all of this information as I can get more board space. Okay, so in order to find this k I'm just going to pick a trial and plug everything in. So I'm going to make it simple, I'm going to do trial 1 so the rate for trial 1 is 2 times 10 to the negative third equals the k which we're looking for times the concentration of a in this case is a 0.1 to the first power times the concentration of b which is 0.1 also squared and so I solve this. So this is 0.001 times k equals 2 times 10 to the negative third so of I divide by 0.001, 0.001 I find that k is equal to 2.
The bigger the k the faster the reaction actually takes place, the smaller the k, the slower the reaction actually takes place. So okay, so then we need to find the units for this guy, because the units actually vary for each rate law. This is the units that are going to change. So let's do that so we're going to have to plug in the rate is in molarity per second equals our k times the concentration is in molarity and it's the first power times the concentration molarity per second power, and I want to get to the overall unit in molarity per second. So what is k well then I'm going to say molarity times seconds equals k times m to the third okay. And so then I'm going to say that k is equal to 1 over molarity squared times seconds, so I wanted to make sure that works. Molarity over seconds equals 1 over molarity squared times seconds times molarity cubed, these will cross out molarity per second so it does the check off perfect okay.
There's actually instead of having to go through this every time there's actually a simple rule that we can actually take place our little trick. So the units for k is k is equal to 1 over molarity to the order exponent minus 1 times seconds. Now this seconds is just because we're using our seconds in, when dealing with this. But this could be any whatever unit of time you're dealing with hours, days, minutes whatever then we just call it seconds because that's what time we're dealing with. But this is how you can find the rate without actually, sorry the units rate, the rate law without actually having to go through this whole mess of Math. Okay so this is essentially how you would use empirical data to determine the rate law and how fast the rate is going.
Please enter your name.
Are you sure you want to delete this comment?
- Collision Theory 24,787 views
- Reaction Rates Factors 24,360 views
- Reaction Mechanism 22,209 views
- Reaction Rate Problems 19,011 views
- Energy Diagrams 24,422 views
- Determining Order of a Reaction Using a Graph 18,671 views
- Factors Affecting Collision Based Reaction Rates 4,035 views
- Factors Affecting Homogeneous Reactions 4,253 views
- Rates of Disappearance and Appearance 24,201 views
- Tips for Figuring Out What a Rate Law Means 3,636 views
- Tips on Differentiating Between a Catalyst and an Intermediate 9,353 views
- Understanding Energy Diagrams 4,113 views