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The equation represents the decomposition of = generic diatomic element in its standard state_{xz(g)X(g)Assume that the standard molar Gibbs energy of formation of ...

Question

The equation represents the decomposition of = generic diatomic element in its standard state_{xz(g)X(g)Assume that the standard molar Gibbs energy of formation of X(g) is 4.41 KJ-mol - at 2000. Kand -46.45 KJ-mol - 3000. K Determine the value of K (the thermodynamic equilibrium constant) each temperature.Kat 2000. K =Xat 3000. KAssuming that 4 Hixn is independent of temperature , determine the value of 4 Hixn from this data:UJ-molAHtxn

The equation represents the decomposition of = generic diatomic element in its standard state_ {xz(g) X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.41 KJ-mol - at 2000. Kand -46.45 KJ-mol - 3000. K Determine the value of K (the thermodynamic equilibrium constant) each temperature. Kat 2000. K = Xat 3000. K Assuming that 4 Hixn is independent of temperature , determine the value of 4 Hixn from this data: UJ-mol AHtxn



Answers

The standard Gibbs energies of formation, $\Delta_{f} G^{\circ},$ for $\mathrm{KO}_{2}(\mathrm{s})$ and $\mathrm{K}_{2} \mathrm{O}(\mathrm{s})$ are $-240.59 \mathrm{kJ} \mathrm{mol}^{-1}$ and $-322.09 \mathrm{kJ}$ $\mathrm{mol}^{-1},$ respectively, at $298 \mathrm{K} .$ Calculate the equilibrium constant for the reaction below at $298 \mathrm{K}$. Is $\mathrm{KO}_{2}(\mathrm{s})$ thermodynamically stable with respect to $\mathrm{K}_{2} \mathrm{O}(\mathrm{s})$ and $\mathrm{O}_{2}(\mathrm{g})$ at $298 \mathrm{K} ?$ $$2 \mathrm{KO}_{2}(\mathrm{s}) \longrightarrow \mathrm{K}_{2} \mathrm{O}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g})$$.

So far are given a chemical reaction. That is two K two in the solid state generates K 20 and 3/2 2. So we have our Delta G the reaction equation because we want to calculate Gibbs Energy. Remember that oxygen and Standard State we have it's free energy is equal to zero. So why Delta G After we consider the other components, there's 159.9 killed Jules Month. So next we're calculating K so we can use the following equation So the G not is equal to negative r t l N k So we can rearrange for lnk. What we get is negative 64.208 and then we simply just So if okay, that is 1.30 times 10 to the negative 28. So k two is the most dynamically stable with respect to cater. I want to

So a continuing work with our main group chemistry, where we're looking at elements across the main groups that are periodic table looking at how they are able to react with one another. And we're expressing this in terms of, like that sort of summer dynamic work as well as chemical equations and how exactly electrons do transfer between species. So jumping straight into what we have here, we have equation for our reaction. We have an 02 oh, to that is in the solid state that generates on boat to, oh, solid state one half 02 in the Gezi estate. So now we're looking at Gibbs Energy of formations of Doubt. The G F is defined as Gibbs Energy off formation. That's our definition. So we have this valley for and 202 So for no Thio, it's negative. 379.79 We'll all starting material. It's negative. Full for nine 0.63 on then for oxygen. It's obviously zero for a pure state. So we have all some Dr G. F not off products, minus reactant. So then family 70 point five. Well, killer jewels, part mall. So the equation that we next need We know that Delta G no equals negative. R t l N k. So therefore, we can rearrange for Alan Kay. Well, im k Andi solve. Okay, so then what we get is K equal to four 0.3 to times 10 to the minus 13. So thermodynamic stability is when a species is at a lower energy state. The entropy change of the reaction from standard heat of formation. We can calculate that as well. So we do have Delta h of formation values. And so what we have is the sum Delta H formation off. Products minus are reactant. Delta H formation off. This reaction is 96 0.67 Keeler jewels. So we have a positive value, meaning that our products are at a higher energy than our reactant. And what I mean here is that if we go back to our reaction at the top, that are starting material is more stable than our product.

So the equation for the reaction is as follows. We have n a 22 in the solid state that generates an A two. Oh, and the solid stay at half two in a gaseous state. So we have our relative gives energies of formation and we have our equation. That is, data do not is able to the son of doubt g formation of the products. Then we subtract some of Delta Jian formation of the reactant. What we get is 17.54 kg jewels from the temperature is 298 Calvin's. We can calculate the equilibrium constant using the following formula that is Delta G is equal to negative r t l N k. So we just rearranged for K So we have after plugging in some values. Yeah, Ellen K is equal to 20 negative 28 0.47 and then we have K is equal to 4.3 two times tent Negative 13. So thermodynamic stability is when a species is at a lower energy state, so we can determine the end will be change of this reaction. So the equation of interest we have Delta H only action is equal to the sum does H formation of products attract with some Delta H formation of reactant. What we get is 96.7 killer jewels. And so, since the value is positive, it means that the products are higher energy than the reactant. So this implies that the reaction is a lower energy than its product. Therefore, any two or two is thorough, dynamically stable with respect to any too o and 02

First we will write the equilibrium constant expression K p. It's going to be equal to the pressure of the product. N O C L squared because of the coefficient divided by the pressure of Eno, also squared And then times the pressure of Cl two. And then we can plug in the values to calculate the value of the constant. So um NOCL is 1.2 atmospheres, So it's 1.2 squared. We omit the units for the K expression you have, N O is Um 0.05. Okay. And we'll square that. Yeah. And then cl two is 0.3. So then we just have to calculate 1.2 Squared, Divided by .05 Squared And then divided by .3 Is 2, 2 significant figures, 1.9 Times 10 to the 3rd. Okay. And that is the value of K p.


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