consider-the-general-reaction-2-mathrmagmathrmbs-rightleftharpoons-mathrmcg3-mathrmdg-quad-delta-h11

Question

Consider the general reaction
$$2 \mathrm{A}(g)+\mathrm{B}(s) \rightleftharpoons \mathrm{C}(g)+3 \mathrm{D}(g) \quad \Delta H=+115 \mathrm{kJ} / \mathrm{mol}$$
which has already come to equilibrium. Predict whether the equilibrium will shift to the left, will
shift to the right, or will not be affected if the changes indicated below are made to the system.
a. Additional B(s) is added to the system.
b. C(g) is removed from the system as it forms.
c. The volume of the system is decreased by a factor of 2.
d. The temperature is increased.

Stephanie L · Accepted Answer

So for this question, were given the following equilibrium reaction. And we&#x27;re told that it&#x27;s already at equilibrium and were asked what will happens into this reaction if we do the following perturbations. So for a what happens if we add B right and being is a solid now because solids and liquids do not show up in the equilibrium, constant expression, there will be no effects, so the reaction will stay at equilibrium. Now, the part b were as what will happen to me equilibrium if we remove see gas as its forms. Now, if we remove, see yes, right which is a product according to lash Achilles principle and the system gonna wanna undo the change. And so because we&#x27;re moving, see, the system is going to shift to produce See again you know, the reestablish equilibrium. So in this case, the reaction is gonna shift to the right and for part C were asked what will happen if the volume of the system is the decrease. So if the container sizes decrease by a factor of two, all right, so if we decrease the volume and in this case the factor of two doesn&#x27;t matter, because this is just a qualitative question. And so if we decreased the buying when you think about what happens to the pressure of the system I and if we decrease the volume, the total pressure increases. These are the pressure of all the gases and the container will increase. And because of that, the Excalibur is gonna want to shift to reduce the pressure. And in order to do that, it will shift to the side with les miles of gas. Since we know that, yes, we know that the moles of gas and the pressure are directly proportional. So in this case, if we look at the left side, we have two moles of gas. This by looking at the Stoke Geometrical efficient uh, the gas one and then on the right side, we have four moles of gas since we have one well coming from C and three moles coming from the deep. So in order to reduce the pressure is going to shift towards the left where we have a less molded gas when we have two moles of gas chefs left and then the final perturbation in part D is what happens if the temperature is increased So if you increase the temperature now to consider that we need to look at the facts that this reaction is Endo thermic nice and still th is confident. And so that means that heat is absorbed so heat is used up as a reaction proceeds. So we want to think about heat as a reactant because it&#x27;s consumed as a reaction and happens now. If we increase the temperature, we want to think about it. As Addie heats, these were increasing the amount of heats. So now, according to the shuttle, it&#x27;s principal write just for, like any of the species analysts, solid or liquid. If we increased the amount of heat, the reaction isn&#x27;t a shift to consume it. Maize is gonna shift over A from it. And so here the reaction is going to shift to the rights

Kyle Angle · Answer

for this problem. We&#x27;re looking at the reaction where two molecules of sulfur dioxide combined with one molecule of 02 to form sulfur Tri ox act and that is in equilibrium. The question is, how do a series of conditions affect our mixture? The first is we&#x27;re adding 02 And when we do that, we&#x27;re going to put positive pressure on the system by increasing 02 to shift some of the reactors to the products. So we see a shift products, by the way, Shadley a principle when we heat the reaction. Well, we&#x27;re told that for this reaction, Delta H. It&#x27;s less than zero that makes it endo thermic in an ex, a thermic reaction heat is kind of like a product. So if it helps you for endo thermic reactions, you can think of it like a starting material. So just like we saw with adding Oh, too, if we heat this reaction, it will absorb that heat and use it to go forward. So once again, you&#x27;ll see a ship to products next doubling the volume. On the reactant side, we have three moles of gas to s 02 and 102 On the product side, we have two moles of gas. When we allow the system to occupy more volume, we have more room for more gases. So this will see a shift to reactors as we open up this value and the reaction can shift back to the side that has more gas to occupy. This for you, adding a catalyst will have no net impact. Old catalysts do is speed up the reaction by lowering the energy barrier so they can more quickly come to equilibrium. But the equilibrium itself remains unperturbed. When we increase pressure, the system is going to favor the side with fewer moles of gas because with more pressure, there&#x27;s not as much freedom for the gas is. So this is going to see a shift Products. Finally, if S 03 is removed, the system&#x27;s going to be under pressure to use s 02 and 02 to replace it. So once again by Les Chateaux Yates principal. When you remove something, you see a shift to that side and we see a shift. Products

David Collins · Answer

to review for any chemical reaction that is excell thermic. When you increase the temperature, you decrease the equilibrium constant for an endo thermic reaction, just the opposite happens. When you increase the temperature, you increase the equilibrium constant. So the first reaction with the positive delta H is endo thermic. So an increase in temperature will result in an increase in the equilibrium constant. The next reaction dealt ages negative, so an increase in temperature will result in a decrease in the equilibrium constant and for the last one, dealt H zero, so there&#x27;s no change.

Kevin Zaborsky · Answer

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&#x27;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&#x27;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&#x27;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&#x27;re left with total energy units and killer jewels, and that comes out to negative 213 0.6 killer JAL&#x27;s. It&#x27;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

David Collins · Answer

to determine the direction of shift. We simply follow the lay shot Elia&#x27;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&#x27;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

Natalie Almond · Answer

So for the falling balance chemical equation, we want to look at two hypotheticals. So for the first we want to look at what happens if we keep a constant temperature for the system. What? We changed the volume. We&#x27;re gonna look at a at A P versus one over the grant and then for the second part here, we don&#x27;t wonder what would happen if we keep a constant pressure, but then changed the temperature. So we&#x27;re gonna look at a volume versus temperature graph here. So, looking first for pressure versus one over volume for an ideal gas, we would see a straight line. This is thanks to what we know is Boyle&#x27;s law, which is that pressure is proportional to one over the volume. We&#x27;re looking at one over the volume here and the pressure on the Y axis here. But when we look at a system we know in terms of increasing one over V, we do not have the same number of moles starting out between the products and the reactions. So we&#x27;re actually going to see a deviation in the system make that a little clearer at low one over V values. But when we look at V vs T volume versus temperature. That means we&#x27;re putting volume on the Y axis in temperature on the X axis. Again, an ideal gas. Be a straight line. However, we know based on this envelope, you value here that since it is positive, we&#x27;re adding we&#x27;re gonna add heat to the reactant sign. So when we increased temperature were increasing, the heat added, which means we&#x27;re going to shift the reaction to the right as to mature increases, which means that the volume is going to decrease. So we&#x27;re going to see a change where the system figures off when the ideal gas constant is a straight line.

Problem

Hydrogen gas and chlorine gas in the presence of …

Problem 36 Medium Difficulty

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