137

DEPENDENCE OF RATE ON TEMPERATURE Experimentally it is found that for most reactions, the reaction rate ____________ as the temperature is raised.

By trying various combinations of plots, it was soon realised that a plot of ln k vs 1/T (or log k versus 1/T) gave a linear plot with a ______________ slope. What does it mean if a plot of ln(something) plotted against 1/T gives a negative slope? What sort of algebraic equation does this imply?

Arrhenius supplied the answer in 1899:

k = A exp{– Ea/RT} where k = rate constant, A = constant and Ea = Activation Energy

If you take natural logarithms of this equation: ln k = ln A – Ea/RT

i.e., a plot of ln k vs 1/T gives a linear plot with intercept ln A, with negative slope = – Ea/R,.

(where have you seen this sort of equation before?)

The Arrhenius equation allows us to do is to calculate the _____________________ for a reaction if we know the rate constants for the reaction at two temperatures.

Thus:

(1) at T2: ln k2 = ln A – (Ea/R)(1/T2)

(2) at T1: ln k1 = ln A – (Ea/R)(1/T1)

Subtract equation (1) from (2): ln (k2/k1) = – (Ea/R)(1/T2 – 1/T1)

By rearranging the equation: ln (k2/k1) = – (Ea/R)(T1 – T2)/(T1 × T2)

OR

ln (k2/k1) = (Ea/R)(ΔT)/(T1 × T2) {Recall ln(K2/K1) = (ΔH/R)(ΔT)/(T1 × T2) – Van’t Hoff Eq’n}

The _____________________________, Ea is the energy that the reactants need to

make the reaction "go". The significance of the temperature, T, is that an increase in

temperature increases the _________________________ and hence kinetic energies

of molecules. The larger the value of Ea, the more ____________________ is the

rate constant k to temperature changes.

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Reference J. Olmsted & G.M. Williams, Chemistry, 3rd ed., John Wiley & Sons, Inc., New York, 2002, p. 682.

We can create plots to show the relationship of Ea and ΔHRxn – the

________________________. These plots display the energy difference between

reactants and products (ΔHRxn ) and the energy changes as ________________ go to

products.

Reference: L. Jones & P. Aikens, Chemistry - Molecules, Matter and

Change, 4th ed., W.H. Freeman and Company, New York, 2000, p. 592

QUESTION: In the diagram provided, the quantity x is:

A) ΔE B) Ea(forward) C) Ea(reverse) D) The collision energy E) The steric factor p

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Collision Theory In collision theory, the potential energy __________________ as reactants approach each other, reaching a maximum as the molecules distort into what is known as a _________________________ or “activated complex”. The energy then ___________________ as the atoms rearrange into the bonding pattern of the product(s). The reaction profile indicates this increase in energy due to the transition state, located at the peak of the profile. The energy difference between the top of the curve and the reactants represents the Ea for the reaction. Only molecules with sufficient energy (E > Ea) can cross the energy barrier and react to form products.

Catalysts

Catalysts _________________ the rate of reaction without being consumed in the reaction. The catalyst provides an ________________________________________ which requires less energy (i.e., Ea (with catalyst) < Ea (no catalyst)). Not all reactions can be accelerated with a catalyst.

Question 1

A reaction has an activation energy of 48 kJ mol–1 at 25°C. A catalyst increases the rate of reaction by a factor of 1000. What is activation energy in the presence of the catalyst, assuming the Arrhenius A–factor is unchanged?

Rate(catalyst)/Rate(no catalyst) = kcat[A]m[B]n /kno cat [A]m[B]n = Aexp{–Ea(cat)/RT}/Aexp{–Ea(no cat)/RT}

∴ kcat/kno cat = exp{–Ea(cat)/RT}/exp{–Ea(no cat)/RT} = 1000/1 ln[exp{–Ea(cat)/RT}/exp{–Ea(no cat)RT] = ln(1000)

–Ea(cat)/RT + Ea(no cat)/RT = ln1000

Ea(cat) =

Ea(cat) =

Ea(cat) =

Question 2 The reaction A → B is 1st order in A. At 25°C, 10.1% reacts in 20.0 minutes. If Ea = 84.5 kJ mol–1, what percentage of A will react during the same period at 50°C?

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If 10.1% of A reacts, then what % remains?

Using the integrated rate law for first order reactions: k = There are two ways of solving this problem: Method I: First use k = A exp{– Ea/RT} to determine A using k1, Ea and T = 298K. Then solve for the new rate constant (k2) at the new temperature (50°C). Method II: Use: ln(k2/k1) = (Ea/R)(T2 – T1)/(T1 × T2) where k1 = 5.32 × 10–3 min–1 at 298K and k2 = ? min–1 at 323K (50°C). Now that the new rate constant is known, the percentage of A that has reacted at 50° C within the same amount of time can be determined. k = at 323K Therefore, [A]/[A]0 = exp{–kt} = This indicates that ?% of A remains, so therefore, ?% has reacted. Question 3 At an altitude of 3000 m on a mountain, water boils at 90°C and it takes 300 minutes to cook a "three–minute–egg". Calculate the activation energy for the coagulation of egg–albumin = denaturing a protein.

[Hint: the rate constant is inversely proportional to cooking time.]

At 3000 m, water boils at 90°C (T1 = 363K) and t = 300 minutes OR k1 = 1/300 = 3.33× 10–3 min–1

At "sea level", water boils at 100°C (T2 = 373K) and t = 3 minutes OR k2 =

Use ln(k2/k1) = (Ea/R)(T2 – T1)/(T1 × T2)

Ea = 500 kJ/mol

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Collision Theory of Gas Phase Bimolecular Reactions Consider the gas phase bimolecular elementary rxn: A(g) + B(g) → C(g) + D(g)

Rate = - d[A]/dt = -d[B]/dt = k[A][B]

Assumption 1: A and B must ______________ to react.

The rate is _________to the frequency of collisions per litre (e.g., collisions L-1 s-1). The frequency of collisions can be calculated from the average velocity of A and B (from the Kinetic Theory of Gases) and their cross-sectional areas.

This frequency is proportional to [A] and [B], i.e., moles of collisions between A & B per litre per second = Z[A][B],

where Z = moles of collisions L-1 s-1 when [A] = [B] = 1 M. Assumption 2: Collisions must have ______________ than some minimum

Kinetic Energy (K.E.) to result in a reaction.

The rate is ____________ to the fraction of collisions having K.E. greater than or equal to some value E, where E depends on the reaction. For gases at STP (1 atm, 273 K) the molar volume is 22.4 L. Therefore, concentration is (1 mole /22.4 L) and there are typically around 108 moles of collisions L-1 s-1. If every collision resulted in a reaction, the initial rate would be ~108 mol L-1 s-1 and the reaction would be 99.9% complete in around 10-7 s. Obviously only a small fraction of the collisions can lead to product in most reactions. Z increases with temperature (Z is proportional to T0.5).

The fraction of collisions having K.E ≥ E can also be calculated as exp{-E/RT}. e.g., if E = 100 kJ/mol, then the fraction of collisions at 298 K is exp{-E/RT} = exp{–100,000 J/mol)/(8.314 J/mol K × 298 K) = 3.0 × 10-18

This fraction increases sharply with temperature and is the main reason that k increases with T.

Assumption 3: Only collisions having the proper orientation of the two reactants

can result in a reaction.

Rate is ________________ to the fraction (p) of the collisions having the proper relative orientation.

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Putting this all together: Rate = k[A][B] = p × Z[A][B] × e-E/RT where, p is the fraction of collisions having the proper orientation for reaction; Z[A][B] is the moles of collisions L-1 s-1 e-E/RT ≥ E

i.e., Rate/([A][B]) = k = pZe-E/RT

Now cast your mind back to the Arrhenius Equation:

k = A exp{-Ea/RT} and compare it to k = pZ exp{-E/RT} i.e., pZ = A and E = Ea Ea is the minimum K.E. for reaction

Transition State Theory OR Activated Complexes Theory

– focuses on the species formed at the top of the potential barrier, e.g., AB‡

Reactions will:

1. Break or weaken bonds in reactants then __________________ in products.

2. Reactants pass over a ____________________ on way to products.

Plot (Exothermic Rxn):

Consider the elementary reaction:

Cl•(g) + CH4(g) → Cl–H(g) + •CH3(g), a reaction for which Ea = 16 kJ mol–1 and ΔUº = ΔHº = +8 kJ mol–1 Reaction is endothermic since ΔHº is positive.

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Energy Profile:

Question A: The bimolecular gas-phase reaction, NO(g) + Cl2(g) → NOCl(g) + Cl(g) has ΔHº +83.7 kJ/mol and Ea +84.9 kJ mol-1. The value of the activation energy for the reverse rxn is:

A) –84.9 B) 1.2 C) 84.9 D) 168.6 E) none of these

Question B: According to the Collision Theory of bimolecular gas phase

reactions, the Activation Energy in the Arrhenius model is associated with A) The potential energy of interaction of two molecules. B) The frequency of collisions. C) The fraction of the collisions having the proper orientation. D) The pressure of the reactants. E) The fraction of the collisions having sufficient kinetic energy for rxn.

REACTION MECHANISMS Consider 2 reactions:

1. H2(g) + I2(g) → 2HI(g); Rate = k[H2][I2]

2. H2(g) + Br2(g) → 2HBr(g); Rate = k[H2][Br2]3/2/{[Br2] + k´[HBr]}

How do we account for complicated rate laws?

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Kinetics and Mechanisms To understand the rates of reactions, to explain the temperature dependence of rates and (ultimately) explain the mechanism of the reaction, we need to consider some theories of reaction rates.

Reaction mechanisms must involve a __________________________, each an “elementary” process producing the overall reaction. An overall reaction is made up of one or more elementary reactions. These are reactions that occur in a SINGLE step i.e., there are no intermediate reactions. Types of Elementary Reactions: 1. Unimolecular Reactions: _______ molecule reacts/decomposes to give products,

e.g., O3(g) → O2(g) + O(g) N2O4(g) → 2NO2(g)

2. Bimolecular Reactions: _________ molecules react to give products, e.g.,

NO(g) + O3(g) → NO2(g) + O2(g)

2NO2(g) → N2O4(g) 3. Termolecular Reactions: ________ molecules react to give products.

These are RARE - WHY??

Elementary Reactions Once you know that a reaction is an elementary reaction (& this must be found out by experiment), the rate law follows from the ______________________ of the rxn. Thus for an elementary reaction, but ONLY for an elementary reaction, the rate equation can be written down just by looking at the balanced chemical equation. Therefore, for Elementary Reactions the overall ____________ is equal to the MOLECULARITY of the elementary reaction. For example, in the case of the elementary rxn: NO(g) + O3(g) → NO2(g) + O2(g)

it is first order with respect to NO and O3, but is overall second order, i.e., Rate = k[NO][ O3].

We can think of a MECHANISM as being a sequence of _____________________ reactions that convert the reactants to products. The __________ of all the elementary rxns (mechanistic steps) will give the overall stoichiometric equation. Also, if you have a postulated mechanism, the overall rate law derived from it ______________ agree with that which is found by experiment or the postulated mechanism is wrong! [Recall that the Rate Law can NOT be predicted from the overall stoichiometric equation.]

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The term molecularity can only be applied to ___________________ reaction. It describes the number of chemical species participating in the elementary reaction. For elementary processes, the molecularity is the same as the overall ___________ ___________. The kinetic order of a reaction is what is determined experimentally from the rate equation. You can talk about the order of the reaction, or the order with respect to any reactant.

Example: The reaction of NO2 in the laboratory: CO(g) + NO2(g) → NO(g) + CO2(g) follows the following rate experimental law: rate = k[NO2]2

(i) Is this an elementary reaction? (ii) Does it involve a bimolecular reaction between NO2 and CO?

Answer: If we write a rate law based on the above chemical equation (assume it was an elementary reaction), we would have:

Rate =

Question A: For: A + B → C + D has Ea = 40 kJ mol-1 and ΔH = -12 kJ mol-1.

The activation energy (kJ mol-1) for the reverse reaction is: A) –40 B) 28 C) 52 D) 12 E) none of these

Question B: For the diagram below, which set of labels is correct? Reactant Transition State Intermediate Product

A) E B,D C A B) A B None C C) C D None E D) A C B,D E E) A B,D C E

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Reaction Mechanisms - the detailed path by which reactants are converted to products.

We need to know or predict or guess: • ______________ and type of elementary reactions

• reaction intermediates (energy ______________________)

• possible transition states (energy ____________________) • energy barriers involved

The overall reaction is obtained by the sequence of elementary reaction steps. The rate law CANNOT be predicted from the overall reaction (stoichiometry) equation. Very often it is found/implied/suggested/proposed that one of the elementary reaction steps is much slower than the others; in this case the Overall Rate = Rate of Slow Step, which brings us to the idea of a Rate Limiting Step.

Rate Limiting Reactions • The reaction mechanism is the dissection of a complicated reaction sequence

into its elementary steps.

• The overall rate of a complex reaction is the rate of the slowest of these elementary steps, which is called the ____________________________ or the rate-limiting step.

• In a chemical reaction, the reaction rate is determined by the events leading up to and including the rate-determining step, but _________ by the (fast) reactions which occur afterwards.

Example 1: The following mechanism is suggested for the reaction:

CO(g) + NO2(g) → NO(g) + CO2(g) having an exp’tal rate law: rate = k[NO2]2

Possible Mechanism:

1. 2NO2(g) → NO(g) + NO3(g) (slow)

2. NO3(g) + CO(g) → NO2(g) + CO2(g) (fast)

rate of slow step =

Thus the rate equation includes _______________ of NO2 (slow step of the rxn), but does not include CO (fast step). Since the rate of the reaction does not depend on the concentration of CO, the

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reaction is said to be ____________________ in CO. We can then conclude: "The proposed mechanism is ___________________ with the experimental data."

If one adds up all the steps in the mechanism, you will arrive at the overall stoichiometric equation and can identify any intermediates in the process:

2NO2(g) → NO(g) + NO3(g) + NO3(g) + CO(g) → NO2(g) + CO2(g) NO2(g) + CO(g) → NO(g) + CO2(g)

Therefore, NO2 and CO are the _________________ while NO and CO2 are the ________________________. A species that is formed and then consumed in a later step is called an ________________________. In this reaction the intermediate is _____________________.

General Energy Profile:

The overall reaction is exothermic, which is indicated by the fact that the products are lower than the reactants (you can determine this from calculating ΔHRxn). Slow steps are slower because there is a larger energy barrier involved in the reaction while fast steps have a lower Ea.

Example 2:

F2(g) + 2NO(g) → 2ONF(g); experimental rate law: rate = k[NO][F2]

Possible Mechanism: 1. NO(g) + F2(g) → ONF(g) + F(g) (slow) 2. F(g) + NO(g) → ONF(g) (fast)

If this mechanism were true, then it would give rise to the rate equation:

We can then conclude: "The proposed mechanism is ___________________ with the experimental data."

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Reactions with a Rapid Equilibrium Step For the reaction, 2NO(g) + O2(g) → 2NO2(g), we find experimentally that:

–d [O2]/dt = k[NO]2[O2], i.e., third order overall

Possible Mechanism: 2NO(g) N2O2(g) rapid equilibrium, K1

N2O2(g) + O2(g) → 2NO2(g), slow, k2

Pictorially:

Based on the slow step: -d[O2]/dt = Rate =

N2O2 is a reaction intermediate.

General Rule: The Rate Law can ONLY include measurable concentrations, e.g., Reactants but NOT Intermediates.

So, do some 'Chemical Algebra' to eliminate the intermediates from the rate eq’n.

Based on the equilibrium reaction, we can define:

K1 =

Using this relationship, we now can substitute for [N2O2] in the rate law: –d[O2]/dt =

From experimental data: –d[O2]/dt = kobs[NO]2[O2] We can then conclude: "The proposed mechanism is _______________________

with the experimental data." Note: Ea for the overall reaction is the energy required to go from reactants to the

top of the energy barrier for the slowest step.

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Alternate Mechanism: NO(g) + O2(g) OONO(g), rapid equilibrium, K1

NO(g) + OONO(g) → 2NO2(g), slow, k2

Rate =

Substituting for [OONO]:

K1 = Substitute for [OONO] in the rate law:

–d[O2]/dt = Rate =

This mechanism produces the same rate law as the previous. Therefore, we would need further evidence to distinguish between the two mechanisms (i.e., which intermediate is correct?).

For a good proposed mechanism, the rate limiting step must be consistent with the rate law. Also, the sum of all the elementary steps must give the correct stoichiometric equation. ALTERNATE SUBSTITION:

Instead of using the equilibrium constant, one can also determine a substitution expression based on the fact that the forward and reverse rates are equal when a reaction is at equilibrium.

For example, in the case of 2NO N2O2 (rapid equilibrium), Rateforward = Ratereverse

where k1 is the rate constant in the forward direction &

k-1 is the rate constant in the reverse direction.

Using this relationship to substitute for [N2O2] in the rate law:

Rate =

Rate = k[NO]2[O2] where k = k2 k1/ k-1

Consider the following: H2(g) + I2(g) → 2HI(g)

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Rateobserved = k[H2][I2] (a) What was taught in high school ~50 year ago:

H2(g) + I2(g) → 2HI(g) - a bimolecular elementary reaction

(b) Current thinking:

I2 2I (fast equilibrium, K1) I + H2 H2I (fast equilibrium, K2) H2I + I → 2HI (slow, k3)

Rate = k3[I][H2I] {both intermediates - need to substitute}

Based on rxn#2: K2 = [H2I] = K2[I][H2]

OR [H2I] = k2/k–2 [I][H2]

which gives: Rate =

Based on rxn#1: K1 = [I]2 = K1[I2]

which gives: Rate =

WHY is this a "better" mechanism?

Chain Reactions Chain reactions have highly reactive intermediates that react to produce another, which produces another, which … etc.

e.g., H2 and O2

CH4 and O2 can be mixed with no effect

H2 and Cl2

However, if a reaction is initiated, it goes very rapidly with a large negative ΔH; i.e., explosion!

For example, in the case of H2 and O2:

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Initiating step: creates a highly reactive species (e.g., free radical)

spark H2 → H• + H• rate limiting, high Ea

Branching step: increases # of reactive species - rapid acceleration of rxn.

H• + O2 → O• + •O• •O• + H2 → HO• + H•

Propagating step: maintains number of reactive species.

HO• + H2 → H2O + H•

Terminating step: reduces number of reactive species.

HO• + H• → H2O H• + H• → H2

Catalysis Previously we have seen that we can speed up a reaction by increasing T, or increasing the concentration of reactants. There are two other ways that this can be achieved:

1. lower the barrier, i.e., _________________________________

2. improve collision efficiency by improving the ______________________ of the molecule(s).

A Catalyst increases the ____________ of a chemical reaction without itself being used up in the process by

1) providing an ___________________ rxn path with a lower activation energy, 2) helping to ________________ bonds, 3) providing a correct steric orientation for the conversion of reactants to

products, etc.

There are two kinds of catalysts: Homogeneous and Heterogeneous

Homogeneous Catalysis: reactants and catalyst are in the _____________ phase.

e.g., the conversion of 2O3(g) → 3O2(g) is catalysed by freons (molecules like CF2Cl2 which are used in air-conditioners, etc.).

Heterogeneous Catalysis: reactants and catalyst are in ______________ phases.

e.g., catalytic hydrogenation of alkenes. Ozone Decomposition:

Uncatalysed mechanism:

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O3 O2 + O rapid equilibrium, K1

O3 + O → 2O2 slow, k2

Overall reaction: 2O3(g) → 3O2(g)

Rate = ½ d[O2]/dt =

K1 = [O2][O]/[O3], where [O] =

Rate =

Rate = kobs.[O3]2/[O2] and this is what is observed by experiment.

Homogenous catalysed mechanism for the slow step in the previous mechanism:

CF2Cl2 → Cl• + •CF2Cl Cl• + O3 → •ClO + O2

•ClO + O → •Cl + O2

Cl• + •CF2Cl → CF2Cl2

Net effect is O3 + O → 2O2, but now no longer "slow".

The catalyst CF2Cl2 emerges unchanged. Compare Reaction Profiles:

Reference J. Olmsted & G.M. Williams, Chemistry, 3rd ed., John Wiley & Sons, Inc., New York, 2002, p. 689.

Note that with the catalyst the overall reaction has a lower activation energy than the uncatalysed reaction, and therefore a larger fraction of molecules will have sufficient energy to exceed Ea and hence participate in the reaction.

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To distinguish between a catalyst and an intermediate: A catalyst is present at the beginning and reacts to form something else. Later the catalyst is reformed unchanged, so it is cancelled out of the overall reaction. An intermediate must be formed, and then is consumed in a following step. Heterogeneous Catalysis:

These types of catalysts are very important to the chemical industry, e.g., Fe(s)

Haber process: 3H2(g) + N2(g) → 2NH3(g) (400ºC and 200 atm)

2H2(g) + CO(g) → CH3OH(g) (in the presence of Al and Cr oxides or Pt)

Catalytic converter in the exhaust system of automobiles provides a catalytic

surface so that the following reactions occur at higher efficiency: Pt(s)

2CO(g) + O2(g) → 2CO2(g)

Pt(s) 2NO(g) → N2(g) + O2(g)

Mechanisms involving catalysts are often complex and are not fully understood.

There is a great deal of interest in metal clusters, organometallic chemistry and surface/surface reactivity for industrial applications.

You have already seen an example of heterogeneous catalysis from CHEM*1040. During hydrogenation of alkenes, hydrogen is adsorbed to the surface of a metal catalyst (Ni, Pt or Pd). The surface of the catalyst allows for the weakening of the H–H bond, and thus the addition of H´s to the alkene.

e.g., H2C=CH2(g) + H2(g) → H3C-CH3(g) (in the presence of a metal catalyst)

Possible mechanism: H2C=CH2(surface) + H(surface) → H2C—CH3(surface) H2C—CH3(surface) + H(surface) → H3C=CH3(g)

Catalysts are also involved in biological processes, such as enzyme catalysis to allow biochemical process to take place that would otherwise be too slow. Enzymes are ______________________ and have active sites where reactions take place allowing for a better orientation or weaken bonds in reactants, which lowers the Ea.

Catalysts can be poisoned. In this case, another species may react with the catalyst to inhibit the intended reaction.

e.g., Lead poisons a catalytic converter in a car Arsenic poisons some enzymes in biological systems.

Problem 1: One pathway for destruction of ozone in the upper atmosphere is:

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O3 → O2 + O (rapid equilibrium) O3 + NO → NO2 + O2 (slow) NO2 + O → NO + O2 (fast) 2O3 → 3O2 (overall reaction)

The catalyst in this reaction is: (a) O (b) NO2 (c) O3 (d) O2 (e) NO Problem 2:

The rate law for the reaction between chlorine and chloroform, CHCl3 (i.e., CHCl3 + Cl2 → HCl + CCl4) was found to be, rate = kobs[CHCl3][Cl2]½ Show that the following mechanism is consistent with this rate law:

Cl2 → 2 Cl (fast) CHCl3 + Cl → CCl3 + HCl (slow)

CCl3 + Cl → CCl4 (fast) Problem 3: Plot the energy profiles {and identify Ea(forward)} for the following mechanisms:

i. a slow step followed by a fast step (endothermic). ii. a fast step followed by a slow step (exothermic rxn).

Answer 1:

A catalyst is used up and then reformed unchanged. Therefore, NO is the catalyst.

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O – intermediate NO2 – intermediate O3 – reactant O2 – product

Answer 2:

Rate is determined by the slow step: rate = k2[CHCl3][Cl] Cl is an intermediate and must be replaced. Use equation one, an equilibrium reaction, to find a relationship between this intermediate and the reactant(s).

K1 = [Cl]2/ [Cl2]; ∴ [Cl] = {K1[Cl2]}½

Substituting into the rate expression: Rate = k2[CHCl3] {K1[Cl2]}½

Rate = k2 K1½[CHCl3] [Cl2]½

Rate = k [CHCl3][Cl2]½

This rate expression is the same as the observe rate law, so the proposed mechanism is consistent with experimental data. Answer 3: i. a slow step followed by a fast step (endothermic)

ii. a fast step followed by a slow step (exothermic rxn).

Note, Ea for the reaction is the energy required to go from reactants to the top of the energy barrier for the slow step.

156

KINETICS REVIEW QUESTIONS I. In the diagram provided, the quantity x is:

A) ΔE B) Ea(forward) C) Ea(reverse) D) The collision energy E) The steric factor p

II. For the diagram provided, identify the one correct answer:

A) The forward reaction is exothermic B) The forward reaction is endothermic C) Ea(forward) < Ea(reverse) D) Ea(forward) = 0 E) Ea(reverse) = 0

III. For the diagram below, which one of the following is true?

Α) ΔE(forward) = C – B

Β) ΔE(forward) = B – A C) Ea(forward) > Ea(reverse) D) Ea(reverse) = C – A E) The potential energy of the transition state

is greater for the forward reaction than for the reverse reaction.

IV. For the diagram below, which set of labels is correct for the forward reaction?

Reactant Transition State Intermediate Product

A) E B,D C A

B) A B None C

C) C D None E

D) A C B,D E

E) A B,D C E

157

V. For a simple two-step mechanism: reactants → intermediates → products,

the potential energy diagram shown below is obtained. Which one of the following statements applies? A) The reactants quickly form intermediates that slowly form products. B) The presence of intermediates is difficult to determine kinetically, although the two-step

reaction is not ultrafast. C) The intermediates are the final products D) The intermediates accumulate in measurable quantities, and the reaction is easily

resolved into two steps. E) None of the above.

VI. To establish first-order kinetics unequivocally by graphical methods, which of the following

graphs {[A] is concentration, t is time} does one draw? A) [A] vs t B) 1/[A] vs t C) ln[A] vs t D) ln[A] vs 1/t E) none of these.

VII. To establish second-order kinetics unequivocally by graphical methods, which of the

following graphs {[A] is concentration, t is time} does one draw? A) [A] vs t B) 1/[A] vs t C) ln[A] vs t D) ln[A] vs 1/t E) none of these.

VIII. For a second-order reaction 2A → B, when the concentration of [A] is doubled the value

of the rate changes by a factor of: A) 0.25 B) 0.5 C) 2 D) 4 E) none of these.

158

IX. The reaction 2NO(g) + O2(g) → 2NO2(g) has the rate equation: Rate = k[NO]2[O2] When an experiment is started with equal concentrations of both reactants, (i) the rate after 25.0% of the NO has been consumed will be what fraction of the initial rate:

A) 0 B) 0.125 C) 0.188 D) 0.413 E) 0.500 F) none of these. (ii) the rate after 25.0% of the O2 has been consumed will be what fraction of the initial rate?

A) 0.0625 B) 0.125 C) 0.188 D) 0.413 E) none of these

X. The reaction: 2NO(g) + O2(g) → 2NO2(g) is 2nd order in NO and 1st order in O2. Starting with equal concentrations of NO and O2, the rate after 50% of the NO has been consumed will be what fraction of the initial rate?

A) 0.125 B) 2. 0 C) 0.188 D) 0.250 E) 0.413 F) None of these

XI. To establish first-order kinetics unequivocally by graphical methods, one can draw a plot of ln[A] vs. t (where [A] is the reactant concentration at time t). The slope is then equal to

A) +k B) –k C) ln(2/k) D) 1/k E) none of these.

XII. To establish simple second-order kinetics unequivocally by graphical methods, one would plot a graph of 1/[A] vs. t (where [A] is the reactant concentration at time t). The slope of this graph is then equal to

A) +k B) –k C) 1/( –k) D) 1/k E) none of these.

XIII. The reaction A → B is simple second-order in A with a rate constant of 2.0×10-2 L mol-1 s-1 at 25ºC. The time (s) required for 75% reaction of A under these conditions is

A) 0.33 B) 3.0 C) 29 D) 14 E) not enough data given

XIV. To determine the Activation Energy, Ea, one measures rate constants (k) at various temperatures (T K) and

i) plots which graph? A) k vs T B) 1/k vs T C) lnk vs T D) lnk vs 1/T E) none of these

ii) The slope of the graph is equal to: A) –Ea B) Ea C) –Ea/R D) Ea/R E) none of these

XV. The bimolecular gas-phase reaction, NO(g) + Cl2(g) → NOCl(g) + Cl(g) has ΔHº +83.7

kJ/mol and Ea +84.9 kJ mol-1. The value of the activation energy for the reverse rxn is: A) –84.9 B) 1.2 C) 84.9 D) 168.6 E) none of these

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XVI. The rxn A → B is 1st order in A and 1st order overall, with a rate constant = 4.00×10–4 s–1 at 323 K. The percentage of A which will react in a 1.00 hour period is:

A) 0.42 B) 14.4 C) 23.7 D) 69.4 E) 76.3

XVII. The reaction: A + B → C + D has Ea = 40 kJ mol-1 and ΔH = -12 kJ mol-1. The activation energy (kJ mol-1) for the reverse reaction is:

A) –40 B) 28 C) 52 D) 12 E) none of these XVIII. According to the Collision Theory of bimolecular gas phase reactions, the Activation

Energy in the Arrhenius model is associated with A) The potential energy of interaction of two molecules. B) The frequency of collisions. C) The fraction of the collisions having the proper orientation. D) The pressure of the reactants. E) The fraction of the collisions having sufficient kinetic energy for reaction. F) None of these.

ANSWERS

Question Answer(s) Question Answer(s) I. B) XI. B) II. B) XII. A) III. B) XIII. E) IV. E) XIV. D); C) V. A) XV. B) VI. C) XVI. E) VII. B) XVII. C) VIII. D) XVIII. E) IX. E); C) X. C)