# Why is the rate of reaction calculated as d[A]/dt?

For example, for a simple chemical reaction like

$$\ce{A -> C}$$

the rate of the reaction could be calculated as the rate of appearance of $$\ce{C}$$ or the disappearance of $$\ce{A}$$, which makes a logical sense. But if the reaction is

$$\ce{A + B -> C}$$

as I read somewhere the rate could be calculated the same way as the first case either as the disappearance of $$\ce{A}$$ or $$\ce{B}$$, or the appearance of $$\ce{C}$$. Is it logical to calculate the rate as the disappearance of $$\ce{A}$$ or $$\ce{B}$$ as the rate?

Think about it, isn't the appearance of $$\ce{C}$$, the combined effect of the disappearance of both $$\ce{A}$$ and $$\ce{B}$$, which mean if the rate of the equation can be calculated as the change in concentration of $$\ce{C}$$ using the product, shouldn't the rate be calculated as the change in concentration of $$\ce{A}$$ plus the change in concentration of $$\ce{B}$$ through some time change $$t$$?

Although the appearance of C is due to the combined disappearance of A and B, the relation is still 1:1 between A and C. Therefore, the rate that A disappears = the rate that B disappears = the rate that C appears.

Let's look at an analogy in stoichiometry. Say I have 1 mole of A and 1 mole of B. Will I make 1 or 2 moles of C? By your logic, I should make 2 moles of C. But I don't.

• but as you said doesn't the rate that C appears have a 1:1 relation ship with B, which means it have a combined relation of 2 with A and B. and does the stoichiometry analogy work for this kind of cases. – Hilea Mar 8 at 23:18
• Consider if I have 10 red legos and 10 blue legos, and I make red-blue legos by sticking 1 red and 1 blue lego together. I will be able to make 10 red-blue legos, and let's say I do it in one minute. The rate of disappearance of the blue legos is 10/minute, as is the rate of disappearance of the red legos. What is the rate of appearance of the red-blue legos? Is it 10/minute, or 20/minute? – jezzo Mar 8 at 23:35

Elementary reaction rates have stoichiometric ratios just like the reaction amounts themselves do. So, in your example, the rate of change in concentration of A, B, and C are all equal in magnitude - For every mole of A you lose, you also lose one mole of B and gain one mole of C. Other stoichiometric ratios would work the same way - you would use the coefficients to figure out how the reaction rates of each species were related.

In practice, you pay attention to whichever compound you care about or is most useful for solving your particular problem. For example, if you could easily measure A but not C, you might use A in your calculations. Or, if you were mostly interested in how much C you would produce, you might track C. They are all related to each other through the reaction equation, though, so you could substitute in a rate equation in terms of any species, as long as you get the coefficient right. This is very useful in cases where one thing is easy to measure but another is not.

Note that for non-elementary reactions the exponents will not be the stoichiometric coefficients, but the amount of any given species will still be related to the amounts of the other via the rate laws and mole balances, and you can still write the rate equations for whichever species or intermediates that you want to.

• RE: "Reaction rates have stoichiometric ratios..." That is true for elementary reactions, but not reactions in general. The statement is thus a tautology since elementary reactions are those reactions where the powers of the coefficients in the reaction equation are the same as the coefficients in the chemical equation. – MaxW Mar 8 at 23:44
• @MaxW that's a good point - I was assuming elementary reactions since they were giving them as examples, but should have made that more clear. – thomij Mar 11 at 14:39