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A reaction A B has the following time dependence for theconcentration of [A] vs time. For t=(0 s, 5 s, 10 s,15 s, 25 s) the concentration of [A]=(30.00 M, 21.13 M,...

Question

A reaction A B has the following time dependence for theconcentration of [A] vs time. For t=(0 s, 5 s, 10 s,15 s, 25 s) the concentration of [A]=(30.00 M, 21.13 M, 16.30 M,13.27 M, 9.68 M). The initial concentration of [A] is the valueat t=0 s.(A)Calculate the values of the rateconstant k assuming that the reaction is firstorder.What is the value of k at 5s? s-1What is the value of k at 10s? s-1What is the value of k at 15s? s-1What is the value of k at 25s? s-1(B)Calculate the valueof k as

A reaction A B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 21.13 M, 16.30 M, 13.27 M, 9.68 M). The initial concentration of [A] is the value at t=0 s. (A)Calculate the values of the rate constant k assuming that the reaction is first order. What is the value of k at 5 s? s-1 What is the value of k at 10 s? s-1 What is the value of k at 15 s? s-1 What is the value of k at 25 s? s-1 (B)Calculate the value of k assuming that the reaction is second order. What is the value of k at 5 s? M-1s-1 What is the value of k at 10 s? M-1s-1 What is the value of k at 15 s? M-1s-1 What is the value of k at 25 s? M-1s-1 (C)Use your results from parts A and B above to decide what the order of the reaction is for your data. (Enter 1 or 2) (D)What is the concentration of [A] at time t=63 s? M



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 concentration of A remaining after 110 s in the (a) zero-order, (b) first-order, (c) second-order reaction?

For this next problem, you need to go back to problem 27. If you haven't done problem 27 you need to do problem. 27 1st in problem 27 you're going to plot the concentration as a function of time went over the concentration and natural log of concentration as a function of time for all three data sets. Once you've done that and you in Excel show the equation of the line, you can also ask Excel to show the r squared value. And I'm not showing all three graphs. You'd have three graphs here. Natural log of concentration, one over concentration and just concentration as a function of time. And you find that for data set, one natural log of concentration gives you the best straight line. You do all three here also, and find out that concentration versus time gives you the best straight line having the r squared value closest to one. That's how you determine if I had all three here. Which one is the best straight line is the one that has an R squared value closest to one, So its concentration is a function of time for the second one and one over. The concentration is a function of time for the third one. So natural law giving you the best straight lines. First order concentration is a function of time. Giving the best straight line is zero order and one over the concentration is second order. Then, with these graphs and the equations of the line, you can identify the the rate constant. The rate constant is the slope unless the slope is negative and then the rate constant is the negative of that negative slope or the positive value of the slope. Great constant is always positive, so the rate constant for the first one for first order is 0.1 Great constant for the second one's also 0.1 Great constant for the experiment three is also 30.1 or more like 0.992

This question will take a little bit of work because the best way to justify the order of a reaction in this case First order is when you just have time and concentration data is to plot the concentration as a function of time, the natural log of the concentration as a function of time and one over the concentration as a function of time. Whichever one of these graphs gives you a straight line that will tell you the order of the reaction. Yes, concentration is a function of time gives you a straight line. It's zero order natural log of concentration as a function of time gives you a straight line. Its first order and one over concentration as a function of time gives you a straight line, then its second order. So it suggests this reaction is first order, so to verify that it's first order, we need to go into excel and plot the natural log of the concentration as a function of time. So you'll need to put in all the time values you'll need to put in all the concentrations as provided in the question, and then calculate the natural log of the concentration at all the time values and then plot the natural log of the concentration as a function of time to validate that you have a straight line in excel. You can ask it to show the equation. The equation will be the best fit, straight line and the better fit you have. The closer the R squared value will be to one. So because we plotted the natural log of the concentration as a function of time and we got in r squared value, that is essentially one, this validates that this reaction follows first order kinetics, which is what the question wanted you to do when it asks, Show that the reaction is first order. Now that we have plotted the data as a function of the natural log of the concentration as a function of time and obtained are straight line and its equation, then the slope allows us to identify the K value or the rate constant for the reaction if its first order will always get a negative slope. So Kay is going to be equal to the negative of the negative slope because K, the rate constant is always positive. So in this case, the rate constant is 0.200185 positive because the negative of negative 0.185 is positive 0.185 Now that we know the rate constant and we have validated that it is first order, we can use first order kinetics in order to calculate the concentration of a at time T Time T in this case is going to be 750 seconds. So it should have a concentration somewhere between 0.1 point 198 and 0.94 because the time is somewhere between here. So we'll set up our first order. Integrated rate like equation. Natural log of concentration at time t over concentration at time. Zero equals negative. K, which was just determined, is 0.185 one over seconds, multiplied by seconds. The seconds is the 7 15. If you remember, your log rules natural log of a value over another value is the natural log of the numerator minus the natural log of the denominator. And then that will be equal to the product of this right here. Well, then, add natural log of 0.6 to both sides and we get natural. Log of X is equal to negative 1.898 Well, then do e to the both sides in order to get out of the natural log in into X and so e to the negative. 1.898 gives us our X value of 0.150 which is definitely between these two values here, as we predicted.

So this problem begins by telling us that won over the concentration of a B first time is a straight line with a slope of negative zero point zero five five more minus one second minus one. So we know that this is a second order reaction with the rake constant of negative zero point zero five five Mueller minus one second, minus one. And so question, eh just wants us to give that rate constant value Question. Be wants us to write the rate law, which is just rate equals k times the concentration of a B squared since its second order. Questions see and I'LL go to a new page for this question. C wants us to find a half life. So the half life of a second order reaction is equal to one over kay times the initial concentration and so we can plug our values in here, or K is negative. Zero point zero five five. Our initial concentration is zero point five five and so a half life is thirty three point zero five eight seconds. Now question dear wants us to find the amount the concentration of A and B gained and so first we want to find the concentration of a B lost. So we start with our second order integrator rate law won over the concentration of a B equals one over the concentration of a bee. Initially plus Katie, we can plug in our values. And so after seventy five seconds, our concentration of a B is zero point one two three Moller. And so that is our new concentration of a B. And so I go to a new page and find our concentration of a B lost, which is our initial concentration, minus our final concentration. And so we know that we lost zero point one two seven moller off a B. And so a B lost is equal to A and B gained. So the concentration of a gained zero point one two seven moller and the concentration of be gained is zero point one two seven Moller. And so those are our final answers

Let's consider the reaction given for E. Called the rate lot is first order in H two second order in you know so let's raise the rate law. It is equal to k. First order with respect to H two second order risk respect. And then also there is our rate lock for B. Were used to calculate the rate we're given the rate constant here which is 6.0 to 10 to the four M -2 S -1. The each concentration is .015 Mueller. Yeah And you know is .035 more squared. The rate would be equal to 1.1 more second or B. For C The rate is equal to our constant 6.02 times 10 to the 4th And -2 S -1. Each concentration is .010 Moller and Oh is .10 moller Squared. And a rate in c works out to 6.0 more per second. And for d rates is equal to the constant 6.010 to the 4th and -2 -1. H two is 030 Moller. Yeah. and Oh is .010 boy Squared. And the d works out 2.18 more per second.


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