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Spontaneous process; AG < 0. For Gibbs free energy change AG is defined as AG = AH - TAS. For = reversible reaction at equilibrium; 4G = The equilibrium constant...

Question

Spontaneous process; AG < 0. For Gibbs free energy change AG is defined as AG = AH - TAS. For = reversible reaction at equilibrium; 4G = The equilibrium constant K of the reversible reaction is relate 4G" =-RTn(K): Symbol refers to thermodynamics standard condition: 298 K and atm: 97" ' Exercise: Consider the following reaction: 2 HNOs(aq) + NO(g) -> 3 NOz(g) + HzO() AH =+136. kJ; AS =+287.5 JIK Below what temperature does the following reaction becomes nonspontaneous?5, `

spontaneous process; AG < 0. For Gibbs free energy change AG is defined as AG = AH - TAS. For = reversible reaction at equilibrium; 4G = The equilibrium constant K of the reversible reaction is relate 4G" =-RTn(K): Symbol refers to thermodynamics standard condition: 298 K and atm: 97" ' Exercise: Consider the following reaction: 2 HNOs(aq) + NO(g) -> 3 NOz(g) + HzO() AH =+136. kJ; AS =+287.5 JIK Below what temperature does the following reaction becomes nonspontaneous? 5, ` {e1_ b: Is this reaction spontaneous at 298 K and atm? equilibrium constant of the reaction at 298 K Calculate the Ia nur OC



Answers

From the values given for $\Delta H^{\circ}$ and $\Delta S^{\circ},$ calculate $\Delta G^{\circ}$ for
each of the following reactions at 298 $\mathrm{K}$ . If the reaction is not spontaneous under standard conditions at 298 $\mathrm{K}$ , at what temperature (if any) would the reaction become
spontaneous?
$$
\begin{array}{l}{\text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)} \\ {\Delta H^{\circ}=-844 \mathrm{kk} ; \Delta S^{\circ}=-165 \mathrm{J} / \mathrm{K}} \\ {\text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)} \\ {\Delta H^{\circ}=572 \mathrm{kJ} ; \Delta S^{\circ}=179 \mathrm{J} / \mathrm{K}}\end{array}
$$

To determine whether reaction is spontaneous or not. We have to know whether the gibbs free energy of the reaction was positive or negative. Looking at the equation delta G equals delta H minus T. Delta S. If it's negative it's spontaneous and if it's positive it's not. And since we have a combination of entropy and entropy, different combinations might give us times where reactions are always spontaneous, not spontaneous ever, or whether there is a high temperature spontaneity or low temperature spontaneity. And so if a reaction is shown as non spontaneous, sometimes we can look to determine the temperature at which the the reaction would become spontaneous. So let's look at a couple examples. We'll figure out whether the reaction is spontaneous or not. If it's non spontaneous, what temperature could it become spontaneous potentially. So let's start out with a reaction where we have lead, sulfide and oxygen and that becomes led to oxide and sulfur dioxide. All right now, if we know the DELTA H. For the process occurring at 298 kelvin, we know that because we have the not. It's under standard conditions, we have negative 844 killer jules is our delta H. And our delta S. Is equal to negative 165 jules per Calvin. Now be careful when you're using the Gibbs free energy equation, because we need everything in the same units, and Delta G. Is normally given in kila jewels, her mole. Now delta H is given. All right. I should say here in killer jules, we're dealing with one mole of each substance. Now here for Delta H were given in killer jewels. But if you notice the esses in jewels per kelvin, where you have to change that to kill a jules to be able to use it in our equation by dividing by 1000 and now we can just plug it into delta G equals delta H minus T. Delta S. So we have our negative 844 killer jewels for delta H. It's done at standard conditions. So we know 298 kelvin is our temperature and our delta S. Is negative 0.165 joules per kelvin. The killer jewels for Calvin, kelvin's cancel out and we wind up with the delta G here as negative 795 killer jewels. So under these conditions this is spontaneous because the delta G is negative. Let's take a look at a second example though, let's take some popsicle and will decompose it and we'll make some phosphorus, try chloride and we'll make some oxygen. And if we do the same thing here and get the entropy values at 298 kelvin. The change in entropy for this process is 572 killers rules and the delta S. Is 100 179 jewels per kelvin. And we're going to divide that by 1000 to get our killer jewel value. Yeah. And we can do the same thing down here. Delta G. Not we have our delta H temperature is still 298 kelvin and our delta S. Is 0.179 Kill the jewels per killing. And when we do the math on this one, we wind up with a value of 519 killing tools. So in this case we have something that's not spontaneous, but can we get to a point where we can make it spontaneous and so we can take a look at what's going to happen with temperature here to see whether we can make this spontaneous or not. So remember when we're dealing with our gifts, free energy, if it's negative then it is going if it's negative then it's going to wind up being spontaneous. So let's take a look at how we can figure this part out. If we have delta jeannot is equal to delta H minus t delta S. What is the temperature at which we can make this delta G negative? So we know that this term is going to have to be overall less than zero. And if that's the case then that means that the delta H not term is going to have to be less than the T delta S. Term in order to make that happen. If we have a larger T delta S, that's going to be more negative than the delta H. We can make that to be less than zero. So if that's the case and we want to know what's going on with temperature, if we divide both sides by temperature, we can see what that means because that means at this point that go to H over cheese, going to have to be less than the change in entropy. Or put another way so we can take a look at temperature, temperature is going to have to be greater than the change in entropy over the change in entropy. So we have those values. So it's going to have to be greater than 570 to kill the jewels over 179 killer joules per kelvin. So t when we do the math, it's going to have to be greater than 3.20 times 10 to the third killed her.

In this question we're going to be doing some analysis on the relation of end to all reacting with an autumn produce green walls and more. So what we're hearing here is we first of all want to determine in this and that keeps energy change delta G. Standard. And we can use to say this is equal to some of the standard gives energy change of formation of the product minus the sum of the standard gives energy change of formation of created. And we always have to remember to include the stock geometric coefficient of the respective species. This reaction, because these values in articles, they are always evaluated pair more of substance. So we have to multiply by the number of morals or rather this documentary coefficients in our reactions so that at the end of the day this energy is reported in units of energy for example in killer joe. So if we are to look at our standard booklet, are you touching standard? This is going to be equal. The products that we are looking at here is you know, So say three multiplied by the standard Gives energy formation of and which is 8 7.6 -1. multiplied by here we're now looking at the Victims one or 3.7 and as one multiplied by the one. Here is the one more story geometric coefficient of the respect you created Which is here we have 5 1.3. So our guilty qi standard. This is going to be equal to 10 7.8 and this is in killer Joe. So moving on for us to determine the partial pressure. We know the partial pressure can be related using our K. P equilibrium constant. And if we are to look this reaction our K. P will be the ratio of the partial pressure of the product to those of the reaction. And here we are going to have the partial pressure of you know divided by the partial operational or into or multiplied by the pressure operation of and or two. And we always have to remember to raise used to their respective documentary coefficients. So we have to raise the pressure pressure of N. O. To the power of correct No. All we need here is to rearrange this asked then determine the partial pressure off and all. But how do we get our K. P. We know that guilt achieve standards is equal to negative. Arab T. Lynn. Okay, he So we can make our K. P subject with the formula here to say K. P. Is equal to the exponent or negative standard gives energy change divided by R A T. And we've already calculated the and it keeps energy change in the initial step here. So all we have to do now is to better mind our K P. K. P. Is equal to the exponents of -10 7.8 times 10 to the poetry. This is our gives energy that we determine them. This is going to be in jos and we divide this by Divide this by our our which is 8.314 Multiplied by the temperature to 98. So the K. P value for this reaction is going to be equal to 1.2 69 seven extend to define negative 19. We have this value and remember K. P equilibrium constant is always evaluated under equilibrium conditions. So these partial pressures here in this formula are going to be the partial pressures at equilibrium and we've been given the initial partial pressures. What we can do here to determine the partial pressures equilibrium is to have an I C. E. Table that is we know we are dealing with And two or we're also dealing with one or 2. We are also dealing with and all. So what we can do is first of all look at the initial we look at the change and we use those to to take your mind the final at equilibrium. The initial pressure pressure here. We've got one. We also have one yet. And before the reaction starts we won't be having any and all. So this is going to be equal to zero. So if this changes by a factor of X, let's say this is negative. We're putting a negative year because the quantity is going to be decreasing, that is the partial pressure is going to be crazy. So if we look at this, we have one more of into or reacting with one more of an auto from three moles of and or so if this end to or decreases by a factor of X. It also means that the N. 02 is going to decrease by a factor of X because we have one more of each. Then if these are decreasing by a factor of X, it means this is going to increase by a factor of three eggs because We are forming three miles off. And so what we're going to have here, this is negative X negative because it is decreasing and here are going to produce so we've got a positive, This is three x. So at equilibrium this is going to be the partial pressure of And to always going to be 1 -6 and 1 -6. And that equilibrium the partial pressure of N. O. Is going to be three X. Now that we have these within substitute these into this formula K P. At equilibrium. So This is going to be three x. Remember we have to raise this to the power of three Minor divided by 1 -6, multiplied by one miners X. And this is you're supposed to be equal to The K. P which is 1.2 1.2 1.2 697 times 10 to the power negative one man. So what we have here, if we simplify this, we're going to have 27 cube Is equal to 1.269 seven Trump's 10 to the Power -19. So what we did here was to assume that this is a very small change X very small is very small. We deduced this because if we look at this This is a very small one Because it is raised to the power of -19. So what this tells us is that at equilibrium november? Yeah K. P. Is given by the ratio of the pressure pressure of the products divided by the partial pressure of the reactor. So if we have this funny being very small, it means this is a small value compared to this and partial pressure of the product is a small value which means the reactant concentration, they didn't change much. That is why we have a larger, a larger denominator than a human later remember this is products divided by creating so if we have a lot of product equilibrium, this Cape is going to be very large. But in our case we have a very small K. P. Meaning we still have a lot of reacted stand product this means that the concentration or the partial pressure of the reactant. They didn't change much from initial conditions. As a result, this change is very small and if X is very small, it means one minor aspect is approximately equal to one because this is a very small vial. What is going to happen if this is a small value, you say 1 -1. The X. If X is very small you're going to get something like zero 999999 min. So it might as well be Approximately equal to one. So 1- takes is approximately equal to one. This is how we simplify this expression. Otherwise it was meant to give us a lot of trouble unless of course software public that can where we can just punch in these values and it does the calculation for us. So at the end of the day we simplify this. This is going to be 27 whether we want to teach my X. X is going to be the cube root of 1.2697 times 10 to the negative 19 devoted five 27. So our X. Here is going to be equal to 1.6 Times 10 to the power negative set up. So we want to deter mined the partial pressure of N. O. Remember we said the question of pressure of an or at equilibrium From our icy tail but the partial pressure of 10 or is three X. So this is equal to three x. So the partial pressure of you know is going to be three multiplied by 1.6 times 10 to the negative servant. So the partial pressure of n is equal trump life Times 10 to the power negative seven. And this is in atmosphere. Moving on to the final part of the question we were supposed to determine the temperature we know for example we have a standard gives energy change being equal job the standard entropy change minus T. It's T talk to us standard. So what we can do here is to determine the standard entropy change and the standard entropy change and then the standard gives energy change. And then we make tea subject of the formula. So if we do this we say delta edge standard. This is going to be remember this is a state function. So we can use this formula to you to say this approach to say the standard entropy change is going to be some of the entropy change of formation of the product. Remember this documentary coefficients minus the standard and the change of formation of created So our daughter our standards or Delta eight standard here is going to be three multiplied by 9 1.3 -1 x 8 1.6 -1 x 3. 3.2. So this is going to be equal to 159 .1. And this isn't a killer job. We do the same for the entropy. This is going to be three multiplied by 2, 10 0.8 minus one by 222 2 to 0 -1 x 24 0.1. And this is 172 three in Johannesburg kelvin. So if we look at this, we know that our due to age Our Delta H is greater than zero. And we also note that our delta is is also greater than zero. So what this tells us is or negative teeth, Delta S. is left on zero. We are looking at this formula right here. So if we reduce the temperature that is as T approaches zero degree socials, what this means is our magnitude of R and r. B. Change is going to be greater than the magnitude of negative T delta S. So this implies that delta ki is Greater than 08 law temperature. Because if we are increasing the temperature, if we increase this temperature, this becomes very large than this portion here. And if this is a large number than this part, if the left hand side of this right hand side, this is larger than this. It means this is possible. So if we if we reduce temperature don't achieve becomes large, delta T becomes large at low temperature. So the reaction is name spontaneous and I'm spontaneous. But if we increase the temperature right up to infinity, we know that this part. If we increase the temperature, this party is going to be smaller than this part. As a result, don't achieve It's less than zero. It high temperature. Therefore the reaction is spontaneous. It's very high temperatures. As a result we can say or do teaching Built achy standard is equal to zero and this is equal to delta H. Standard minus T. Daughter is so if this is equal to zero, it means our delta edge standard is equal to T. Delta is standard. So as a result, if we are to cooperate the team T. is equal to Delta eight standards divided by doctor. Yes. Thunder. So the temperature here, This is going to be 15 9.1 to understand to the power three divided by 172 .3. So the temperature t here is going to be equal to 29 right there. This is 9 to 3.4 and this isn't killed.

Yeah. For this question, we have a chemical reaction of one mall, The hydrogen sulfide reacting with sulphur dioxide, producing three moles, gaseous sulphur and two moles liquid water. Delta H is negative at -233 killer jewels. And Delta S. is also negative at negative for 24 jewels per Kelvin. Because both of these are negative. That indicates that this reaction is spontaneous at low temperatures. If they were both positive, it would suggest that reaction is spontaneous at high temperatures. So let's calculate delta G at 393 Kelvin to do that will take DELTA G is equal to delta H minus T. The temperature 3 93, multiplied by delta S delta S. However, we will convert from jules per kelvin to kill jules per kelvin. So the units are the same when we sum them up to get delta G And we get Delta G equal to negative 66.4 killer jewels. To determine at what temperature is it spontaneous? We first need to determine the temperature at which it said equilibrium, and delta G is equal to zero. To do that will set DELTA G equal to zero, put in our delta H and then minus T delta S. Solve for T. We get 5 50 kelvin, And because both of these are negative, we know that it's spontaneous at low temperatures, so it must be spontaneous at temperatures less than 550 Calvin.


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