# What is meant by the rate of a reaction?

##### 2 Answers

#### Answer:

Refer to this video, and I hope you will find your answer.

#### Explanation:

Refer to this video, and I hope you will find your answer.

This video is one of a series of six videos that explains in details the **Chemical Kinetics** topic.

**Chemical Kinetics | Reaction Rates & Rate Law.**

Well, think about what a rate is. It's how fast something happens. Remember how **velocity is the change in distance over time**? Yeah, it's the rate at which one's travel distance changes over time.

*So if reactions involve concentrations changing over time, shouldn't that mean a rate is the change in concentration over the change in time?*

I think you can answer that question.

Once you do, it would make sense to say that for the following reaction:

#"A" + 2"B" -> "C" + 2"D"#

the rate becomes:

#r = r(t) = -(d[A])/(dt) = -1/2(d[B])/(dt) = (d[C])/(dt) = 1/2(d[D])/(dt)# since evidently,

reactantsget(used up) andreactedproductsget... well,.produced

And you might also see that the stoichiometric coefficients (i.e. the number of *relative* to those of the others) are found as **their reciprocals** in the rate.

It makes sense that if there is twice as much **act on twice as much** **as it expects to act upon.**

So to be efficient, ** half** and has each half deal with one equivalent of

**is consumed half as quickly.**

Had there been twice as much

#r = r(t) = -1/2(d[A])/(dt) = -1/2(d[B])/(dt)#

*What does that change about the above reasoning, then? What does it mean for both rates to be halved?*

It just says that now, there is a **twice as much** **and twice as much** **than as if there were only one equivalent of each.**