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Ch.19 - Chemical Thermodynamics

Chapter 19, Problem 68

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: CH3COOH(l) → CH3OH(g) + CO(g) By using data from Appendix C, calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that ΔH° and ΔS° do not vary with temperature.

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Welcome back everyone in this example. We're told that Dick our box elation reactions are those in which a car boxes group is removed and releases carbon dioxide gas. We're told that the following reaction shows the Dick our box realization. Sorry the dick. Our box elation of moronic acid. So we have our maloney acid which decomposes into acetic acid and carbon dioxide gas. We need to assume that our change in entropy and entropy do not vary with temperature and to determine the minimum temperature at which this reaction will occur spontaneously under standard conditions. So below were given our standard entropy of formation of each of our re agents as well as our standard entropy of formation for each of our re agents. So we want to recall our formula where we have that our free energy change or change in gibbs. Free energy is equal to our standard entropy change subtracted from the temperature in kelvin multiplied by our standard entropy change. And we want to recognize that we're not given any information on our free energy change for the reaction. But we are told that we have a spontaneously occurring reaction because our reaction is spontaneous that already tells us that we should recall that our free energy change should be less than zero for the reaction to be spontaneous. And so that means that since we don't know our free energy change but we do know that it should be less than zero or a negative value to be a spontaneous reaction. We would say that We can say zero is equal to our entropy change minus our product of our temperature and our entropy change. And if we figure out our values for standard entropy and standard entropy, we can plug them in to solve for the minimum temperature at which the reaction occurs spontaneously at standard conditions. So we're going to begin by finding our entropy change of our reaction. So recall that we can calculate that by taking our entropy a formation of our products subtracted from the entropy of formation of our reactant And so given from the info and the prompt above. We're just gonna plug in what we know. So we'll say that the entropy change of our reaction are standard entropy change of our reaction is equal to the entropy of formation of our product. Where we have our one mole of acetic acid Which is multiplied by acetic acids entropy of formation given in the prompt as -487.0 with units of kilo jewels per mole. We're then going to add onto this our second reactant or sorry, our second product which is our one mole of carbon dioxide, multiplied by its Entropy of formation given the prompt as -393.5 with units of killing jewels Permal. This is going to complete the some of the entropy of formation of our products. And now we're subtracting from in brackets to some of our entropy of formation of our reactant. So we have on our react inside our malic acid, we have one mole of our malic acid which is multiplied by malic acids, Ethiopia formation given in the prompt as negative 891. killer jewels per mole. And that completes the entropy of formation of our reactant. Now we want to simplify our brackets. We can say that our standard and they'll be change of Our reaction is equal to Our first bracket. Should simplify to negative 880.5. Recognise that we can cancel out our units of moles leaving us with killer jewels as our final unit for both brackets. So we have units of kill jules here Which is subtracted from our brackets in red which should simplify to negative 891.1 kill a jewels. And taking the difference here we're going to get that are standard entropy of our reaction is equal to a value of 10.6 kg. So we're going to note this for later and we want to continue to find our standard entropy change for the reaction using the values given in the prompt. So we're going to scroll down for more room and we see that our standard entropy change for our reaction we want to recall can be found by taking the entropy change of our reactant subtracted from the entropy change the standard entropy change of our products. I'm sorry. This needs to be corrected to say products minus reactant. We want to follow that same process. So we would say that our entropy change of our reaction is equal to beginning and we'll use the color red for our reactant again. So beginning with our products we have our first product which is our one mole of acetic acid multiplied by its standard entropy given in the prompt as negative or sorry positive 159.8. We have units of not killing jewels but rather jewels divided by moles times kelvin. This is then added to our second product where we have one mole of carbon dioxide multiplied by its standard entropy We'll put the multiplication symbol there it's standard entropy of carbon dioxide is given in the prompt as 213.6 jewels divided by Moles Times Kelvin. This completes the some of the standard entropy of our products. Now we're going to subtract our standard entropy of our reactant where we have our malic acid, we have one mole of malic acid Multiplied by its standard entropy given in the prompt as or sorry, 149 point oh jules divided by moles times Calvin And so now we want to simplify our brackets just as before. So we would say that our standard entropy of our reaction is equal to our first brackets. Should simplify to 373.4. And as far as our units, we can cancel out moles leaving us with jewels per kelvin. We're going to subtract this from our brackets in red which would simplify to 149.0 Jules divided by Kelvin because we can cancel out moles here too. And so taking the difference, we would get that our standard entropy change of our reaction is equal to a value of 224.4 joules per kelvin. But we want to recall that our entropy change should be in kila jewels per kelvin. So we're going to multiply by the conversion factor to go from jewels to kill a jewels by recalling that are prefix kilo tells us that we have 10 to the third jewels for one kg jule. And this allows us to cancel out jules leaving us with killer joules per kelvin as our final unit. And giving us a standard entropy change of our reaction equal to 0.2244 kg joules per kelvin. Now that we have this value, we're gonna note it to the side and because we know that our Gibbs free energy change should be less than zero to be or for our reaction to be spontaneous and we know our values for now entropy and standard entropy of our reaction. We're gonna plug in these values to solve for that minimum temperature. So we can say that zero for our free energy change for a spontaneous reaction is equal to our standard entropy which we calculated above to be a value of We'll say 10.6 kg jewels which is I'm sorry actually before we just plug things in, let's make this clear. We know that our free energy change. Delta G Should be less than 0 to be spontaneous. And so we would say that zero is equal to our entropy change minus our temperature times our standard entropy change. And if we simplify since we know that the left hand side is zero, we would say that our product of temperature times the entropy change is equal to our entropy change because we would subtract delta age from both sides. Or sorry not subtract delta H. But we would add the product of temperature times entropy to both sides. And so with that being simplified to that point, we want to isolate for temperature and to isolate for temperature. We're going to divide both sides by our standard entropy change, meaning that we can say that our temperature is equal to the standard entropy change divided by our standard entropy change. So plugging in our values we can say that our minimum temperature is equal to in our numerator, We calculated above that. Our standard entropy change of our reaction is 10.6 kg jewels. Whereas in our denominator we just calculated above that our standard entropy change of our reaction is 0.2244 kg joules per kelvin. And as you can see we can cancel our units of killing jewels leaving us with kelvin as our final unit for temperature where we would say that our minimum temperature is going to equal positive 47.23 kelvin. But because we want this to be an absolutely minimum temperature, we're going to subtract 2 73.15 kelvin to give us our temperature in Celsius. And so what we would get is that our temperature is equal to negative 226 degrees Celsius. And we can confirm that our reaction is spontaneous at temperatures That are greater than 200 than negative, 226°C. And so our final answer is going to be this statement here. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.
Related Practice
Textbook Question

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontaneous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T. (c) N2F4(g) ⟶ 2  NF2(g) ΔH° = 85  kJ;  ΔS° = 198  J/K

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Textbook Question

From the values given for ΔH° and ΔS°, calculate ΔG° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous? a. 2  PbS(s) + 3  O2(g) → 2  PbO(s) + 2  SO2(g) ΔH° = −844  kJ;  ΔS° = −165  J/K

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Textbook Question

A certain constant-pressure reaction is barely nonspontaneous at 45 °C. The entropy change for the reaction is 72 J/K. Estimate ΔH.

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Textbook Question

Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g) → 3 NO(g) (a) Use data in Appendix C to predict how ΔG for the reaction varies with increasing temperature.

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Textbook Question

Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g) → 3 NO(g) (b) Calculate ΔG at 800 K, assuming that ΔH° and ΔS° do not change with temperature. Under standard conditions is the reaction spontaneous at 800 K?

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Textbook Question

Consider the following reaction between oxides of nitrogen: NO2(g) + N2O(g) → 3 NO(g) (c) Calculate ΔG at 1000 K. Is the reaction spontaneous under standard conditions at this temperature?

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