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Ozone (0s) in the atmosphere can reacl with nitric oxide (NOJ: Og) NO(g) NOxg) 0x(g). Calculate the AG? for this reaction at 259€. (AlF 199 kJmol, 45" 41...

Question

Ozone (0s) in the atmosphere can reacl with nitric oxide (NOJ: Og) NO(g) NOxg) 0x(g). Calculate the AG? for this reaction at 259€. (AlF 199 kJmol, 45" 41 WK-mol)

Ozone (0s) in the atmosphere can reacl with nitric oxide (NOJ: Og) NO(g) NOxg) 0x(g). Calculate the AG? for this reaction at 259€. (AlF 199 kJmol, 45" 41 WK-mol)



Answers

One of the steps involved in the depletion of ozone in the stratosphere by nitric oxide may be represented as $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ From the data in Appendix $2,$ calculate $\Delta G^{\circ}, K_{P},$ and $K_{c}$ for the reaction at $25^{\circ} \mathrm{C}$

So we need to figure out the theoretical yield of this if we have 4.55 grounds and to you, change this into molds. Divide by the molar mass off 13 close extreme times two. It's 46 grands that is equal to zero point non nine Mose, and we have that. It's a 1 to 1 ratio between, ah, nitrogen dioxide and ozone. So there would be 0.99 moles of Boat three. And that is equal to how many grounds? Let's above. About 48 4.7 for eight grams of 03 That's a theoretical yield. So to get the percent yield, we simply have to divide 4.58 grams, which was are given I 4.748 grams. We get 96 0.5% their yield

Mhm. Before calculating the equilibrium constant for this reaction, we must first calculate delta G standard. Do you see the thermodynamic values found in appendix for to calculate delta G standard will take the delta G of formation of no two Added to the Delta Geo formation of oxygen, which is zero. Then subtract off the delta geo formation of eno and the delta geo formation standard of ozone. And we get a delta G standard for this reaction of -198 Killer Jewels. The equation used to solve for K. From delta G standard is the equilibrium constant equals E. To the negative delta G standard divided by R. T. However, when using this equation, delta G standard needs to have units of jewels, not kill the jewels. So we'll convert our negative 198 killer jewels into jewels which is negative 198,000 jewels, divide by R. N. T. And we get 51 times 10 to the 34 as our equilibrium constant.

In this problem, we want to find the amount of the product that is formed and the living you reacted in this problem. So first of all, let's start with the ozone. We're going to divide the mass of the ozone given by the molar mass of the ozone. Then we see from the balanced chemical reaction, One mole of ozone gives us one mole of nitrogen dioxide. They were going to convert to the mass of nitrogen dioxide. Using the Mueller mass of nitrogen dioxide, We find that we get what 709g of the nitrogen dioxide for the next part we have to find the massive nitrogen dioxide we get from the mass of nitrogen monoxide. So divided by the molar mass of nitrogen monoxide to get moles of nitrogen monoxide. One mall of nitrogen monoxide gives us one more of nitrogen dioxide. Then multiply by the molar mass of the nitrogen dioxide to get mass of nitrogen dioxide formed from the night stream monoxide from this, we can see ozone is a limiting react. It produces less moles of nitrogen dioxide. Now we want to know the excess reaction. Used up the exits reaction is going to react with the amount of the limiting react to that as president. So we determined that the eliminating reaction was ozone. Therefore, let's take the mass of the ozone and figure out how much of the nitrogen monoxide is going to react with this massive pose. Um so We take .740 g of ozone divided by the molar mass for the ozone. We end up getting molds of ozone, one mole of ozone reaction, one mole of nitrogen oxide from the balanced chemical reaction. So we see that we have .01542 moles of nitrogen monoxide. You stop. Let's determine the amount of the nitrogen monoxide that is originally present. So take the mass of nitrogen monoxide divided by the molar mass to get the moles of the nitrogen monoxide originally present. Subtract the amount that has reacted with the other reacted. Yeah. And then we are able to get the moles of nitrogen monoxide that are left over. So this is how you would approach this problem.

They were looking at a reaction off. 03 ad Oh, generates 202 So here we're just estimating oxygen oxygen during bond, energy and Ozo. So we need to look at the bonds that are present in 03 So we have a single Bondo on a double bond. Oh, these are interchange balls through residence. So the bond energy for a single bond oxygen oxygen bond is born for two, where as the bond energy for oxygen oxygen double bond is 498 kg joules per mole, so are changing entropy for this reaction for 03 is negative. 394 kg jewels pummel. So next we need to compare this value with the values for O single bond. Oh, on a double Bondo. So the value that we've calculated here is an intermediate off the values off the O single bond in O double bond values that aren't just listed above this. This is because the bonds and ozone have a bundled off 1.5, and this is due to the residence that is possible within the structure


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