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Main content
Current time:0:00Total duration:9:14
AP.Chem:
TRA‑3 (EU)
,
TRA‑3.A (LO)
,
TRA‑3.A.1 (EK)
,
TRA‑3.A.2 (EK)

Video transcript

the rate of a chemical reaction is defined as the change in the concentration of a reactant or product over the change in time and concentration is in moles per liter or molar and time is in seconds so we express the rate of a chemical reaction in molar per second molar per second sounds a lot like meters per second and that if you remember your physics is our unit for velocity so average velocity is equal to the change in X over the change in time and so thinking about average velocity helps you understand the definition for rate of reaction in chemistry if we look at this applied to a very very simple reaction all right so we have one reactant a turning into one product B and let's say at time is equal to zero we're starting with an initial concentration of a of one molar and a hasn't turned into B yet so at time is equal to zero the concentration of B is zero let's say we wait two seconds so we wait two seconds and then we measure the concentration of a obviously the concentration of a is going to go down because a is turning into B let's say the concentration of a turns out to be 0.98 molar so we lost point zero two molar for the concentration of a so that turns into since a turns into B after two seconds the concentration of B is 0.02 molar right because a turned into B so this is our concentration of B after 2 seconds if we want to know the average rate of reaction here we could plug in to you our definition for rate of reaction change in concentration let's do change the concentration of our product over the change in time so the rate is equal to the change in the concentration of our product that's final concentration minus initial concentration so the final concentration is point zero two so we write in here point zero two and from that we subtract the initial concentration of our product which is zero so point zero 2-0 that's all over the change in time that's the final time minus the initial times that's two minus zero so the rate of reaction the average rate of reaction would be equal to 0.02 divided by two which is point zero one molar per second so that's our average rate of reaction from time is equal to zero to x equal to two seconds we could do the same thing for a right so we could instead of instead of defining our rate of reaction as the appearance of B we could define our rate of reaction as the disappearance of a so the rate would be equal to all right the change in the concentration of a that's the final concentration of a which is 0.98 minus the initial concentration of a and the initial concentration of a is 1 so 0.9 H minus 1 and this is all over the final time minus the initial times this is over 2-0 now this would give us this would give us a negative point zero two negative point zero two here over two and that would give us a negative rate of reaction but in chemistry the rate of reaction is defined as a positive quantity so we need a negative sign we need to put a negative sign in here because a negative sign gives us a positive value for the rate so now we get point zero two divided by two which of course is point zero one molar per seconds so we get a positive value for the rate of reaction all right so we calculate the average rate of reaction using the disappearance of a and the formation of B and we could make this a little bit more general all right we could say that our rate is equal to this would be the change in the concentration of a over the change in time but we need to make sure to put in our negative sign right we put in our negative sign to give us a positive value for the rate so the rate is equal to the negative change in the concentration of a over the change of time and that's equal to all right the change in the concentration of B over the change in time and we don't need a negative sign because we already saw in the calculation right we get a positive value for the rate so here's two different ways to express the rate of our reaction so here I just wrote it a little bit more general terms let's look at a more complicated reaction here we have the balanced equation for the decomposition of dinitrogen pentoxide into nitrogen dioxide and oxygen and let's say that oxygen forms at a rate of 9 times 10 to the negative 6 molar per second so what is the rate of formation of nitrogen dioxides well if you look at the balanced equation for every 1 mole of oxygen that forms 4 moles of nitrogen dioxide form so we just need to multiply the rate of formation of oxygen by 4 and so that gives us that gives us 3 point 6 times 10 to the negative 5 molar per second so no.2 forms at four times the rate of OH - what about dinitrogen pentoxide so n2o5 look at your mole ratios for every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide so if we're starting with the rate of formation of oxygen right because our mole ratio is one to two here we need to multiply this by two and since we're losing dinitrogen pentoxide we put a negative sign here so this gives us negative 1.8 times 10 to the negative 5 molar per second so dinitrogen pentoxide disappears at twice the rate that oxygen appears all right let's think about the rate of our reaction right so the rate of our reaction is equal to well we could just say it's equal to the to the appearance of oxygen right we could say it's equal to 9 times 10 to the negative 6 molar per second so we could write that down here the rate is equal to the change in the concentration of oxygen over the change in time all right what about if we wanted to express this in terms of the formation of nitrogen dioxide well the formation of nitrogen dioxide was three point six times 10 to the negative 5 all right so that's three point six times 10 to the negative 5 so you need to think to yourself what do I need to multiply this number by in order to get this number since this number is 4 times the number on the Left I need to multiply by 1/4 all right so down here down here if we're talking about the change in the concentration of nitrogen dioxide over the change in time to get the rate to be the same we'd have to multiply this by 1/4 all right finally let's think about let's think about dinitrogen pentoxide all right so we said that that was disappearing at negative 1.8 times 10 to the negative 5 so once again what do I need to multiply this number by in order to get 9 times 10 to the negative 6 well this number right in terms of magnitude was twice this number so I need to multiply by 1/2 I need to get rid of the negative sign because rates of reaction are defined as a positive quantity so I need a negative here all right so that would give me right that gives me 9 times 10 to the negative 6 so 4 I could express my rate if I want to express my rate in terms of the disappearance of dinitrogen pentoxide I'd write the change in n 2 this would be the change in n 2 O 5 over the change in time and I need to put a negative 1/2 here as well all right so now that we figured out how to how to express our rate we can look at our balanced equation right so over here we had a 2/4 dinitrogen pentoxide and notice where the 2 goes here we're expressing our rate for nitrogen dioxide right we had a 4 for our coefficient right so the 4 goes in here and for oxygen for oxygen over here let's use green we had a 1 all right so I could have written I could have written 1 over 1 just to show you the pattern of how to express your rate
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