Chem 1310: Introduction to physical chemistry Part 2: Chemical Kinetics Exercises

From concentrations to rates and rate laws t[H 2 O 2 ] (s)(mol/L) H 2 O 2 → 2 H 2 O + O 2

From concentrations to rates and rate laws t[H 2 O 2 ]av rate (s)(mol/L)(mol L -1 s -1 )

From concentrations to rates and rate laws t[H 2 O 2 ]av [H 2 O 2 ]av rate (s)(mol/L) (mol L -1 s -1 ) k = slope

From concentrations to rates and rate laws t[H 2 O 2 ]av [H 2 O 2 ]av rate (s)(mol/L) (mol L -1 s -1 ) Numerical order check: conc(600)/conc(2200) = 3.25 rate(600)/rate(2200) = 2.80 conc(200)/conc(1400) = 2.36 rate(200)/rate(1400) = 2.40 Reasonable (not great) for first-order.

From concentrations to rates and rate laws t[H 2 O 2 ]av [H 2 O 2 ]av rate"k" (s)(mol/L) (mol L -1 s -1 )(s -1 ) Average rate constant: s -1

From concentrations to rates and rate laws t[H 2 O 2 ]ln [H 2 O 2 ] (s)(mol/L) k = - slope

From concentrations to rates and rate laws t[H 2 O 2 ]ln [H 2 O 2 ] (s)(mol/L) k from data at two data points: k = (ln [H 2 O 2 ] t1 - ln [H 2 O 2 ] t2 )/(t 2 -t 1 ) so from data at 400 and 2000 s: k = s -1

Using rate laws What is the rate when [H 2 O 2 ] = 2.00 M? rate = k [H 2 O 2 ] = mol L -1 s -1 What are [H 2 O 2 ] and rate a t = 500 s? [H 2 O 2 ] = [H 2 O 2 ] 0 exp(- k t) = 1.59 M rate = k [H 2 O 2 ] = mol L -1 s -1 What is the half-life? t ½ = ln(2)/k = 925 s

From rate constants to activation energies tree cricket chirping frequency interpreted as a rate constant chirping freqT (chirps/min)°C

From rate constants to activation energies chirping freqln freq1/TTT (chirps/min)K -1 K°C E a = -R * slope = 55 kJ/mol

From rate constants to activation energies chirping freqln freq1/TTT (chirps/min)K -1 K°C Calculate A, E a from rate constants at 16 and 26°C E a = R (ln k 1 - ln k 2 )/(1/T 2 - 1/T 1 ) = 58 kJ/mol A = k 1 exp(E a /R T 1 ) = 2.9*10 12

Using activation energies What will be the frequency at 0°C? k = A exp(-E a /RT) = 24 (chirps/min) At what temperature will the frequency be 250 (chirps/min)? T = (E a /R)/(ln A - ln k) = 302K = 29°C

Rate laws and mechanisms An "elementary step" or "elementary reaction" is one step that happens "as a single step", without any intermediates. A unimolecular elementary step can happen spontaneously, provided enough energy is available; it does not require anything else. A bimolecular elementary step can happen spontaneously on collision of the two reaction partners, if enough energy is available; it does not require anything else.

Rate laws and mechanisms Rate laws for elementary steps can be written down immediately. Unimolecular, A → X: rate = k [A] Bimolecular, A + B → X: rate = k [A][B] If a reaction consists of several elementary steps, its overall rate law cannot be written down so easily. If there are several steps, the slow or rate-limiting elementary step determines the reaction rate.

Rate laws and mechanisms ex: the observed rate law for 4 HBr + O 2 → 2 Br H 2 O is rate = k [HBr] [O 2 ] Is this compatible with the proposed mechanism HBr + O 2 → HOOBrk 1 HOOBr + HBr → 2 HOBrk 2 HOBr + HBr → H 2 O + Br 2 k 3 If so, what must be the rate-limiting step?

Rate laws and mechanisms Assume k 1 is rate-limiting: rate = k 1 [HBr][O 2 ]OK! Assume k 2 is rate-limiting (then k 1 is faster and we also have k -1 ) steady-state

Rate laws and mechanisms Assume k 3 is rate-limiting: (then both k 1 and k 2 are faster and we also have k -1, k -2 )