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39) Consider the following reaction at equilibrium. Whatwill happen if O2 is removed from the reaction?4 FeS2(s) + 11 O2(g) ⇌ 2Fe2O3(s) + 8 SO2(g)...

Question

39) Consider the following reaction at equilibrium. Whatwill happen if O2 is removed from the reaction?4 FeS2(s) + 11 O2(g) ⇌ 2Fe2O3(s) + 8 SO2(g)

39) Consider the following reaction at equilibrium. What will happen if O2 is removed from the reaction? 4 FeS2(s) + 11 O2(g) ⇌ 2 Fe2O3(s) + 8 SO2(g)



Answers

What happens to the following equilibrium reaction when $\mathrm{Br}_{2}$ gas is added to the equilibrium mixture? $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$

For this question. According to Lucia Taylor principle, we know that after a reaction has reached equilibrium, when there's an increase causing treasure on one side, the system would shift the equilibrium to the other side. And if there is a decrease, concentration on one side they call the room would be shifted to that side. So in this case, if we have an equation here and the reaction has reached equilibrium and we add BR to, then the equilibrium be shift to the other side of the R two, which is the product side order right inside. And the second question asked, if there's a changed a Colombian constant. We know that's the equilibrium. Constant is only dependent on the temperature that the reaction is running out. And since we're not changing the temperature here, then they live in. Causton will not change in this case.

For this question where asked to write an equation that fits this equilibrium expression. We know that in the nickel liberum expression, the exponents is the number of molecules present in the equation. So on the product side we have n 02 and H 20 and their respective number is four and six according to the exponents nd equation in the equilibrium expression and on the reaction site, we have four and H three and 702 according to the equation, because the four is two noted asked the exponents. So the foreigner seven arcs are denoted asked the exponents for an eight story and old two, respectively. And four and six are to know that, respectively for and No. Two and h 20

So we have the reaction at equilibrium. The reaction between sulfur dioxide gas and oxygen gas to produce sulfur Try oxide. Ah, the delta H in the reaction was given as a negative 1 98 killer jewels. That's an exo thermic reaction. Heat is given off. He is produced it. You want to write it in the reaction on the product side when Delta H is negative and you want to write it on the left hand side with the reactant when Delta H is positive. So what changes in condition favor the reactant. So the idea here is we want to increase the amounts of reactant since either the sulfur dioxide or the oxygen or both. So one way to do that would be to ADM or of the sulfur try oxide that would, uh, increase the relative amount that would that would increase the amounts of both eso to and 02 Ah, and then another thing that we could do in terms of pressure volume effects. We have three moles of gas on the left, two of the S 02 and one of the 02 arm, adding the coefficients two and one and then two on the right. So there's more moles of gas on the right. We would need to increase the volume to get the reaction to shift to the left. Uh, now we wanted, ah, more eso to than the other thing that we could do would be to change the temperature and what we would do in terms of heat to push the reaction to the left. Uh, we would want to put heat in heat is on the right. If we add something on the right, the reaction will move to the left. The reverse reaction. And so we want to increase the temperature, and that will cause the reaction to go to the left.

Proposing that we have a decomposition reaction, so the decomposition reaction involves a gaseous product. So knowing this information where identifying what must be done for the reaction to reach equilibrium. So if we have a decomposition product where one of the products is in the gay c'est form, we must make sure that the reaction is carried out in a closed our vessel. If the reaction was not to have been carried out in a closed our vessel, we would lose thes Gacy's product on the reaction will never be able to reach equilibrium, and instead we would continue with the forward reaction until all of the reactant have been used up.


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