Kinetics & Collision Theory

BASIC IDEA - Before a rxn can occur the particles must hit each other AND this collision must have enough energy to cause a change.

To summarize reaction rates, they depend on TWO factors:

OR Reaction = number of x percent of collisions with rate collisions enough energy to react

FACTORS EFFECTING REACTION RATES:

1.) NATURE OF REACTANTS

2+ 4+ 3+ 3+ a) Fe (aq) + Ce (aq) -----> Fe (aq) + Ce (aq)

b) CH4 (g) + 2 O2 (g) ---> CO2 (g) + 2 H2O (g) + 2+ c) Mg (s) + 2 H (aq) ---> Mg (aq) + H2 (g)

2.) CONCENTRATION

3.) TEMPERATURE

4.) PRESENCE OF A CATALYST

1 Ch. 12 - Rates of Reactions

• The rate of a reaction is a positive quantity that tells us how the concentration of a reactant or product changes with time.

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

Unit for rate is:

Express the rate of reaction in terms of reactants/products

(-) ∆ conc. NH3 or ∆ conc. N2 or ∆ conc. H2 time time time

∆ [NH3] ≠ ∆ [N2] ≠ ∆ [H2]

If rate of formation of N2 is 0.040 M/min, what is the rate in terms of a. ∆ conc. NH3 b. ∆ conc. H2

• Rate may be average rate OR instantaneous (see graph)

• For instantaneous find slope of tangent on rate curve for that time. ------

2 N2O5 (g) -----> 4 NO2 (g) + O2 (g)

Time (min) 0 1 2 3 4 conc. N2O5 (M) 0.160 0.113 0.080 0.056 0.040 Inst. rate (M/min) 0.056 0.039 0.028 0.020 0.014

Q - What is the average rate at:

a. 3 min

b. 4 min

Differential rate law -

Integrated rate law –

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

n Rate Expression: Rate = k [ NH3]

0 If Zeroth Order n = 0 Rate = k [NH3] k = Rate 0 [NH3]

1 If First Order n = 1 Rate = k [NH3] k = Rate 1 [NH3]

2 If Second Order n= 2 Rate = k [NH3] k = Rate 2 [NH3] ______• Sometimes rates depend on more than once substance. X + 2 Y + Z ----> Q + 2 R Rate = k [X]1 [Y]0 [Z]2

Rxn is 1st order with respect to X Zero order with respect to Y Reaction is 3rd order overall 2nd order with respect to Z

How will rate change if: A) [X] is doubled B) [Y] is tripled C) [Z] is tripled D) If ALL of the above occur ______To Determine Order: Compare rate and concentration changes. Concen. Δ Effect on Rate Order [A] triples Rate triples [B] doubles Rate unchanged [C] doubles Rate quadruples

Rate Law:

NOTE: 1) The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

3 2) The value of the exponent n must be determined by experiment (i.e. the reaction mechanism must be determined); it cannot be written from the net equation.

4 Find Rate Order from the Following Data

Ex. 1 CH3CHO (g) --> CH4 (g) + CO (g)

Conc. CH3CHO (M) 0.10 0.20 0.30 0.40 Inst. rate (M/min) 0.085 0.34 0.76 1.4 n Rate 2 = Conc. 2 log Rate 2 = n log Conc. 2 Rate 1 Conc. 1 Rate 1 Conc. 1

Ex. 2 Conc. Rate 0.010 M 0.0082 M/min 0.015 M 0.0277 M/min

Another Method: + - NH4 (aq) + NO2 (aq)  N2 (g) + 2 H2O (l)

+ - Initial Rate Experiment Initial [NH4 ] Initial [NO2 ] (mol/L•s) 1 0.100 M 0.0050 M 1.35 x 10-7 2 0.100 M 0.010 M 2.70 x 10-7 3 0.200 M 0.010 M 5.40 x 10-7

-7 n m Rate 2 = 2.70 x 10 M/s = k (0.100 M) (0.010 M) -7 n m Rate 1 1.35 x 10 M/s = k (0.100 M) (0.0050 M)

Rate 2 = Rate 1

-7 n m Rate 3 = 5.40 x 10 M/s = k (0.200 M) (0.010 M) -7 n m Rate 2 2.70 x 10 M/s = k (0.100 M) (0.010 M)

Rate 3 = Rate 2

5 Rate Law:

6 Integrated rate laws – another method to find rate order • 1st order rate law: Rate = k [A]1 from calculus ln Ao = k t where Ao = original conc. A A = Conc. @ time (t) Rearrange and solve for ln A:

Zero First Second

Rate Law

Integrated Rate Law Ao - A = k t ln (Ao/A) = k t 1/A - 1/Ao = k t

Solve for y = mx + b

Plot needed to give straight line

Slope of line

Solve for Half-life (T1/2) (A = 1/2 Ao) OR (2A = Ao)

7 Examples - Use Table 12.6 1) For a 1st order reaction it is found that 20% of the sample (X) has decomposed in 100 seconds. (A) Find the value for the rate constant. (units?)

(B) What is the half-life of this reaction?

2) Redo assuming a 2nd order reaction and an initial concentration of 0.200 M. (A) Rate constant?

(B) Half-life?

What is the order for radioactive decay?

Activation Energy and Potential Energy Diagrams

• Activation Energy - the energy required to form the activated complex

• The greater the activation energy the ______the reaction.

POTENTIAL ENERGY DIAGRAMS:

• Three key points: (1) Start ∆ H rxn = (2) Activated Complex Ea = (3) End

PE PE CO + NO CO + NO 2 2

CO + NO CO + NO 2 2

Reaction Coordinate Reaction Coordinate

8 CATALYSTS - Change rate of reaction without being consumed.

•They do this by forming a different ______which has a ______activation energy. • a catalyst is often a solid on whose surface a reactant particle can be adsorbed, or held. • Enzymes are biological catalysts.

Homogeneous Catalyst – all reacting species are in the same phase

Heterogeneous Catalyst –reacting species are in different phases 1) Adsorption and activation of the reactants 2) Migration of the adsorbed reactants on the surface 3) Reaction of the adsorbed substances 4) Escape, or desorption, or the products

Rates and Temperature

Kinetic Energy Distribution plot

# molec. OR # collisions

Kinetic Energy

KEY IDEAS:

1) At a given temp. (T1) not all molecules have the same kinetic energy. Only a few have high energy.

2) When the temperature is increased (T2) the number of high energy molecules increases greatly, even though the average increases only slightly. 3) A threshold energy line (E) is used to relate to chemical reaction. Only the molecules beyond the threshold energy can react.

9 Summary of Diagrams and applications

POTENTIAL ENERGY KINETIC ENERGY DISTRIBUTIONS: DIAGRAMS: A) Draw a kinetic energy distribution. A) Draw a PE diagram. B) In the same diagram can you draw the effect of adding a catalyst. If so, do so. C) In the same diagram, can you show the effect of increasing temperature? If so, do so.

======1) Which of the two reactions 1) Which of the two rxns below is faster? below would be faster? 2) Show how the graphs below would change if a catalyst were added. 3) Which graph below would have a larger increase in rate if… i) temp increased? PE PE ii) a catalyst were added?

Reaction Coordinate Reaction Coordinate cles cles

parti parti

10 Rate and Temperature Cont. Consider Collision Theory - • Arrhenius proposed the idea of threshold (activation) energy. Function of Molecules with energy needed to react. • Rate and # collisions both increase exponentially with temperature. • Reaction rate is much smaller than the calculated collision frequency. • Arrhenius postulated that the # collisions having an energy greater than or equal to the Ea is given by:

J R = gas law constant (8.31 /mol.K)

# collisions = tot. # (-Ea/RT) T = Temperature (K)

w/ Ea collisions e e = base for natural log Ea = Activation Energy (J)

• Experiments show that the observed reaction rate is considerably smaller than the rate of collisions w/ enough energy to surmount the barrier.

k = (z p) e-Ea/RT

k = A e-Ea/RT (Arrhenius Equation)

a more useful form is: ln k = - E a 1 + ln (A) Thus slope (m) = R T or Ea =

Another useful form ~ “Two Point Form”

- Use T2 as higher temp to avoid negative logs

Examples - • Find Ea if rate doubles when T changes 25 °C --> 35 °C.

• Given k = 0.0251 s-1 @ 0 °C, Ea = 116 kJ. Find a) Ratio of rate increase at 100 °C

11 b) value of k at 100 °C

12 Reaction Mechanism • Most reactions take place in a series of steps, each step usually involving two particles. • To find the rate expression for a multi-step mechanism: 1) Focus on the slow step (rate determining step - R.D.S.) 2) Coefficients for each reactant in R.D.S. are the order of reaction with respect to each reactant 3) Eliminate any unstable intermediates from rate expression. a. Write the equilibrium expression for the previous reaction. b. Rearrange the equation and solve for the intermediate. c. Substitute this equation in for the [intermediate] in the original rate expression.

Find Rate Expression

(1) HBr + O2 <==> HOOBr  If the slow step is a) Step 1? (2) HOOBr + HBr <==> 2 HOBr

(3) HOBr + HBr <==> H2O + Br2 b) Step 2?

(4) HOBr + HBr <==> H2O + Br2 ______NET:

Another Example:

1) X2 (g) <==> 2 X (g) FAST

2) X (g) + A2 (g) ---> AX + A (g) SLOW

3) A (g) + X2 (g) ---> AX + X (g) FAST ______NET:

Find Rate Expression: