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For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant(a) becomes one-fourth(b) is doubled(c) is halved(d) remains...

Question

For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant(a) becomes one-fourth(b) is doubled(c) is halved(d) remains same

For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant (a) becomes one-fourth (b) is doubled (c) is halved (d) remains same



Answers

A reaction in which A, B, and C react to form products is zero
order in A, one-half order in B, and second order in C.
a. Write a rate law for the reaction.
b. What is the overall order of the reaction?
c. By what factor does the reaction rate change if [A] is doubled
(and the other reactant concentrations are held constant)?
d. By what factor does the reaction rate change if [B] is doubled
(and the other reactant concentrations are held constant)?
e. By what factor does the reaction rate change if [C] is doubled
(and the other reactant concentrations are held constant)?
f. By what factor does the reaction rate change if the concentrations
of all three reactants are doubled?

Thiss problem gives us the equation. See Solid plus H two o gas Reversible E reacts to form C O gas plus age to gas and then gives us several different scenarios and asks if they were which direction The reaction Well moved too, or if there will be no change on the first one is adding, see to the reaction reaction mixed here. This will result in no change and the reason for this is because C is solid and equilibrium is on ly measured by gases and so adding si will not affect the equilibrium. Let her be has h two o being removed. And so since h two I was removed, the reaction will move to the I'm sorry left to compensate for that for that removal and it will shift towards the react answer to the left for letter C C O carbon monoxide is added. And so since a product has been added, the equilibrium will shift to the left ah or to the reactant ce to compensate. And now for letter d. H two is removed and so since a product is being removed, the reaction will move towards the products or to the right in order to compensate for that. So these are all of the answers for this question

Chapter to Problem number five from Campo, asked which of the following statements correctly describes any chemical reaction that has reached equilibrium. A. The concentrations of products and reactions are equal Be the reaction is now irreversible. See both forward and reverse reactions have halted or d The rates of the forward and reverse reactions are equal. So remember that when a reaction is in chemical equilibrium, that means that the concentrations of the reactant and products remained stable. And for the reactions to remain stable, there has to be an equal number of units going towards be as there are units of B going towards a. So does this mean that the concentration of the products and reactions are equal? Not necessarily. So say we have sixty units of A and A is transitioning to be, with a third of units transitioning to be permitted. So times one third ah, per minute. So each minute you're going to get twenty more units of B at the same time we have Ah, let's say forty units of B that are transitioning towards a at half of all of B is going towards a per minute. So in each circumstance, so At this point, you're going to get twenty units of a going towards B sixty units of a times Ah, one third of it changing per minute. So that's twenty units. There's going to be, Ah, forty units of B times, one half of all units transitioning back towards a per minute. So that's also going to be twenty units. So you're getting twenty more units of B per minute, but you're also getting twenty more units of April minute. So this is an equilibrium. Basically, everything's just going back and forth at an equal rate, but you can still possible for you to have less units of B of your reactant. Then you do love your I'm sorry for. You can have less units of your product. Then you can have your reacted. So is not necessarily the case. The concentration of cryonics and reactions aren't necessarily the same. The reaction is now irreversible. A lot of times you're constantly going in reverse. You're going from a to B and then you're going frumpy today, so it definitely is reversible. So that's not the case. Both forward and reverse reactions have halted. That's not really a reaction, because yeah, how? Yeah. Um so that's not the case, either. The rates of the Ford and reverse reactions are equal. Yeah, that's the keys. So A and B are going back and forth at the same rate. And so the concentrations of the same of the of the react instead, products aren't changing through time, so that's what it means to be an equilibrium. So it's going to be D three eights of forward and reverse reactions are equal.

In this question, we are given a reaction and the rate constance for its forward and reverse reactions being told that these air elementary steps, the first thing that we're asked to do is compute the equilibrium constant. We can remember that the equilibrium constant when we have elementary steps like this is equal to K F over K r. The units of both of these rate constance that have been provided to us match. So we don't need to do any further conversions, and we can plug our numbers directly in here. When you do that calculation, you get 1.5 times 10 to the minus 39. We then ask if this favorites the products are the reactions. You can see that Kay is much, much smaller than one. It's a very, very small number. Times 10 to the minus 39 is its order of magnitude. And so this heavily favors the reactant Sze. Since Kei is products over reactant, the fact that it's a small number must mean the ratio of products to react in CE is very, very, very small

So this question has two parts to it. The first part says, um, doubling the concentration of reacting increases the rate of a reaction four times. Then it says, What is the order of the reaction in respect to that reactant, and to figure out how to answer this first part, Um, we can just do some like, basic logic and put in some, um, random numbers to see how they work out. Until we can say that, let's say our concentration, um, to begin with is, let's say one, and it's to the power of X because we don't know what that ISS and the reaction rate is. Why. And so let's say we double the reaction or double the concentration, like the questions says. So I'll say it's now two, then it has the same order. But our, um, reaction rate is now four times that. So now we have four. Why? And so now to figure out how to pay, uh, what are value of X is what our order is. We can just plug in some numbers and see if they work out, and so one to the power of one is just one, and so are. Why would equal, um one? So one to the power of one equals one. And so we double that two to the power of one just equals two. And so that is not correct. And so now we can try to so one to the power of two. That's just one. And let's see two to the power of two that equals four. So this one checks out. So for the first part of our question, um, it would be, Let's see what? It wasn't asking. What is the order? So now we can say that the order is second order. So for the second part of the question, it says tripling the concentration of a different reactant increases the reaction rate three times. What is the order of the reaction in respect to that reactive. So by the same logic, we can say one to the power of X equals y. And what does it say? Tripling it three to the power of X equals three y. And so we start by plugging in our random values again, Um, we can see that one to the power of one equals one and three to the power of one equals three. And so this checks out. And so we know for that react in the order is first order. Um, and so we go back. This isn't in the question, but just two. Add this in if we're rating. If we were writing our great expression, we could write it as rate equals our rate constant. And let's say this was, um a then our concentration of a to the second power. And then if this was be, then it would be our concentration of B to the first power.


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