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Student obtained the following data for the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 *HzOz(aq) ~HzO() 'h 02(g)[[HzO2k. M time , miu4...

Question

Student obtained the following data for the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 *HzOz(aq) ~HzO() 'h 02(g)[[HzO2k. M time , miu422*10-22.Uxl0-2 73406*10-2 14.7[s.30*10- 22.0(1) What is the hall-life for the reaetion starting a (0 min? 7.34minWhat is the hall-lile for the reaction stanting a(=7.34 mmin"IuinDves the half-lile increase, deerease Or remain constant aS the reaction proceeds" increase(2) [s the reaction zero, first, or second order? second

student obtained the following data for the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 * HzOz(aq) ~HzO() 'h 02(g) [[HzO2k. M time , miu 422*10-2 2.Uxl0-2 734 06*10-2 14.7 [s.30*10- 22.0 (1) What is the hall-life for the reaetion starting a (0 min? 7.34 min What is the hall-lile for the reaction stanting a(=7.34 mmin" Iuin Dves the half-lile increase, deerease Or remain constant aS the reaction proceeds" increase (2) [s the reaction zero, first, or second order? second (3) Based on these datz, what is thc rale constant for Ilie: reaction?



Answers

The decomposition of hydrogen peroxide was studied at a particular temperature. The following data were obtained, where $$ \text { Rate }=-\frac{d\left[\mathbf{H}_{2} \mathbf{O}_{2}\right]}{d t} $$ $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right](\mathrm{mol} / \mathrm{L})} \\ \hline 0 & 1.00 \\ 120 \pm 1 & 0.91 \\ 300 \pm 1 & 0.78 \\ 600 \pm 1 & 0.59 \\ 1200 \pm 1 & 0.37 \\ 1800 \pm 1 & 0.22 \\ 2400 \pm 1 & 0.13 \\ 3000 \pm 1 & 0.082 \\ 3600 \pm 1 & 0.050 \end{array}$$ Determine the integrated rate law, the differential rate law, and the value of the rate constant. Calculate the $\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]$ at $4000 .$ s after the start of the reaction.

Mhm. Given it was all over the country, uh hydrogen peroxide decomposes to give each tool and half more or to gas. You don't want to mm. It is the first order reaction. Yeah. Look at that. Mhm. Given mhm half life frequency. I'm gonna leave you. Mhm denoted by T. Have is equal to 17.0 minutes. In part A. We have to calculate rate constant for the reaction rate constant. Mm hmm is denoted by K. It is equal to Ellen to divide by T. Half. For first order reaction, Which is equal to 0.693, divide by 17.0 minutes. And we get K equals to zero 041 minute universe. Mhm. Mhm. This is the value of rate constant. Okay. In part B, we have to calculate dying to decompose 86.0% of It's two or 2 late. The indicates. Yeah. Mhm. Time required. Four 86.0%. I still think mhm decomposition of H two or 2 may not indicate initial concentration. Mhm of H202 which is equal to 100 is wounded and 80 indicates final concentration. The Republican president of H 208 H two or two after fine. The Here 86% of extra. Or to decompose it at 90. So final concentration is remaining concentration Which is equal 200 86%. Both 100 Which is equal to 14. For faster reaction. T. Is equal to one by rate constant Cape Times and then initial concentration divide by final concentration. Now substitute the value of K. A. note and 80 in this expression to innovate the value D. Second. Mhm. We get the equals two. Yeah. Mhm. Approximately 48 minutes. Okay. In part C initial concentration He's given 0.1 juro juro Mueller time. T. Is given 15.0 minutes. We have to find concentration after time. T. Which yeah we know that. Ellen A not divide by 80 is equal to Katie. Four. First order the accent good. Uh huh. I'm looking for that one. Mhm. Mhm. We can get in this expression A not divide by 80 is equal to eat to the Power Katie. Therefore 80 is equal to in our times. Here is to the bubble minus Katie this afternoon. The value of a note K. Anti this expression To get eight equals 2 0.100 Mueller into Areas to the Power -0.041 minute universe into 15.0 minute. We get 18 was 2 0.0 five four mona. Yeah so required concentration christian off It's two or 2 after 15.0 Minutes is equal to 0.0 five four moller. Mhm. This is our answer for but see. Mhm.

The first thing that we need to do Recognizing this is the first order reaction is used. The first order half life equation to solve for K K is gonna be equal to natural log of two divided by 17.3 minutes. We get 0.0 for 011 over minutes. Now that we know the K value, we can use a form of the first order integrated rate law and solve for the concentration after 10 minutes have passed natural log of concentration at time T minus natural log of concentration. Well, natural concentration time zero minus natural. Lack of concentration at Time T equals Katie. So if I subtract this from both sides, I'll get this expression here. And then I will change my negative sign over to this side and then take the anti natural audible sides and X equals 0.293 Moeller

So we have this mechanism for Thebe composition of hydrogen peroxide. And the data that we had in number 37 showed that the plot of the log of the concentration of the peroxide against time was linear. We determined that it was first order in Htoo too, and, uh, it's a little puzzling in that, uh, the only possibilities were to show that it was zero first or second order in h 202 It wouldn't pick up the possibility, for example, of something else Ah, some other reactant having a an effect on the rate. But, um, actually, we don't see any other Ah reactant on the left hand side, Those substances that we see on the left hand side other than peroxide are all reaction intermediates. And so the, uh uh, conclusion about the slow step is that when the ah order of the reaction ah is the same as the coefficient in an elementary step that is likely the slow step. So the reaction is first order in peroxide and there's one peroxide on the left In the first step. I'm not sure how we would have detected hydroxide radicals with the data that we had. But anyway, as I said, being first order in peroxide indicated that the slow step is one that has one molecule of peroxide and nothing else on the left side. And so the first step would be the slow step. Now, in obtaining the overall balanced equation, we simply cancel anything identical on both sides. So, for example, we have this H 02 one of each on both sides, and so we can cancel one of each on both sides. And then we have 20 H is on the right and 20 H is on the left so we can cancel those two on each side with each other. And, uh, nothing else seems to cancel. We've got to H 202 on the left, 200 peroxide and then two waters and an oxygen on the right.

So chemical equations must be balanced that the quantities of reactant and products match that is, for either side off our equilibrium our or are just simple reaction arrow. So for an equation to be balanced, there must be an equal number of atoms on each side. So the rate determining step is off the slowest step of a chemical reaction. So we have hte to go to, which is an equilibrium with 20 H. This is the rate determining step in this equation. So we know this because rate equals K concentration H 202 So then we have our three reactions written out and all we've done is gone through and canceled out what we are able Thio. Which leaves us with our overall reaction on our overall reaction. Here is two h 20 is an equilibrium with two h 20 ad 02 This is our balanced final equation.


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