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249 HSet 7 Q3 Homework AnsweredFill in the BlanksTypeyour answers in all of the blanks and submit Calculate the appropriate [H;O H'Jor [OH-!]for the conditions...

Question

249 HSet 7 Q3 Homework AnsweredFill in the BlanksTypeyour answers in all of the blanks and submit Calculate the appropriate [H;O H'Jor [OH-!]for the conditions shown below at 25 "€.4.2E-8If the [OH-1]=2.37E-07 M,then the [H;o--=You are incorrect0.0000012If the [H;0+1]= 8.33E-09 M, then the [OH" J= You are correct Submit your answer using notation. For examplel.2x 10 would be 1.2e-02HintThe equilibrium constant Kw = [HsO+1[OH

249 HSet 7 Q3 Homework Answered Fill in the Blanks Typeyour answers in all of the blanks and submit Calculate the appropriate [H;O H'Jor [OH-!]for the conditions shown below at 25 "€. 4.2E-8 If the [OH-1]=2.37E-07 M,then the [H;o--= You are incorrect 0.0000012 If the [H;0+1]= 8.33E-09 M, then the [OH" J= You are correct Submit your answer using notation. For examplel.2x 10 would be 1.2e-02 Hint The equilibrium constant Kw = [HsO+1[OH



Answers

Calculate the equilibrium constant $K_{\text {eq }}^{\prime}$ for each of the following reactions at $p$ H 7.0 and $25^{\circ} \mathrm{C},$ using the $\Delta G^{\prime \circ}$ values in Table 13-4. a.(FIGURE CAN'T COPY) b.(FIGURE CAN'T COPY) c.(FIGURE CAN'T COPY)

In this video, we're going to work backwards from the equilibrium plus an expression to write the balanced equation, we'll start with the bottom that gives the reactant and then the exponents turn into coefficients. That's two s. 0. two Plus two H 20. Put a double arrow since this is an equilibrium. And then we have two H 2 s Plus 302. And then let's just check are we missing any solids or liquids or is everything balanced to suffers for Hodgins? Six oxygen's, yep. Alright. For part B. We have I. F. And then one half. F. Two plus one half I too. And that's our complete equation for part C. We have cl was to be ar minus should be a cl two, shouldn't it? Yeah. And then it's too see all my nests and it's going to be br too because it's not included in this expression because brahman is a liquid. So it wouldn't show up. But you you can't just have growing on one side and nothing on the other side. That's how you know, it's missing. All right. And then we have I know three And then eight h plus. Yeah. And then on the top of the products to n. o. and four h. 20. And three See you two plus. Uh So nitrogen is balanced, water, oxygen's and um yeah, oxygen and high jeans are balanced. The only thing is that the right side has a copper um three copper specifically. So the left side needs to have three copper that we're just going to surmise are solids that end up being oxidized. So the equation is balanced and we're done.

To estimate the value of the equilibrium constant for each of the following reactions. In order to do so we'll have to use the expression Lawn K casar equilibrium constant is equal to delta, each knots minus delta H. Not reaction. They are one over T plus delta s not reaction over our for a equilibrium reaction is to C E. O. Gas. Also to gas will produce To co two gas. We'll have to find the delta H. Not reaction and Delta asked not reaction before calculating the equilibrium constant. Dr. Each not reaction is equal to be some resulting age, not formation of the products minus the sum got each formation of the reactions and this is equal to two of the H not formation CO two gas on the product side minus to build a judge, not formation ceo plus without the H not formation oh two Equals two times minus 3 93.5. Yeah. You know jules per mole minus you Times -110 .5 Killer Jewels Per Mole. Yeah. Plus zero. And this would give us value here equal to negative 566. Yeah. Okay. Kill a jules per mole. Which is equivalent to negative 566,000 jules Permal. Now let's calculate delta S not and delta S not here is equal to Mhm. Reaction is some D delta? S not information of the products. Mhm minus the sum delta S. No information of the reactant. Mhm. This is equal to two times the Delta S. Not formation of CO. Two gas minus two times he dealt s not formation C E. O. Gas plus delta S. Not formation or to gas. Mhm. This is equal to two Times they'll face not formation is 2 13.8 fuels per more. Yeah minus two Times 1 97. 7. Yeah, jules per mole Plus two. 05 22 jules promote. Mhm. This is equal to 427.6 tools promote minus 4 to 9. Mhm fuels promote the delta S. Not reaction works out to be Mhm 24.7 fuels per mole. Now we'll plug this in and solve for our equilibrium constant. Lawn K is equal to minus. Don't each not reaction over are one over T plus delta. S. Not reaction over our lawn of K is equal to minus and then delta H. Not. We solved was minus 560,000 jules per more for over jules per mole. Sorry, this is jules per mole. This juice from kelvin here for the adult S. Not and jules promote Calvin, add that into each one of these here. Yeah. In the denominator Are as 8314 jules per more Calvin and temperature is 525 Kelvin and plus 24.7 jules per mole. Tell them over 8.314 tools Permal. Tell them find that lawn K. Here is equal to This is 128 point three plus 297 Yeah. For lawn K. Z. Equal to 131 points through 331.27. Mhm. And taking the inverse. Here we find that. Okay, equilibrium constant is equal to 1.02 Times 10 to the 57. For part A. For part B. Our equilibrium reaction is two H 2 s. Yes, in equilibrium with two h. 2 gas & S. two gas. So we'll have to again find delta each not reaction. Delta. S not reaction. And then the equilibrium constant about the H not reaction. Think of the sun got a transformation the products when is his son? You felt the original information of the reactant. And this is equal to you. To belt. It's not formation of each to gas. Yeah. Plus we built the age not formation S. Two gap minus the of age. Not too times adult age, not formation H. two s. Yes. Mhm. And substitute numbers here about the national formation of Each two is 0 Felt. Additional information. S two Is 128 6 cuba jewels promote -2 times a delta. Additional information of Each two SS -20.6 killer jewels promote. And this gives me adult each not Reaction. That's equal to 160 9.8 cured joules per mole. Or this would be for two, 169 1008 100 jules per more. Delta S. Not reaction is equal to the sum you delta. S. Not information of the products minus the sum the delta S. Not formation of the reactant. This is equal to two times the Delta S. Not formation http gas plus. You built the S not formation of as two guests minus. You dealt it two times we built? S not formation of bench to us. Yeah, sure. And substituted numbers in here and I have to ask the information of H two Is 130 0.7 jules per roll kelvin plus. Delta's new information of S. two is 2 28.2. Mhm jules promote Calvin and the delta S confirmation of H. Two S. Is 205.8. Yeah, This would give me a delta is astronaut reaction of 78 jules per mole kelvin. Now to calculate the equilibrium constant lawn of K is equal to minus delta H. Not reaction over our one over T. Plus delta. S. Not reaction over our. Therefore lawn over K is equal to minus. And we found that Delta H not is equal to 169,000 800 jewels per mole. Our is 8.314 jewels promote Calvin Temperature is one over 525. Calvin. Delta S not is equal to 78 jewels per mole Calvin over 8.314 fuels promote Calvin. And here lawn of K Is equal to 38-9 plus 9.3. 8 Lot of K is equal to -29.5. Mhm. Yeah. And using the inverse natural log of both sides, we find that the equilibrium constant has 1.51 Times 10 to the -13.

So for this problem, we've been actually write down the expression for KP for all these cases. In the first case, we have, um all three terms are gases, so all of them will be included in the KP occasion field and no, to the bar to, um he parks was not been to on prosecution of oxygen. In the second case, you have any two or four, um, disassociating to two motives and not two. And both of them are gases. So the most appear in the KP equation and Cape Easy goes to the part of Russia and to square divided by the pressure pressure of mental four. Uh, in the third case, all of these are gases. So all of these will make an appearance in the final equation. The pasha pressure roaming que posh Russia PCL three square Bartra pressure, PVR tree square and partial pressure seal too cute. And I think I skipped one. So in the third part again, all of these are gases. All of these will feature and KP. We have partial pressure of silicon chloride hydrogen. The part too, um, to the can hide right and turning to the part, too. Um, as according to the band, it's kind of deliberation

So here's the question. 84 in Chapter 19 part a right, the chemical equation for the Taliban that was correspond to KP. First, we need to understand the term of K B, which means the association constant for bass. And in this case, is that the association off Ch three image too. So the reaction will be ch three in age, too. Chris Plus Water Liquid and the Associates into ch three Inish three plus hey Chris, Plus oh H minus also equals. And then part B using the value ofthe K B Cat Coy Delta zero. The relationship between Delta zero and KP is that Delta zero in for two negative Artie Long baby. And this is equal to negative. A point 314 times 10 to the minus. Three times 2 98 Kelvin Time's too long. 4.4 times 10 to the minus four and this is equal to 19.1 killed Jules. So that will be part B part. See, what is the value ofthe Delta G at equilibrium? So remember that as a hologram, the value ofthe Delta ji will always be zero now, the last one. What is the value ofthe Delta G when given the following conditions. So the relationship between Delta G and Delta zero is that Delta G is equal to zero plus Artie Long. Q. So the question tells you that the concentration of age plus is 6.7 times 10 to the minus nine. Moller, however, in the question, we were required to know the concentration of Polish miners, so the concentration of which minus and H plus there's a relationship between them that is the concentration of each plus times. The concentration of Polish minus will be equal to 10 to the minus 14. So in this case, for have the concentration off wish minus equal to 10 to the minus 14 divided by 6.7 times 10 to the minus night, and this will give you a number of 1.5 times 10 to the minus six. So now plugging the numbers to Q would have Delta G equal to Delta zero, which is 19.1 plus Artie. That is a 0.314 times 10 to the minus three temperature is 2 98 lawn of Q Q Is the concentration of the products divided by the concentration of the reactant CE. So that is 2.4 times 10 to the minus. Three times 1.5 times 10 to the minus six, divided by 0.98 on this will give you a number off. Negative 23 point three Killer Jules.


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