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CHEM125 ILLINOIS INS itute Experiment 3: TECHNOLOG Chemical Kinetics Rate of Reaction Post-lab Questions ( [ Opts)It has been found that in acid solution (grealer t...

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CHEM125 ILLINOIS INS itute Experiment 3: TECHNOLOG Chemical Kinetics Rate of Reaction Post-lab Questions ( [ Opts)It has been found that in acid solution (grealer than MH ) the reaction of H,Oz with rale expression for the is ofthe form rate k [HzOzI[I ] where k, 0.25 L? mol 25"C. Therefore the rate law over the whole range of acidity from very acid solutions to neutral solution is given by two-term rate expression:rate k[HzOz][T-] + kz[HzOzl[-J[H+]where k, has the value of 0.0115 Lrmol &#x

CHEM125 ILLINOIS INS itute Experiment 3: TECHNOLOG Chemical Kinetics Rate of Reaction Post-lab Questions ( [ Opts) It has been found that in acid solution (grealer than MH ) the reaction of H,Oz with rale expression for the is ofthe form rate k [HzOzI[I ] where k, 0.25 L? mol 25"C. Therefore the rate law over the whole range of acidity from very acid solutions to neutral solution is given by two-term rate expression: rate k[HzOz][T-] + kz[HzOzl[-J[H+] where k, has the value of 0.0115 Lrmol ' $"' at 25*C. Compare the relative magnitudes of these = [WO terms under the conditions employed in your rate measurements for mixture in beaker # (You should find that the second term makes negligible contribution to the reaction rate at Ht concentrations smaller than 10 5



Answers

The reaction between propionaldehyde and hydrocyanic acid has been studied at $25^{\circ} \mathrm{C}$. In a certain aqueous solution at $25^{\circ} \mathrm{C}$ the concentrations at various times were as follows:
$$\begin{array}{llll}t / \min & 2.78 & 5.33 & 8.17 \\{[\mathrm{HCN}] / \mathrm{mol} \mathrm{L}^{-1}} & 0.0990 & 0.0906 & 0.0830 \\{\left[\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{CHO}\right] / \mathrm{mol} \mathrm{L}^{-1}} & 0.0566 & 0.0482 & 0.040660\end{array}$$ $$\begin{array}{llll}t / \min & 15.13 & 19.80 & \infty \\{[\mathrm{HCN}] / \mathrm{mol} \mathrm{L}^{-1}} & 0.0706 & 0.0653 & 0.0424 \\{\left[\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{CHO}\right] / \mathrm{mol} \mathrm{L}^{-1}} & 0.0282 & 0.0229 & 0.0000\end{array}$$ What is the order of the reaction, and what is the value of the rate constant $k ?$

Is this the sore question? Where you really need to know about what? Linear plots, plots result with what order? If they're linear And so on this sizzling hot into with you integrate rate law, you know? Know what? I'm time. Well, you should definitely read your book on this. Anyways, if we take the concentration of a and plotted versus time and it ends up linear, we know this is first order with respect to a And with that, we can just pump out a bunch of information that counts from the integrated rate law. For instance. We know that what the integrate ray law is gonna be It's natural log of a equals negative. Katie, were case the rate constant plus the natural log of the initial concentration? We also can know what the half life is. Jiggles the natural log of two over K that's derived from that integrated ray lock. And we can also say pretty quickly here if the rate law is always first rate equals k, the rate constant and times the concentration of a the first power. So that Cindy is concentration of a because it's first order with respect to a and there's no other reactions. So this is our rate lock. And so that answers the first question. Mostly, um, now to figure out the value of the rate constant. We also can use the fact that for US plot like this in its first order, that slope is equal to negative K. And so we're told what the slope is, and so we can easily pump out cages by copying it from the problem. It's 2.97 times, 10 to the negative, second in verse minutes. And so we're in business. Certainly we want to figure out the half life well. It's legal to the outline of two over K, and that equals 23.3 minutes plugging k from right up here. And, uh, finally we want to figure out how much time it's gonna take to get to a certain concentration. Well, we're gonna use the integrated rate law. It's so that concentration working, looking for is 2.5 times tense, and I give third our initial concentration. There's two times 10 today in a second we have K. It's still 2.97 times 10 Judy negative second and all of that. We consult for the time, which will come out of 69.9 minutes. Fantastic

So if we know the half life, we can use toe half life equation for first order kinetics because we're told to assume that it's a first order process and to calculate the K value K will be equal to attach a log of two divided by the half. Life 4.99 times 10 The negative 31 over minutes, but you can convert toe one over seconds if you would like. Then we use the first order. Integrated rate lie equation Natural log of concentration at time. Zero minus natural log of concentration at time T equals 4.99 times. Tend to negative 31 over minutes, multiplied by the time the time is five hours, which we will convert two minutes by multiplying by 16. Well, then go ahead and multiply everything over here together and then subtract off this value and will get negative. Natural log of X will be equal to some number. 0.484 will then move the negative sign from the left hand side to the right hand side and then take the anti laudable sides E to the natural log of X is X and then e to the negative. 4.84 is 0.616 Mohler

This one also is similar to some of the previous problems. We need to choose to experiments where we have a changing concentration of one reactant and the other reactant concentrations. Staying constant chemical reaction is an 02 and CEO going to an O and Co. Two. So the rate law is going to be rated sequel to K, multiplied by the concentrations of each reactant raised to some unknown power that we will determine So enoh to raise two X and CEO raised toe. Why? Well, then go to the data and we'll look at the two experiments where we have one concentration changing and one staying constant. I'll take experiments one and two and from experiments one into I can write. The rate is equal decay multiplied by the concentrations given raised to the unknown powers, which I'm attempting to determine. You'll see that the case will cancel if I divide these two rate laws for the first two experiments by each other, and in addition to the case canceling, so will the concentrations of n 02 and the exponents X. So now we just have one variable, which is why the ratio of the rates is 11 is going to be equal to this ratio is too to the why. So how can I get one by taking two and raising it to some number? Well, that number needs to be zero. Any number raised to zero will give me one. So I know it. Zero order with respect to this particular reactant that has the changing concentration with respect to CEO. Well, then take two additional experiments where I have a changing concentration of n 02 with a constant concentration of CEO. Divide these two expressions and I get the K values cancel and the concentrations associated with CEO to cancel because the wise will cancel. But they didn't. I could plug in. Why? I just calculated it. Zero. And all of this is going to become one anyway, because it zero order so then I have left. The ratio of the rates is equal to the ratio of these concentrations raised to the x 0.25 is equal 2.50 Raise two x. So how do I get 0.25 from 0.5? Well, if I square 0.5, I'm going to get 0.25 so X is equal to two. So my rate then is equal to the concentration of n 02 Squared X is too multiplied by the concentration of CEO raised to zero, or I could have just simply written it as this. Then I can choose any experiment to determine my K value the data from any experiment to determine my K value. I'll take the rate from experiment one equal to K, multiplied by the concentration of n 02 squared, multiplied by the concentration of CEO raised to zero or again, I don't have to include this because anything raised to the zero is just one. So all of this could be excluded. Then I have K is equal to the rate divided by the concentration of an 02 squared. And I get K is equal to 2.8 times tend to negative four. But what about the units? While the overall order of this reaction is to so the units are going to be won over polarity seconds

In discussion are critical to kurdish the extreme the all which C two H five B 25 which is able to be a dish th three the Alright, going scared we have is equal to 20 by here, dash, etc. Feel. Or it not staging right on what? Please return to one by And uh tennis for Minor street 204 through OK. is equal to Pander next hear me neck interests.


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