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QUESTION 20Consider the equilibrium system:2iCl(s) ? Z 12(s) + Cl 2lg)Which of the following changes will increase the total amount ofof Clz that can be produced?re...

Question

QUESTION 20Consider the equilibrium system:2iCl(s) ? Z 12(s) + Cl 2lg)Which of the following changes will increase the total amount ofof Clz that can be produced?removing some of the (2(s)adding more ICI(s)removing the Clz as it is formedd.decreasing the volume of the container

QUESTION 20 Consider the equilibrium system: 2iCl(s) ? Z 12(s) + Cl 2lg) Which of the following changes will increase the total amount ofof Clz that can be produced? removing some of the (2(s) adding more ICI(s) removing the Clz as it is formed d.decreasing the volume of the container



Answers

Consider the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ When this system is at equilibrium at $700^{\circ} \mathrm{C}$ in a 2.0 - $\mathrm{L}$ container, $0.10 \mathrm{~mol}$ $\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}$, and $0.40 \mathrm{~mol} \mathrm{C}$ are present. When the system is cooled to $600^{\circ} \mathrm{C},$ an additional $0.040 \mathrm{~mol} \mathrm{C}(s)$ forms. Calculate $K$ at $700^{\circ} \mathrm{C}$ and again at $600^{\circ} \mathrm{C} .$

So here we are given information about essentially a certain system and were given information that the entropy for this chemical reaction is greater than zero, which is going to be important. And were given information that we have to uh no BR push reacts to form to nitric oxide in its gaseous form. And bro mean in liquid form, essentially we're going to stress the system to see how it affects the equilibrium. So in our first case we're going to be adding more bro ming. But in this case since bro ming is in its liquid form and not in its gaseous form, brahmins liquid form isn't included in the equilibrium expression since liquids and solids aren't included since they don't have a significant change in concentration over the course of the reaction. So as a result of this, since there's not a significant change in concentration and since it's not included in the equilibrium expression, this shift will have no effect at all on the system, so no change on the system. So in the second case, we're going to be removing an O B E R. So the system will shift in the direction to oppose this change. So the main way to oppose this change would have the reaction shift to the left, which would increase the amount of N O. B. R. The reaction will shift to the left, and in our third case were given information that we're lowering the temperature. So for an end to thermic reaction, we have to have a certain amount of heat in order for the forward reaction to occur. However, in this case, since we are losing that amount of heat, the system cannot have the forward reaction occurred or it has the forward reaction occur at much lesser extent. So the reaction would shift to the left in this case, and these are our final answers.

To determine the direction of shift. We simply follow the lay shot Elia's principal concepts for the 1st 1 If we decrease the volume, we now have less room for the gas molecules, so it shifts to the side with the fewest moles of gas, which in this case is to the right. If we were to increase the iodine concentration with now, we now have too much. Uh, I, too, so to compensate, it needs to shift to the left if we were to decrease the temperature in order to correctly predict the shift, we need to know whether or not this is an endo thermic or an exo thermic reaction. And because iodine wants to exist as I to it naturally wants to be a die atomic species. Then it's highly likely that this is exo thermic. If it is excellent, thermic than a decrease to temperatures going to result in an increase in the Casey value, and it will shift right

So here we have an example, specific chemical reaction. We have to N. O. B. R. Gashes form leading to the formation of two nitric oxide gas in brahmi in liquid. And were given that for this reaction, the essentially entropy spirited than zero, which means it's end of thermic and were asked how certain shifts would affect the reaction. In our first case we're going to increase the amount of protein in the system. However, it's important to remember that roaming is essentially a liquid and it's not included directly in our equilibrium expression. As a result, this would not have any direct impact upon the system. In our second case, essentially were essentially removing some N O B. R. So the system, according to these principles shift in the direction to replenish that N O B. R. As a result, the system would have to shift to the left. And in our last case we're going to decrease the temperature to decreasing the temperature provides less energy for the foreign reaction to occur. And as a result, since this is an end to thermic reaction, the system would essentially have to shift to the left and this gives our final answers

Were given this reaction and we want to determine how the equilibrium position will change when we impose the various disturbances on the system outlined in parts A through C important A. We want to determine how the system will respond if we increase the total pressure by decreasing the volume according to a shat layers principle. If the stress that we imposed on the system is an increase in pressure in the system will respond to get back to equilibrium by decreasing the pressure the way that we decrease the pressure for a gas phase reaction is to form the fewest number of moles of gas is possible, and the way that we do that is by favoring the side that has the fewer number of moles. We see that we have two moles of gas on the reactant side and one mole of gas on the product side. And so, in order to decrease the number of moles of gas, we have to favor the right side, the product I to in order to decrease the overall pressure because when we have fewer number of moles of gas than we have fewer concentrate, we have fewer contributions to the total pressure based on the partial pressures of those gases. And so that is why important A. The equilibrium position will shift to the right and in part B, we went to determine the effect of adding I, too, so increasing concentration of I two guests for you, the effect of concentration. We should consider the equilibrium expression for Casey that would be equilibrium concentration of I to divided by the equilibrium concentration of I squared. And this is also this expression is also equal to queue. But these are general concentrations where when we take the ratio, it no longer comes out to the value of the equilibrium constant. Because these both of these concentrations concentrations no longer correspond to their equilibrium concentrations because we changed the concentration of one or both of the species to get it out of equilibrium, an important peewee specifically increased the concentration of I to gas from its equilibrium value. It's just since it appears on the numerator. When we take their ratio for Q, we would find that Q is greater than K, and whenever Q is greater in K, the reaction will respond by shifting to the left 24 more reactors. So when we add I to increase its concentration and we have to four more reactors to get back to equilibrium and finally import see, we want to determine what happens when we decrease the temperature. The effective temperature is related to whether this is an endo, thermic or exit thermic reaction. So we need to begin by calculating Delta H using thermodynamic data in the appendix. We know that Delta H at standard conditions is equal to the total change in entropy of the products minus a total change in entropy of the reactive. We see that the only product is I to gas and that is a naturally forming diatonic molecule. And so it's still teach a formation valued standard conditions. Zero. So that means that the total change in entropy of the products zero killer jewels and then we subtract the total and will be changed of the reactions. On the reactant side, we see that we have two moles of i gas and we multiply those two moles by its value, for it's to each of formation at senior conditions value. And when we looked at up, it is 100 six point eight killer jewels per mole. So we see that the units of most cancel out. We're left with total energy units and killer jewels, and that comes out to negative 213 0.6 killer JAL's. It's a Since Delta H comes out to a negative value, we know that this is an exo, thermic reaction. And so Delta H can be written as a product and any time that we have an exo thermic reaction when we decrease, the temperature system will respond by shifting to the right to four more products when we decrease the temperature. Then, according to the chandeliers principle, the system will respond by trying to increase the temperature. We know that when we four more products, he is given off as one of those products. Sue, we are producing more heat to raise that temperature. And so that is why, since this is an exo thermic reaction, where we lower the temperature assistant will respond by shifting to the right


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