1. Chapter 15 Kinetics

2. Kinetics Deals with the rate of chemical reactions Deals with the rate of chemical reactions Reaction mechanism – steps that a reaction takes Reaction mechanism – steps that a reaction takes Haber Process: uses iron oxide as a catalyst Haber Process: uses iron oxide as a catalyst N 2 + 3H 2  2NH 3 N 2 + 3H 2  2NH 3 Spontaneous – will occur on its own – not necessarily fast Spontaneous – will occur on its own – not necessarily fast

3. Reaction Rate Change in the concentration of a reactant or product over time Change in the concentration of a reactant or product over time Rate = Δ[A]/Δt Rate = Δ[A]/Δt A- Molarity (final – initial) A- Molarity (final – initial) Units – moles/L/s or M/s Units – moles/L/s or M/s Reactants can have neg. rates – change to a + Reactants can have neg. rates – change to a +

4. Average rates – M changes over a given time period Average rates – M changes over a given time period Instantaneous rate – found at any given second. Instantaneous rate – found at any given second. Rate is affected by 2 factors: Rate is affected by 2 factors: 1. Coefficients in the balanced equation. 1. Coefficients in the balanced equation. 2. Time 2. Time

5. Rate Laws Concerned only with the concentration of the reactants Concerned only with the concentration of the reactants Rate = k[A] n Rate = k[A] n k = rate constant; n = order – must be determined by experimental data k = rate constant; n = order – must be determined by experimental data Rate is directly related to concentration Rate is directly related to concentration

6. 2 types of rate laws: 1. differential - (rate law) – shows how reaction depends on concentration. 1. differential - (rate law) – shows how reaction depends on concentration. 2. Integrated – shows how concentration depends on time 2. Integrated – shows how concentration depends on time

7. Determining the order of the differential rate law: Order is not the same as the coefficient in the balanced equation Order is not the same as the coefficient in the balanced equation Following method only works for differential rates (concentration and rate are given) Following method only works for differential rates (concentration and rate are given) First you must find the order First you must find the order

8. To determine the order: If there are 2 or more reactants: If there are 2 or more reactants: Find where one of the reactants concentrations remains the same will the other reactant’s concentration changes. Find where one of the reactants concentrations remains the same will the other reactant’s concentration changes. Compare rates by the division method demonstrated. Compare rates by the division method demonstrated. It is easier if the larger concentration goes on top. It is easier if the larger concentration goes on top. But whichever experiment goes on top for the concentration, the rate for the same experiment must be on top in the other side. But whichever experiment goes on top for the concentration, the rate for the same experiment must be on top in the other side.

9. Order: 1 st order – concentration doubles – rate doubles 1 st order – concentration doubles – rate doubles 2 nd order – concentration doubles – rate quadruples 2 nd order – concentration doubles – rate quadruples 3 rd order – concentration doubles – rate increases by a factor of 8 3 rd order – concentration doubles – rate increases by a factor of 8 0 order – concentration doubles – rate stays the same 0 order – concentration doubles – rate stays the same

10. Do the same thing for each reactant in the table. Then put them together in one rate law. Do the same thing for each reactant in the table. Then put them together in one rate law. Overall reaction order – sum of all the orders Overall reaction order – sum of all the orders General pattern for units: General pattern for units: L (n-1) /mol (n-1) *s L (n-1) /mol (n-1) *s N is the overall reaction order N is the overall reaction order To calculate the rate constant – plug #’s of any row back in the equation and solve for k To calculate the rate constant – plug #’s of any row back in the equation and solve for k

11. Integrated rate laws: Use this method when concentration and time are given Use this method when concentration and time are given Must graph integrated rate laws to determine the order. Must graph integrated rate laws to determine the order. Slope is related to the rate constant. Slope is related to the rate constant.

12. To determine the concentration at a certain time, plug back into the appropriate equation for the rate law to solve. The initial concentration is the M at time = 0. To determine the concentration at a certain time, plug back into the appropriate equation for the rate law to solve. The initial concentration is the M at time = 0. To determine the rate, plug back into the rate law to solve. To determine the rate, plug back into the rate law to solve. Pseudo- first order rate law – if more than 1 reactant Pseudo- first order rate law – if more than 1 reactant Can determine the rate law by graphing one reactant at a time. Can determine the rate law by graphing one reactant at a time. Then put them together in one rate law. Then put them together in one rate law.

13. Factors affecting reaction rates: 1. Concentration – (from rate laws) 1. Concentration – (from rate laws) 2. Temperature – speed up when temp. is increased 2. Temperature – speed up when temp. is increased 3. Catalyst – speeds up a reaction without being used up itself 3. Catalyst – speeds up a reaction without being used up itself

14. How does temperature affect the reaction rate? Collision model – molecules must collide to react Collision model – molecules must collide to react Higher the temp – higher the KE – more collisions Higher the temp – higher the KE – more collisions

15. Two requirements for reactants to collide: 1. The molecules must be oriented correctly to allow bond formation 1. The molecules must be oriented correctly to allow bond formation 2. The collisions must have enough energy to produce a reaction 2. The collisions must have enough energy to produce a reaction

16. Activation Energy - E a Energy required to break a chemical bond and produce a new one in a chemical reaction Energy required to break a chemical bond and produce a new one in a chemical reaction To be broken the KE must be high enough to overcome its bond and convert it to PE in the new product To be broken the KE must be high enough to overcome its bond and convert it to PE in the new product If the E a is low – reaction happens fast If the E a is low – reaction happens fast If the E a is high – reaction happens slow If the E a is high – reaction happens slow

17. Arrhenius Equation Ln(k) = -E a /R(1/T) +ln(A) Ln(k) = -E a /R(1/T) +ln(A) Ln(k) – y axis; (1/T) – x axis Ln(k) – y axis; (1/T) – x axis When graphed a straight line should form. When graphed a straight line should form. R= 8.31J/mol*K R= 8.31J/mol*K Units for E a = J/mol Units for E a = J/mol Slope = -E a /R Slope = -E a /R

18. Ln(k 2 /k 1 ) = E a /R(1/T 1 – 1/T 2 ) Ln(k 2 /k 1 ) = E a /R(1/T 1 – 1/T 2 ) **Use this form if temperature changes** **Use this form if temperature changes** Catalyst – lower the E a Catalyst – lower the E a Does not affect the overall energy difference of the reaction Does not affect the overall energy difference of the reaction 2 types of catalyst: 2 types of catalyst: A. Homogeneous – same phase as the reacting molecule A. Homogeneous – same phase as the reacting molecule B. Heterogenous – different phase (usually 2 gases being absorbed on the surface of a solid) B. Heterogenous – different phase (usually 2 gases being absorbed on the surface of a solid)

19. Reaction Mechanisms: Series of steps for a chemical reaction Series of steps for a chemical reaction Must meet 2 requirements: Must meet 2 requirements: 1. Sum of the elementary steps must give overall balanced equation for the reaction 1. Sum of the elementary steps must give overall balanced equation for the reaction 2. The mechanism must agree with the experimentally determined rate law 2. The mechanism must agree with the experimentally determined rate law

20. Elementary steps – reaction whose rate law can be written by looking at the # of species colliding Elementary steps – reaction whose rate law can be written by looking at the # of species colliding 3 types of molecularity (same as the order) 3 types of molecularity (same as the order) A. Unimolecular – 1 molecule – 1 st order A. Unimolecular – 1 molecule – 1 st order B. Bimolecular – 2 species – 2 nd order B. Bimolecular – 2 species – 2 nd order C. Termolecular – 3 species – 3 rd order – rare since all have to hit at the same time C. Termolecular – 3 species – 3 rd order – rare since all have to hit at the same time

21. Intermediate – a species that is neither in the overall reactant or product but is used up during the reaction Intermediate – a species that is neither in the overall reactant or product but is used up during the reaction To determine the rate law look at the rate determining step. To determine the rate law look at the rate determining step. Rate determining step – in multi- step reaction it is the slowest step Rate determining step – in multi- step reaction it is the slowest step