Chemical Kinetics Flashcards

Question

describes the multi step mechanism

Answer 1

1. The elementary reactions in a mechanism add up to give the overall chemical equation. 2. Intermediates do not appear in overall chemical equation or overall rate law. Intermediates are species produced in one elementary reaction and consumed in another 3. The reaction mechanism explains the experimentally determined rate law. 4. Reaction mechanism includes only elementary steps. Usually only unimolecular or bimolecular elementary reactions

Answer 2

When mechanisms have a slow initial step, overall rate law depends on the first step Rate law = k[A]

Answer 3

If the first elementary reaction is not the slow step, a later reaction is the bottleneck The fast initial reaction produces the intermediate needed for the slow, rate determining step Fast initial reaction often establishes an equilibrium (a powerful simplifying assumption). Generally: the reverse of the first reaction is faster than the slow second forward reaction, leading to equilibrium condition. Rate1(forward rxn) = rate-1(reverse rxn)

Answer 4

The rate is determined by the slow step. Slow step rate law refers to [intermediate] Assume fast equilibrium in step 1: Rate 1 = Rate -1 Obtain the expression for intermediate concentration Use this expression in rate law for slow step Ex: 2A ←→ B: Rate1=K1[A]^2 Rate-1=K-1[B] B + C → D + G: Rate2=K2[B][C] D + C → E + G: Rate3=K3[D][C] Overall equation: 2A + 2C → E + 2G Experimental rate k=[A]^2[C] The slow step is step 2, and the rate is dictated by Rate2=K2[B][C] We can then manipulate rate1 and rate-1 to isolate B K1[A]^2=K-1[B] [B]=(k1/k-1)[A]^2 Then plug it in to the rate 2 equation to get Rate2=k2(k1/k-1)[A]^2[C] k2(k1/k-1) = K (the overall rate constant) Elementary reactions add up to give overall chemical reaction and only involve bimolecular steps (feasible)

Answer 5

A more general approximation for cases when no single step is slower than the others. The overall reaction rate depends on a combination of steps The assumption here is that the concentration of the reactive intermediates remain nearly constant throughout the reaction. Ex: 2A ←→ B: Rate1=K1[A]^2 Rate-1=K-1[B] B + C → 2D Rate2=K2[B][C] Overall equation: 2A + C → 2D Experimental rate k=[A]^2[C] 1. The reactive intermediate is B. If it’s concentration doesn’t change, then ∆[B]/∆t = 0 ∆[B]/∆t=k1[A]^2-k-1[B]-k2[B][C]=0 2. We have an equation that allows the determination of the constant concentration of B. [B](k-1+k2[C])=k1[A]^2 [B]=k1[A]^2/(k-1+k2[C]) 3. Write the rate equation for the overall reaction, based on the product concentration, and observe that the product is obtained in step 2. Rate=∆[D]/∆t=k2[B][C] 4. Substitute the expression for the intermediate based on the steady-state approximation: rate=k1k2[A]^2[C]/(k-1+k2[C]) Compare with rate derived assuming step 2 was slow: Rate=k2(k1/k-1)*[A]^2[C] If step 2 is slow then k-1 >> k2[C] so: k-1 + k2[C] = k-1 So the two rate laws agree! Note: the steady state approximation is a much more general assumption.

Answer 6

Only a fraction of the molecules will have enough energy to react Fraction = e^-Ea/RT There are more molecules with KE>Ea at higher temperature ~ Temperature For a given value of T, molecules have a specific distribution of energies.

Answer 7

K = Ae^-Ea/RT K is rate constant, Ea is activation energy in J, R is gas constant (J/K), T temperature on Kelvin A is frequency factor: accounts for the fraction of sufficiently energetic collisions that lead to product formation considering how many have the correct orientation

Answer 8

Taking the ln of both sides of the equation ln(k)=lnA-Ea/RT For k1 and k2 at T1 and T2: ln(k1)=lnA-Ea/RT1 ln(k2)=lnA-Ea/RT2 Since lnA is constant ln(k1)+Ea/RT1=ln(k2)+Ea/RT2 We assume that the activation energy (Ea) and the frequency factor (A) do not depend on the temperature. But the rate constant (k) does

Answer 9

Measuring rates at different temperatures allows the determination of k at different T ln(k1/k2)=Ea/R(1/T2-1/T1) two methods 1. measure k1 and k2 at two different temperatures, T1 and T2 and use above equation to solve for activation energy 2. Or is we have a data set with k verses T, we can plot lnK as a function of 1/T lnK=-(Ea/R)(1/T)+ln(A) Method two is better for determining Ea since it has more data and will be more accurate.

Answer 10

A catalyst increases the rate of the reaction Provides an alternative pathway (alternative reaction mechanism) with lower Ea Changes the value of k: Catalysts usually do not undergo a permanent chemical change.

Answer 11

Homogeneous Catalysis Catalyst is in the same phase as the reaction molecules Catalyst is consumed in 1st reaction and produced in second reaction

Answer 12

The Catalyst is in a different phase from reacting molecules, usually a solid. e.g. Hydrogenation of unsaturated hydrocarbons such as ethylene or vegetable oils. Where a solid Ni surface acts as a catalyst. Automobile Catalytic Converter Converts incomplete combustion products like Hydrocarbons, CO, and NO buy combining with O2 to produce CO2, H2O, N2, and O2 Catalysts are Pt, Pd, Rh and transition metal oxides.

Answer 13

Enzymes catalyze biological reactions so that they can occur at a reasonable rate under physiological conditions. Enzymes have a region that specifically binds the reactant molecule. This is called the active site. The substrate (reactant) binds to the active site to form a complex. Once bound, substrate is activated and react to form products Inhibitors bind to active sites and block entry of substrate Some enzymes are very efficient: turnover of 10^3 - 10^7 per second

Answer 14

Ozone (O3) absorbs photons (240-310 nm) protects us from damaging UV radiation UV-A < UV-B < UV-C O3 formed between 30 km and 90 km Reaction: 3O2 ← → 2 O3

Answer 15

At higher altitude - high energy photons dissociate O2: O2 + hv → O + O The right balance between the amounts of O2 and O is found in the stratosphere. Stratosphere - sufficient O2 and O 1. O2 + O → O3 (exothermic) 2. O3 + M → O3 + M* + heat 3. O3 + hv → O2 + O 4. O + O + M → O2 + M* + heat M is an inert gas, the * means the molecule is excited Ozone is formed and decomposed naturally in the atmosphere. 3 x 10^8 tons per day Photons with wave length 240-310 nm (UV-B) are absorbed by the ozone

Answer 16

1. traces of inactive molecules containing Cl are present in the atmosphere over Antarctica ClONO2 and HCl 2. Cold dark winter forms polar stratospheric clouds (PSCs), which are made of ice crystals 3. Ice crystals (heterogeneously) catalyze the formation of Cl2, which remains trapped in clouds while cold and dark 4. In the southern spring (sept to Dec) sunlight dissociated the Cl2: Cl2 + hv → Cl* + Cl* Cl* results in catalytic destruction of ozone until PSCs melt in summer Dobson unit: one DU is 0.4462 millimoles of ozone/m^2

Answer 17

Ozone layer depletion Monitoring of ozone from 1978 onwards revealed that O3 in the stratosphere was depleting

Answer 18

Rowland (noble prize 1995) showed that naturally occurring nitrogen oxides depleted ozone. N2O + O → 2NO NO + O3 → NO2 + O2 N2O is also a by-product of fertilization, fossil fuel combustion and biomass and biofuel burning

Answer 19

Molina and cruzen (also noble prize in 1995) showed that chlorine from chlorofluorocarbons (CFCs) that react the stratosphere react to produce chlorine atoms. CF2Cl2 + hv → CF2Cl + Cl* (reacts with high energy UV < 220 nm CFCs (trade name: freon) persist in stratosphere for 120 years. Ozone decomposition is catalyzed by Cl 2 Cl* + 2 O3 → 2 ClO + 2O2 2ClO + hv → 2 Cl + O2 Overall: 2 O3 → 3 O2 Second order rate law depends on [Cl*} and [O3] k = 7.2 x 10^9 /MS at 298K CFCs in spray cans were banned in north America in 1978, in world by Montreal protocol in 1987.