Rate Law represents an equation for a chemical reaction that connects the reaction rate with the concentrations or pressures of the reactants and the rate constant.
Understanding Rate Law
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Rate Law Concept 1
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Now the rate law is just an expression that relates the rate of a reaction to its react change in reactant concentrations, rate constant, and reaction orders. Now your rate constant is a proportionality constant that links the rate to these reactant concentrations. And your reaction orders, well, these are just the exponent component for the given concentrations. Now they're determined mathematically with a chart, which with which has given values, or from a series of steps called a reaction mechanism. Here, we're not gonna go into detail on reaction mechanisms just yet because pretty abstract. Later on, we'll go in greater detail on what reaction mechanisms are and how we can use them to figure out our reaction orders. Now also what's important is that rate law ignores products, because notice I only talked about the concentration of the reactants. I never mentioned concentrations of products, because when it comes to rate law, we ignore products altogether. Now our rate law has its rate law expression, which is rate equals k, which is your rate constant, times a to the x, b to the y. Now a and b represent your reaction concentration or reactant concentrations. Here, if your chemical equation had more than 2 reactants, then we continue. We have c to the z. X and y, these are your exponents or reaction orders. So just remember your rate law is trying to connect the ideas of changes in concentration for your reactants, rate constant, and reaction orders, in order to find the overall rate of any given chemical reaction.
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Rate Law Example 1
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Chemical reaction has a rate law of k a to the 3, b to the 1, c to the 0. By what factor would the rate increase if the concentration of a were tripled, the concentration of b was cut by half, and the concentration of c was increased by half, and the rate constant k was kept constant. Alright. So here we're going to say rate equals k a to the 3, b to the 1, c to the 0. They're telling me that our rate constant k is staying the same, it's not changing, so it's not going to affect any change in rate, so we can ignore it for now. And here we're talking about tripling, cutting in half, increasing by half, each one of these reactant concentrations. To make it easy for us, they don't give us their initial amounts, so just make them all 1 molar. K? Just to make it easy. Because proportionally, it wouldn't matter. I can make them all 3 molar. Doing what it tells me to do, I'd still get the same answer. So here we're gonna assume that they're all starting at 1 molar, and here I'm going to triple a, so it's gonna go from 1 molar to 3 molar, and it's still to the 3. B is getting cut by half, so you start at 1 molar and you cut it by half, so now you only have 0.5 left, still to the 1. And c, we're still gonna do c, but it's not gonna really matter. We're increasing it by half, so now it becomes 1.5. C blue doesn't matter because any number to the zeroth power is equal to 1. So I can make this concentration inside for c a 1000000. It won't matter because a1000000 to 0 is still equal to 1. So really all that is important to figure out the change in rate is a and b. So 3 to the 3 is 27 times 0.5 is 13.5. This means that my rate would be expected to increase by 13.5 fold. This is how much faster my rate would go, because increasing the concentration of my reactants would cause an increase in the rate. Here we're seeing it mathematically being shown to us. So here, option c would be the correct answer.
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Rate Law Concept 2
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Now the units for the rate constant k can be determined by first calculating the overall order. Now our overall order uses the variable n. It's a numerical value calculated from the addition of all reaction orders. And here, the formula to figure out the units for your rate constant k is k equals m 2 n+one times time inverse. So remember k is your rate constant, capital m is your molarity, and n again is your overall order. So adding up all the orders of your rate law helps you to figure out n, take that and plug it into this formula, and you'll know what the units for k will be.
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Rate Law Example 2
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What is the overall order and the units for the rate constant k for the following chemical reaction shown below that has a rate equal to k n o 2 to the 2cl2 to the 1. Alright. So remember, 2 and 1 are your reaction orders. To find your overall order, you just add them together. So this would be 2 +1, so your reaction would be 3rd order overall. Now for k, we'd say, remember, k is m to the negative n plus 1 times time inverse. Here, time could be in units of seconds, minutes, years, whatever. So this would be m to the negative 3 plus 1. Here, let's just say seconds are the units of time that we're looking at, and so this would be m to the negative two times seconds inverse. So whatever the actual value of k would be, these would be the units that follow it. So it would be some value m to the negative two times s to the negative one. Okay? So we need to be 3rd overall, and these would be the units for our rate constant k.
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Rate Law Concept 3
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The rate constant and reaction orders of the rate law are determined mathematically when given our reactant concentrations and initial rates. Now this is the sequence that you have to use in order to solve all of the components of the rate law. The first thing that we can solve are our reaction orders. Now once you have your reaction orders determined, you can use that information to determine your rate constant k. And then once you have that, you can then determine your new rate. So if you're given a question where they're giving you information on the reacting concentration and initial rates, and they want you to find your rate constant k, You can't just skip to rate constant k. You have to first find your reaction orders. Once you do that, then you can find k. If they give you reacting concentrations and initial rates and ask you to find your new rate, you can't just skip all the way to new rate. You have to do reaction orders to find your rate constant, and then you can find your new rate. Right? So keep this in mind anytime we're asked to deal with any calculations dealing with rate law determination.
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Rate Law Example 3
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The initial rates of reaction 2 n l gas plus c l two gas, giving us 2 n l c l gas are given below. Here we have experiments 123, and each experiment has attached to it reacting concentrations and an initial rate. Now here it says determine the new rate if given new initial concentrations for n o and c l 2. Now remember our sequence for solving is we first have to determine our reaction orders. Once we do that, then we can figure out our rate constant k, and then finally we can determine our new rate. Right. So for step 1, we're gonna choose a reactant and look at 2 experiments where its concentration changes, but the other reactants stay the same. These other reactants we're going to ignore. So you're gonna ignore the reactants whose concentrations remain the same. Here we're gonna look for the reaction order of n o, so let's make it x, c l 2 would have y. You can you can make them any variables you want, a, b, c, whatever. Here I just do x and y. We have to look for 2 experiments where n o is changing, but c l two is staying the same. That occurs with experiments 23. With experiments 23, we see n o is changing and the concentrations of c l two remain the same. Once we figure that out, we go to step 2 where we create a pair of ratios for the reactant that sets react rates equal to concentrations of the reactants. This is important. You place the larger rate value on top of the smaller rate value to get whole numbers when solving. And here you're gonna raise the reacting concentrations to an unknown power or variable for the reaction order and solve. Alright. So if we take a look here, experiments 23, we can see that rate 3, it's rate initial rate 3 because it's connected to experiment 3, it's larger than initial rate 2. So that would mean that we're gonna have rate 3 over rate 2 equals the concentrations of no to the x. Now we're going to plug in the number, so 18.2 divided by 9 point 08 equals 0.0500 divided by 0.0250, still to the x. 18.2 divided by 9.08 is approximately 2. And then 0.05100 divided by 0.0250 is 2, and it's still to the x. Now you say 2 to what number gives you 2? The answer would be 1. So the reaction order for n o is 1. Now we're gonna do the same thing for step 3. We're gonna repeat the process for any remaining reactants until all reaction orders are determined. Right? Right? So now we have to look for c l 2. We look for 2 experiments where c l 2 is changing, but n o is staying the same. We see that happen with experiments 12. Now here initial rate 1 is larger than initial rate 2. So if we come down here, we're gonna say rate 1 over rate 2 equals c l two divided by c l two to the y. So let's see. That was 18.2, 18.2 divided by 9.08. And then here we're gonna plug in the concentrations for c l 2. So here for c l 2, it's 0.0 510 and this is 0.0255 to the y. This also comes out to 2 equals 2 to the y. 2 to what number gives us 2? The answer is 1. So it's first order for both of our reactants. Now step 4, if necessary. Here it is necessary because we need to find our new rate. And to find our new rate, we have to figure out our react our rate constant. So here we're gonna say, if necessary, to solve for the rate constant k, plug in the reactant concentration and reactant orders into the rate law. Alright. So here we're going to say that our rate, we're gonna say rate 1. You can use any of the rates, but just to keep it simple always go with rate 1. So rate 1 equals k, and we're going to say here n o to the 1, c l 2 to the 1. Okay? And then we're gonna start plugging the numbers. So 18.2, we're looking for k, we don't know what k is. Come back up here. Let's use a concentration for so since we're using rate 1, we have to use the concentrations of experiment 1. So that's 0.0250 and 0.0510. To the 1 and 0.0510 to the 1. All right, so here, when we multiply these together, that's gonna give me 0.001275, still multiplying with k. Rate 1, again, is 18.2. Divide both sides by 0.001275 and we'll have our k. Here, k equals 14117.6 47. Now, here we could technically give the units for k because remember, to find the units of k, it's m to the negative n+one times time inverse. And here's our overall order which is 2, and it comes from adding up your reaction orders together, so that's negative 2+1. Here, let's just use seconds inverse since our initial rates are molarities per second. So that would be m to the negative one times seconds inverse. Okay. Finally, we can figure out our new rate. And again, if necessary, we're gonna say to solve for the new rate, plug in the k, the reaction orders, and the additional reaction concentration given. So here we're going to say that our new rate equals k and we're gonna plug in the new concentrations given to us in the very beginning. So let's come up here. So here we have 0.0730 for our n o, and then we have 0.0510 for our c l 2. So those are the new concentrations we're gonna plug into here. So let's come down here. So this is 0.0730. Actually, and that's again for n o to the 1 and c l 2 to the 1. So new rate, so k we got is 14117 0.647, n o is 0.730 to the 1, c l 2 is point 0510 to the 1. When we plug all that in, that gives us 52.6 molarities per second for our new rate. So we can see that if they're asking us for the new rate, it is quite in ordeal. You have to go through a lot of things to get to your final answer. Remember, the sequence we're solving is important. 1st, figure out reaction orders before you can figure out your rate constant k, before you can figure out your final new rate.
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Problem
Problem
Given the following chemical reaction, A → B. If the concentration of A is doubled the rate increases by a factor of 2.83, what is the order of the reaction with respect to A.
A
1
B
0.5
C
1.5
D
0
E
2
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Problem
Problem
The data below were collected for the following reaction:CH3Cl (g) + 3 Cl2 (g) → CCl4 (g) + 3 HCl (g)
Calculate the value and units for the rate constant k.