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# Tips for Determining Direction Shifts in Reactions - Concept

###### Jonathan Fong

###### Jonathan Fong

**U.C.Berkeley**

M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

Here's some tips and tricks for determining the direction your reaction will shift, to reach chemical equilibrium. I'll show you the simple way first because everyone likes the simple way.

So say if I have one reactant and then it changes into a product. Say when I start or initially I have one molar of my reactant and I have none of by-product B. Well, just by common sense, it should be pretty easy to know that the direction of the reaction would be to the right, because it needs to make products. So in this case it would go to the right. So the reaction would produce products, so to the right to make products.

The other simple way is say if we have A and B again, and in this case, we have no Molar of A, but we have say 2 Molar of B and this is at the start or initial. So which way is the system going to shift? Well, it's going to shift to the left. And it should make sense because, we're going to make reactants because we had none at all. That's the easy way, the simple way. Here we go, ready for the hard stuff? Well, it's not that hard.

So we have something called Q and something called K. Q is known as the reaction quotient and it happens with initial concentrations. K is the equilibrium constant and that's with equilibrium concentrations. Now how do you calculate them? Well, say for example if we had say a pure Q expression, Q would equal, well, let's use our equation up above. So we would have the concentration of B and I'll put a little 0 or not at the bottom to signify intitial concentration over the concentration A to the initial. K would equal the concentration of B at equilibrium, over the concentration of A at equilibrium. So that's the only difference. So if you calculate those you can actually figure out which way is the reaction going to shift, in order to basically achieve equilibrium.

So if Q equals K, then you have no shift at all, because basically you're already at equilibrium if Q equals K because they would be the same number, already at equilibrium.

Now if you calculate Q and Q equals 0, then the reaction is going to shift to the right, or towards the forward direction. And the reason why is, it's because it's going to make more products to try to get that Q. So it's a greater number than 0. And then if Q is equal to infinity, and how you get that is when you get basically some number over 0, which means you have no reactants at all. So what happens is that, of course if you have no reactants, the reaction is going to shift to the left.

And if you take a look for Q equals 0, take a look at our very first example, what we started off with, one molar of A and 0 molar of B. So basically you would have zero over one for Q and that's equal to 0. Whereas if Q equals infinity, then take a look at our second example where we have A is 0 molar and B is 2 molar. Two divided by zero that would be infinity. So the reaction would shift to the left just like we talked about before.

And then if Q is greater than K, the reaction will shift to the left, and then that's because Q is going to try to decrease, to get to be equal to K.

And then the last one if Q is less than K, the reaction will shift to the right. And then for these last two, think about the reaction shifting to the left, look at where the greater than sign is and look at where it's trying to eat. So, it's trying to eat to the left. Whereas on Q is less than K, the less than sign is eating to the right and so it's like it's eating to the right. So it's going to shift to the right. Maybe that's the way to help out.

So you have your simple method on top, and you have your little more complicated method with Q versus K. But hopefully these tips for determining the direction of a reaction will help you out and make things a little easier for you. Have a good one.

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###### Jonathan Fong

U.C.Berkeley

M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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