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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Rates of Disappearance and Appearance - 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.

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Here's some tips and tricks for calculating rates of disappearance of reactants and appearance of products. Here we have an equation where the lower case letters represent the coefficients, and then the capital letters represent either an element, or a compound.

So if you take a look, on the left side we have A and B they are reactants. If we take a look at the reaction rate expression that we have here. We have reaction rate which is the over all reaction rate and that's equal to -1 over the coefficient and it's negative because your reactants get used up, times delta concentration A over delta time. Because remember, rate is something per unit at a time. So here it's concentration per unit of time.

If we know this then for reactant B, there's also a negative in front of that. -1 over the coefficient B, and then times delta concentration to B over delta time. You take a look at your products, your products are similar, except they are positive because they are being produced.

Now you can use this equation to help you figure it out. Say for example, if we have the reaction of N2 gas plus H2 gas, yields NH3. So the formation of Ammonia gas. We're given that the overall reaction rate equals; let's make up a number so let's make up a 10 Molars per second. We want to find the rate of disappearance of our reactants and the rate of appearance of our products.

Here I'll show you a short cut which will actually give us the same answers as if we plugged it in to that complicated equation that we have here, where it says; reaction rate equals -1/8 et cetera. The first thing you always want to do is balance the equation. Don't forget, balance, balance that's what I always tell my students. If you balance your equation, then you end with coefficients, a 2 and a 3 here. I'll show you a short cut now. So since the overall reaction rate is 10 molars per second, that would be equal to the same thing as whatever's being produced with 1 mole or used up at 1 mole.

N2 is being used up at 1 mole, because it has a coefficient. So since it's a reactant, I always take a negative in front and then I'll use -10 molars per second. Now we'll notice a pattern here.

Now let's take a look at the H2. Now I can use my Ng because I have those ratios here. The ratio is 1:3 and so since H2 is a reactant, it gets used up so I write a negative. And then since the ration is 3:1 Hydrogen gas to Nitrogen gas, then this will be -30 molars per second. I'll show you here how you can calculate that.

I'll take the N2, so I'll have -10 molars per second for N2, times, and then I'll take my H2. I have H2 over N2, because I want those units to cancel out. I'll use my moles ratio, so I have my three here and 1 here. So I'll write Mole ratios just so you remember.

I use my mole ratios and all I do is, that is how I end up with -30 molars per second for H2. I do the same thing for NH3. So I can choose NH 3 to H2. Let's use that since that one is not easy to compute in your head. Say if I had -30 molars per second for H2, because that's the rate we had from up above, times, you just use our molar shifts. H2 goes on the bottom, because I want to cancel out those H2's and NH3 goes on the top. 2 over 3 and then I do the Math, and then I end up with 20 Molars per second for the NH3.

Yeah you might wonder, hey where did the negative sign go? Well notice how this is a product, so this we'll just automatically put a positive here. So this will be positive 20 Molars per second. If you take a look here, it would have been easy to use the N2 and the NH3 because the ratio would be 1:2 from N2 to NH3. Since 2 is greater, then you just double it so that's how you get 20 Molars per second from the 10.

You can use the equation up above and it will still work and you'll get the same answers, where you'll be solving for this part, for the concentration A. The rate of concentration of A over time. Then basically this will be the rate of disappearance. This will be the rate of appearance of C and this is will be the rate of appearance of D.

If you use your mole ratios, you can actually figure them out. And it should make sense that, the larger the mole ratio the faster a reactant gets used up or the faster a product is made, if it has a larger coefficient.

Hopefully these tips and tricks and maybe this easy short-cut if you like it, you can go ahead and use it, will help you in calculating the rates of disappearance and appearance in a chemical reaction of reactants and products respectively. Have a good one.
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