Three rate laws and order of reaction, uh here, if you look at this reaction, two and no plus o2 give you two nodes. The rate of reaction is depends on the reactant, uh, let's say, the general, then you have ax plus b, y x and y is your reactant a and b is a coefficient. Then the rate is proportional to the concentration of the x and concentration of the y, basically your reactant.

And the power of this concentration is m and n. We find this m and n. Experimentally. So look at that is proportional to that. But r actually equal to k concentration of x power, m, multiplication, y power. So v, k, equal k. The rate constant, uh. And it depends on the temperature. Okay, looking at that. Then we have this, um reactions.

So the r is equal to k. No2 concentration of the no2 and caution of f2, uh, experimentally, we will realize that the power of these brackets are one and one. It means the order of reaction with respect to no2 is one. And the order of reaction with respect to f2 is one. And the overall is 2 1. Plus 1 is the. Overall rate of reaction order of reaction.

Here is another example. We do experiment. We realize that the r is about k. X power.

One concentration of y power, two because we have the power is e, rho. It means uh, the rate of reaction with respect to x is the order of the reaction respect x is one respect to y is two respective. Z is zero. It doesn't then is like is that zero means that does not have any effect under ready because any number of powers that is one of the initial conditions that is.

Double then the rate of that is multiplied by one will be unchanged. If the concentration of the y becomes double two to the power of two is 0 or 8 become 4 times bigger. If the constitution of the x becomes 2 times, bigger 2 to the power 1 is 2. So the rate will be 2 times bigger. If the concentration y becomes 3 times bigger. How does it affect the rate of reaction, so it's going to be 3 to the power of 2?

This is going to be 9 times. The rate of reaction will be 0. But if this that becomes 2 times.

Bigger 2 to the power 0 is 1 or three times a two three directors is one. So means it won't impact the rate of reaction means overall rate of reaction is just k. X power. One y power, two. Because if that power is one doesn't have an effect all right here. Look at this question, the decomposition of the dinitrogen pentoxide is given the first order with respect to n2o5 means the rate of reaction is k and 205 the power of one because it's the respect to n2 pi. The order is one.

The first order means the. Power is one of the initial rate concentration is given. So the rate is given to this r. And the initial concentration of n2o5 is 0.4 concentration of and 205 is given first equal would be. If the another experiment were performed in initial concentration is 0.8.

Now knowing the rate of reaction and concentration of n2o5, you can find k after we can find check, and you can put no 205 association 0.8 to find what is the rate of reaction, okay, so the dimension or okay. So this is an example, the. Dimensionalization reaction of 1 3, butane butadiene as two c4h6 gives you c8h12 that in second order with respect to c4h6 and second order, which is basically that is c or a six to the power of two. Second order r is k multiplied.

By that I said, if the initial rate is given so r is given 32 millimoles, uh per liter per minute, make sure, uh, pay attention to this. Um units at a given initial concentration of the c for h, 6, what would be the initial rate of reaction if the initial concentrations were. Doubled so if this is doubled, of course, 2 power 2, it will give you 4 times bigger, right? If this become four times, the game is just simply multiply this one by four to get the um, the rate of the reaction here is another example when aqua promote and by sulphide ion react, produce boron mine. The overall equation is given. So here is that consider this theorist of experiment recorded in table two in which initial reactant concentration are very varied and read are compared and the evidence provided. Determined a rate equation.

So the thing is that here is the rate your concentration of hso3 because I'm going to be on road three. Okay, uh. Overall. I have this r is equal to k. Bro3 that's, the version of that power of n. Because the p orbital is your reactant hso3, concentrating the power of n here. It just so theory is your reactant now I'm using one of the two of that.

So I use the one that they have the same concentrations. These six and six, I would say concentration. So I use one and two. So let me. Just put that here or equal to k b or o3 power of m and h. So three the power of n so r for the first one is 1.6. Here is k. A b or root 3 is 4 4 to the power of m. X, 3, 3 is 6 as you see 6 to the power of n. And the second data is 0.

Zero point, eight that's, two to the power of m. Here, 6 to the power of n. If we divide both sides so 1.6 divided by 0.8 is 2, k, divided by k. It will cancel here. You have 4 divided by 2 to the power of m. Also, n. N, 6, power and x 1 are divided by each other. It is one. So from here. Two is equal to two to the power of m. Both sides have the same base. So the power should be the same. So m is equal to one.

Okay. Now we do the same thing, but this time I'm going to find something that the bro3 is the same because I use this two for so theory to business thing. So the second one, and the third one is good, you have a saying of that. So again, r of that is zero point.

Eight zero point, eight is equal to k euro to resist two to the power of m. And here is 6 power of n. Here is the rate is 0.2. Equal to k, uh 2, power, m. Here 2 power and then 3 to the power of n. So 0.8, divided by zero point. Four zero point two is four k. Divided by k is one two m, divided by twelve is one six power and divided by three power is six divided by three to the power of m. So 4, you can write it as a 2 to the power of 2. 6, divided by 3 is 2. So it's 2 power n since the bases from both are the same. The powers should be the same. So n is equal to 2.

Power is the same. So I got m and n now I'm going to find k. I can. Use one of this information just over okay, maybe last one so 0.2 is your rate equal to 2 to the power of m. You got m to be 1 and then 3 to the power of n. We got n to be 2. Here. I forgot to put k so k here.

So k is equal to 0.2 divided by 2, divided by 3 power 2 is just used your calculator, find the k to be 1.1, 10 power 4. And then you have to uh write down the velocity instead of k. You put that number v or three to the power of one that's, just so three to the power of two. So the overall rate of. Reaction is one plus two is three. Here is another question, mixing an aesthetic solution, containing IDATE, Io theory ion with another solution containing.

So this is the reaction we have. So we have a theory reactant when you want to write the rate laws, that's going to be OK Io 3 to the power of a negative to power BH, the positive power c. So basically, these three the power of ABC. Now we need to find out the rates rate laws. So the same thing I did in the previous examples. We find the one that me. Have the same see where these two and these two are the same.

So I take one and two to find the power a here I'm going to do it I'm taking these two. So the rate is 5, 10 power, negative 4 equal to here. 0.1 is k. We have k 0.1 to the power of an um 0.1 to the power of b and 0.1 to the power of c. The second one, we have 1 10 power negative 3 equal to k 0.2 to the power of a 0.1 to the power of b 0.1 to the power of again. We divide this 5 divided by 1 is 5, 10 power, negative 4 divided by 10 bonding. So. The overall of these divided by that is 0.5 so k, divided by k will go away.

Zero point, divided by that's going to be one over two to the power of an is will cancel this divide by the cancel. So as you see 0.5 is 1 over 2 equal to 1 over 2 to the power of a both sides, the base are the same so equal the powers. So here is more. I got a then I can choose, uh, the other one see this two are the same this.

And this is the same and these are different. So if I use these two, I do the same thing I divided, I can. Find b, right, also if I use this, and this is the same this.

And this is the same then these two are different. So if I is you use the 3 and 4 do the same thing, I can solve for c and the same way when I have a b and c, I use one of the information to solve for k. The same thing I did in periods. Previous example, I want to also go over that solution and do it and check the answer. So for praxis, you do quotient number three and six. So relating reaction rates to time. Okay, when we're talking about the half-. Life what is the half-life is if you have the decaying?

So it can start like that at the beginning, you are 100 the time it read the time when your material decay to the half of the original 50 percent. We call it t of half t of half life, or sometimes we call it h. So the equation for the half-life is an amount at time, t, equal to a zero. A zero is initial amount half t, divided by half life. I instead of the t, one over two, it would itch. It is a period that or the time that it takes for.

Material to decay to half of its original. So you have that you have here is another one that related the rate constant into the thing. So k, the rate constant multiply by half life is equal to 0.693. The radius, the isotope that has a half-life of 10.6 hour. What is the rate constant easy chemotherapy half is that? So you have half life, which is 10.6 hours. You just stop that end.

And you will have a rate constant here. If the mass of an antibiotic in a patient, what mass antibiotic will remain at 36. Hours if the half-life is 2 and no further drug. So what can you do?

You can use the equation an is equal to a zero one over two. T, divided by h is half life. So the mass of dendrite is the initial mass is two point. Um, what mass interval will remain after the six-hour means I'm looking for an amount after time? T, a zero is two point.

Four, six, four, like we have the time is two hours. Sorry. The time is six hours. Half life is two is 1 over 2 to the power of 3. Okay. This 2.464 multiplied by 1 over 2 with.

The power of 3 you use your calculator. You'll get thirty-one gram because this isn't gram. So that answer will be in gram.

Okay. So to practice, you can do question number, seven.