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Question 31 ptsWhat is the initial rate of change of [A] in the following experiment?Enter your answer in units of Mls.0,200,15[A] 0.10 m810.050,00100200Time. sec ....

Question

Question 31 ptsWhat is the initial rate of change of [A] in the following experiment?Enter your answer in units of Mls.0,200,15[A] 0.10 m810.050,00100200Time. sec .

Question 3 1 pts What is the initial rate of change of [A] in the following experiment? Enter your answer in units of Mls. 0,20 0,15 [A] 0.10 m81 0.05 0,00 100 200 Time. sec .



Answers

Three different sets of data of $[\mathrm{A}]$ versus time are giv the following table for the reaction $A \longrightarrow$ prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ What is the approximate initial rate of the second-order reaction?

Hey, guys. So in this question were given you reaction between A and B forming C and D and were asked to find the rate law for this reaction. So the first thing I'm going to do is find the order with respect to A by using experiments one and two. This is because the concentration with respect to be is constant between these two and the constant concentration with respect to a is changing between experiments one and two. So to do that, I'm going to use my rate law equation that I have here and write some proportionality ease. So I have experiment one an experiment to, and I'm gonna put experiment to over experiment one. Since experiment twos, values are larger, and I won't have to worry about fraction when I divide all of this out. So what I'm doing now is writing out my rates. So I have that for my rate for Experiment two and I have this for my rate for experiment, one for the other side. I have the rest of my rate law. So my rate constant multiplied by my concentration of aid to the Empower Times. My concentration of B to the end power. This is going to be over the same thing K times my concentration of a for experiment one to the empower over my concentration of B to the end power. When I simplify this my case cancel out and the concentration for be cancels out. So I'm left with three on the left side and three to the M on the right side solving for m. I find that em is equal to one. And this is my order for the concentration A. Now that I have this, I can go ahead and do the same thing to find the order with respect to concentration. Be. So when I do that, I'm gonna use experiments two and three this time. And even though there's no concentrations that are held constant, I can go ahead and just use these threes. And since I already know the order with respect to a already, I'm putting Experiment three over experiment, too, for the same reason as last time. And I'm going ahead and writing down my rates for both Experiment three and experiment to now the how that I have that I can go ahead and write the rest of my rate law. So K times my concentration of a to end power multiplied by my concentration of B to the end power. Same thing on the bottom, this time 1.5 for both A and B. Nothing cancels out besides our case for this case And when we simplify this, we find that this is equal to four. This concentration for a is equal to two. And since m is you know the one I have a cough and exponents of one and this is going to be multiplied by two to the end. Power dividing the two over this is two is equal to two to the end. And when I saw for N and is also equal to one making be the first order. Now that I have this, I can go ahead and sulphuric a high rate constant. So go ahead and say my raid, which I'll abbreviate toe R is equal to k times my concentration of a times my concentration of B I left the exponents out because they're both one. So now I can go ahead and self okay, I can use any of the experiments that are given to me, but I chose to use experienced one. So I just have my rate, which is 4.2 to the negative three. Power to the Moeller over minute, and this is equal to K. Times 0.50 Moeller multiplied by 1.50 Ruler. When I saw for K, I find that K is equal to 0.56 with units of inverse Moeller and multiplied right, inverse minutes. Now that I have that, I can go ahead and write the rest of my rate law out for this reaction. So I have My rate is equal to 0.56 times the concentration of A to the one power times the concentration of be the one power. And this is our final rate law for this reaction.

So we're continuing our with our kinetic theory of So this is where we have molecules that are very small in comparison to the distances between them. They will collide frequently with one another with random motion as well as the container off the box. So what we're looking for here is the great constant. So in chemical kinetics, a right constant or a reaction coefficient K will quantify the right and direction off a chemical equation. It's often expressed in terms of K. And so what we find here is that we have 1.0 plus or minus, not point. No, No. Five. Don't forget units. We have moles per liter, two minus one hours to the minus one, right?

So initial rate. This is given by rate constant and product of the rate constant and the concentration, the rate constant. Okay and the concentration is initial concentration here. This is given in the question that the red constant is one into 10 ways to power -2. This is the rate constant which is given to us and the initial concentration is one murder. This is also given to us integration. So that's why the initial rate will be terroristic AR -2. And this will be Wanting to generously 4 -2. Moon for later per second. So this is the initial read now the final the b part. So that be part or be part for foster and reaction or fast dollar ricin. So we have a equation that is the rate constant. K is equal to 2.303 divided by a T long divided by a minus X. So is the initial concentration and a minus X. Is the final concentration here. So Their content is known to us. That is standards of AR -2. We have sold it 2.303 divided by 30 times the time is here. One minute. This is also given to us one minute time and this is to be converted into second. So we will multiply it by 60 because we all know that one minute If they call to 60 seconds, one minute is equal to 60 seconds. So log they're the initial concentration. So this is the same. Right? So we can cancel also this will be one minus one minus 61 divided by one minus x. Okay so solving this equation we will have one minus X. Is equal to 0.549. So this is the final concentration here more later. So now we have to find the rate after one minute. Right? Great after bandwagon is equal to Yeah Into one -X. This is what This is the great constant. And this is the final concentration here. 1 -1. With the final concentration, right? So they will be given by Rate constant is similar. That is standards department to or you can also say one into tennis department to 1 -6 is 0.549. That is the funny concentration what we have sold here. So solving it will be 5.49 multiply by Can raise the bar -3. Born for later. What a second. So this will be the rate For the reaction after one minute.


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