Reaction Rate The rate of appearance of a product The rate of appearance of a product or disappearance of a reactant or disappearance of a reactant units: M/s Collision Theory- reaction rate depends on the rate of “effective” collisions between reactants. See “collision Model” video

Factors that affect reaction rate Nature of reactants Nature of reactants Ionic reactions tend to be fast (double displacement) Ionic reactions tend to be fast (double displacement) Molecular reactions are slower (bonds rearrange) Molecular reactions are slower (bonds rearrange) Concentration Concentration Rate = k[A] m [B] n Rate = k[A] m [B] n (rate constant x product of concentration of reactants) Increase pressure on a gas Increase pressure on a gas Temperature Temperature Directly proportional Directly proportional Surface area Surface area Catalyst present Catalyst present

activated complex- a transitional structure created from an effective collision. activated complex- a transitional structure created from an effective collision. activation energy- the energy required to form the activated complex activation energy- the energy required to form the activated complex

E f = activation energy for forward reaction = 30kJ E f = activation energy for forward reaction = 30kJ E r = activation energy for the reverse reaction = 60kJ E r = activation energy for the reverse reaction = 60kJ

catalyst a substance that speeds up a chemical reaction without being changed itself. It lowers the activation energy needed for a reaction to occur.

Exothermic- gives off heat Exothermic- gives off heat Products have less energy than reactants Products have less energy than reactants Tend to be spontaneous Tend to be spontaneous Endothermic- absorbs heat Endothermic- absorbs heat Products have more energy than reactants Products have more energy than reactants

N 2(g) + 3H 2(g)  2NH 3(g) If the rate of disappearance of H 2 is.048 M/s, what would be the rate of formation of ammonia?.048M/s x 2 mol NH 3 =.032 M/s 3 mol H 2 3 mol H 2 Use stoichiometry to relate the rate of one species to the rate of another species in the same reaction (under the same conditions)

Predict Draw a graph of concentration vs. time for Draw a graph of concentration vs. time for A + 2B  3C A + 2B  3C Time Concentration (M) legend [A] …….[B] ___ [C]

Reaction rate varies directly as the product of the concentrations of the reactants in the rate determining step. Rate Law Rate Law Rate = k[A] n [B] m K is the rate constant, increases with temp. [A], [B] are molar concentrations of reactants n, m – reaction order, determined experimentally Overall reaction order = n + m

0 order: Rate = k 1 st order: Rate = k [A] If [A] doubles, rate doubles If [A] triples, rate triples 2 nd order: rate = k [A] 2 If [A] doubles, rate quadruples (2 2 ) If [A] triples, rate is 9x (3 2 ) 3 rd order: rate = k [A] 3 If [A] doubles, rate is 8x (2 3 ) If [A] triples, rate is 27x (3 3 )

A + B  C Trial[A][B]Rate(M/s) 10.1M0.1M M0.2M M0.1M Determine the rate expression for the above reaction. Rate = k[A] 2 Determine the rate constant, k. K =.0076/(.1) 2 =.76 M -1 s -1

C 2 H 4 Br 2 (l) + 3I - (aq)  C 2 H 4 (l) + 2Br - (aq) + I 3 - (aq) Three particles colliding with another is unlikely to occur in one step. Three particles colliding with another is unlikely to occur in one step. Actual reaction mechanism: Actual reaction mechanism: C 2 H 4 Br 2 + I -  C 2 H 4 Br - + IBr (slow) C 2 H 4 Br -  C 2 H 4 + Br - (fast) IBr + I -  Br - + I 2 (fast) I 2 + I -  I 3 - (fast) Rate = k[C 2 H 4 Br 2 ][I - ] overall order of a reaction = two

Reaction Mechanisms NO 2 + F 2  NO 2 F + F slow F + NO 2  NO 2 F fast 2NO 2 + F 2  2NO 2 F overall The rate law should have the reactants from the slow (rate determining) step in it. The rate law should have the reactants from the slow (rate determining) step in it. Rate = k [NO 2 ][F 2 ] Rate = k [NO 2 ][F 2 ] Catalyst- appears as a reactant and a product Catalyst- appears as a reactant and a product Intermediate- is created then used up (not a reactant or product. Intermediate- is created then used up (not a reactant or product.

molecularityElementary Reaction or slow step Rate Law unimolecularA  productsRate = k[A] bimolecularA + A  productsRate = k[A] 2 bimolecularA + B  productsRate = k[A][B] TermolecularA + A + A  productsRate = k[A] 3 termolecularA + A + B  productsRate = k[A] 2 [B] termolecularA + B + C  productsRate = k[A][B][C] You can determine the rate law from a balanced equation IF it is an elementary reaction (one step) or if you know the slow step. Write the rate law for the single elementary reaction below. H 2 (g) + Br 2 (g)  2HBr(g) Rate = k [H 2 ][Br 2 ]

Half life T 1/2 is the time required for the concentration of a reactant to drop by one half. T 1/2 is the time required for the concentration of a reactant to drop by one half. The longer the half life, the more stable the compound. The longer the half life, the more stable the compound.

N 2 O 5(sol)  2NO 2(sol) + ½O 2(g) What is the half life in this reaction? 2.5 hours What concentration of N 2 O 5 will be left after two half lives (5 hour)?.37M

orderRate equationK unitsStraight line graph Half life 0Rate = kM/s[A] vs time 1Rate = k[A]S -1 ln[A] vs time.693/k 2Rate = k[A][B] = k[A] 2 M -1 s -1 1/[A] t vs time1/k[A] 0 Summary

Based on these graphs of concentration of hydrogen peroxide vs. time, what order reaction is this? Which graph is most linear?

Activation energy k = Ae (-Ea/RT) Arrhenius equation k = Ae (-Ea/RT) Arrhenius equation A is a constant related to the number, orientation and frequency of collisions occurring between the particles in the reaction, called frequency factor A is a constant related to the number, orientation and frequency of collisions occurring between the particles in the reaction, called frequency factor k is rate constant k is rate constant R is universal gas constant R is universal gas constant T is absolute temperature T is absolute temperature E a is activation energy (J/mol) E a is activation energy (J/mol) e is e is As activation energy increases, reaction rate (and k) decreases- a smaller fraction of the molecules will have enough energy to react

As seen on the AP Chem equations sheet A graph of lnk vs 1/T will be a line with slope –E a /R and a y-intercept of lnA. Y = mx + b

Sample problem Temp (°C)K (s -1 ) x x x x From this data, calculate the activation energy. Temp (K) 1/T (K -1 )ln k x x x x

Slope = -6 – (-10.4) = -1.9 x 10 4 = -1.9 x 10 4 Slope = - E a R E a = -(slope)(R) = -(-1.9x10 4 K)(8.31J/mol∙K)(1kJ/1000J) = -(-1.9x10 4 K)(8.31J/mol∙K)(1kJ/1000J) =1.6 x 10 2 kJ/mol =1.6 x 10 2 kJ/mol