SCH 4U - Rate Laws

Rate Laws & Order of Reaction

Typically chemists wish to quantify the affect of changing factors on reaction rate. The rates of reactions change when the reactant concentrations change. This relationship can be expressed in an equation called the rate law:

 For some general reaction A  +  B  →  products     

                                                                rate ∝ [A]^x[B]^y

                                                                rate = k[A]^x[B]^y

  

     *note:  ∝ is a symbol meaning "proportional to" and indicates a relationship between rate and the reactant concentrations.  Mathematically, if we wish to remove ∝ and replace it with = (to create an equation), we must include a constant.  Chemists have chosen the constant k.*

The values of exponents x and y are determined by experiment (which you will see below). k is the rate constant (it is temperature and reaction dependent).

Order of Reaction

The order of reaction is the exponent used to describe the relationship between the initial concentration of a particular reactant and the rate of the reaction.

ex. The reaction, BrO₃^-  +  3SO₃^2-  →  Br^-  +  3SO₄^2-, has the rate law, 

rate = k[BrO₃^-][SO₃^2-].   

What is the order of the reaction wrt each reactant?  What is the overall order of the reaction?

Answer

Just look at the exponent values:

👉 wrt BrO₃^- 1^st order because the exponent on [BrO₃^-] is 1

👉 wrt SO₃^2- 1^st order because the exponent on [SO₃^2-] is 1

👉 2^nd order overall because 1 + 1 = 2

Determining Exponents of the Rate Law

 

To determine the exponents, we study how changes in concentration affect the reaction rate

ex. Given the table of values, determine the rate law for the reaction of A  +  B  →  products.  Calculate the value of the rate constant.

Initial [A] (molL-1)	Initial [B] (molL-1)	Initial Rate of Formation of Products (molL-1s-1)
0.10	0.10	0.20
0.20	0.10	0.40
0.30	0.10	0.60
0.30	0.20	2.40
0.30	0.30	5.40

 

Answer

Finding the rate law requires the use of the table above.  Watch this video and follow along.

rate = k[A]^x[B]^y

rate = k[A]¹[B]²

Using the data from any of the trials (I used trial #1), you can find the value and unit of the rate constant. Watch this video and follow along.

rate = k[A]¹[B]²

0.20 Ms^-1 = k(0.10 M)(0.10 M)²

k = 200 M^-2s^-1

      If k were calculated using any other trial, would the value of k change?

No way dude, k is the rate constant, therefore it will not change.  Give it a try, if you are skeptical.

P.S. For the above reaction, wrt to A the order is 1^st, wrt to B the order is 2^nd and the overall order is 3^rd.

Chemical Kinetics & Half-Life

The rate of a reaction of many processes is a first order process (rate ∝ [A]¹).  

Nuclear decay is one example of a first order process, thus, if the initial concentration of a reactant doubles then the rate doubles. 

When U-238 decays, ²³⁸₉₂U  →  ⁴₂He  +  ²³⁴₉₀Th, so the rate law is rate = k[U-238].  Since this is a first order process, the quantity of radioactive sample remaining follows a predictable pattern – the half life is the time required for half of the sample to decay. 

Different isotopes have different half lives. 

Many chemical systems may have a half life as well (for instance the concentration in the body of toxic lead or cadmium, from cigarette smoke, follows a first order process – so we can calculate how much lead is left in the body after a certain number of half lives). 

Recall that we learned:

• rate = - Δ[A]/Δt and rate = k[A] (from above)
• thus, since we can equate the rates, we can write -Δ[A]/Δt = k[A]
• using integral calculus (which you don't know yet, so just trust me) yields 
  
• ln ([A]_o/[A]) = kt where [A] is the concentration at any time, [A]_o is the initial concentration
• after one half life, [A] = ½[A_o], so ln ([A]_o/½[A]_o) = kt

So, ln 2 = kt_1/2   ↜this brand new equation (the derivation of which you.do.not.need.to.know. can be used alone or in combination with m₂ = m₁(½)^x (where x = t/t_1/2) when needed. 

***Looking for more info?  This video is quite good.  So, is this video.  And this one, too (but don't go beyond the 3 min mark for this one).***

 

 

Homework # 11-18

Answers to Student Questions:

1. The exponents in the rate law are dependent on experimental evidence only.  Thus, the coefficients in the balanced equation having no bearing on the exponents.

In HW question #12 it reads,  "If the concentration of B is doubled, the rate is doubled."

So the [B] x 2 and rate x 2.  So, 2 to what exponent = 2?  Answer: 2¹ and therefore, B’s exponent in the rate law will be 1 (r = [A]² [B]¹ [C]⁰).