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Chemistry 13 Online
OpenStudy (superhelp101):

The progress of a reversible reaction can be illustrated graphically to show how the rates of the forward and reverse reaction compare. Which of the following graphs represents a reversible reaction that has reached equilibrium?

OpenStudy (superhelp101):

OpenStudy (superhelp101):

@mathmate ?

OpenStudy (mathmate):

Please tell me something about the forward and backward reaction when they are in equilibrium.

OpenStudy (mathmate):

Especially the rates of reactions.

OpenStudy (superhelp101):

The rates of the forward and reverse reactions are equal.

OpenStudy (mathmate):

Very good!

OpenStudy (superhelp101):

:)

OpenStudy (mathmate):

Would the rates be constant or varying with time?

OpenStudy (superhelp101):

i am not sure but i am thinking varying ??

OpenStudy (mathmate):

Think of the word "equilibrium" and try to answer (and explain) again.

OpenStudy (superhelp101):

ok well equilibrium means balanced so if it is balanced it would not be varying but constant. is that right??

OpenStudy (mathmate):

Yes. The explanation would be that if both reactions are varying, but equal, there must be some external cause to make it vary, so it is not really in equilibrium. Try to look up the definition of "dynamic equilibrium" of chemical reactions. You might need it any way for other questions, if not just for your own understanding.

OpenStudy (mathmate):

Now, are you in a position to examine the four graphs, and make your choice, with explanations?

OpenStudy (superhelp101):

dynamic equilibrium is a state of balance between continuing processes. i know that the rates are constant but i don't know how to determine which ones are constant on the graphs

OpenStudy (mathmate):

Just like in math, if you have y=x, it's a line with a slope. When you have y=k, where k is a constant, then you get a horizontal line. |dw:1406850817662:dw|

OpenStudy (superhelp101):

oh ok. so i was thinking the answer could be the last option because the rate of forward reaction is the same as the rate of reverse reaction.

OpenStudy (mathmate):

So what do you choose now?

OpenStudy (superhelp101):

the last option

OpenStudy (mathmate):

Doesn't the last option mean that the rates of reaction have not stabilized? It would be rare to find the forward and backward reaction have varying but perfectly equal rates. One of the requirements of dynamic equilibrium is that the rates must be equal and constant. If they vary, equilibrium has not been reached.

OpenStudy (superhelp101):

sorry but i am kinda confused isn't stabilizing mean equilibrium ?

OpenStudy (mathmate):

Option 4 has both reaction rates in a curve (not a straight line). If you observe the two graph I just drew, the one with a horizontal line represents constant rates. If the rates of reaction are still changing (as in a curve), the reactions have not stabilized.

OpenStudy (mathmate):

Equilibrium is like the water surface in a glass. If you shake the glass, water surface will go up and down, and it's not in equilibrium.

OpenStudy (mathmate):

After a certain time, the water surface will be perfectly plane (flat), then we conclude that the surface has a constant height, so it is (macroscopically) in equilibrium.

OpenStudy (superhelp101):

okay so i don't think it could be option 3 because that looks like it is varying. is that right? But option A and B don't have a straight line either.

OpenStudy (mathmate):

The independent variable is time. We are looking at the time from a particular moment. |dw:1406851692744:dw| We see that the first part of graph 1 is not equilibrium, but they are merging together to have an equal (but opposite) rate, which stabilizes after t=t0. So after t=t0, the reactions are in dynamic equilibrium, because the quantity of products formed (forward reaction) is exactly equal to the quantity of reactants formed by the reverse reaction. Hence they are in dynamic equilibrium.

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