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Calculatc thc rate constant; k for reaction at 66.0 %C that has an aclivation energy of 82.1 kJlmol and frcquency factor of 7.79 * [0"...

Question

Calculatc thc rate constant; k for reaction at 66.0 %C that has an aclivation energy of 82.1 kJlmol and frcquency factor of 7.79 * [0"

Calculatc thc rate constant; k for reaction at 66.0 %C that has an aclivation energy of 82.1 kJlmol and frcquency factor of 7.79 * [0"



Answers

Consider the first-order reaction $$ \mathrm{CH}_{3} \mathrm{NC}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(g) $$ Given that the frequency factor and activation energy for the reaction are $3.98 \times 10^{13} \mathrm{s}^{-1}$ and $161 \mathrm{kJ} / \mathrm{mol}$ respectively, calculate the rate constant at $600^{\circ} \mathrm{C}$.

Here were given information for a specific chemical reaction. We have phosphorus Penta oxide plus water yields essentially phosphoric acid and we're essentially given a table for the different quantities and we're asked to find essentially the delta the the standard free energy of the reaction. So in this case we have to remember that delta jeannot equals delta H. Now of the reaction minus T delta S not of the reaction. And were given that this reaction occurs At 298 Kelvin. And we have a long list of a data table. So r N trapeze and jewels per mole kelvin. So we have to remember to convert it to kill jules since our and therapies are given in kilograms promote And were given that the entropy for the first component 2984, -285.8 -1281 Or entropy is 228.969.91. 110.5. So essentially in this case we basically for delta H not of reaction, we take the summation of the basically heats of formation for the the entropy of formation of the products minus delta H, nods half of the heat of formation of the reactions. So here we're seeing that we're making four moles of phosphoric acid, so that's our only product. So this would be four times negative 1,281. While for the reactant we essentially see that we're making reactant, we have basically six miles. We're consuming six bowls of water, So six times negative 285.8. and we're essentially consuming one mole of phosphorus oxide. And essentially for the first term We have negative 5,124, I'll add units later -2698. And basically adding everything together. We can find that essentially the Uh the entropy of the reaction is -425. About negative 425 killer goals per mole. And for the uh the entropy of reaction is the same process. We're going to summit the entropy of the products and minus the entropy of the reactant. So once again we have to look at their data table again, we're making four moles of essentially phosphoric acid. Then we're losing essentially six moles of water and we're losing one mole of phosphorus Penta oxide And you see that's 440- -648.4. When we find that the entropy of the reaction is negative 206 jaws prime or kelvin. However, this has to be intelligible since we have to agree, we have need units agreement with a century and kill people. So this is negative 0.206 killer jewels promote Calvin. So now let's plug everything back into our equation And remember that we're running this at 298 Kelvin and essentially we can find by plugging our values in Delta G. Not for this reaction. At 298 Kelvin is negative. 364 killer goals promote and this is our final answer.

Let's start this problem by remembering the Arrhenius equation, which states that we have some value k r rate constant, which equals a E to the negative e a over r t. Okay, so when we're working at different temperatures, will remember that this value a doesn't change, but values K and T will change E a r and literally, of course, are all constant. So let's go ahead and divide these two expressions. So let's take K one over que two, which is going to be equal to a over a the A's cancel e to the negative e a over our tea one plus yeah, over our t two. Okay, so let's work through this now and take the natural log of both sides. So natural logarithms gets us k one over que two e drops away and we are left with Yea, there should be an equal sign in here equals e a over our one over t to minus one over t one. Now let's go ahead and substitute in our values and we find that we get an e A of 133 0.76 killer jewels. Permal That's kilo jewels Permal. Okay, next up were asked to figure out the rate constant at 750 k. So this is just a matter of doing the reverse process now. So we're going to say Okay, we know everything else, but we want to find some K one. So in this case, K one is going to equal some K two which we can use as being either of the values that we started with. So let's take our rate constant at 700 Kelvin, which is 1.3 per moller per second, as our rate constant times eat the negative e a or rather, we can take E to the EA over our. So that's e to the 134 hello Jewels per mole over our times. One over t to remember t to in this case is 700 Kelvin minus one over t one and t one in this case is 750. Calvin note that even though we've gone ahead and figured this out, we didn't actually need to calculate a We could have calculated a after we found e a, but we don't actually need to know that value to figure out the answer to this. So if we go ahead and calculate this out, we get a value of 6.0 17 per moller per second.

Problem 66. Poor calculating E and a first. We need to calculate LNK and value of one. But after that we need to draw up. Sure this value is given question. We need to calculate this and this value. So alan Kay Alan 0.02 equals two -3.58. Ellen 0.0.22 Equals two -1.51. Alan 1.3 equals two 0.26. Ellen six equals two. 1.79. Allen 23 equals two 3.14. Now one divide by 600 equals two. 1.67 into 10. to the power three one divide by 6 50 equals to one point Hi put one divided by 700 equals to one 43 One divided way 7 50 equals to one point 33125. equals to 1.25. So now we need to draw a graph between Ellen K and one by T. So there's a grab between LNK and won by T. This graph is state line. Using the we draw this grabbed by using relationship Ellen K equals two L N E minus. Yeah, divided by RT Therefore slope of this graph calls to -15.94 into 10 to the power three. That is equals two minus. Yeah. Bye. Are therefore we can write as equals two minus 15 point Sorry plus 15 point nine pour into 10 to the power three multiplied by gas constant. That is 8.31 port. That it equals two. 1.3 into 10. to the power two kilo jule Farmall. So this is the value of Yeah. Now we need to calculate value of a To calculate a we will use the rate data at 700 Calvin from the equation that is L N E. Sorry that equation is alan key equals two L N E minus. Yeah. Babe. Rt sure. Hello. LNK it's zero point 262 So we use this value okay at 700 Calvin At 700 Calvin value of Alan Alan Kay is called the 0.26. So we can write as zero point 262 equals two. Ln a minus E value P already calculated. That is Yeah but sorry sorry sorry actually we also simplify this question because this they have already calculated here. So directly used this show that is minus 15 point 94 multiply by 10 to the power three and value of Okay little bit mistake this started already given. Okay should that this year party so used directly here it is 15 point 94 into 10 to the Power three divided by value of temperatures 700. After calculating we can write us Ln equals two 0.262. Bless 22.7 71 therefore equals two. One in two, tend to the power and this is final and support this question where you up? Is this value of? Is this value of it? Yes?

For the equation off decomposition off hydrogen iodide, so give hydrogen and iodine. The activation energy is given at temperature 555. Galvin. This is the rate constant for the reaction we have to find out the rate constant when the temperature is 645 Calvin. For this, we will use the Arnie's equation and, assuming the earnest vector a Toby temperature independent, the question can be written for two conditions. As on substituting all the values in this equation, we can further calculate the value off key to the value off Kato. At 645 Calvin comes out Toby 9.5 into 10 days to power minus five later Permal per second.


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