Activity 1. Simple Activity on Factors Affecting the Rate of Reaction Objective:At the end of this activity, you will be able to determine the factors affecting rea...

The use of iodine crystals is a popular way of making small quantities of water safe to drink. Crystals placed in a 1 -ounce bottle of water will dissolve until the solution is saturated. After saturation, half of the solution is poured into a quart container of water, and after about an hour, the water is usually safe to drink. The half-empty 1-ounce bottle is then refilled, to be used again in the same way. Suppose that the concentration of iodine in the 1 -ounce bottle $t$ minutes after the crystals are introduced can be approximated by $C(t)=250\left(1-e^{-t}\right) \quad t \geq 0$ where $C(t)$ is the concentration of iodine in micrograms per milliliter. (A) What is the rate of change of the concentration after 1 minute? After 4 minutes? (B) Graph $C$ for $0 \leq t \leq 5$.

Yeah. Then why? Be equal to f f T. NFL T denotes the amount of iodine and the blood at time. T. So given the idea answers and since the blood a rate of 4%/h, the Andean answers to your the your brain Um at a rate of 10 per hour. So the rate of change of the rate of change of iodine iodine levels of blood per hour will be dy over DT. Which is equal to negative for 100. Yeah. Tom's wide minus 10 over 100 times. Why, jesus? This becomes negative 14 over 100 times. Why? And the the initial condition Y. of zero. Yeah, Becomes f of zero mm. Here the negative son indicates that the iodine levels decreasing as time progresses. Yeah.

All right. So, in this question, it involves the analysis of a tablet of potassium iodide H Right. You want to determine the mess of potassium iodide date in the tablet. So, first the tablet was dissolved in water and acidified and it was converted to this tri iodide iron. Then they try Knight Iron. I was reacted with this amounts of it by yourself. A solution. All right. So, We want to know their mass of potassium iodide eight in the tablet. So France. We need to know the most of potassium my date, but we need to start with the face off it. All right. So, we will start with that. And if you have zero the most of thai software, we must apply the malacca city by the volume of the by yourself eat in liters. 0.02261. Yeah. Yes. All right. So, we will find the most of the try try how you died. I am by using a more issue. Uh So, we have one more of the it's a date I am and to most of the thyself feet. Yeah. All right. So that would be the most of the tri iodide. Now, there's more asia between the tri iodide and the idea is this. Yeah. So, uh we'll have one more Yeah, of the a date and three malls of Mhm. Uh The miners. All right. So, that's give us the most of the uh I did in that tablet. Okay, so that gives us 3.967/10 Temper minus four moles of Yeah. Mhm. Okay, so, we will convert this to mess by using the miller mass of potassium. Are you dates? Which is 21 4.001g of Yeah. For more. Yeah. All right. So That gives us zero 08 Just 61 grabs in Milligram. Will multiply by 1000 to get 80.61 milligram of protests in. Yeah, I did. All right. So this is the mass of potassium a date in the tablets. Yeah. All right. Yeah.

In this question were given some data on the reaction of di ethyl hydrazine and iodine. First, we want to find out what the great expression for this uh reaction is. So we know that the rate should be equal to K times the concentration of dimethyl hydrazine. Okay. Yeah. Some experts at A times the concentration of iodine to some extent, be to solve for A B and K. We can take the data from the experimental, from the experiments to make the systems of equations. So we know From experience at one at 1.08 Times 10 to the negative 4th Should be equal to K. 0.15 To be a time 0.25 to the B. Some experience to we know that 1.56 times 10 to negative west equal K Times 0.15 to the A. Time 0.362 to the B. and from experiment three, we know that 230 Times 10 to the negative fourth should equal okay, Times 0.2 to the A. Time 0.4 to the B. Solving the systems of equations. We get that A. Should be equal to one point oh 05 which is according to one Be should be equal to zero 99 which is a client one. And K. Should be equal to 0.00 two 88 leaders per mole per hour. So now that we know the orders of both diable hydrazine and iodine, we can come up with the overall order. Which should be to which is the son of them and also the overall rate expression which is great equals K. Times the concentration of diagonal hydrazine. Yeah. Mhm. To be a. Which is one time is the concentration of iron to the A. To the B. Which is also. So that's a great expression. Finally, we want to determine what the concentration of dimethyl hydrazine has to be so that the rate of the reaction is five times 10 to the negative fourth moles per liter per hour. When the concentration of iodine is 100.5 Mueller. So using the right expression we just came up with we plug in what we know and solve for a concentration that we won't divide. Yeah. Mhm. Yeah. Mhm. Mhm. We find back to the concentration. Mhm. Of diet to hydrazine necessary should be 0.347 bowler

Let's go over how to determine the rate law for the reaction In this problem, we need to find the experiments in which one of the reacting concentrations remains the same and one of them changes. So if we look at reaction one and three, we can see that hydrogen peroxide concentration remains the same and the iodide concentration changes. So let's take a look at how the iodide concentration changes and how this changes the eight of the reaction. So if we divide the concentrations of iodide, then we see that the concentration goes up by a factor of four. Now we're going to do the same with the rates. So we're going to take The rate from reaction three. The boy by The rate from reaction one. We also get about four. So from this we can see since the concentration and the reaction rate go up by the same factor. This reaction is first order in the aisle diet ion. Now, we're going to look at experiment two and four. So in these two experiments, you can see that the iodide ion concentration say the same, but the hydrogen peroxide concentration is changing. So let's determine how the hydrogen peroxide concentration changes. So if we divide the concentrations, We see that the concentration goes up by a factor of two, and they were going to do the same for the rates. And we see that is also a factor of two. So, because the concentration and the rate is going up by the same factor, it's also a first order in hydrogen peroxide. So this is going to be the expression for the Great Law. Now we are asked to calculate the value of K. To get the value of K. We're going to plug in the numbers from any experiments. So, I'm just going to do experiment one. So now we're going to plug in the numbers Mhm. And now, So for cake. So this is the value that we get for K. Now we're giving some reaction conditions and we add K I. As well as a solution of hydrogen peroxide. In this case, one more of KI contributes one more of the I have died ion. So since we know this, if we calculate the moles of K I, we have also calculated the moles of iodide ion. So what we're going to do is we're going to take the mole aren t of the solution, multiply it by the volume, which we need to convert two leaders first And then we're going to put that over the total volume. And we're told that we have 25 ml at its 25 million. So the total volume is 50 Bill leaders. Then this will give us the polarity of the iodide ion. So now we have to determine the modularity of the hydrogen peroxide. So we're going to use the density and the volume to find the mass. The density is one g over one millimeter And we have 25 ml of the solution. So that means that we have 25 g of the solution. 10% of this is a hydrogen peroxide. So we're going to convert the percentage to a decimal, multiply it by the mass of the solution and that will give us a massive hydrogen peroxide. Now that we have the mass of the hydrogen peroxide, we can convert two moles of hydrogen peroxide using the molar mass. So from a periodic table we get the molar mass. Well the masses 34.01 g. And that loves to convert two moles. And then we're going to put this over the total volume, which is as many leaders. And then we're going to end up with the polarity of ha jin peroxide. Now that we have the more clarity of both of the reactant. We are going to solve for the rate using the rate law. So we calculated the value of K earlier and we can use that. And we're just going to plug in the numbers and recall that this reaction is first order in both the iodide ion and the hydrogen peroxide. So then we're just going to multiply everything together. So we get .36 polarity per minute as the rate.