Daniel Evans: So we covered rates and rate stoichiometry.
00:15 - Daniel Evans: method of initial rates. Daniel Evans: Now we’re going to look at three orders so we can get Daniel Evans: Different exponent on our Daniel Evans: Reactions for a rate all Daniel Evans: The most common one is going to be zero order first quarter second quarter. There are other uncommon ones fractional order negative orders, but these are the common ones.
00:50 - Daniel Evans: So they’re read law. Daniel Evans: For zero order reaction is rate equals cost of times to castration zero order at zero order makes the concentration disappear, and we get rate is equal to a constant.
01:07 - Daniel Evans: For first order rate, rate is equal to a custom times the class situation to power one so rate.
01:17 - Daniel Evans: cost in terms of concentration for second order rate law you have rate equals constant times a concentration to the power to Daniel Evans: So these are the three most common right laws and we’re going to Daniel Evans: integrate those right boss.
01:37 - Daniel Evans: So these rate laws and middle column, we can call them differential rate loss.
01:44 - Daniel Evans: Weight loss in the right column are the integrated raid boss.
01:49 - Daniel Evans: So when we have zero order reaction.
01:53 - Daniel Evans: Concentration equals minus Katie Daniel Evans: Log of plus the initial concentration Daniel Evans: So that has integrated rate loft for Daniel Evans: A zero order reaction.
02:10 - Daniel Evans: So the rate does not depend upon concentration we subtract that from the initial concentration to wash the concentration in time.
02:20 - Daniel Evans: This is a form of y equals mx plus b. So, Daniel Evans: Wise concentration M is native case, the slope is a negative rate constant x is time and the intercept is the initial concentration Daniel Evans: So that is a zero orders.
02:41 - Daniel Evans: Per first order reaction, we end up with lager, the concentration equals minus k t plus lager, the initial concentration Daniel Evans: So again it has form of a y equals mx plus b. Why is log of a slope is negative k x. This time he interceptors is log natural log of initial concentration Daniel Evans: For a second order reaction, the integrated rate laws one orchestration equals Katie plus one over the initial concentration is still has a form of y equals mx plus b.
03:24 - Daniel Evans: So why is one over a slope is k and the intercept is one over the new construction.
03:33 - Daniel Evans: So each of these orders has a different form for its linear equation. So your order is concentration versus time is linear.
03:44 - Daniel Evans: First orders lot of concentration versus time is linear and second order is one orchestration person time he’s linear. So we can use these differences.
03:56 - Daniel Evans: In what is the most linear graph to determine the order of the reaction we of course need sufficient data to comply graph, if we have that data, we can get the order of the reaction.
04:12 - Daniel Evans: So we take data forgot concentration log castration in one overconcentration and plot all three of these Daniel Evans: And determine which ones. Most linear. So in this example here, the one overconcentration is the most linear car is deviation for longer castration obviously deviation for concentration. So that would mean that we have a second order reaction. So that’s how this Daniel Evans: Graphing method works. We have to do all three graphs and determine which one is the most linear Daniel Evans: And what’s not shown on the graphs here is really good to get the correlation coefficient, along with the equation, the correlation coefficient, the closer it is to one more negative one, depending on your software.
05:07 - Daniel Evans: Makes it linear and if software gives both positive, negative, positive for positive so negative is for native stuff.
05:19 - Daniel Evans: So we can use this graphing method with the integrated weight loss to determine the order of the reaction.
05:31 - Daniel Evans: So just a little bit more Daniel Evans: About how the half life behaves will give you the equation for the fights that momentarily. So for his order reaction that rate is costing just drops down and it has a hard Daniel Evans: Endpoint. So a racist toward that zero Daniel Evans: So it says the straight line from the initial classification down to zero point Daniel Evans: So our first half life is half the time.
06:05 - Daniel Evans: Second half life is a quarter of the time. Third, half life.
06:10 - Daniel Evans: eighth of the time. So our half life escapes getting shorter and shorter and shorter and shorter and Nicholas microscopic in size but we’re racing for hard in point on this one.
06:24 - Daniel Evans: So the Daniel Evans: Slope has a negative slope. So that means slope is negative k the right customer is always positive.
06:32 - Daniel Evans: And the slope for all three of these is Daniel Evans: Just positive and negative. Right, so the rate cost is always the absolute value as a slope.
06:49 - Daniel Evans: For first quarter we actually approved pot concentration first time is not linear has zero or was Daniel Evans: But it’s getting less and less deep as we go along.
07:00 - Daniel Evans: But what we’ll find out. Daniel Evans: Momentarily with the equations that they have life is a fixed time Daniel Evans: So we have the same time. So in this case I have 500 seconds 200 seconds be to half life 363 half life. So the half life is a constant. So our rate keeps getting slower slower and we can’t point toward a hard endpoints. This keep the body until those funding nothing left.
07:29 - Daniel Evans: But they have a physical constant for first order reaction.
07:41 - Daniel Evans: For a second order reaction. Daniel Evans: Is slows down faster than does for a first order reaction. So what we find that half life keeps increasing.
07:54 - Daniel Evans: For Success of half lives here. It keeps doubling Daniel Evans: So we have a half life next half life is twice the time the next half life is twice again. So we keep doubling the time of the half lives for a second. What are we actually so Daniel Evans: We have our Daniel Evans: Particular plot that will be linear. We also see that the have five behave differently. So this is our summary here. So for zero order reaction. This rate equals a custom Daniel Evans: The integrator rate was causing this concentration equals minus Katie plus initial concentration Daniel Evans: So that means our most linear plots because filtration versus time.
08:41 - Daniel Evans: slope is negative k. So, okay, will be a Daniel Evans: negative slope or absolute value is slow.
08:53 - Daniel Evans: So for zero reaction in the rate does not depend upon concentration. So if we double the concentration rate means the same are half like we saw it from the integrator rate law we get that the half life is initial concentration and do it by two k Daniel Evans: So the bigger the initial concentration. The larger are half like the smaller the initial concentration smaller Daniel Evans: So if we double our concentration we double our half Daniel Evans: As we go to smaller concentrations success of half lives.
We have the time of the initial Daniel Evans: Body.
09:37 - Daniel Evans: For first order reaction. Daniel Evans: rate equals k times castration first power, sorry I don’t have my powers up there right spot.
09:49 - Daniel Evans: The integrator rate laws natural log of concentration is equal to minus k t plus the natural log of initial concentration. So that means log of castration first time will be our most linear plot our slope is negative k. So, k is the absolute value slope.
10:08 - Daniel Evans: If we double our concentration Daniel Evans: Her rate is going to double. So we look at this rate law we double here to to the first power is still a doubles. We double our rate.
10:26 - Daniel Evans: The half life for a first order reaction is a constant is the natural log two which is 0. 693 divided by the rate cost.
10:39 - Daniel Evans: So the half life always remains the same amount of time for first reaction.
10:46 - Daniel Evans: So if we double concentration the half life remains the same and the success of half life remain the same.
10:55 - Daniel Evans: So second quarter reaction will be rate was Kate have concentration squared.
11:00 - Daniel Evans: The integrated rate. Last one overconcentration equals k t plus one over the initial concentration in the most linear line should be one of orchestration versus Daniel Evans: The slope will be equal to the right constant Daniel Evans: And if we double our concentration look up in the first rate law here. We double A that to square before. So the rate quadruples so we double our concentration great quadruples. This is what we’re using in the method initial rates to determine or raise the rates.
11:42 - Daniel Evans: Are half life. Daniel Evans: Is equal to one over k times initial concentration. So for doubling our concentration. We’re cutting our half life and a half. So double our concentration we cut our half life and a half.
11:58 - Daniel Evans: So our success of half lives as is a keeps dividing by to the half life double so success of halfway, the doubles.
12:09 - Daniel Evans: So by looking at how half lights behaves. We can also help determine the order reaction. In addition to what we use for method initial rates to concentration Daniel Evans: So let’s Daniel Evans: Go to the whiteboard and try a couple problems using these right loss.
12:36 - Daniel Evans: Okay, this makes Okay.
12:45 - Daniel Evans: So we have Daniel Evans: One. Here we have an initial concentration Daniel Evans: Of 2. 67 times in my three were told that the plot of concentration first time is the most linear with a slope of minus 0. 00132 minutes investments. So we’re asked what our great customer is our reconnaissance is always the opposite value of the slope.
13:19 - Daniel Evans: So I will this quarter. This to scientific 1. 32 times Daniel Evans: 10 to the minus three investments.
13:32 - Daniel Evans: So that’s our rate constant. Daniel Evans: Were asked for the rate law.
13:40 - Daniel Evans: Well castration first time is zero order rate.
13:46 - Daniel Evans: Right out Daniel Evans: Long form.
13:51 - Daniel Evans: And condense it down to its final form.
14:00 - Daniel Evans: So rate is degree. Okay. That case 1. 32 Thompson my three and it’s Daniel Evans: So the integrated rate law.
14:20 - Daniel Evans: P Daniel Evans: Equals minus k t plus Daniel Evans: A zero Daniel Evans: Power numbers in there. So we have a minus Daniel Evans: 1. 32 times 10 to the minus three.
14:43 - Daniel Evans: T plus zero. The 2. 67 Daniel Evans: Times 10 to the minus three.
15:02 - Daniel Evans: So we know this is your order reactions we put in the order rate law and then fill in the numbers that we have for that.
15:12 - Daniel Evans: Half Life. Daniel Evans: Is the initial customer tree as we feel in our numbers.
15:42 - Daniel Evans: And then we reduce it down. Daniel Evans: And cut out to be one minutes Daniel Evans: Last question dealing with this but information that we start off with is while we’re adding one more. So what is the concentration at 15. 1 minutes Daniel Evans: So use our integrated rate law.
16:10 - Daniel Evans: So our concentration Daniel Evans: Is r minus k.
16:25 - Daniel Evans: Times time plus us Daniel Evans: Initial Daniel Evans: Or time we have 15. 1 Daniel Evans: And we just make sure the use of matching. So, we have 15 minutes, the slope is inverse minutes so the universal castle off.
16:55 - Daniel Evans: So we run this through our Daniel Evans: Final answer. But now, Daniel Evans: Concentration at 15. 1 minutes Daniel Evans: Away.
17:54 - Daniel Evans: Equals a 6. 77 times 10 to the minus three.
18:13 - Daniel Evans: So we can try logic check our half life was 10 minutes we’re Daniel Evans: Beyond the first half life. We are half my cousin half who are close to our second half. I Daniel Evans: Put that in half. We had a 1. 3 kind of half we get between a six or seven so yes he does seem like he we have Daniel Evans: We’re Daniel Evans: Logic is not right. So, Daniel Evans: Let’s accept 2. 67 minus negotiator smaller number wise not sure on the smaller number here.
18:59 - Daniel Evans: Oh, six, seven mystery I typed in wrong. Okay.
19:31 - Daniel Evans: Cognitive Daniel Evans: Right here. This is some some Daniel Evans: Right.
19:43 - Daniel Evans: 15 half life. Daniel Evans: It’s 10 we’re not true half lives.
19:57 - Daniel Evans: Slope. Daniel Evans: Says, Daniel Evans: I’m sorry for that. But my logic check. I did their call you stay ahead.
20:26 - Daniel Evans: Hours working on it working off a piece of paper head tension lines to appear not to my street so I missed copy of my road appear Daniel Evans: So it didn’t affect our very constant a great law integrated great law that affected the half life. So I had to my Daniel Evans: To my notes and Timothy down here. Give me 10 minutes, but it’s really 1. 01 minute for the half life. So I had to give us a different time to work with this. I gave us a point six five minutes.
21:01 - Daniel Evans: So putting that into the integrated rate law.
21:06 - Daniel Evans: So we’re going to have our Daniel Evans: Point six five minutes for the time and again minutes we’ll catch off with him, first, miss. Nurse slope, then that gives us a value of Daniel Evans: 1. 81 times 10 to the minus three similarity Daniel Evans: At that point six five minutes, so we’re Daniel Evans: Not up to the half life yet so we haven’t used up half of the class integration yet.
21:41 - Daniel Evans: So another problem. Okay, I won’t have mistake on this phone.
21:47 - Daniel Evans: We have initial concentration of 3. 25 times too much to clarity.
21:52 - Daniel Evans: And our log of castration versa time ism most linear graph, it gives us a slope of minus 1. 57 times too much to inverse minutes. So what’s our HR a customer a customer is always the absolute value slow Daniel Evans: So it’s 1. 57 times 10 to the minus two inverse minutes Daniel Evans: Error rate law. Well, the natural log plot. He said, we have a first Daniel Evans: order rate law.
22:26 - Daniel Evans: So, Daniel Evans: Our rate is going to be a Daniel Evans: Constant times the concentration to the first power will be either Constance will put it in 1. 57 Daniel Evans: Times 10 to the minus two times concentration Daniel Evans: So the integrated root ball is now based on our natural log of a Daniel Evans: And it would be a minus, Katie.
23:06 - Daniel Evans: Binary face that K Daniel Evans: With the numerical value.
23:14 - Daniel Evans: 1. 57 times 10 to Daniel Evans: The minus k t times t plus the natural log of the initial concentration and we have that. So I’ll put that number in and the Daniel Evans: Right out the two steps here. So we’re doing Daniel Evans: This number Daniel Evans: Natural log of the 3. 25 times 10 to the minus two.
23:49 - Daniel Evans: So we’ll put the actual number in and that is a Daniel Evans: Minus 3. 427 so that’d be our intercept.
24:01 - Daniel Evans: The half life is natural log of to home for our a custom Daniel Evans: Natural log of two is a 0. 693 Daniel Evans: over k 1. 57 times 10 to the minus to Daniel Evans: reduce that number and we end up with a half by 44. 1 minutes Daniel Evans: So now we’re asking for the castration at 60 minutes time. So we want to use our Daniel Evans: Integrated rate loft solve this, we have our el en have a Daniel Evans: So sorry minus k 1. 57 times 10 to the minus two times time.
We’re looking at 60 minutes, two minutes inverse Minister our units are kissing off.
25:05 - Daniel Evans: Then minus 3. 47 to Daniel Evans: reduce this down.
25:14 - Daniel Evans: For national over eight comes out to be Daniel Evans: Minus 4. 369 Daniel Evans: So that are a Daniel Evans: Will be Daniel Evans: Equal to Daniel Evans: He raised the power of negative 4. 369 Daniel Evans: run that through our calculator.
25:44 - Daniel Evans: Point 0127 modularity. Daniel Evans: So we start off with that point 03 we’re down to a point 01 on this.
26:01 - Daniel Evans: So let’s do one more of these. Let me swap the board.
26:25 - Daniel Evans: Concentration of 5. 65 times in my district, we find that the plot of one overcome concentration versus time is the most linear, we have a slope, that’s Daniel Evans: Actually 4. 75 times to my free inverse seconds.
26:46 - Daniel Evans: Point. Daniel Evans: Number one here.
26:54 - Daniel Evans: Okay, sorry. Daniel Evans: So our great customers always the absolute value. So 4. 75 times 10 to the minus three and four seconds.
27:06 - Daniel Evans: So resolve the one over concentration first time means that we have second order rate lot so it’s Ray cousin times concentration squared. And of course, we can write this with the value in here for break constant 4. 75 Daniel Evans: Times 10 to the minus three.
27:28 - Daniel Evans: Times concentration, squirt. Daniel Evans: So integrated rate law.
27:36 - Daniel Evans: That would be the one overconcentration Daniel Evans: Can we see a positive slope here. So it’s a passive Katie Daniel Evans: Plus one over the initial concentration Daniel Evans: can stick in our values for that. So we have a Daniel Evans: Okay, of 4. 75 times 10 to the minus three and verse seconds.
28:08 - Daniel Evans: Time. Time. Daniel Evans: Plus Daniel Evans: Up here, so one over the initial concentration 5. 65 times 10 to the minus three and that older down here and it’s Daniel Evans: 170 7. 0 Daniel Evans: So our integrated rate law one over castration equals 4. 75 times two times three times time plus 1770 so our half life.
28:53 - Daniel Evans: Air half life is one over there. Ray constant times the initial concentration Daniel Evans: Hello bit so Daniel Evans: Our numbers in there.
29:18 - Daniel Evans: Are re conason 4. 75 times 10 to the minus three.
29:26 - Daniel Evans: Our initial concentration 5. 65 times 10 to the minus three, we run this through our calculator, we find that our half life.
29:38 - Daniel Evans: Is Daniel Evans: 37,000 to 61 Daniel Evans: So we can take that to F3 significant is we have a 3. 73 Daniel Evans: Times 10 to the fourth seconds for our top five.
30:02 - Daniel Evans: So now we’re asking for the castration at time of 2. 5 times center for 2. 5 so we’re not quite at half life. So I have more than half the compound left. So we’re going to use our integrated reply here. So we have one over the custom curation.
30:22 - Daniel Evans: equals the rate constant 4. 75 times 10 to the minus three.
30:30 - Daniel Evans: Times a time Daniel Evans: Are 2. 5 times 10 to the fourth seconds.
30:39 - Daniel Evans: Plus 177 Daniel Evans: So we run that through our calculator.
30:49 - Daniel Evans: We end up with a 290 5. 75 Daniel Evans: But this is for one over the concentration Daniel Evans: So it’s not concentration yet so to get the concentration we had to do one over this. So our concentration Daniel Evans: Will be equal to one over the 290 5. 75 which equals Daniel Evans: 3. 38 times Daniel Evans: 10 to the minus three with terms of malarkey. So that would be our final concentration at Daniel Evans: 2. 5 times 10 to the fourth seconds.
So that’s how we can use our grass, we can determine the order reaction. And then from that we can get a constant rate law half life in concentration at any time that we want. .