Deriving the rate laws for first and second order reactions

In summary, the conversation discusses the average and instantaneous rates for a reaction of type A --> product, which can be represented by the equations \text{rate} = -\frac{\Delta A}{\Delta t} and \text{rate} = k \cdot \text{A}, respectively. The speaker also mentions that the reaction rate can be written as a separable differential equation, \ln A = -kt + C, and explains how to obtain the equation \ln{\frac{A}{A_0}} = -kt from the derivation by using definite integration. The conversation ends with a comment about the use of calculus in deriving the equation.
  • #1
erik-the-red
89
1
So, the average rate for a reaction of type A --> product is given by [tex]\text{rate} = -\frac{\Delta A}{\Delta t}[/tex]. Also, [tex]\text{rate} = k \cdot \text{A}[/tex].

The instantaneous rate for a reaction of that type is [tex]\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}[/tex].

Setting the instantaneous rate for a reaction equal to the second equation, there is [tex]-\frac{dA}{dt} = k \cdot \text{A}[/tex].

Well, this is a very friendly separable differential equation. I get [tex]\ln A = -kt + C[/tex].

How do I get [tex]\ln{\frac{A}{A_o}} = -kt[/tex] from my derivation? Definite integration?
 
Chemistry news on Phys.org
  • #2
erik-the-red said:
So, the average rate for a reaction of type A --> product is given by [tex]\text{rate} = -\frac{\Delta A}{\Delta t}[/tex]. Also, [tex]\text{rate} = k \cdot \text{A}[/tex].
The instantaneous rate for a reaction of that type is [tex]\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}[/tex].
Setting the instantaneous rate for a reaction equal to the second equation, there is [tex]-\frac{dA}{dt} = k \cdot \text{A}[/tex].
Well, this is a very friendly separable differential equation. I get [tex]\ln A = -kt + C[/tex].
How do I get [tex]\ln{\frac{A}{A_o}} = -kt[/tex] from my derivation? Definite integration?



At time =0 C=Ao
 
  • #3
yes, definite integration, you'll merely have one term on the right since initial time is considered zero. Also remember to always derive the equation for a particular reaction...if you're going to write the rate equation in terms of a reactant/product with a unique coefficient, they'll be multiples to account for.
 
  • #4
Thanks! It's like every intro chem book always uses the exact phrase, "Using calculus, we can derive..."

I mean at the intro chem level most students haven't taken calculus, so to show the derivation is unnecessary. I was just curious :)
 

1. What is a rate law for a chemical reaction?

A rate law is an equation that describes the relationship between the concentration of reactants and the rate of the chemical reaction. It shows how the concentration of reactants affects the speed at which the reaction takes place.

2. What is a first order reaction?

A first order reaction is a chemical reaction in which the rate of the reaction is directly proportional to the concentration of one of the reactants. This means that as the concentration of the reactant increases, the rate of the reaction also increases.

3. What is a second order reaction?

A second order reaction is a chemical reaction in which the rate of the reaction is directly proportional to the square of the concentration of one of the reactants. This means that as the concentration of the reactant increases, the rate of the reaction increases at a faster rate.

4. How do you derive the rate law for a first order reaction?

To derive the rate law for a first order reaction, you must plot the natural log of the reactant concentration versus time. The slope of this graph is equal to the rate constant, which can then be used to write the rate law equation.

5. How do you derive the rate law for a second order reaction?

To derive the rate law for a second order reaction, you must plot the inverse of the reactant concentration versus time. The slope of this graph is equal to the rate constant, which can then be used to write the rate law equation.

Similar threads

Replies
2
Views
1K
Replies
131
Views
4K
  • Chemistry
Replies
7
Views
1K
  • Biology and Chemistry Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
334
  • Chemistry
Replies
7
Views
2K
Replies
7
Views
9K
  • Chemistry
Replies
4
Views
3K
Replies
9
Views
3K
Replies
10
Views
578
Back
Top