Chemical Kinetics

What is Kinetics?• The study of the rate at which a chemical

process occurs.

• In chemical equations, we see the starting reactants and the ending products, but what about the middle?

• Important because kinetics is involved in how quickly medicine

reacts in the body. Industrial applications on how to make a

reaction more efficient, and quicker.

Factors That Affect Reaction Rates• Physical State and Surface Area

If reactants are in gaseous forms, there is a greater ability to react because of a greater kinetic energy. Medicine in the form of a powder enters the bloodstream more quickly than a solid tablet.

• Concentration of ReactantsAs the concentration of reactants

increases, so does the likelihood that molecules will collide and cause an increase reaction speed.

Factors That Affect Reaction Rates• Temperature

At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

• Catalysts Speeds up a reaction by changing the mechanism. It

decreasing the activation energy required for the reaction to take place.

Reaction Rates

The speed at which the reaction occurs is solved for by dividing the change in concentrations over the time the reaction takes place. Reaction Ex: A B

rate = - [A]

t

rate = [B]

t

Reaction Rates

• In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times during the experiment.

• It makes sense that the concentration [C4H9Cl] is decreasing. The reactant is being used up.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

[C4H9Cl] M

Average Rates The average rate of the reaction is determined by subtracting the change in concentration divided by the change in time:

Average Rate, M/s

(0.0905 – 0.1000) (50.0 – 0)

- 1.9 x 10-4 M/s

Negative value b/c it is decreasing in amount.

Reaction Rates • Note that the

concentration and the average rate are both decreasing as the reaction proceeds.

• This is because as the reaction goes forward, there are fewer collisions between reactant molecules, thus a slower rate.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Practice Problem• In an experiment, the concentration of

substance D was measured at different times. Determine the average rate of the reaction between 3.0 and 7.0 seconds.

Time (S) [D] M

0 0.500

3.0 0.470

7.0 0.360

• Rate = (.360-.470) (7.0 – 3.0)

• Rate = -0.028 M/s

• Is “D” a reactant or a product?

Reaction Rates and Stoichiometry

• In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. For every mole of C4H9Cl lost, there was a mole of C4H9OH created.

• Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. They are directly related.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Rate =-[C4H9Cl]t

= [C4H9OH]t

Reaction Rates and Stoichiometry

• What if the ratio is not 1:1? For every 1 mole of H2 that react, 2 moles of HI are produced.

• In this reaction, the rate that Hydroiodic acid is produced is twice the rate Hydrogen gas is being depleted.

• In order to get one correct reaction rate for all parts of the equation, we must account for reaction stoichiometry.

H2(g) + I2(g) 2 HI(g)

Reaction Rates and Stoichiometry

• So that the reaction rates for all chemical compounds will be the same, the rate for Hydroiodic acid needs to be divided by 2.

• (Remember it produced twice as much, so dividing it by two will make the reaction rate the same for all compounds in the reaction)

H2(g) + I2(g) 2 HI(g)

Reaction Rates and Stoichiometry

• So how would we write the rates for the reaction of nitrogen and hydrogen, in the production of ammonia (NH3)?

• Rate = -[N2] = -1 [H2] = 1 [NH3] t 3 t 2 t

N2(g) + 3 H2(g) 2 NH3(g)

Reactants are negative, Products are positive.

Generalized Rate Expression• Use this generalized equation to write rates for a reaction

aA + bB cC + dD

Reactants decrease(Negative rates)

Products increase(Positive rates)

Practice Problem: • Write the rate expression for all

molecules in the following equation:

2 N2O 2 N2 + O2

Rate = -1 [N2O] = 1 [N2] = [O2] 2 t 2 t t

Practice problem with calculations

• The disappearance of dinitrogen monoxide occurs at a rate of -3.25 x 106. Determine the rate of nitrogen and oxygen gas appearance.

-1 [-3.25 x 106] = 1 [N2] N2 = 3.25 x 106 2 2 t

-1 [-3.25 x 106] = [O2] O2 = 1.63 x 106

2 t

Rate = -1 [N2O] = 1 [N2] = [O2] 2 t 2 t t

Rate Expression and Calculations Practice