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QUESTION 7Kinetics is 'hinder; availability 2availability, hinder equilibrium; rate rale equilibriumcontrol and thermodynamics iscontrol:QUESTION 8Visible regi...

Question

QUESTION 7Kinetics is 'hinder; availability 2availability, hinder equilibrium; rate rale equilibriumcontrol and thermodynamics iscontrol:QUESTION 8Visible region in the electromagnetic spectrum has range of what? 400 - 700 nm 1000 1500 nm 200 400 nm 700 1OO0 nm

QUESTION 7 Kinetics is 'hinder; availability 2availability, hinder equilibrium; rate rale equilibrium control and thermodynamics is control: QUESTION 8 Visible region in the electromagnetic spectrum has range of what? 400 - 700 nm 1000 1500 nm 200 400 nm 700 1OO0 nm



Answers

For each reaction in Question 17 , write the equilibrium constant expression for $K_{n}$

Yeah. All right. So, we have a chemical reaction. Let's go ahead and write out the equilibrium law expression for this reaction. And remember that only gases and Aquarius substances go into equilibrium law expressions. Pure solids and pure liquids are omitted. This reaction is being done at 25°C. That makes water a liquid at 25°C. So it's actually not going to be part of our expression. So we're going to write out our law of mass action. We have our products raised to the power of their coefficients in the numerator. Now iodine here is written out as Aquarius, so it is going to be a part of our expression a lot of times you'll find iodine is a solid at room temperature and it wouldn't be. But here it's a quiz. So we're going to keep it in and then we're gonna have our H five I. 06 That's gonna be raised to the 5th power because coefficients five. And then on the denominator only the I. 03 and H plus. We'll go in there. Aquarius. The water is a liquid that stays omitted, but the H plus and the i +03 minus are both to the seventh power, And that is equal to the value of our equilibrium constant, which were given as one times 10 to the minus 85.

So this problem is a little different than other problems leading up to this one in the book. And we want to remember Van Hoffs equation here, and we're working with the following balanced chemical equation and we want to find and full p. But to find empathy in full piece, we'll need to actually doing graph so that we can find the slope of a line given all of our different points. But we can't just graph are KP versus R T Looking at this, you know that we're gonna need the Ln of K, and we're also gonna need one over tea. So we're definitely gonna dio Ellen of K versus one over the temperature. So I'm gonna make a little chart here that is going to take the values in the book and I'm going to for the KP values. I'm going to take the Ln of them. Then for the one over t, we're going to zero point 00167 0.0 143 0.0 1 to 5. 0.0 one of 10 and I'm just gonna add the zero so that we can see, we all have the same six figures. So I highly suggest that you use either a graphing calculator or put these into excel. I put them into a cell personally. So you knew Get the graph itself here is moving town so we can see a little bit better. So I have thesis local line listed here and that this is everything we need to know to find the infill p. So the slope value is a value that comes before the X When you're looking at the form, why equals m X plus B So this right here it's a slope. And we know that this slope is equivalent to negative Well, p over our gas constant. So we found here I'm just going to go ahead and round it to 1.3 times 14th a little bit easier to deal with, and we're gonna know that equals and, you know, p over our gas constant. So since we know are, we can go ahead. I calculate this out and we find that the inflow p equals negative 1.1 five times 10 to the fifth. But this is in jewels per mole because that is what our or ask constant is in our gas constant is 8.314 So we know that if we converts to kill jewels per mole, this is 1 15 killing jewels promote, which is what we usually see and feel P values in.

I want My name is now I will answer this question by using the appropriate value off KSB. We will use KCB value from table. You have table and table for key air. Uses his value for finds equilibrium for these equations. But right thing is, he's a question here. If we want equilibrium off, eh? We will divide it. The concentration off Bruder see, And six naked before times about selfies Negative too divided by the concentration off season I our c But here p f if he s no booted in the clever so now using the appropriate value for Casey in Kiev to find this. So from Kiev, we have relations. Sees that if he lost too when re act with six and I you see forms of complex if e see and six negative four So every rides this k f it will be the concentration off complex for May, divided by the concentration off iron and plus two times by the concentration off scene I barber six the K f in the table is 1.5 times party in 35. This we can use from Pierre may be related to this case also we can use from Kiesa bees is if we have f p s using Kinkade's be it form F B plus toe last seen night Negative too. And the variety ksb off these value. It will be the concentration off our plus two times of isa concentration off, sir. Fire And the key is be value is 3.7 times fighting negative nine. So how can we lied to relation toe finds? Is he found? If we times is his relation by these relation we get this how a few times his pies is some fine will be a path Foolery is the iron were repeated in bull sign up and down so he can remove two Bizerte Very meaning is a complex and 6 to 9 and some fine So we can see that to get his relation with time Kiev boy K s B. So it will be one boy five times by teen certify times for sleep on seven to times fighting negative 19 and the whole result is fine 0.6 times by then Barber 16. Thank you

In this question I've been asked to determine the equilibrium question K. And since we are dealing with state functions, thermodynamic state functions we can use the expression standard keeps energy change is equal to negative Paracin Lynn K. And this is our equilibrium constant that we're looking. So making K substitute formula we get cape being equal to the exponent of negative and it gives energy change divided by our routine. So the main goal here in the first part of the question is to just remind this standard gives energy change and we know this to be and that gives energy change can be calculated from standard and they will be changed minus T. Multiplied by standard entropy change. So since we are dealing with state functions for example for these two we can use this expression to say for example the standard entropy change is equal to the sum of the entropy changes of formation of the reactions of the product minus the sum of and that entropy change of formation of the reactions. And we always have to make sure that we multiply by the respective strict geometric coefficient because these expressions or these parameters are always given in Permal of substance. So we always have to make sure that we multiply by the respective documentary coefficient. Now applying this, we can use the same expression on the standard entropy change. So are playing this would get this from our data booklets all appendices. This is going to be one, one More of product, multiplied by 9.16 Multiple miners to multiplied by three 3.2. So this gives us a standard entropy change of negative 57 point 24 and the season tell a joke. Then we can do the same for the standard entropy change. This is equal to one more. Multiplied by The entropy change which is equal to 304 0.4 minus two moles. Multiplied by 240.1. And this is going to be equal to negative 175 .8. And this is in Joel's pair. No. Now that we have our standard entropy change and our standard entropy change. We can then use this formula to take your mind there. Standard keeps energy change and we take the standard gives energy change and plug it in here to finally get our equilibrium constant. So the standard keeps energy change to achieve standard is equal to negative five 7.25. This is in college also. We multiply by 10 to the poetry to get it in jewels because our standard entropy changes also in job. So this multiplied by there should be minus our temperature which is 655 Kelvin Multiplied by the entropy change which is negative. one point Okay, so our deal turkey standard. It's going to be equal to 5.7 Times 10 to the powerful. Sure what? This tells us. We have a standard mm Then that keeps energy change. That is greater than zero. So it tells us that under standard conditions this reaction is not spontaneous. It needs an external force or external source of energy for it to kicks that all to initially get it, get a drink. So now that we have our delta G standard. We then use this formula, this initial formula here to determine our equilibrium constant K. So our K. is going to be equal to the exponent of negative 5.7 to extend to the powerful divided by Uh which is 8.314. Multiplied by the temperature of 655 Children. So our K here is going to be equal to 2.41 multiplied by 10 to the power negative what? So we do the same for remaining part of the question that is we calculate the standard and then change this is equal to two, multiplied by wonderful 26 -1. 30.9 When there's one multiplied by zero, this is zero. Because understand that conditions for a pure element this standard and they'll be changing formation is zero because all of these are referenced under Few elements. eight standard condition This is going to be equal to negative 1.7. So for this one to calculate our standard, let's first of all calculate our standard standard gives energy change. If we use this formula we're going to get to Want to play it by -1 one, multiplied by 3.1 -1, multiplied by zero To get a negative 5.1 killer job. And for us to then get delta is the standard entropy change. This is going to be the entropy change -1.7 times 20 poetry Miners. The standard keeps energy change off negative 5.1 times 10 to the power three divided by two nine. was These are standard conditions at 25° sources here. What we're doing we are using this formula. Yeah. To data mine our standard entropy chain. So now that we have our standard entropy change, this is equal to standard entropy change is equal trump 11.404 in Joel's pair. Sure. So now that we have this you can then use this formula again to say to achieve is equal to delta age standard minus t daughter is standard. Now this is now going to be evaluated at a temperature of six. Quite fine. Remember all this what we're doing here. We're evaluating these under standard conditions of 25 Degrees sources. So now we want to account for the temperature that we've been given Which is 255 kills. So we can then say it all to change standard is equal to -1.7. to understand to the poetry -655 multiplied by 1 1.40. Sure. So our delta chief standard he is going to be equal to negative 9.1 from stand to the power trade in job. Now that we have our don't Turkey, we then use this formula to get our equilibrium constant, so our equilibrium constant K is going to be the exponents of negative Gibbs, which is negative 9.1 times 10 to the power train, divided by our great point 314 Times 655. So our K is going to be equal to five point then.


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