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At equilibrium, the concentrations do not change with time. So, is it true that the rates of the chemical reactions are zero at equilibrium? Wikipedia says that they are not zero. Why is this?

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    $\begingroup$ Who say these two r zero?? $\endgroup$ – Vidyanshu Mishra Nov 22 '16 at 18:27
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In equilibrium, the forward and backward rates are equal to each other. The net is zero, but the individual rates are not zero.

Consider something as simple as water — we know that the water molecule can dissociate in to an $\ce{OH-}$ and $\ce{H+}$, and the pH is a direct measure of the amount of $\ce{H+}$. Once equilibrium is achieved, it isn't like $\ce{H2O}$ stops splitting, or $\ce{H+}$ and $\ce{OH-}$ get back together — it is just at a macro level nothing changes.

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The magic word here is "dynamic equilibrium". Let's look into that.

Let's say we got a simple reaction like: $$\ce{A <--> B}$$

Now how fast does A react to B? For this case we assume first order in both directions, which means the reaction rate would be:

$$r_1= [\ce{A}] \times k_1$$

and of course for the reaction back it's

$$r_2=[\ce{B}]\times k_2$$

So in both cases the reaction rate is dependend on the concentration of the reactant, that's very important.

Now let's say we start off with only A, in this case $r_1$ will be large and $r_2$ will be 0, since there's no B there. Now A gets less and B is getting more, so $r_1$ is going down and $r_2$ is going up. At some point $r_1$ will be equal to $r_2$, but this means there will be as much of A produced from B, than A will react to B. And the same for B. So over all the reaction is still going in both directions, but it's in a dynamic equilibrium.

Here's an example which describes this pretty well:

Let's imagine you got a backyard and right on the boarder to the next one is an apple tree and a lot of old apples are laying on the ground. You don't like those in your garden, so you go around, pick them up and throw them into your neighbours garden. He doesn't like that and starts throwing them back. Now he is old and is moving much slower than you are, so what is happening?

At first there are a lot of apples on your side and you might need to walk some meters but then you can pick one up and throw them. So you will throw a lot of apples very fast. On the other hand there aren't many apples on the other side and the old man is slow, so he needs much longer to grab one. So he isn't throwing them back to you very rapidely. But here's the problem: the more apples you throw the less apples there are, and you need to move longer distances to grab one. So the rate in which you are throwing apples decreases. On the other side there are more and more apples, so the old man doesn't need to move far to grab one, so his rate in throwing apples increases.

After some time there will be a point where for every apple you throw to your neighbour he will throw one back. And if you count the total number of apples on each side it will pretty much stays the same, even if there are apples flying in both directions the whole time.

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Correction to above statement which assumes that there must be equal concentrations of products and reactants in an equilibrium reaction.

In equilibrium there is no net change of products or reactants. The forward and backward rates are only equal to eachother if there are equal amounts of products and reactants which is not always the case.

A reaction can be 99% reactants, 1% products and be in equilibrium as long as those concentrations are not changing, and in that case the forward and reverse reaction rates are not equal.

Consider for example the disassociation of a weak acid: HA <> H+ + A- since it is a weak acid you already know the equilibrium will lie to the left with most of the acid not disassociated. The forward rate is smaller than the reverse rate. If the [HA], and [H+][A-] are not changing over time, the reaction has reached equilibrium. There is always HA disassociating, and H+ and A- combining, but the overall concentrations do not change.

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    $\begingroup$ I don't see anywhere that it says that the concentrations are equal. $\endgroup$ – hBy2Py Nov 21 '16 at 19:11
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So, is it true that the rates of the chemical reactions are zero at equilibrium?

Here you are making a mistake. Just take a simple look at the definition of equilibrium you will find that:

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction.

What the last line says is the state of equilibrium results when the forward reaction proceeds at the same rate as the reverse reaction i.e rate if reactions (forward and backward) become equal not each zero.

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