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QUESTIONKc is the equilibrium constant for forward reaction what is Kc' for the reverse reaction?(Kc)" none of theseQUESTION 2The equilibrium constant, Kp...

Question

QUESTIONKc is the equilibrium constant for forward reaction what is Kc' for the reverse reaction?(Kc)" none of theseQUESTION 2The equilibrium constant, Kp: equals 3.40 at 25"C for the isomerization reaction: cis-2-bulene trans-2-bulene If a flask initially contains .00 atm of each gas in what direction will the system shift to reach equilibrium? It will shift left: It will shift right: The system is already at equilibrium: The system is not at equilibrium and will remain in an un

QUESTION Kc is the equilibrium constant for forward reaction what is Kc' for the reverse reaction? (Kc)" none of these QUESTION 2 The equilibrium constant, Kp: equals 3.40 at 25"C for the isomerization reaction: cis-2-bulene trans-2-bulene If a flask initially contains .00 atm of each gas in what direction will the system shift to reach equilibrium? It will shift left: It will shift right: The system is already at equilibrium: The system is not at equilibrium and will remain in an unequilibrated state _



Answers

Consider the reaction $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$ for which $K_{\mathrm{c}}=278 \mathrm{M}^{-1} .0 .001 \mathrm{~mole}$ of each of the reagents $\mathrm{SO}_{2}(g), \mathrm{O}_{2}(g)$ and $\mathrm{SO}_{3}(g)$ are mixed in a $1.0 \mathrm{~L}$ flask. Determine the reaction quotient of the system and the spontaneous direction of the system : (a) $Q_{c}=1000$, the equilibrium shifts to the right (b) $Q_{c}=1000 ;$ the equilibrium shifts to the left (c) $Q_{c}=0.001$; the equilibrium shifts to the left (d) $Q_{c}=0.001$; the equilibrium shifts to the right

Okay, So for this question, we have this following chemical question right here and then the Chelsea K at 300 degrees Kelvin is 3.9 times 10 to the negative three. And to give us the concentrations of the reaction reactant and the products and Bob. So they wanted to determine whether the reaction is an equally real and if not, which way will a shift to reach equilibrium. So the first thing that we do is we got to calculate the cue of the reaction. So cue is the concentration of the products over the concentration of the reacting. Right. So we see the concentration of I know is in the product Gotta square because of the two. Then we have C l two. Now, let's go to the reactant we have N o c. L gotta square this because of the two. And now we've got to do is just plug in the values given to us from the question. We got 2.5 times. Censor the negative three square and we got two point. Oh, time Centrally negative. Three over 5.0 times 10 to the negative three square. And we should get Q. As five times sensory negative for now. When we compare Q and K, we see that Q is smaller than K. This means that the reaction is not at equilibrium. The only time a reaction is that equilibrium is when Q equal K reaction reaction. That's equilibrium now, since we have Q is smaller than K, which will will a shift to reach equilibrium were one. Q Is Smalling K it must perceive to the right or two D product side to reach equilibrium?

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.

In order to predict whether this reaction is in equilibrium, we need to cut that are quotient which will be end. Oh, squared C two over and O. C. L squared and were given these values so it's in end. Old is 2.5 times 10 to the negative three. Let's look up seal to 2.0 times 10 to the negative three. And you know see Ellis 5.0 times 10 to the negative Eat squared was calculated quotient. Here, get a caution on. This would be 5.0 times 10 26 Let's compare that to the equilibrium constant. Our quotient here is 5.0 times 10 to the sixth. An equilibrium constant is 3.9 times 10 to the negative three. So it is much larger or not of equilibrium. Therefore, the reaction must shift to the reactant since to reach equilibrium

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


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