Integrated Rate Law...

Added: 2026-05-12

[00:00:00]So, hello guys, good.

[00:00:03]So, let's quickly treat on chem 101 integrated rate law.

[00:00:08]Integrated rate law. I believe at the end of the video you are going to understand.

[00:00:13]And don't forget to subscribe to the channel, drop a comment and like the video. Now, let's start with zero order reaction.

[00:00:23]Zero order zero order reaction. So, um we have this is the formula. We have AT equals to minus KT plus A naught.

[00:00:42]This is the concentration at time T.

[00:00:45]This K This K is the rate constant.

[00:00:51]This K is the rate constant, this time, and this is the the initial concentration, initial concentration, concentration at time T, the rate constant, and the time.

[00:01:01]Now, we can liken this to the normal equation of a straight line that we know, Y equals to MX plus C. So, where we can liken AT as Y, um minus K as M, minus K as M, while AO as what? As C.

[00:01:18]Sorry for something.

[00:01:27]Yeah. So, AT can be Y. As in you are plotting a graph for this, on the Y axis you plot your AT.

[00:01:34]On the on the X axis you plot your time.

[00:01:37]So, um your slope will be minus K. Your slope will be your minus K, while um your intercept be your initial concentration. So, that's the zero order.

[00:01:51]Where's this duster?

[00:01:52]So, that's the zero order. Now, for the first order for the first for the first order reaction, we are still going to solve questions involving calculations on this.

[00:02:03]So, for the first order reaction, for the first order, we have our ln AT equals to - KT + ln A0.

[00:02:16]Where um this is the ln of the um of the concentration at a particular time T. This is the rate constant. This is the ln of the concept of the initial concentration. So, this is the formula of which you can also liken to the um equation of a straight line Y equals to MX MX + C. So, ln A0 is C, the intercept. - K is the slope. T is the on the x-axis, while ln AT on the y-axis if you are to plot the graph. So, that's the formula for the first order. Now, for the second order, for the second order, we have 1 over concentration of AT equals to KT + 1 over um A0.

[00:03:07]So, there's no negative here for the second order, and there's no ln. We only have the inverse of the of It's just like the zero order, but the inverse, except this aspect, and this one does not have negative. So, you can also liken it to the equation of a straight line Y equals to MX + C.

[00:03:25]1 over A0 is your intercept C. Your K is your M, your T is your X, while the whole this on the y-axis. So, that's that formula now.

[00:03:35]That's everything you need to know about the formula. Now, let's solve a few questions on what you have done so far.

[00:03:41]So, now, this question on is on zero order. And and there is something you need to know.

[00:03:46]For zero order, let me write all the three formulas that I wrote initially.

[00:03:51]It is very, very important.

[00:03:53]It's very, very important. And something you can be tricked by it. Now, let me say the the zero order the formula is what?

[00:04:03]Concentration at time t equals to what?

[00:04:04]- kt plus what? Concentration initial concentration. For the first order we have ln of the concentration at time t equals to - kt plus what? ln of the what? Initial what? Concentration.

[00:04:18]All this can be These are the linear form of integrated rate law. Like that's why we liking them to be equals to y equals to mx plus c and the one I did earlier, so. For the second order, this second order I don't think you have been taught about, let me just add it. One over concentration of concentration at time t equals to kt plus what? The the reciprocal of the initial concentration.

[00:04:41]Now, one thing you need to know is that for here this is concentration. You know the unit it will be in mole. This one too, the unit will be in mole.

[00:04:53]It >> [snorts] >> So, this one too, the unit will be in mole.

[00:04:57]Now, the unit here there is no unit here. Why? Because you are taking a natural log of something.

[00:05:03]ln of whatever whatever it is, the whole result here it will just be like a number. There is no Do we Okay, let me let me tell us what I'm saying. I just want to show us about this unit.

[00:05:16]Now, here if I'm to make k the subject of the formula, this one will be from this one you know AO will come to the other side it will become AT conc of AT minus what? conc of AO, right? Equals to what?

[00:05:30]Okay, sorry, sorry. Let take AT to the other side. If KT comes to the other side, you know it's minus it will become what?

[00:05:37]Plus I mean it will become KT.

[00:05:40]So, there's no minus again. KT will give us what?

[00:05:43]conc of AO then this one will come minus what? Conk of what? AT, right?

[00:05:49]Now, this is conk. The unit is mole. The unit is mole. If you subtract them, you still get mole, are we?

[00:05:55]So, if you make K the subject of the formula, D here will be mole over time.

[00:05:58]That's mole per second. So, the unit for K here will be mole per second. Do we get?

[00:06:03]You subtract this from the D, unit is still mole. Because this is conk, this is conk. If you subtract two force from each other, the unit will still be Newton. So, just like this. So, if you make K the subject of the formula, that means you divide this place by T, you have mole per second as the unit for rate constant. Now, for this, that's for zero order. For zero order, the unit for the rate constant is mole per second.

[00:06:24]So, if they mention zero order, and you you have a a different unit for your rate constant, then that question is wrong.

[00:06:32]It's either you leave it it it's it's not a mistake, it can be a trap. So, just know that. Now, for first order, for first order, the unit is just per second. How? Now, let's make this formula. If this comes to the other side, it will become KT equals to what?

[00:06:50]Lin what? Lin AO.

[00:06:52]If this one comes to the other side, minus what? Lin what?

[00:06:55]AT, right?

[00:06:57]All right. This will transform this to this. So, this one becomes KT equals to what? Lin.

[00:07:03]Now, if you have Lin A minus Lin B, same thing as Lin A over B. This one give us what?

[00:07:09]AO over what? AT. Now, whatever you have here, because it's inside the logarithmic function, Lin of whatever it's here, whether it has unit or not, the whole answer will not have a unit because you have attached it to a logarithmic function.

[00:07:24]Just like you are saying sign.

[00:07:27]Since it's inside an angle, whether it's even force that is inside the angle, since it's already an angle, everything will takes the identity of degree. So, this one now, since there's no unit for logarithm, Lin is natural logarithm. So, Lin of whatever we have in here, even though the unit is concentration like mole.

[00:07:45]Do we get? So, not mole amount in mole.

[00:07:48]I mean big M like um mole per dm cube or mole per cm cube. Since the unit is big M mole per dm cube, since it's attached to logarithm logarithm function, then it won't have unit at all anymore because you are taking learn of it. Do we get? So, now whatever we get here, there will not be unit. So, if you are to make it the subject of the formula here, then the unit for K will now be what?

[00:08:10]per seconds alone. You see, since whatever we have here doesn't have unit, so divided by T, then you just have per seconds. For first order, for zero order, the unit for the rate constant is mole per like that big M, not mole not this mole.

[00:08:26]It's like this big M which represent which can represent mole per dm cube.

[00:08:29]So, I will just use the word mole, but you understand what I'm saying.

[00:08:32]So, M per seconds for zero order. For first order, it is just per seconds. Do we get? Now, for second order, you can you can do it yourself. But, for second order, it is um per M per second for second order.

[00:08:47]Please take note of those unit. They are very important. Don't take them for granted. Now, let's quickly solve this question.

[00:08:53]A zero order reaction has rate constant of this. So, our K equals to 0.02 mole M per seconds and initial conc of 0.75 M.

[00:09:04]Initial conc that's A0, right?

[00:09:08]Equals to 0.75 what? M. Calculate the conc after 25 seconds. T equals to what?

[00:09:14]25 what?

[00:09:15]seconds.

[00:09:17]Now, what is the formula for zero order?

[00:09:20]Don't mix formulas together. The question is talking about zero order.

[00:09:23]Even though if the question does not mention zero order, you can see the unit of your rate constant. That's another thing you need to know. Please take note of that.

[00:09:31]So, here we have um So, looking at the end I'm not even looking whether it's zero order, but looking at the unit mole per second in big M per seconds.

[00:09:40]No, mole is actually different. This capital letter M represents mole per dm cube. Take me, but don't worry. So, the formula for the initial for zero order is um AT conc of concentration at the time T equals minus KT plus what? Concentration the initial concentration, right?

[00:10:00]So, you know, what are we asked to find?

[00:10:02]Calculate the conc after after So, we are looking for the conc at a particular time T after 25 seconds.

[00:10:11]At a particular time.

[00:10:14]So, the formula directly is just what you are looking for. Do we get that? So, equals to minus This is what we are looking for. K was our K? 0.02.

[00:10:22]What's our time? 25.

[00:10:24]Plus what's our initial What's our initial um condition? 0.75.

[00:10:31]So, are we there? So, now 0.02 * 25 0.02 * 25 That's 0.5. So, this don't give us minus 0. 0.5 plus 0.75.

[00:10:45]So, what do we get at the end of the day? This don't give us 0.25 big M.

[00:10:51]This This This unit represents concentration. So, that's that for that now.

[00:10:58]For the second question So, um this is another question.

[00:11:02]A first order reaction took 500 seconds for it to decrease to 25% of its initial value. What is the rate constant for the reaction? Now, our time equals to 500 seconds.

[00:11:17]Um It It took 500 seconds for it to decrease to 25% of its initial value.

[00:11:26]So, the initial concentration Let's assume that it is one.

[00:11:33]Or you can also represent it with 100%.

[00:11:37]Whether one. So, if you want to write the 25% of one now, you say 25 over 100 * 1. It gives you 1/4. You see that you write 100% and you leave 25% for your 80. Or you write one and you write 1/4 for your 80. Either way, you still get your answer.

[00:11:53]So, okay, let's just use 100%.

[00:11:56]So, like this is the This is the um the initial conc. Like nothing has happened. Um do we get? No decomposition here. This the initial concentration.

[00:12:07]100% now. A first-order reaction took 500 seconds for it to decrease to 25% of its initial. Now, that means A at a particular time T.

[00:12:20]Now, listen. What we need here now is um how can I put it now?

[00:12:24]Sorry.

[00:12:25]What we need here is um the conc at a particular time T.

[00:12:32]The conc at a particular time T is that part that is yet to be used.

[00:12:38]In a sense now, you know, the thing decreased to 25% of its initial value.

[00:12:44]This 25% is no longer conc.

[00:12:47]It has been removed. You know, the the initial concentration is 100.

[00:12:51]So, and the when the reaction took place, it decreases it decreases to 25% of this. That is, we still have um Now, this just it. How to find that conc at a particular time T is going to be 100, the initial one, minus the one it decreased to.

[00:13:11]100 - 25, which is what? 75 percent.

[00:13:18]Which is 75%.

[00:13:22]Oh, sorry. It's not It's not this. Sorry. sorry. That's That's a mix-up.

[00:13:31]This just 25%.

[00:13:34]Yeah, yeah.

[00:13:35]That's a mix-up. Don't mind me. A first order reaction took Now, let's understand the question very well. So, a first order reaction took 500 seconds for it to decrease to 25% of its initial value.

[00:13:52]Now, this is it. Initial value is 100%.

[00:13:56]At a time 500 seconds, this already this concentration already decreased. So, the concentration at a particular time is 25%. Just get it right. There's no confusion there. Sorry for the initial.

[00:14:10]Let me repeat again. Let me repeat myself. A first order reaction took 500 seconds for it to decrease to 25% of its initial value. Isn't that what you're using one here and one over four here?

[00:14:24]You're using 100 here and you're using 25 here.

[00:14:27]Sorry for the initial. When I get the one we do something like that that I did earlier, then I will I will refer back to this and show you the differences.

[00:14:35]There's actually different. The differences are obvious. Sorry about that.

[00:14:39]So, this is the initial condition initial concentration. Now, at a particular time T, this 100% is already reduced to 25% according to the question. Decreased to 25% of the initial.

[00:14:51]Do we get it? So, at a particular time T, we already have this 100 to have reduced to 25. So, what is the rate constant from the reaction? So, this just what we want to do this. So, here we have conc of uh our AT, you know, equals to minus KT plus what?

[00:15:10]A not, right? So, what's our AT?

[00:15:13]Our AT is 25 equals to minus KT, 500 seconds. That's 500 K plus what?

[00:15:23]100. Is it it?

[00:15:26]Oh, we are told that first first order, learn.

[00:15:30]Yeah, that's learn.

[00:15:32]Learn A not.

[00:15:34]Don't mix the formula together. The initial formula I use is for zero order.

[00:15:38]So, this is first order.

[00:15:40]Please take note of that. So, first order equation is this learn learn.

[00:15:44]So, we have learn 25 this learn 100.

[00:15:48]Learn 100. So, you have to be careful during your exam, please. So, yeah, if you bring this 500k to the other side, it becomes positive. 500k equals to what? Learn 100. If this one comes, it becomes minus what? Learn what? Learn 25. Isn't it? So, this one here we have 500k equals If you have learn A minus learn B, it's the same thing as learn A over B.

[00:16:09]100 over what? 20 what? 25.

[00:16:12]Right? So, here we have 500k equals to what? Learn four. 100 / 25 is four.

[00:16:19]So, K equals to learn four over what?

[00:16:24]500.

[00:16:25]K equals to learn four over 500. Now, what's learn four?

[00:16:30]Learn four.

[00:16:31]Learn four equals to 1. 1.3863.

[00:16:37]So, we have K equals to 1.3863 over 500.

[00:16:43]Divided by 500.

[00:16:46]So, our our rate constant equals to 2.773 * 10 raised to the power minus three per seconds, not mole per seconds now.

[00:17:01]This is just per seconds. Now, the this and that trick there, in your option, you can see M per second. I'm mentioning mole. Mole is different from that big M.

[00:17:10]This small M is it stands for number of amount in mole. But big M is mole per dm cube. Take note of that. So, if you if you hear me mentioning mol, I mean mol per dmq.

[00:17:20]Because I use just big M. So, now if in your option you have something like this, for rate constant unit for first order is wrong. It's per seconds and I've explained it earlier in the video. This unit for rate constant is for zero order. This one is for first order. The one for second order is per M per S.

[00:17:42]And this M this M is same thing as small M per dmq.

[00:17:49]This capital big M is equivalent to small M per dmq.

[00:17:54]It can also be M per centimeter cube.

[00:17:56]Small M per centimeter cube. So, take note of that. Now, the next question.

[00:18:03]So, this the next question that seems to be mixed up with the one we just finished. Now, a first order reaction is 50% complete in 30 seconds. So, our time is 30 seconds.

[00:18:16]Now, our concentration at a particular time T.

[00:18:21]Now, concentration is equivalent to and that's the mix-up I made earlier.

[00:18:27]So, listen now. That one we just we grabbed the one we just finished. Now, for this one, concentration is equivalent to um No, the first one is decreased to 25%.

[00:18:40]Is that 25% that has not been used? So, that one is still equal to the conc at a particular time T. So, that 25% is what we should use earlier. That was correct, what we did. Now, this one now the amount of um the the the incomplete now.

[00:18:55]The amount of the incomplete part is equivalent to concentration. So, now a first order reaction is 50% complete.

[00:19:03]If it is 50% complete, that means there's a 50% that is incomplete. So, that incomplete is equivalent to conc to concentration that we should have. So, now our concentration at the particular time T will give us 100% - 50% like the whole the initial concentration which must be 100% minus the the complete 50% to give us incomplete 50%. So, that incomplete is our concentration at the particular time T. Assuming this is this is 30% complete, it will be 100 - 30. So, our concentration at the particular time T will now be 70.

[00:19:41]If it is 30% complete, then we have 70 incomplete. That incomplete is equal to our concentration at the particular time T. So, which is what?

[00:19:50]50%. So, our concentration at the particular time T is 50%.

[00:19:54]So, what is our initial concentration?

[00:19:57]100%. We are not given the value, but it's 100% because at that time nothing has happened.

[00:20:04]So, how long will it take for the reaction to be 75% complete? Let's leave that for now.

[00:20:09]Let's leave that for now. So, what before we can get that question, the question we are asked, we need to first of all find the rate constant.

[00:20:18]So, this is the first order. The formula for first order is ln AT equals to what? - KT plus what? ln what?

[00:20:28]ln A0.

[00:20:30]It's concentration at the time T, rate constant, time, and initial concentration.

[00:20:36]So, what's our AT? 50. What's our A0?

[00:20:38]100.

[00:20:39]So, this will give us ln 50 equals to minus our time is 30 seconds.

[00:20:47]30K plus ln what? ln 100. So, here we have 30K. If it comes to the other side, it becomes positive. It equals to what?

[00:20:55]ln 100 minus what? ln what? ln 50, right?

[00:20:59]So, here we have 30 30K equals to what?

[00:21:05]ln ln A minus ln is Lean A over B. 100 over what? 50.

[00:21:11]So, K equals to Lean 100 over 50, that's Lean 2. Divided by what? 30.

[00:21:17]Isn't it? So, I think Lean 2 should be 0.693 divided by 30 divided by 30 That is 0.023.

[00:21:28]So, our K equals to this 0.693 divided by 30 equals to 0.

[00:21:37]0.023.

[00:21:40]This first order, that will be what? Per seconds, not M per seconds. Let me write it here. K equals to 0.0.023 per seconds, which is same thing as 2.3 * 10 raised power minus 2 per seconds.

[00:21:54]Please take note of that. Per seconds of first order. For zero order M per seconds. For second order, per M per seconds.

[00:22:02]Where's this duster?

[00:22:05]So, now this is not yet what the question is asking us to do. The question says, "How long will it take for the reaction to be 75% incomplete?"

[00:22:14]You know this rate constant, this one is always constant. Whether whatever is happening, this constant. Whether the time is very Whether the time varies, this constant. Whether more varies, this one is constant. So, since it's called rate constant we still leave it that way, not touched. 2 times 2.3 * 10 raised power minus 2 per seconds.

[00:22:33]So, What was our initial condition, initial concentration rather? It's still 100%.

[00:22:39]So, we we asked to find time when it is 75% complete. So, if it is 75% complete, that means it will be 25% incomplete. 100 minus 75. So, we have 25% incomplete. That incomplete is equivalent to the concentration at a particular time T.

[00:23:01]So, since this is first order Sorry, what's going So, here we have learn learn AT equals to minus KT plus what?

[00:23:10]Learn what? Learn A not, right? So, here we have KT equals to since we have our If this come to that side, it become positive. Since we have our K Our K is 2.3 which is the same as 0.

[00:23:23]023 T equals to learn this 100, right?

[00:23:29]Learn 100 minus learn 25 learn 25 So, this will give us 0.023T equals to That's learn 100 over 25. Learn A minus learn B is learn A over B. So, this will give us learn four.

[00:23:45]Right? So, here we have T equals to learn four over what? 0.023 So, learn four over 0.023 Learn four divided by 0.023 So, that's 60 equals to 60.27 seconds. So, at time 60.27 seconds it will be 75% complete, which means it will be conquered that time will be 25%.

[00:24:12]That is 25% incomplete. Now, let's solve the second to the last question before we call it off.

[00:24:18]So, now before I solve this question, I will refer back to second to the last one I did that they said decrease to 25% of its initial value. So, that AT there is equivalent to 25% It's different from the one we just finished.

[00:24:36]Complete incomplete. Please get the difference. So, here now we have our The rate constant for the first order reaction at time T is So, our K equals to 2 * 10 solution 2 * 10 raised power minus one per seconds If the initial conc is That's our initial concentration equals to three times 10 raised to power minus three.

[00:25:01]3.0 is three. So, what is the conc after three half life?

[00:25:07]So, we need this, right? Unknown. [snorts] The concentration at the particular time t. And what's our time? Our time is three half life.

[00:25:18]Three half life. This is the time that we are given, three half life. So, because this um first order and even the unit of the rate constant will make me to know that that's actually first order. Please, those units, they are trap. Take note, please.

[00:25:34]So, here we have ln At equals to minus kt plus what? ln what?

[00:25:42]A not.

[00:25:43]Are we together?

[00:25:45]So, now what are we asked to So, we need to find our t. Okay, let's write it first. ln Are we given Now, here we have ln Okay, okay, okay.

[00:26:02]Um Okay.

[00:26:05]ln At equals to um minus minus two times 10 raised to power minus one t.

[00:26:17]Now, our t is three half life plus ln ln What's our A not? Three times 10 raised to power minus three.

[00:26:26]Right?

[00:26:27]Yeah, we are given A not. What we are given is not ln A not all together. We are given A not. It's the initial concentration, initial conc That is this. And the formula is ln of the initial conc. Take note of that. So, we are asked to find the the conc, this At, not ln of the At.

[00:26:47]It's ln At conc conc that we are looking for.

[00:26:50]So, now let's find our T first. Now, half-life equals to 0.693 over K.

[00:26:59]Over K. And what's our K?

[00:27:01]2 * 10 raised to the power -1. That's formula for half-life.

[00:27:06]Do you get you know in normal physics that we know we say over lambda.

[00:27:10]This is like a lambda there now. 0.693 over lambda 0.693 over K. So, over your rate constant. So, here now we have um 2 * 10 raised to the power -1.

[00:27:22]That's 0.2. So, it's 0.693 divided by 0.2.

[00:27:27]That's 3 point So, our half-life is 3.465. If you divide this, you get 3.

[00:27:33]465.

[00:27:35]So, since our T equals to three half-life, so it's going to give us 3 * 3.465.

[00:27:45]* 3. So, our time is what? 10.

[00:27:49]So, our time if you multiply it, our time now three half-life three times this we have 10.4.

[00:27:55]10.4 seconds.

[00:27:57]10.4 seconds. So, that's our half-life.

[00:28:03]So, now here now we're going to have ln AT equals to 10.4 times this times times 2 * 10 raised to the power -1.

[00:28:14]10.4 times 2 times 10 raised to the power -1.

[00:28:19]That give us 2.08.

[00:28:22]So, here we have minus 2.08.

[00:28:26]This time we've looked for our time plus what? ln 3 * 10 raised to the power -3.

[00:28:32]ln of this, let's type it in our calculator. 3 * 10 raised to the power -3.

[00:28:40]ln of the answer.

[00:28:42]So, that's that's negative negative 5.81.

[00:28:50]So, here we have So, here we have um let me clean all these parts so that we can have more space.

[00:28:58]So, here we have learn 80 equals to minus this minus this. So, minus 2.08 minus 5.81.

[00:29:08]Minus 2.08 minus 5.81.

[00:29:13]That give us minus 7.89.

[00:29:15]So, here we have minus 7.89.

[00:29:21]So, how do we find 80 now? Here we have learn 80 equals to this. So, you take exponent of both side. In mathematics, if you have e raised power learn x is equal to x.

[00:29:32]It's just like this learn we cancel e.

[00:29:34]They are like opposite to each other, something like that. It's just like you are saying y plus minus y. So, they cancel. That's what happened to here and learn. It will remain x. So, how do we have 80 here is by saying e raised power learn 80.

[00:29:51]Since you say e raised power this side, you must also say e raised power this side so that you not alter this this structure the the question. So, since you are taking exponent of both side, then we have not done anything but since what you apply here we also apply it here. So, this side since this will go like I said earlier, that's not the way it cancel but me I'm just but it goes you have 80 left. So, here we are taking exponent of this.

[00:30:17]So, that's if you in your calculator, how do you press e? That's shift learn or arc learn, whatever. So, shift this shift and this learn.

[00:30:29]Shift learn cancel it will turn to e.

[00:30:30]Here you have shift learn. That's you can see e raised to power minus 7.89.

[00:30:41]So, I'm coming.

[00:30:46]It's showing error here. This is how we are going to do it then equals to 1 over e raised power 7.89.

[00:30:54]Maybe it's this type of calculator. You know, if you take it up, it will become negative. So, let's do it this way.

[00:30:59]I think this calculator can manage this or not. Shift link 7.89.

[00:31:09]Okay, okay, okay. Sorry, sorry. We don't need to put power.

[00:31:13]That's the um sorry.

[00:31:15]If you want to use this kind of calculator to press it, don't put power.

[00:31:19]Just type it directly. Don't put power.

[00:31:21]That's shift link shift link minus 7.89. Don't put power. If you want to type it exponent in your calculator, don't put Don't put this power. Don't put it. Just say e then after your shift link, then you type minus 7.89.

[00:31:41]So, we can see now. If you put power, whether it's positive or negative, you put power, it will bring syntax error. I don't know, maybe that's how it is for every other calculator.

[00:31:50]But for this calculator, don't put power.

[00:31:52]So, here we have our conc at that time is what? At a particular time at three half life is 3.7 times 10 raised power minus four mole per dm cube, which can be written as big M.

[00:32:10]So, that's all about zero and first order.

[00:32:14]Now, for second order, well, let me just For second order, let's say if k equals to 0.02 per m per s. Yeah, this is the unit for second order. And our t equals to I just want to brief it. Our t equals to 10 seconds.

[00:32:32]And we have our initial condition equals to 0.5 mol per dm cube and we have to find initial concentration rather. We have to find the concentration at a particular time t. So, for second order we know it's 1 over this equals to um kt. That one is positive, no ne- there's no negative there plus what? 1 over this.

[00:32:58]That is the formula. Right now, here we have 1 over this at equals to what's our k? 0.02. What's our t? Times 10 equals to 1 over what's our initial concentration?

[00:33:11]Sorry, plus 1 over 0.5.

[00:33:15]Do we get? Just input. It's not there's nothing there's no big deal there.

[00:33:20]1 over at equals to 0.02 * 10 that's 0.2 plus 1 divided by 0.5 that's 2.

[00:33:29]Right? So, here we have 1 over at equals 0.2 plus 2 that's 2.2. So, how do we find at? Cross multiply. So, here we have 1 equals to what? 2.2 multiplied by at, right? So, make at the subject of the formula. That means at equals to 1 over what? 2.2 equals to what?

[00:33:51]1 divided by 2.2 So, equals to 0.45 45.

[00:33:58]So, even though you are given a second order question, but they are not going to I mean for under level four level one, they are not going to but I just brief it even though you come across it. So, thanks. That's the end of the video. Don't forget to like and subscribe and drop your comment in the comment section. See you in the next video.