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Using any data YOu can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 %C for the following reaaicn:Fe,0,(s) + 3H,(g) 2Fe(s) 3H,O()Rou...

Question

Using any data YOu can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 %C for the following reaaicn:Fe,0,(s) + 3H,(g) 2Fe(s) 3H,O()Round your answer t0 significant digits.K-0D,0

Using any data YOu can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 %C for the following reaaicn: Fe,0,(s) + 3H,(g) 2Fe(s) 3H,O() Round your answer t0 significant digits. K-0 D,0



Answers

$K_{\eta}=32$ for the following equilibrium at $298 \mathrm{K}$ $$ \mathrm{A}(z)+\mathrm{B}(g) \rightleftharpoons \mathrm{AB}(g) $$ What is the value of $K_{c}$ for this same equilibrium at $298 \mathrm{K} ?$

So since we're comparing K at different temperatures, we're going to use the Van Hoff Equation, which is Eleanor K two over of K one equals the heat of reaction over art. One over T one minus one over T. Two. Well, we don't actually have delta H for this reaction. So we're gonna need to calculate it. So heat of reaction is equal to all the heats of formation of the products minus the reactant. So if we start with our product or CH four, We've got negative 74 0.8 and then we subtract our reactions but they're both elements. So they're heats of formation are zero. So this is just equal to negative 74.8 killer jewels. Mhm. So we know that K one is .26, and T one Is 1,000°C.. So we'll need to add to 73 to that. So we'll get 1273 Kelvin, thanks. And K two is what we're looking for and our new temperature Is 750°C.. So again we need that to be in kelvin So that will be 1023. So now we can go ahead and plug in what we know. So the Ln of K two over K one is there heat of reaction? Now I'm going to multiply this by 1000 to get it into jewels. Okay, Because our is given to us and as 8.31 jules per mole Kelvin, So negative 74,800 And then 8.31. So this is jules per mole kelvin. So we have to make sure that this isn't jules Okay? And then one over 12 73 -1 over 1023. And we'll get Ellen of K two minus the Ellen of 0.26 And that's all going to equal 172. So that element okay, is going to equal 0.38 mm E to the X. We'll get K two here is going to be 1.5

And for the synthesis of Yuria from ammonia and carbon dioxide. So in order to find our K value here, R k value we need to recall our Gibbs free Energy equation. Here we have our delta Gee, standard is equal to the negative. Universal gas, constant times, temperature times the natural log of our equilibrium constant. This is gonna be a K p. Here. So what we're going to do first is we want to solve this equation for our okay, because that's what we're trying to find. So what we're going to do first is divide this side by the negative Universal gas, constant and temperature divide both sides by that. So after some algebra here we have that are standard energy of reaction over universal gas. Constant temperature is equal to the natural log of okay. And then we can go ahead and raise both sides thio e to cancel out that natural log. So we get that e to the negative delta G standard over universal gas. Constant temperature is equal to K. So in the problem, we were given that our delta G standard value is equal to negative 13.6 killed goals. Permal our our value. We know since we're using killed jewels going to be 0.8314 killed joules per mole. Kelvin R T value is going to be equal to 298 Calvin, because we're at standard temperature. And so we have our G. We have our and we have our tea. Let's go ahead and sell four k. So here we're going to dio e to the negative negative. 13.6 kg per mole. Divided by 0.8314 times 298 Calvin. A better zero there and not from equals. Okay, so when you put this into our calculators, we get a K value of 242.1. There's no unit on K because it is a equilibrium constant, but we get 242.1

Reaction of sulfur dioxide and oxygen gas reacting to forms sulfur try oxide. So in order to find the equilibrium constant, which is our KP, we need to use the equation that are standard free energy of reaction is equal to the negative of the universal gas. Constant times the temperature that this reaction has run out times the natural log of our equilibrium constant. So let's go ahead and do some algebra here because we're solving for art. Okay, so the first thing that we can do is we can divide both sides by the negative are times t and that will give us that Delta G standard negative doctor to standard over our times T is equal to the natural log of Okay, so we can raise both sides to the E to allow this natural log to cancel and we're left with okay is equal to e to the negative Delta G standard divided by our times t. So let's go ahead and plug in some numbers. We know that mm to the negative and then from the previous problem, we know that our delta G standard of reaction is negative. 141 0.8 killer jewels divided by are but his 0.0 8314 Jules Permal Calvin Times 298. Calvin, there we go. Okay, then we can go ahead and put that in our calculators, and we find that our K of this reaction is equal to 7.1 times 10 to the 24th power. All right, so okay does not have any units because it is a constant and a couple of tips and tricks for solving these types of problems. It's always good to make sure you keep track of your negative signs here. So you see that our delta G standard of reaction here is negative, but this entire term is also negative. So make sure not to lose your negative signs here. And then you could easily plug this into your calculator and get an answer for your KP

Of our values for K one uh, K two t one and t two. So then what we need to find is through the one over t 11 of the t two lnk one and L N K two. So these values are as follows. One of the T one is 3.35 times 10 to the minus 31 over t two is equal to 2.87 times 10 to the minus three l N K one is equal to approximately two Ln k two is approximately negative. 3.42 So the next thing we need to consider is the equation. That is O N K is equal to negative. Delta h not over our t Delta s no over our So we saw for Delta h, not on Delta s no. So Delta h nought is negative. 97.279 kg jewels, part mall. Also astronaut is 33 344.73 jewels part Calvin. So we substitute these values ah, into the following while we're solving 40. So we're plugging these values into this equation here that we looked at in the previous slide in solving for temperature. So we have zero Ln k is equal to nine, 7279 joules per mole. I went right on the units just to save space multiplied by 8.314 multiplied by temperature. We add 344.7 jewels for Calvin, divided by 8.314 Jules for Calvin per mole, where temperature is equal to 9.2 degrees Celsius once we re arrange for temperature.


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