1st order, 2nd order Rate reactions HELP

[brandon1]

1st order, 2nd order Rate reactions HELP!

I have read the section of my book over and over and studied the practice problems, but I still do not understand 1st order, 2nd order, or 0 order rate reactions. What does it mean to be of any particular order?

 

[algebra_chem]

what are you talking about?

what lesson are you reading?

you information is incomplete, I'm sorry... that's why i cannot understand the order that you're referring to...

 

[andrewchang]

the order is just the relationship between the molarity of a reactant and the amount that it affects the rate of the reaction.

 

[algebra_chem]

oh! so that's what brandon meant... lol! i didn't get it first...

 

[Gokul43201]

If the rate of a reaction is given by :

[tex]rate= dP/dt = k*[A]^a \cdot ^b \cdot [C]^c \dots [/tex]

then the order of the reaction is said to be a+b+c+...

So, a first order reaction has a rate that is linearly proportional to the concentration of one of the reactants. A second order reaction can be one of the two following kinds :

$$rate = k*[A]^2$$
or
[tex]rate = k*[A] [/tex]

A zero order reaction has a rate that is independent of the concentration of any of the reactants.

 

[ksinclair13]

Let's say you have the following data:

A + B ---> C

[A]-----------reaction rate
.010----.020-----.050
.020----.020-----.100
.020----.040-----.400

Look at the first and second lines; [A] increases by a factor of 2 as does the reaction rate ( is held constant). 2^x = 2? x = 1, so [A] is a first order reactant. Now look at lines 2 and 3; [A] is held constant, increases by a factor of 2, and the reaction rate increases by a factor of 4. 2^x = 4? x = 2, so is a second order reactant.

Now you right the differential rate law like this: rate = k[A]^2. The overall order of the reaction is 3.