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Equation: NO) 3020) 2Ni. 42 2 KJ N26) +3h4e ZNH4o -92,20) B) Which lemperature wauld give greater yield of ammonia, room temperature 100"C? Lm Teth Pua+2 62 au...

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Equation: NO) 3020) 2Ni. 42 2 KJ N26) +3h4e ZNH4o -92,20) B) Which lemperature wauld give greater yield of ammonia, room temperature 100"C? Lm Teth Pua+2 62 ause Toc Oi lous remciro Oor_ GuyoredExplain your answer Deca~>2 jl Ra Laln erhlr < Rury R Fuvo-&ble 2 5 ' or UC decre +L p4_ ~ulre iOcA6*e > JAn Pts pack Jercl olcr 50 tn Icbrium 8hi Yec d 2 7 hin VA Nkz 474 Az b 0itre Lnere 1/? 2 ) A) An equilibrum constant (Kc) constant: does nol change Ahen chemicals are added eq

equation: NO) 3020) 2Ni. 42 2 KJ N26) +3h4e ZNH4o -92,20) B) Which lemperature wauld give greater yield of ammonia, room temperature 100"C? Lm Teth Pua+2 62 ause Toc Oi lous remciro Oor_ Guyored Explain your answer Deca~>2 jl Ra Laln erhlr < Rury R Fuvo-&ble 2 5 ' or UC decre +L p4_ ~ulre iOcA6*e > JAn Pts pack Jercl olcr 50 tn Icbrium 8hi Yec d 2 7 hin VA Nkz 474 Az b 0itre Lnere 1/? 2 ) A) An equilibrum constant (Kc) constant: does nol change Ahen chemicals are added equilibrium: the concentrations change response the stress placed on the system t0 keep the. constant Ihe same value. There exception t0 this: situation where the VALUE of the Kc WILL change. thinking about what you saw the videos, what is the situation where the value of the Kc will change? B). Explain your answer:



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For reactions carried out under standard-state conditions, Equation 18.11 takes the form $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ (a) Assuming $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where $K_{1}$ and $K_{2}$ are the equilibrium constants at $T_{1}$ and $T_{2}$ respectively. (b) Given that at $25^{\circ} \mathrm{C} K_{\mathrm{c}}$ is $4.63 \times 10^{-3}$ for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{kJ} / \mathrm{mol}$$ calculate the equilibrium constant at $65^{\circ} \mathrm{C}$

Part a serpent calculating the partial pressure of B R two initial, which is I defined us. We used the ideal gas law and Artie over V plugging her values. 0.974 contrary to six meters per atmospheres. Mole Kelvin temperature is 1000 Calvin over 1.0 leaders and a pressure would be 79.9 atmospheres initially, but also for the partial pressure of H two. And we're told that starting with 1.22 more, 28 to 06 leaders per atmospheres. Mole Kelvin 1000 Calvin 1.0 is your leaders that would be equal to 100 atmospheres Using this, let's set up a nice table each to guests, beer to guests, equilibrium with to h p r. We use their partial pressures initial from up above. We found that each to was 100 br to a 79.90 computer rice table. Let's define our kee p expression KP expression. Partial pressure of h B. R squared over the partial pressure of each two times a partial pressure br two. We're told that the liberal constant is 2.1 times 10 to the sixth two x squared over, uh, 3 100 minus x 79.9 minus X. Solving this quadratic equation will yield to roots. X would be equal to 100 and 79.9. We're gonna reject this route here. And if we go back to our ice table, the partial pressure of H two at equilibrium is going to be 100 minus 79.9, which is going to give us 20.1 atmospheres. The partial pressure of B R. Two at equilibrium would be 79.9 minus acts which 79.9, which would be zero. In the partial pressure, HBR at equilibrium would be two times X, which will yield 159.8 atmospheres at equilibrium for part B. Let's look at what one of the following changes will I for the equal of impart a Each of the following changes will increase the partial pressure of HBR. Choose the change that will cause the greatest increase on expenditure choice. So for part B, all three of these increase the partial pressure of h p. R. So if we great start by writing or equilibrium and we're told that it is XO thermic and XO thermic to each PR, I guess. And excellent. Eric, by back to the original question. 107.7. Kill a Jules. So we've got three changes that are going to take place. Uh, we could either, um, if we increase so change one is add one more 10.1 mole of each to so 0.10 Moeller of each to going up, or that's changed. One change too, is 0.10 Moeller of beer to increase and change three is to decrease the temperature, which would cause a shift. Now, all three of these cause a shift here temperature goes down. Uh, from initial temperature is 1000 to 700 a decrease of 300. Calvin, uh, decrease. So going from 1000 kelvin to 700 Calvin the one that's going to cause the greatest increase of HBR because we're focusing on this one here would be changed. Number three. A decrease in temperature would really causes equilibrium shift towards the products, adding an equal 30.1 miles of age to our adding 0.1 moles of B R two that would cause the seam increase towards the rate. So the one that's gonna cause the greatest increase towards the rate would be the decrease in temperature with which would cause the increase in.

Of the reaction for the dissociation of Die Atomic A into two singular a molecules. And we're going to discuss some of the entropy and the P and energy properties of this reaction. So our first step here is to look at the signs our standard adults H and R standard. So the standard Delta US is a little bit easier to find. You can look right at this reaction equation and look at the number of moles on either side. Remember that if there's more moles of gas, the entropy of that reaction mixture will be higher. So let's look here. We have one more gas reacting on the reacting side, and then we have two moles of gas on the product side. Therefore, we have more Mel's on the product side, and the entropy of this forward direction reaction is going to be positive. To find Delta H. We want to look at these two reaction mixtures. Here we have initial state one with eight Die Atomic A's and two singular A's, and we have reaction. We have this reaction make sure equilibrium that has six Die Atomic A's and six singular A molecules so visually here you can see that majority of the die atomic A molecules are still in the reaction at equilibrium. This must mean that the reaction is endo thermic and requires an input of energy in order to fully dissociate die atomic A into singular A. And because this reaction is an author, Mick, that means our Delta H is greater than zero. It is positive. So for our next part here, we're going thio distinguish between what Delta s means versus Delta s standard. So our delta s standard so we can conceptually understand the difference between these two just from the idea of standards alone. So our Delta s standard is the amount of entropy in the entire reaction when associating a 21 mold A to completely converting it into two moles off singular A. So that is a standard for this reaction. Whereas this Delta H with the note or this Excuse me, this dealt, asked with the no standard is the entropy for the reaction at its current progress. So this number can change. This number cannot change. So for our third part of the reaction or third part of our question here we are talking about the sign of our Delta G standard. So when you think about Delta G standard, we can use our equation that Delta G standard is equal to Delta H Standard minus the temperature times the standard entropy of the reaction. So we ran into a little bit of a problem here. We have adult H. That's positive. We have adults, us that's positive. And that makes it almost impossible for us to determine the sign of our Delta G standard here so we can call up a few scenarios. So at points when our temperature is very high, that will cause our negative T Delta s term to increase. This in turn, will cause our Delta G standard to be negative at lower temperatures are negative. T Delta s term here will decrease. Therefore, our delta G standard will increase will be greater than zero. So now that we know are signs for Delta G, we can go ahead and look into our K p. So our KP is our equilibrium constant here in our reaction. So when we have our equation of Delta G standard is equal to the negative. Rt l n of Cape er Gibbs Free energy equation as we increase our temperature from up here. Remember, as we increase our temperature, our Delta G standard will become less than zero. It will become a negative number. Therefore, this entire term here, we'll need to be larger. So we can say that as temperature that was a Delta, uh, temperature increases R K p will also increase. So then, for our final part here, we're thinking about Delta G not standard. Just normal. Delta G. So far a delta Gee, at equilibrium when we're at equilibrium are Delta G is going to be equal to zero because there is no change in free energy at equilibrium. There's no change. This Delta G is going to be equal to zero, even though our adult that you standard or at different points of the reaction this Delta G value can vary when we're at equilibrium. Furthest association reaction are Delta G will be equal to zero

For this question, we have several different K values for the generic reaction A. Goes to be. And first well actually have one K value, we have different coefficient values. If both coefficients are one, then it's going to be the concentration of C. A. B divided by the concentration of a. Both raised to the one power. If we're starting out with one moller of a, then it's going to have to shift to the right will increase be by X and will decrease A by X. We solve for X&X 0.80 Mueller. So that means a is going to be 1 -18 or point to Mueller For the next one. They're both too. So we're gonna have to square both of them. We solve for X annex ends up being The concentration of be still in its .67 Mueller. And then a is going to be still one -X. So it'll be .33 Mueller For the last one. It's a little bit more challenging. We have two on B. I'm sorry, Yeah, two on B and a one on A. So as this shift to the right B is going to increase at twice the rate as a will decrease. So if we're defining the increase in B is X, which we're going to have to square because there will be a two here on little B, then the decrease in A is going to be one half X. So it will be one point oh minus one half X and the coefficient here is one. So we have a superscript one here, then we do a little bit of algebra and solve for X. And we get 1.24 Mueller for be for X. And then A is going to be one Mueller minus one half X or 10.38 Mueller.

Using data from appendix four. Let's calculate the values for Delta H, not Delta s not. And the equilibrium constant k for the haber process. At 25 degrees Celsius or 2 98 Calvin felt h not is equal to some moles. The empathy of the products minus some the moles, the M V p of the react. It's Yeah, this is equal to mm to negative 46 killer jewels per mole minus zero plus zero Delta h not is negative. 92 killer jewels per more Delta s knots is equal to the sum of the malls. The M Papi entropy of the product minus the sum of the moles of the trippy of the reactions this is equal to Yeah. Two tons. 193 Jules per covered. More minus three times. 131. Jules per Calvin More? Yeah, plus two 130 one. Yeah, Yeah. Is there one 90? Hugh? Mhm. Yeah. Okay. Mhm. This works out to negative. 199. Yeah. Jules. Per Calvin Mole. Yeah, and the equilibrium constant. We'll have to calculate Delta G not first. Before we calculate the equilibrium Constant Delta G not is equal to Delta H not minus T Delta s not. This is equal to negative. 92 individuals per mole minus 2 98. Calvin minus 1 99. Tools for Calvin Mole. Mhm one killer jewel over 1000. Jules and Delta. Gina works out to minus 33. Killah jewels Promote. Yeah. Now, using this, we can calculate the equilibrium constant. The equilibrium constant is equal to e to the minus. Delta G not over, Artie. It was equal to e to the minus 33. Your jewels per mole over 8.314 Jules per Kelvin. Mole, your jewel for 1000 jewels. 2 98 Calvin. And the equilibrium constant here works out to 61 times 10 to the fifth. Yeah, well, now, calculate Delta G for this reaction under the following conditions for a we have, uh, Delta G not from above. Delta G, not at 25 degrees Celsius. People. The negative 33 killer jewels per mall. Delta Jean, Delta G is equal to Delta G not plus R t lawn. Okay is equal to Delta G, not bus r t lawn. Partial pressure of NH three squared over the partial pressure and two partial pressure of H two cubed. It was negative. 33 Hewlett jewels per mole plus 8.314 tools. Calvin Mole one Killer jewel over 1000 jewels 2 98 Calvin Lawn of 50 squared over 200. This is 200 cubed and solving for adults. Aggie. Here we get negative. 66 killer jewels. Her more for B. Since we are at the same temperature. The delta value for Delta G not is equal to negative. 33 killer jewels perm all from up above. Okay, Delta cheese is equal to Delta. Do not plus r T lan que people the Doctor Genomics plus R T lawn. Partial pressure of N H three squared partial pressure and two partial pressure of H two cubed. This is equal to negative. 33 killah jewels per mole plus 8.314 Rules for Calvin Mole. Mhm killer jewels grown 2000 jewels. 2 98 Calvin lawn of few 100 square, over 206 100 cubed. Okay, this is equal to negative. 67 killer jewels. Her more for C. Since we're changing the temperature the value of belted Jeannot is going to change, so we'll have to calculate a new value for Delta G knots. Think for the negative 92 killer jewels from all minus temperature of 100 Calvin minus 1 99. Mhm covered more to a tool 1000 jewels. This is equal to negative 72 killer joules per mole. Delta G is equal to adults had she not plus r T lan que the top that do not plus r T lawn. Partial pressure of NH three squared pressure and two partial pressure of age two cubed Delta G knots. We just calculated as at 72 Killed Jules per mole at 100 degrees Kelvin 8.314 Cool spring. Tell them all Jewel over 1000 jewels Temperature is 100 Calvin lawn and scared over 50 and 200 cubed Delta G works out to minus 85. Kill Jules per mole and for D again, we are changing the temperature so we have to calculate a value for Delta. Do not is equal. The Delta h not minus Delta S Scott. Yeah, negative. 92 individuals per mole minus temperature of 700 Calvins minus 1 99. Jules for carbon mole. Mhm one killer jewels for 1000 jewels. Delta G not works out to 47 your jewels per mole. Delta G is equal to Delta G not plus r T lan que sequel Adult to Jeannot. Yeah, R T lawn. Partial pressure of any B squared. And two. Which two? Cubed 47. You're joules per mole are mhm. Temperature is 700. Calvin Lawn of 10 squared over 50 and 200 cube and Delta G works out to negative. 41 Q A. Jules per mole.


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