5

2a(8) D(g) C(e) AC(g) AB(e)The following datu wcre collected for Ihe above reactionIAiIBIIc) 0 20Rale; Mxs-"0.200 20240 400.J00.209 60.200 J00.202 40,200 400, ...

Question

2a(8) D(g) C(e) AC(g) AB(e)The following datu wcre collected for Ihe above reactionIAiIBIIc) 0 20Rale; Mxs-"0.200 20240 400.J00.209 60.200 J00.202 40,200 400, 404 &What /s the rate law for this reaction?rlc MA]" (B] mle MAJ[DJIC) Nlc MA] (C] nic K[B] [C] rale M[A] [C]"

2a(8) D(g) C(e) AC(g) AB(e) The following datu wcre collected for Ihe above reaction IAi IBI Ic) 0 20 Rale; Mxs-" 0.20 0 20 24 0 40 0.J0 0.20 9 6 0.20 0 J0 0.20 2 4 0,20 0 40 0, 40 4 & What /s the rate law for this reaction? rlc MA]" (B] mle MAJ[DJIC) Nlc MA] (C] nic K[B] [C] rale M[A] [C]"



Answers

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+$ $\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results: \begin{tabular}{llll} \hline Experiment & {$\left[\mathrm{ClO}_{2}\right](M)$} & {$\left[\mathrm{OH}^{-}\right](M)$} & Rate $(M / \mathrm{s})$ \\ \hline 1 & $0.060$ & $0.030$ & $0.0248$ \\ 2 & $0.020$ & $0.030$ & $0.00276$ \\ 3 & $0.020$ & $0.090$ & $0.00828$ \\ \hline \end{tabular} (a) Determine the rate law for the reaction. (b) Calculate the rate constant. (c) Calculate the rate when $\left[\mathrm{ClO}_{2}\right]=$ $0.100 \mathrm{M}$ and $\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}$

Based on the initial rate data. So when we keep ohh constant and we change the cielo to by a factor of one third, the rate goes down. If you divide 10.2485 point oh 276 It goes down by a factor of nine. So that means that it's going to be second order in case c l 02 Because three squared is equal tonight. And then for the O H minus, let's see. So we keep C L 02 constant and then we're going to multiply this by a factor of three and this goes up as well by a factor of three. So that's a first order in hydroxide which will omit the one. All right now we got to calculate the rate constant so that was part A. So we can use any of the experiments. Let's just use the first one since it's the first one. So .0248 polarity per second is equal to k Times .060. Let's put the units to as well. So we we know well r squared Times .03 Mueller. So then we're gonna divide By .060 And the .030 to get K by itself. We'll do that on the other side. Yeah. Okay. So when we do that out, we get .0248, divided by .06, divided by .06 again. And then divided by .03 comes out to to 29.6. So 2 30 with a decimal point And then we've more clarity divided by the 3rd. So it's gonna be one over a square polarity uh times seconds. And that makes sense. Since is the third order reaction? Okay. All right. And then for part C we got a calculator rate so we'll plug in our newly discovered rate constant 230. And then we're told, Let's see Cielo to his .1 and OH- is .05. So .100 moller squared 905 General Mueller. And we'll just multiply it all out and then the polarities cancel out. And except for one of them, I'm sorry. Yeah. And then we'll get more clarity per second for a rate. So 2 30 Times .1 times .1 times .5 Is going to be 0.115 polarity over seconds.

In this question were given the following data and we want to determine the rate law for the reaction. So okay we know that the rate law is going to be of the general form rate equals K times A. To sound a. Be to some B. See to some see so solve this. We can just make a systems of equations using the data given. So we know that rate X. Supposed to equal K. Times A. To some A. Times B to some B times C. To some see from the second line on the state of table we get the eight X. Should equal okay times the concentration of A. To the A times the concentration of B. To be to the concentration of C. To the sea. You do this for the other two lines as well. So you get the X. Is equal to K. Times point to the A. Times point to to the B. Times point to to the sea. And that four x equals k. times .4. The A. Times .4 to be times 0.1 to the sea, solve the systems of equations and you get that A Should be equal to two, Be equal to one, And C is equal to one. So you got a great law of rape equals K times a square times B. Time. See

Problem 34 party from their data given when concentration up CLO two increased by Pachter three. In experiment two. To experiment one. That rate is increased by a factor nine Increased word factor nine. And when concentration of which negative Increased by a factor three. an experiment to to experiment three, the rate is increased by By a factor three. The reaction in second the reaction, it's second order in concentration of C L 02 And first ordering always therefore really equals two came to concentration of cl or to the power to into concentration of which negative this is. And support party. No, in part B. Using data from experiments too, K equals two. Rate divided by concentration of cielo too. To the power to into concentration of which negative. Now put the values rate given in question. That is 0.0 0276. And per second. And concentration of Cielo to it's 0.020 mm. To the power to multiply by zero Jiro 30 And after calculating we get 2.3 into 10 to the power To into empty. The power -2. And as to the power -1 this is answer part Part B. No in part C. Red equals to use this formula. Okay, value calculated here that is 2.3 into 10. to the power to I. M. To the power minus two. And as to the power minus one into Concentration of cl or two to the power to that is 0.100 M. To the power to into 0.0 pipe zero M. After calculating we get 0.115 30 approximately equals two 0.12 M per second. This answer part part C.

This is a method of initial rates with two reactant. To determine the order of the reaction with respect to each reactant, we need to treat each reactant separately, namely, if we look at experiments one and two, we see that the concentration of N. 02 is changing, but the concentration of F two is not. So let's just look at what a change to the concentration of N. 02 does to the rate. By looking at experiments one and 2 where F two stays constant. So with these two experiments a doubling of N 02 Only a doubling of eno to keeping F to constant, results in a doubling of the rate going from 20.262 point 51 This is indicative of a first order reaction with respect to N. 02 Now let's look at experiments two and three where the concentration of CO two is stained constant and the concentration of F two is being doubled from 20.12 point two. In this process we see that the rate is doubled from .512.103 Doubling the concentration resulting in doubling the rate is indicative of a first order reaction. So it's also first order with respect to F two. So the rate laws rate is equal to K, multiplied by each concentration raised to the first power. To solve for K will simply plug in the rate for any experiment and the two corresponding concentrations and we'll calculate a K value of 2.61 over Mueller seconds, one over Mueller seconds. Because the overall order is second order. First order with respect to each reactant. Or we could do this for all four experiments, Get four K values and then calculate the average. So the differential rate law is rate is equal to 2.61 over moller seconds, multiplied by each concentration raised to the first power giving us a second order overall


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