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Ch.14 - Chemical Kinetics

Chapter 14, Problem 105

At 28 C, raw milk sours in 4.0 h but takes 48 h to sour in a refrigerator at 5 C. Estimate the activation energy in kJ>mol for the reaction that leads to the souring of milk.

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Hi everyone for this problem. It reads an ester hydraulics. This reaction takes 6.80 hours to complete at 45°C but takes 2.25 hours to complete when the temperature is raised to 85°C, calculate the activation energy in joules per mole for the reaction. So our goal here is to calculate activation energy and the way that we're going to do that is we'll use the two point form of the Iranians equation to solve this problem. And that is Ellen of K two over K one is equal to negative activation energy over constant. R times one over the second temperature minus one over the first temperature. Okay, so let's just take a look at the problem to see what we're given. So we know what the ratio of K to two K one is because it's equal to K two over K one is equal to T one over T two. So that means we have 6.80 hours Over 2.25 hours. So this gives us a ratio of 3.02. Okay, R is r gas constant. This is a value we should know. It's not going to be given in the problem and that value is 8.314 jewels over mole times kelvin. Okay. Our first temperature is 45 degrees Celsius but if we take a look at our our constant, we see that temperature is in the unit of kelvin. So that means we need to convert this degree Celsius to kelvin and we do that by adding 273.15. So that gives us T one is equal to 318.15 kelvin. T two is equal to 85 degrees Celsius and will do the same thing converted to kelvin. And so we'll get T two is equal to 358.15 kelvin. So now we have everything that we need to solve this problem. So let's go ahead and plug in our values. So we have L. N. Of K two over K one. So we said that value is 3.2 is equal to negative activation energy Over our so that's 8.314 jewels over more times Kelvin Times one over T one. RT one is and 18 . Kelvin -, Sorry one over T two. So 300 n 58.15 kelvin minus one over 318.15 kelvin. Okay, so let's go ahead and continue. So we're going to get on the left side, 1.105 is equal to our activation energy times 4. times 10 to the negative five Jules per mole. So now remember we're solving for activation energy here. So we need to isolate our activation energy by dividing both sides of our equation by 4. two Times 10 to the negative 5th jewels caramel. Okay. All right, so now we're going to get when we simplify this even more, our activation energy Is equal to 26,172 jewels Permal. The question specifically asked us for it and kill a jewel. So we need to convert jewels to kill a jewels. And that conversion is and one killer jewel, We have 1000 jewels So our unit of jewels cancel and will be left with joules per mole. And so when we solve this, we're going to get our activation energy is equal to 26.2 kg jewels Permal. And this is our final answer. Okay, this is the activation and kill a jules Permal for the reaction. That's the end of this problem. I hope this was helpful.
Related Practice
Textbook Question

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 103 M-1 cm-1 at 520 nm. (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction.

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Textbook Question

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 103 M-1 cm-1 at 520 nm. (c) Calculate the half-life of the reaction.

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Textbook Question

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 103 M-1 cm-1 at 520 nm. (d) How long does it take for the absorbance to fall to 0.100?

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Textbook Question

The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O: NO1g2 + NO1g2¡N2O21g2 N2O21g2 + H21g2¡N2O1g2 + H2O1g2 (d) The observed rate law is rate = k3NO423H24. If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

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Textbook Question

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: Cl1g2 + O31g2¡ClO1g2 + O21g2 ClO1g2 + O1g2¡Cl1g2 + O21g2 (b) What is the catalyst in the reaction?

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Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O3(g) ⇌ O2(g) + O(g) (fast)

Step 2: O(g) + O3(g) → 2 O2 (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

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