Chemical Kinetics Part - 2

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Section - 3

REACTION MECHANISMS

Collision Theory of Reaction Rates The fundamental notion of the collision theory of reaction rates is that for reaction to occur, molecules, atoms, or ions must first collide. Increased concentrations of reacting species result in greater number of collisions per unit time. However, not all collisions result in reaction; i.e., not all collisions are effective collisions. For a collision to be effective, the reacting species must (1) possess at least a certain minimum energy necessary to rearrange outer electrons in breaking bonds and forming new ones and (2) have the proper orientations toward each other at the time of collision. Collisions must occur in order for a chemical reaction to proceed, but they do not guarantee that a reaction will occur.

A collision between atoms, molecules, or ions is not like one between two hard billiard balls. Whether or not chemical species “collide” depends on the distance at which they can interact with one another. For instance, the gas phase ion-molecule reaction CH 4 + + CH 4 → CH5+ + CH3 can occur with a fairly long-range contact. This is because the interactions between ions and induced dipoles are effective over a relatively long distance. By contrast, the reacting species in the gas reaction CH3 + CH3 → C2 H6 are both neutral. They interact appreciably only through very short-range forces between induced dipoles, so they must approach one another very closely before we could say that they “collide.” Recall that the average kinetic energy of a collection of molecules proportional to the absolute temperature. At higher temperatures, more of the molecules possess sufficient energy to react. If colliding molecules have improper orientations, they do not react even though they may possess sufficient energy. Figure 8 depicts collisions between molecules of NO and N2O. We assume that each possesses sufficient energy to react according to

NO + N 2O → NO2 + N 2 Effective orientation of collision

N N

N

N

N

N

O (a)

O N O

N O

O N O Product

Ineffective orientation of collision

(b)

O N N N O

O N N N O

O N N

N O No reaction

Chemistry / Chemical Kinetics

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Ineffective orientation of collision

O N N

(c)

O N N

O N N O N

O N

O N No reaction

Figure 8: Some possible collisions between N2O and NO molecules in the gas phase. (a) A collision that could be effective in producing the reaction. (b, c) Collision would be ineffective. The molecules must have the proper orientations relative to each other and have sufficient energy to react.

Transition State Theory Chemical reactions involve the making and breaking of chemical bonds. The energy associated with a chemical bond is a form of potential energy. Reactions are accompanied by changes in potential energy. Consider the following hypothetical, one-step exothermic reaction at a certain temperature.

A + B2 → AB + B + heat Figure 9 shows a plot of potential energy versus reaction coordinate. In Figure 9(a) the ground state energy of the reactants, A and B2, is higher than the ground state energy of the products, AB and B. The energy released in the reaction is the difference between these two energies, ∆E. It is related to the change in enthalpy or heat content that you have studies in thermodynamics. Quite often, for reaction to occur, some covalent bonds must be broken so that others can be formed. This can occur only if the molecules collide with enough kinetic energy to overcome the potential energy stabilisation of the bonds. According to the transition state theory, the reactants pass through a short-lived, high-energy intermediate state, called a transition state, before the product are formed. A + B — B  → A ! B! B  → A—B + B transition state AB2

reactants A + B2

A

B

B

Ea forward reaction Reactants A + B2

Endothermic reaction A B B

Ea reverse reaction ∆Ereaction

Reaction coordinate (a)

Potential energy

Potential energy

Exothermic reaction

products AB + B

Ea forward reaction

Ea reverse reaction Products AB + B

Reactants A + B2

∆Ereaction

Products AB + B

Reaction coordinate (b)

The "reaction coordinate" represents Figure 9: A potential energy diagram. (a) A reaction that releases the progress along the pathway energy (exothermic). An example of an exothermic gas-phase leading from reactants to products. reacion is This coordinate is sometimes labeled H + I2 HI + I "progress of reaction." (b) A reaction that absorbs energy (endothermic). An example of an endothermic gas-phase reaction is I + H2 HI + I

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The activation energy, Ea, is the additional energy that must be absorbed by the reactants in their ground states to allow them to reach the transition state. If A and B2 molecules do not possess the necessary amount of energy, Ea, above their ground states when they collide, reaction cannot occur. If they do possess sufficient energy to “climb the energy barrier” to reach the transition state, the reaction can proceed. When the atoms go from the transition state arrangement to the product molecules, energy is released. If the reaction results in a net release of energy (Figure 9(a)), more energy than the activation energy is returned to the surroundings and the reaction is exothermic. If the reaction results in a net absorption of energy (Figure 9(b)), an amount less than Ea is given off when the transition state is converted to products and the reaction is endothermic. Thus, the activation energy must be supplied to the system from its environment, but some of that energy is subsequently released to the surroundings. The net release of energy is ∆E . When the reverse reaction occurs, an increase in energy equal to the reverse activation energy, Ea reverse, is required to convert the AB product molecules to the transition state. As you can see from the potential energy diagrams in Figure 9, E a forward – E a reverse = ∆E reaction

As we shall see later in this unit, increasing the temperature changes the rate by altering the fraction of molecules that can get over a given energy barrier. Introducing a catalyst changes the rate by lowering the barrier. As a specific example that illustrates the ideas of collision theory and transition state theory, consider the reaction of iodine ions with methyl chloride. I – + CH 3Cl  → CH 3I + Cl –

Many studies have established that this reaction proceeds as shown in Figure 10(a). The I – ion must approach the CH3Cl from the “back side” of the C—Cl bond, through the any other angle would not lead to reaction. But a collision with the appropriate orientation could allow the new I—C bond to form at the same time that the C—Cl bond is breaking. This collection of atoms, which we represent as

H

H I

C

We can view this transition state as though carbon is only partially bonded to I and only partially bonded to Cl.

Cl

H is what we call the transition state of this reaction (Figure 10(b)). From this state, either of two things could happen: (1) the I—C bond could finish forming and the C—Cl bond could finish breaking with Cl – leaving, leading to products, or (2) the I—C bond could fall apart with I– leaving, and the C—Cl bond could re-form, leading back to reactants. – H

H

H

H

H

H

(a)

I

C

–

H Before collision Chemistry / Chemical Kinetics

Cl

I

C H Transition state

Cl

I

Cl

C H After reaction

–

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–

–

I

I

H (b)

H

H

H C

C

Cl

H

Cl

H

Figure 10: (a) A collision that could lead to reaction of I – + CH3Cl to give CH3I + Cl – . The I – must approach along the “back side” of the C—Cl bond. (b) Two collisions that are not in the “correct” orientation.

REACTION MECHANISMS AND THE RATE-LAW EXPRESSION The step-by-step pathway by which a reaction occurs is called its mechanism. Some reactions take place in a single step, but most reactions occur in a series of elementary steps.

The reaction orders for any single elementary step are equal to the coefficients for that step. In many mechanisms, however, one step is much slower than the others.

A reaction can never occur faster than its slowest step. This slow step is called the rate-determining step. The speed at which the slow step occurs limits the rate at which the overall reaction occurs. The balanced equation for the overall reaction is equal to the sum of all the individual steps, including any steps that might follow the rate-determining step. We emphasise again that the rate-law exponents do not necessarily match the coefficients of the overall balanced equation. For the general overall reaction aA + bB  → cC + d D

the experimentally determined rate-law expression has the form rate = k[A]x[B]y The values of x and y are related to the coefficients of the reactants in the slowest step, influenced in some cases by earlier steps. Using a combination of experimental data and chemical intuition, we can postulate a mechanism by which a reaction could occur. We can never prove absolutely that a proposed mechanism is correct. All we can do is postulate a mechanism that is consistent with experimental data. We might later detect reaction-intermediate species that are not part of the proposed mechanism. We must then modify the mechanism or discard it and propose a new one. Chemistry / Chemical Kinetics

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As an example, the reaction of nitrogen dioxide and carbon monoxide has been found to be second order with respect to NO2 ad zero order with respect to CO below 225° C. NO 2 (g) + CO(g)  → NO(g) + CO 2 (g)

rate = k[NO2]2

The balanced equation for the overall reaction shows the stoichiometry but does not necessarily mean that the reaction simply occurs by one molecule of NO2 colliding with one molecule of CO. If the reaction really took place in that one step, then the rate would be first order in NO2 and first order in CO, or rate = k[NO2][CO]. The fact that the experimentally determined orders do not match the coefficients in the overall balanced equation tells us that the reaction does not take place in one step. The following proposed two-step mechanism is consistent with the observed rate-law expression. (1) (2)

NO 2 + NO 2  → N 2O 4 N 2 O 4 + CO  → NO + CO 2 + NO 2 NO 2 + CO  → NO + CO 2

(slow) (fast) overall

The rate-determining step of this mechanism involves a bimolecular collision between two NO2 molecules. This is consistent with the rate expression involving [NO2]2. Because the CO is involved only after the slow step has occurred, the reaction rate would not depend on [CO] (that is, the reaction would be zero order in CO) if this were the actual mechanism. In this proposed mechanism, N2O4 is formed in one step and is completely consumed in a later step. Such a species is called a reaction intermediate. However, in other studies of this reaction, nitrogen trioxide, NO3, has been detected as a transient (short-lived) intermediate. The mechanism now thought to be correct is (1) (2)

NO 2 + NO 2  → NO3 + NO NO3 + CO  → NO + CO 2 NO 2 + CO  → NO + CO 2

(slow) (fast) overall

In this proposed mechanism two molecules of NO2 collide to produce one molecule each of NO3 and NO. The reaction intermediate NO3 then collides with one molecule of CO and reacts very rapidly to produce one molecule each of NO2 and CO2. Even though two NO2 molecules are consumed in the first step, one is produced in the second step. The net result is that only the NO2 molecule is consumed in the overall reaction. Each of these proposed mechanisms meets both criteria for a plausible mechanism: (1) The steps add to give the equation for the overall reaction, and (2) the mechanism is consistent with the experimentally determined rate-law expression (in that two NO2 molecules and no CO molecules are reactants in the slow step). The NO3 that has been detected is evidence in favour of the second mechanism, but this does not unequivocally prove that mechanism; it may be possible to think of other mechanisms that would involve NO3 as an intermediate and would also be consistent with the observed rate law. Important Points 1. In the reaction discussed above, the slowest (rate determining) step in the mechanism involves a collision of two molecules. Such a step is said to be bimolecular. Likewise, depending upon the number of molecules, ions or atoms taking part in the rate determining step, we classify the elementary reactions according to: unimolecular : when only one species takes part bimolecular : when two species take part termolecular : when three species take part The probability of a reaction where 4 molecules take part in a single step is almost nil. Chemistry / Chemical Kinetics

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Molecularity (like order) cannot be predicted from the stoichiometric coefficients of a balanced equation. It is determined theoretically after proposing the reaction mechanism (whereas order as we’ve seen is determined experimentally).

Please Answer this Question before seeing its solution. •

You are a chemist of a research laboratory that is trying to increase the reaction rate for the balanced chemical reaction: X + 2Y → Z. a. One of your researchers comes into your office and states that she has found a material that significantly lowers the activation energy of the reaction. Explain the effect this will have on the rate of the reaction. b. Another researcher states that after doing some experiments, he has determined that the rate law is rate = k[X][Y]. Is this possible? c. Yet another person in the lab reports that the mechanism for the reaction is:

2Y → I

(slow)

(fast) X+I→ Z Is the rate law from part b. consistent with this mechanism? If not, what should the rate law be? Solution: a. Her finding should increase the rate since the activation energy, Ea, is inversely related to the rate constant, k; a decrease in Ea results in an increase in the value of k. b. This is possible because the rate law does not have to reflect the overall stoichiometry of the reaction. c. No. Since the rate law is based on the slow step of the mechanism, it should be Rate = k[Y]2.

Pseudo first order reactions. Consider the following reaction: +

t=0

H CH3 COO C2H5 + H2O  → CH3 COOH 0.02 M 100 M 0M

t

0M

99.98 M

0.02 M

+

C2 H5 OH 0M 0.02 M

This reaction is a bimolecular elementary reaction. Therefore, its rate law can be written as: Rate = K [CH3 COO C2H5] [H2O]. But observing the data above, we see that H2O is in large excess and its concentration has almost remained constant in the reaction. So, if we consider [H2O] as a constant, then, the rate law may be rewritten as: Rate = {K [H2O]} [CH3 COO C2H5] ⇒

Rate = K ' [CH3 COO C2H5]

We see that this rate law is same as the one for a first order reaction. This is an example of a pseudo-first order reactions. They are defined as: the reactions in which the molecularity of the reaction is 2 or more but they follow first order kinetics are called as pseudo first order reactions. Chemistry / Chemical Kinetics

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Section - 4

THE ARRHENIUS EQUATION

Temperature: The Arrhenius Equation The average kinetic energy of a collection of molecules is proportional to the absolute temperature. At a particular temperature, T1, a definite fraction of the reactant molecules have sufficient energy, Ea, to react to form product molecules upon collision. At a higher temperature, T2 a greater fraction of the molecules possess the necessary activation energy, and the reaction proceeds at a faster rate. This is depicted in Figure 11. From experimental observations, Svante Arrhenius developed the mathematical relationship among activation energy, absolute temperature, ad the specific rate constant of a reaction, k, at that temperature. The relationship, called the Arrhenius equation, is k = Ae – Ea / RT or, in logarithmic form, ln k = ln A –

Ea RT

or

log k = log A –

Ea 2.303 RT

Fraction of molecules with a given kinetic energy

In this expression, A is a constant having the same units as the rate constant. It is proportional to the frequency of collisions between reacting molecules. Actually A (called frequency factor or pre-exponential factor) is equal to Z ρ i.e. A = Z ρ where Z is the number of collisions of the molecules per second in a unit volume and ρ is the steric factor. The necessity of introducing the factor ρ in the Arrhenius equation is explained by the fact that the collisions even between active molecules (i.e. molecules with sufficient energy to bring about reaction) do not always result in a reaction, but only when the molecules have a definite orientation. (Refer figure – 8 of section – 3 for an example). The factor ρ is proportional to the ratio of the ratio of the number of ways of the mutual orientation of the molecules favourable for proceeding of a reaction to the total number of possible ways of orientation: the greater this ratio, the more rapidly will a reaction proceed. The steric factor ρ is usually much smaller than unity; it has an especially great influence on the rate of reactions proceeding with the participation of complex molecules (for example, glucose and proteins), when the total number of various possible orientations is very great, and the number of orientations favourable for proceeding of a reaction is very limited. R is the universal gas constant, expressed with the same energy units in its numerator as are used for Ea. For instance, when Ea is known in J/mol, the value R = 8.314 J/mol ⋅ K is appropriate. Here the unit “mol” is interpreted as “mole of reaction,” as described the unit on thermodynamics. One important point is the following: The greater the value of Ea, the smaller the value of and the slower the reaction rate (other factors being equal). This is because fewer collisions take place with sufficient energy to get over a high energy barrier. The Arrhenius equation predicts that increasing T results in faster reaction for the same Ea and concentrations. T2 > T1 T1 T2

The area between the distribution curve and the horizontal axis in Figure is proportional to the total number of molecules present. The total area is the same at T1 and T2. The shaded areas represent the number of particles that exceed the energy of activation, Ea.

Kinetic energy Ea Figure 11: The effect of temperature on the number of molecules that have kinetic energies greater than Ea. At T2, a higher fraction of molecules posses at least Ea, the activation energy. Chemistry / Chemical Kinetics

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– Ea / RT k If T ⇒ E a / RT ⇒ – E a / Rt ⇒ e ⇒ ⇒ Reaction increases increases speeds up decreases increases increases

Let’s look at how the rate constant varies with temperature for a given single reaction. Assume that the activation energy and the factor A do not vary with temperature. We can write the Arrhenius equation for two different temperatures. Then we subtract one equation from the other, and rearrange the result to obtain, in natural logarithm (ln) form,

ln

k 2 Ea  1 1 =  –  k1 R  T1 T2 

In base-10 logarithm (log) form this equation is written as

log

k2 Ea  1 1 =  –  k1 2.303R  T1 T2 

Let’s substitute some typical values into this equation. The activation energy for many reactions that occur near room temperature is about 50 kJ/mol (or 12 kcal/mol). For such a reaction, a temperature increase from 300 K to 310 K would result in ln

k2 50, 000 J / mol  1 1  = –   = 0.647 k1 (8.314 J / mol ⋅ K)  300 K 310 K  k2 = 1.91 ≈ 2 k1

Chemists sometimes use the rule of thumb that near room temperature the rate of a reaction approximately doubles with a 10° C rise in temperature. Such a “rule” must be used with care, however, because it obviously depends on the activation energy. Example – 13 The specific rate constant, k, for the following first-order reaction is 9.16 × 10– 3 s– 1 at 0.0° C. The activation energy of this reaction is 88.0 kJ/mol. Determine the value of k at 2.0° C. N 2O5 → NO 2 + NO3 Critical thinking First we tabulate the values, remembering to convert temperature to the Kelvin scale. Ea = 88,000 J/mol

R = 8.314 J/mol.K

k1 = 9.16 × 10–3 s–1

at

T1 = 0.0° C + 273 = 273 K

k2 = ?

at

T2 = 2.0° C + 273 = 275 K

We use these values in the "two-temperature" form of the Arrhenius equation. Chemistry / Chemical Kinetics

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ln

Solution:

k 2 Ea  1 1  =  −  k1 R  T1 T2 

  88, 000 J / mol  1 k2 1  ln  – =   = 0.282 – 3 –1   9.16 × 10 s  8.314 J  273 K 275 K  mol ⋅ K

Taking inverse (natural) logarithms of both sides. k2 = 1.32 9.16 × 10 – 3 s –1 k 2 = 1.32 (9.16 × 10 – 3 s –1 ) = 1.21 × 10 – 2 s –1

We see that a very small temperature difference, only 2°C, causes an increase in the rate constant (and hence in the reaction rate for the same concentrations) of about 32%. Such sensitivity of rate to temperature change makes the control and measurement of temperature extremely important in chemical reactions. Example – 14 The gas-phase decomposition of ethyl iodide to give ethylene and hydrogen iodide is a first-order reaction. C2 H 5 I  → C 2 H 4 + HI

At 600 K, the value of k is 1.60 × 10– 5 s– 1. When the temperature is raised to 700 K, the value of k increases to 6.36 × 10– 3 s– 1. What is the activation energy for this reaction? Critical thinking We know k at two different temperatures. We solve the two-temperature form of the Arrhenius equation for Ea and evaluate

Solution:

k1 = 1.60 × 10 – 5 s –1 at T1 = 600 K

k2 = 6.36 × 10 – 3 s –1 at T2 = 700 K

R = 8.314 J / mol ⋅ K

Ea = ?

We arrange the Arrhenius equation for Ea.

ln

Chemistry / Chemical Kinetics

k2 E a  1 1  =  −  k1 R  T1 T2 

k2 k1 Ea = 1 1   −   T1 T2  R ln

so

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Substituting,

J   6.36 × 10 – 3 s –1    8.314  ln   mol ⋅ K   1.60 × 10 – 5 s –1   Ea = =  1 1  –    600 K 700 K  Ea = 2.09 × 105 J/mol

J    8.314  (5.98) mol ⋅ K   2.38 × 10 – 4 K –1

( i.e. 209 kJ/mol)

____________________________________________________________________________ The determination of Ea in the manner illustrated in the last Example may be subject to considerable error, because it depends on the measurement of k at only two temperatures. Any error in either of these k values would greatly affect the resulting value of Ea. A more reliable method that uses many measured values for the same reaction is based on a graphical approach. Let us rearrange the single-temperature logarithmic form of the Arrhenius equation and compare it with the equation for a straight line. E ln k = – a R

1 T

+ ln A

y

m x + b = The value of the collision frequency factor, A, is very nearly constant over moderate temperature changes. Thus, ln A can be interpreted as the constant term in the equation (the intercept). The slope of the straight line obtained by plotting ln k versus 1/T can be interpreted as – Ea/R. This allows us to determine the value of the activation energy from the slope (Figure 12). Ea R

ln k

Slope =

1/T Figure 12: A graphical method for determining activation energy, Ea. At each of several different temperatures, the rate constant, k, is determined by methods such as those in Section 2. A plot of ln k versus 1/T gives a straight line with negative slope. The slope of this straight line is – Ea/R. Alternatively, a plot of log k versus 1/T gives a straight line whose slope is – Ea/2.303 R. Use of this graphical method is often desirable, because it partially compensates for experimental errors in individual k and T values.

Please attempt the following problem before seeing its solution: •

The chemical reaction A → B + C has a rate constant that obeys the Arrhenius equation. Predict what happens to both the rate constant k and the rate of the reaction if the following were to occur: a. a decrease in temperature. b. an increase in the activation energy of the forward and reverese reactions. c. an increase in both activation energy and temperature.

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Solution: The Arrhenius equation is k = Ae − E a / RT a. When the temperature is decreased, the rate constant, k, will also decrease. When k decreases, the rate also decreases. b. When the activation energy is increased, the rate constant, k, also decreases. When k decreases, the rate also decreases. c. Since the activation energy is in the numerator and the temperature is in the denominator, you cannot predict the effect without knowing the magnitude of the changes. Catalysts Catalysts are substances that can be added to reacting systems to increase the rate of reaction. They allow reactions to occur via alternative pathways that increase reaction rates by lowering activation energies. The catalyst causes unstable intermediates (called activated complexes) to appear as a new transition state which has a lower energy as compared to the transition state corresponding to activated complexes of the uncatalysed reactions. Since, the new transition state has lower energy, a much greater number of reacting species have the required energy to reach this new transition state. This increases the rate of the reaction. Once this transition state is attained, its decomposition leads to the formation of the products. The activation energy is lowered in all catalysed reactions, as depicted in Figures 13 and 14. A catalyst does take part in the reaction, but all of it is regenerated in later steps. Thus a catalyst does not appear in the balanced equation for the reaction. For constant T and the same concentrations, If Ea ⇒ E a / RT ⇒ – E a / RT ⇒ e – Ea /RT ⇒ k ⇒ Reaction increases speeds up decreases decreases increases increases

Ea forward

Ea reverse ∆E

Reaction coordinate for uncatalysed reaction

Potential energy

Potential energy

E′a < Ea E′a forward

E′a reverse ∆E

Reaction coordinate for catalysed reaction

Figure 13: Potential energy diagrams showing the effect of a catalyst. The catalyst provides a different, lower-energy mechanism for the formation of the products. A catalysed reaction typically occurs in several steps, each with its own barrier, but the overall energy barrier is lower than the uncatalysed reaction. ∆E has the same value for each path. The value of ∆E depends only on the states of the reactants and products. Chemistry / Chemical Kinetics

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Minimum kinetic energy for catalysed reaction Minimum kinetic energy for uncatalysed reaction Kinetic energy

Figure 14: When a catalyst is present, the energy barrier is lowered. Thus, more molecules possess the minimum kinetic energy necessary for reaction. This is analogous to allowing more students to pass a course by lowering the requirements.

Please attempt the following problem before seeing its solution:

E+F G+H Progress of reaction E+F G+H

Energy per mol

Considering the potential energy curves for two different reactions:

Energy per mol

•

A+ B C+D Progress of reaction A+ B C+D

a. Which reaction has higher activation energy for the forward reaction? b. If both reactions were run at the same temperature and have the same orientation requirements to react, which one would have the larger rate constant? c. Are these reactions exothermic or endothermic? Solution: a. Since the “hump” is larger, the A + B reaction has a higher activation energy. b. Since the activation energy is lower, the E + F reaction would have the larger rate constant. Keep in mind the inverse relationship between the activation energy, Ea, and the rate constant, k. c. Since in both cases energy per mole of the reactants is greater than the products, both reactions are exothermic. Chemistry / Chemical Kinetics

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Example – 15 The activation energy of a reaction is 75.24 kJ/mol in the absence of a catalyst, and 50.14 kJ/mol with a catalyst. How many times will the rate of the reaction grow in the presence of a catalyst if the reaction proceeds at 25° C? Critical thinking When the reaction is catalysed, the rate constant of the reaction increases and Ea decreases. However A remains constant. We can use this information to calculate the ratio of rate constants between two alternative paths. This ratio shall give us the relative rates quite successfully. Solution: Let the activation energy of the reaction without a catalyst be Ea, and with one, E′a , let k and k' be the respective rate constants of the reaction. Using the Arrhenius equation, we find: k ′ exp ( – E′a / RT )  E − E′a  = = exp  a  k exp ( – E a / RT )  RT  hence ln

k′ E − E′a k′ E − E′a k′ = 2.303 log = a , and log = a k k RT k 2.303RT

Introducting the data of the example into the last equation, expressing the activation energy in joules, and taking into account that T = 298 K, we get: log

3 k ′ ( 75.24 – 50.14 ) × 10 25.1 × 103 = = = 4.40 k 2.30 × 8.314 × 298 2.30 × 8.314 × 298

We finally obtain k′ / k = 2.5 × 104. Hence, a decrease in the activation energy by 25.1 kJ resulted in the reaction rate growing 25000 times.

Example – 16 From the following data for the reaction between A and B, calculate (i) the order of the reaction with respect to A and with respect to B, (ii) the rate constant at 300 K, (iii) the energy of activation, and (iv) the pre-exponential factor. [A]/mol L– 1 –4

2.5 × 10 5.0 × 10– 4 1.0 × 10– 3

Chemistry / Chemical Kinetics

[B]/mol L– 1 –5

3.0 × 10 6.0 × 10– 5 6.0 × 10– 2

Initial rate/mol L– 1

s– 1 at

300 K

320 K –4

5.0 × 10 4.0 × 10– 3 1.6 × 10– 2

2.0 × 10– 3 – –

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Critical thinking Sub-questions (i) and (ii) are left for you as an exercise. The answer are provided for cross-checking. In sub-question (iii), we cannot calculate Ea by the “one-temperature form” of the Arrhenius equation as we don’t have A (called some times as pre-exponential factor). So, the other alternative is to see for the possibility of the two temperature form of Arrhenius equation to give us the solution. In such a case we would need ratio of the rate constants of the same reaction at two different temperatures. But so far in (i) and (ii) we have only got the value of rate constant at one particular temperature (300 K). So, what do we do? Please go back to Example – 12 and note that in cases where ratio of two quantities are involved we can substitute it with ratio of two other quantities which are directly proportional to the ratio of the first two quantities. Here, we know that the rate of a reaction at a particular temperature is proportional to the rate constant at that particular temperature. Note also that the data given in table has two rates of reaction at two different temperatures for first choice of concentrations of A and B. Can you use k1 this information to find your substitute for k ? 2

Solution: The two temperature form of the Arrhenius equation is:

log

k2 Ea  1 1  =  −  k1 2.303 R  T1 T2 

... (i)

But, we know that,

(rate)2 = k 2 [A ] [B] 2

1

... (ii)

and

(rate)1 = k1 [A ] [B] 2

1

... (iii)

Dividing (ii) by (iii) to get: (rate) 2 k 2 [A ] [B] = (rate)1 k1 [A ]2 [B]1 2

1

... (iv)

Hence, if we substitute A and B at the same instant, we get:

( rate )2 ( rate )1 ⇒

Chemistry / Chemical Kinetics

=

k2 k1

k 2 2.0 × 10 – 3 = k1 5.0 × 10 – 4

(for the first choice of concentrations of A and B)

... (v)

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Substituting (v) in (i), we get,  2.0 × 10 – 3   1 Ea 1  = − log     –4   5.0 × 10  ( 2.303)(8.314 J / K / mol )  300 K 320 K 

⇒

Ea =

( 2.303)(8.314 )(300 )(320 ) log 4 (320 − 300 )

= 5.53 × 104 J/mol = 55.33 kJ/mol. (iv) To calculate the pre-exponential factor A, it is easier if we take logarithm of the Arrhenius equation. Hence, log A = log k +

Ea 2.303 RT

... (A)

Here, k = 2.67 × 108 (at 300 K) (You should have calculated it in sub-question (ii) above) Ea = 5.53 × 104 J/mol R = 8.314 J K– 1 mol– 1 T = 300 K Substituting these values in (A) we get log A = log ( 2.67 × 108 ) + ⇒

log A = 8.43 + 9.63 = 18.06

⇒

A = 1.145 × 108 M– 2 s– 1.

Answer (i) order w.r.t. A = 2

5.53 × 104 J mol –1 ( 2.303)(8.314 J / K / mol )(300 K )

order w.r.t. B = 1

Answer (ii) k = 2.67 × 108 M– 2 s– 1.

Example – 17 A first-order reaction, A → B , requires activation energy of 70 kJ mol– 1. When a 20% solution of A was kept at 25° C for 20 minutes, 25% decomposition took place. What will be percent decomposition in the same time in a 30% solution maintained at 40° C? Assume that activation energy remains constant in this range of temperature.

Chemistry / Chemical Kinetics

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16

Critical thinking First of all we need to understand an important characteristic of first order reactions. Consider the reactions: A→B

...(i)

and

(first order in A)

A→C

...(ii)

and

(sec ond order in A)

A→D

...(iii)

(zero order in A)

We have seen that the integrated rate equation for the reaction (i), (ii) and (iii) are: ln

[A ]0 [A ]t

= akt

(first order)

... (iv)

1 1 − = akt [A ]t [A]0

(second order)

... (v)

[A ]0 − [A ]t = akt

(zero order)

... (vi)

Suppose we have two initial concentrations for A, say, 2M and 5M respectively. And we want to know the time after which a certain percentage of the reactants have reacted (say, 25%). Then in the first order case, we will see: ln

2M 5M = akt and ln = akt 1.5M 3.75M

(1.5 M and 3.75M is what is left after 25% decomposition in the respective cases) and ln1.33 = akt ⇒ ln1.33 = akt ln (1.33) ln (1.33) and t = ak ak We see that, in both the cases, the time required is same and observing closely we can realise, that the extent of reaction in a first order reaction is independent of the initial concentration. You may use the same two concentrations in second order and zero order case to see that the extent of reaction is dependent upon the initial concentration. Zero order case is shown: ⇒

t=

2M − 1.5M = akt and

5M − 3.75M = akt

0.5M 1.25M and t = ak ak Since, the time required for 25% decomposition is different in both the cases, we can see that the extent of reaction in the same time period is different too and is dependent upon the initial concentration in this case.

⇒

t=

Solution: We know that the fraction of reactants reacted in a first order reaction is independent of initial concentration. So, we need not worry about the terms 20% solution and 30% solution. Now, it is given that 25% of A is decomposed in 20 minutes at 25° C. And we need to find the percent decomposition in the same time at 40° C for the same reaction. Quite obviously, the rate constants of this reaction at two temperatures are different. We can also see that the rate constant for the first reaction can be calculated using the integrated rate equation and then this can be used in the “two-temperature” form to get the rate constant of the reaction at 40ºC. This rate constant at 40ºC can then be used in the integrated rate equation to obtain the required data. We illustrate the above methodology we have planned over here: Chemistry / Chemical Kinetics

LOCUS

17

We know that, log

[A]0 akt = [A]t 2.303 ...(vii)

Here, at 25ºC, a=1 k1 = ? t = 20 min. [A]0 100 = [A]t 275

(as 25% decomposition has taken place)

Substituting these values in (vii), we get, log

⇒

100 1. k1 .(20 min) = 75 2.303

k1 = 1.44 × 10–2 min–1

Now, letting k2 be the rate constant of this reaction at 40ºC, we can use the two temperature form of Arrhenius equation as: log

1 1  k2 Ea =  −  k1 2.303 R  T1 T2 

 1 k2 70 × 103 J / mol 1  = − ⇒  −2 −1 −1  (1.44 × 10 ) 2.303 × 8.314 JK mol  298 K 313 K  Solving for k2, we will get, log

k 2 = 5.56 × 10−2 min −1 Finally,

log

[A]t  k  = − 2  t [A]0  2.303 

 5.56 ×10−2 min −1  = −  (20 min) 2.303   ⇒

This remains 20 min. because we have been asked to find the percentage decomposition at 40ºC in the same time interval as the first reaction (at 25ºC) during which 25% decomposition took place.

[A]t 33 = 0.33 = [A]0 100

i.e., percentage of A remaining at (t = 20 min) = 33% percentage of A reacted till (t = 20 min) = (100 – 33) = 67%

Chemistry / Chemical Kinetics

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TRY YOURSELF - III Q. 1

For a first-order reaction (a) the degree of dissociation is equal to (1 – e– kt) (b) a plot of reciprocal concentration of the reactant versus time gives a straight line (c) the time taken for the completion of 75% reaction is thrice that of t1/2 of the reaction (d) the pre-exponential factor in the Arrhenius equation has the dimension of time T– 1

Q. 2

In the Arrhenius equation, k = A exp (– Ea/RT), A may be termed as the rate constant at............. .

Q. 3

A catalyst is a substance which (a) increases the equilibrium concentration of the product (b) changes the equilibrium constant of the reaction (c) shortens the time to reach equilibrium (d) supplies energy to the reaction

Q. 4

A catalyst (a) increases the average kinetic energy of reacting molecules (b) decrease the activation energy (c) alters the reaction mechanism (d) increases the frequency of collisions of reacting species

Q. 5

For an endothermic reaction where ∆H represents the enthalpy of the reaction, the minimum value for the energy of activation will be (b) zero (a) less than ∆H (c) more than ∆H (d) equal to ∆H

TRUE or FALSE (Q. 6 - Q. 8) Q. 6

Catalyst makes a reaction more exothermic.

Q. 7

Catalyst does not affect the energy of activation in a chemical reaction.

Q. 8

The rate of an exothermic reaction increases with increasing temperature.

Q. 9

In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 1013 s–1 and 98.6 kJ mol– 1, respectively. If the reaction is of first-order, at what temperature will its half-life period be ten minutes?

Q. 10 A first-order reaction is 50 percent completed in 30 min at 27° C and in 10 min at 47° C. Calculate the reaction rate constant at 27° C and the energy of activation of the reaction. Q. 11 The trans → cis isomerisation of 1, 2-dichloroethylene proceeds with Ea = 231 kJ/mol and ∆H = 4.2 kJ/mol. What is Ea for the cis → trans isomerisation?

Chemistry / Chemical Kinetics

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Q. 12 Will the value of the rate constant of a reaction change (a) when one catalyst is replaced with another one, and (b) when the concentrations of the reactants and products are changed? Q. 13 Does the heat effect of a reaction depend on its activation energy? Substantiate your answer. Q. 14 For which reaction—the forward or the reverse one—is the activation energy greater if the forward reaction proceeds with the liberation of heat? Q. 15 How many times will the rate of a reaction proceeding at 298 K grow if its activation energy is lowered by 4 kJ/mol? Q. 16 What is the activation energy of a reaction if its rate doubles when the temperature is raised from 290 K to 300 K? Q. 17 What is the value of the activation energy for a reaction whose rate at 300 K is ten times greater than at 280 K? Q. 18 Which of the following procedures will lead to a change in the rate constant of a reaction? (a) A change in the pressure; (b) a change in the temperature; (c) a change in the volume of the reaction vessel; (d) the introduction of a catalyst into the system; (e) a change in the concentration of the reactants and products. Q. 19 The increase in a reaction rate with temperature is due chiefly to: (a) an increase in the average kinetic energy of the molecules; (b) a growth in the number of active molecules; (c) a growth in the number of collisions. Q. 20 The first-order gaseous decomposition of N2O4 into NO2 has a k value of 4.5 × 103 s– 1 at 1° C and an energy of activation of 58 kJ mol– 1. At what temperature its half-life would be 6.93 × 10– 5 s?

Chemistry / Chemical Kinetics

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MISCELLANEOUS EXAMPLES

Example – 18 Under certain conditions the reaction between sulphate (S O 24 − ) ions and hydrogen ions is first order in each of them, i.e., rate = k[SO 24− ] [H+]. A student measured the initial rate of the reaction four times using the combinations of solutions shown in the following table. Write down the actual concentrations of both ions in each experiment and hence determine the value of the rate in terms of k. _____________________________________________________________________________ Experiment

Solution of sulphate ions Solution of hydrogen ions Concentration volume Concentration volume –3 3 –3 /mol dm /cm /mol dm /cm3 _____________________________________________________________________________ A 0.5 125 0.05 125 B 0.5 250 0.05 250 C 0.5 500 0.05 500 D 0.5 500 0.05 125 _____________________________________________________________________________

Critical thinking (1) You will have to think carefully about the difference between concentration and volume. (2) In the rate law for a reaction, the concentration used are in terms of the number of moles of each reacting species w.r.t. the total concentration of the solution. Here the concentrations given are of the reactants before they have been brought together for reaction. We have been given the volume of each of the reactants’ solution. When they are mixed, the total volume will change and hence, the concentration of the reacting species will change too. It is these new concentrations of the reacting species that should be put in the rate law to get the respective rate of reaction. Solution: As pointed out the key thing to remember is that the rate will change if the concentrations of the reagents change. In experiment A when the two solutions are mixed, the total volume becomes 250 cm3. This is twice the volume of each of the separate solutions, so their concentrations are halved just as they are brought together to react. 0.5 M in a 250 cm3 2 0.05 solution. And the concentration of H+ ions (0.05 M in a 125 cm3 solution) becomes in a 250 cm3 2 solution.

Hence, the concentration of SO 24 − (0.5 m in a 125 cm3 solution) becomes

Chemistry / Chemical Kinetics

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21

So, rate of the reaction in this case will be given by, 0.5   0.05  rate = k    [substituting nto the given rate law]  2  2  rate = k × 6.25 × 10–3

You are requested to provide the solution for experiments B, C and D yourself. The answers are given here: (B) (C) (D)

rate = k × 6.25 × 10–3 rate = k × 6.25 × 10–3 rate = k × 4.00 × 10–2

Example – 19 A certain compound A, was found the undergo two parallel first order rearrangements, forming B and C respectively in the two arrangements. At 25° C, the first order rate constant for the formation of B was measured as 1.0 × 10– 5 s– 1 and for the formation of C as 2.0 × 10– 6 s– 1. What is the percentage distribution of the rearrangement products? Critical thinking What the question says is that the same compound A has tendency to form two different products by two different paths and both these different reactions can occur simultaneously: –1 –5 s B 0 .0 × 1 1 = k1 A k2 = 2. 0 × 1 –6 – 0 s 1 C Here, we can write the various rate laws as: d [A ] – = k 1 [A ] + k 2 [A ] ... (i) dt d [B] = k1 [ A ] ... (ii) dt d [C ] = k 2 [A ] ... (iii) dt Dividing (ii) by (i), we get, d [B] k1 [A ] dt = d [ A ] k1 [ A ] + k 2 [ A ] − dt k1 [ A ] d [B] = ⇒ −d [ A ] k +k A

(

1

2

)[ ]

Hence LHS gives us the fraction of amount of B formed to amount of A reacted and this is equal to k1 k1 + k 2 . Can you see that you have all the information and analysis you need to get the answers for

this question? Chemistry / Chemical Kinetics

LOCUS

22

k1

Solution:

A

B

k2

C  k1  percentage of B formed can be obtained by =  × 100   k1 + k 2 

  1.0 × 10 – 5 = ×100 –5 –6   1.0 × 10 + 2.0 × 10    1 × 10 – 5 = ×100 –5 –5   1.0 × 10 + 0.2 × 10  = 83.33 %  k2   × 100 percentage of A formed can be obtained by =  k + k  1 2 

  2.0 × 10 – 6 = ×100 –5 –6   1.0 × 10 + 2.0 × 10    0.2 × 10 – 5 = ×100 –5 –5  1.0 × 10 + 0.2 × 10   = 16.67 % .

Example – 20 Two reactions (i) A → products and (ii) B → products follow first-order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from 300 K to 310 K. The half-life for the reaction at 310 K is 30 min. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300 K. Critical thinking The problem is actually very simple as all you have to do is certain calculations in a logical order. Arrange all the data in a diagram and it will be immediately clear to you so as to how you can solve this question. In the solution to this problem, you will be provided with the rough draft of the data and then how it can be reorganised to see for yourself so as to how solution can be obtained. It is expected of you that you make the rough draft and its reorganisation yourself first and then compare it with the one given here. It is expected then that you provide the complete solution yourself. The answer is provided for cross-checking. Chemistry / Chemical Kinetics

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23

Solution: Rough Draft (data collection from Question) 2k A (300 K ) = k A (310 K )

... (i)

t 1 (for (i) at 310 K ) = 30 min

... (ii)

2

k B (310 K ) = 2 k A (310 K )

... (iii)

1 E 2 a (i )

... (iv)

E a (ii ) =

k B (300 K ) = ?

... (v)

Reorganised draft

2kA (300 K) = kA (310 K) and 30 min =

0.693 kA (310 K)

Use in two temperature form of Arrhenius Equation to get Ea (i) ⇒ we have Ea (ii) (using (iv))

get value of kA at both temperatures

get value of kB (310 K) (using (iii))

Use both these informations in the two temperature form of Arrhenius equation to get kB (300 K)

Answer

1 0.0462 min –1 k B (300 K ) = k B (310 K ) = = 0.0327 min –1 2 2

Chemistry / Chemical Kinetics

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24

EXERCISE Rate of Reaction and Rate Law Q. 1

→ 2NO 2 (g ) change if the volume of the reaction How will the rate of reaction 2 NO ( g ) + O 2 ( g )  vessel is deminished to one third of its initial value? (The rate law for this reaction involves all reactants and their respective orders are same as the stoichiometric coefficients.)

Q. 2

One mole of gas A and two moles of gas B are introduced into one vessel, and two moles of gas A and one mole of gas B into a second vessel having the same capacity. The temperature is the same in both vessels. Will the rate of the reaction between gases A and B in these vessels differ if it is expressed by the equation (a) rate1 = k1[A] [B]; (b) rate2 = k2 [A]2[B]?

Q. 3

The reaction between substances A and B is expressed by the equation A + 2B → C. The initial concentrations of the reactants are [A]0 = 0.03 mol/l and [B]0 = 0.05 mol/l. The rate constant of the reaction is 0.4. Find the initial rate of the reaction and the rate after a certain time when the concentration of substance A diminishes by 0.01 mol/l. (rate = k[A][B]2)

Q. 4

How will the rate of the reaction 2NO(g) + O2(g) → 2NO2(g) change if (a) the pressure in the system is increased three times; (b) the volume of the system is diminished to one-third of its initial value; and (c) the concentration of the NO is increased three times? (rate = [NO]2[O2] )

Q. 5

Rate of a reaction A + B → products, is given below as a function of different initial concentrations of A and B: [A] (mol/l) [B](mol/l) Initial rate (mol/l/min) 0.01 0.01 0.005 0.02 0.01 0.010 0.01 0.02 0.005 Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the reaction?

Q. 6

For the given reaction, A + B → products, following data were obtained.

1. 2. 3.

[A] (mol/l)

[B](mol/l)

Initial rate (mol/l/min)

[A0] 0.1 0.2 0.1

[B0] 0.2 0.2 0.1

R0(mol L– 1 s– 1) 0.05 0.10 0.05

(a) Write the rate law expression

(b) Find the rate constant

Q. 7

The reaction, H2(g) + Br2(g) → 2 HBr(g) has a rate given by the rate law, Rate = k [H2][Br2]1/2. (a) What are the units of the rate constant k? (b) If the volume of the gas mixture is halved, by what factor is the rate changed?

Q. 8

The reaction v1A + v2B → products is first-order with respect to A and zero-order with respect to B. If the reaction is started with [A]0 and [B]0, the integrated rate expression of this reaction would be (a) ln

[A ]0 =k t [A ]0 − x 1

Chemistry / Chemical Kinetics

(b) ln

[A ]0 =k t [A ]0 − v1x 1

(c) ln

[A ]0 [A ]0 = v1k1t (d) ln = –v1k1 t [A ]0 − v1x [A ]0 − v1x

LOCUS

Q. 9

25

Rate of a reaction, A + B → products is given as a function of different initial concentrations of A and B. [A]/mol L– 1

[B]/mol L– 1

r0/mol L– 1 min– 1

__________________________________________________________________________________

0.01 0.01 0.005 0.02 0.01 0.010 0.01 0.02 0.005 Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the reaction? Q. 10 For a first order reaction A → B, the concentration of B at two instant is given: [B] t = 10 sec = 1.0 M and [B] t = 25 sec = 4.0 M. Using this information can you calculate the instantaneous rate of the reaction at at least one instant in the interval from t = 10 sec to t = 25 sec. Q. 11 If the steady-state concentration of O3 in a polluted atmosphere is 2.0 × 10– 8 mol/L, the rate of production of O3 is 7.2 × 10–13 M/hr, and O3 is destroyed by the reaction, 2 O3 → 3 O2, what is the rate constant for the reaction, assuming a rate law, Rate = – (1/2) ∆ [O3]/ ∆ t = k [O3]2? Q. 12 In the enzymatic fermentation of sugar, the sugar concentration decreased from 0.12 M to 0.06 M in 10 hours, and to 0.03 M in 20 hours. What is the order of the reaction? What is the rate constant k? Q. 13 The rate law for the reaction, Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq), is: Rate = (1.0 × 103 M–1s–1) [Ce4+][Fe2+]. If 0.500 L of 0.0020 M Ce(SO4) 2 is rapidly mixed with 0.500 L of 0.0020 M FeSO4, how long does it take for [Fe2+] to decrease to 1.0 × 10– 4 M? Integrated Rate Equation Q. 14 The following statement(s) is (are) correct: (a) A plot of log Kp versus 1/T is linear (b) A plot of log [X] versus time is linear for a first order reaction, X → P (c) A plot of p versus 1/T is linear at constant volume (d) A plot of p versus 1/V is linear at constant temperature Q. 15 At constant temperature and volume, X decomposes as 2X(g)  → 3Y(g) + 2Z(g); Px is the partial pressure of X.

Observation No. 1 2 3

Time (in minute) 0 100 200

(i) What is the order of reaction with respect to X? (ii) Find the rate constant. (iii) Find the time for 75% completion of the reaction (iv) Find the total pressure when pressure of X is 700 mm of Hg.

Chemistry / Chemical Kinetics

Px (in mm of Hg) 800 400 200

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26

Q. 16 While studying the decomposition of gaseous N2O5 it is observed that a plot of logarithm of its partial pressure versus time is linear. What kinetic parameters can be obtained from this observation? Q. 17 The rate constant for the first-order decomposition of N2O5(g) to NO2(g) and O2(g) is 7.48 × 10– 3 s– 1 at a given temperature. (a) Determine the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.1 atm to rise to 0.145 atm. (b) Find the total pressure after 100 s of the reaction. The Arrhenius Equation. Q. 18 The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25° C are 3.0 × 10– 4s– 1, 104.4 kJ mol– 1 and 6.0 × 1014s– 1 respectively. The value of the rate constant as T → ∞ is, (a) 2.0 × 1018 s– 1 (b) 6.0 × 1014 s– 1 (c) infinity (d) 3.6 × 1030 s– 1 Q. 19 At 380° C, the half-life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ mol– 1. Calculate the time required for 75% decomposition at 450° C. Q. 20 The rate constant for the first order decomposition of a certain reaction is described by the equation log(k/s– 1) = 14.34 – (1.25 × 104 K)/T (i) What is the energy of activation of this reaction? (ii) At what temperature will its half-life be 256 minutes? Q. 21 The rate constant of a reaction is 1.5 × 107 s– 1 at 50° C and 4.5 × 107 s– 1 at 100° C. Evaluate the Arrhenius parameters A and Ea. Q. 22 A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 kJ mol– 1. Q. 23 The activation energy of a certain reaction is 15 kJ/mol. The reaction is exothermic, yielding 19 kJ/mol. What is the activation energy of the reverse reaction? Q. 24 The activation energy of the reaction O3(g) + NO(g) → O2(g) + NO2(g) is 10 kJ/mol. How many times will the rate of the reaction change when the temperature is raised from 27 to 37° C? Q. 25 The Ea of the reaction M + N → O + P is 80 kJ/mol. At 50° C, the products are formed at the rate of 0.15 mol/L/min. What will be rate of formation of products at 100° C? Q. 26 The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the pre-exponential factor for the reaction is 3.56 × 109 s–1, calculate its rate constant at 318 K and also the energy of activation. Q. 27 A substance A was found to undergo two parallel first-order rearrangements A → B and A → C with rate constants 1.26 × 10– 4 s– 1 and 3.8 × 10– 5 s– 1, respectively. What were the percentage distribution of B and C?

Chemistry / Chemical Kinetics

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27

TRY YOURSELF

[ ANSWERS KEY ] TRY YOURSELF - I 1.

(a)

2.

(d)

3.

(b)(c)

4.

(d)

5.

(c)

6.

(d)

7.

16 times

8.

[A]0 = 0.042 mol/l [B]0 = 0.014 mol/l

9.

rate of reaction

zeroth order second order

first order concentration of reactant

10.

rate = k [conc.] for a first order reaction. We can see that the slope is given by the value of k here.

TRY YOURSELF - II 1.

(i) 5.0 × 10– 5 mol L– 1 s– 1

2.

(c)

4.

(i) 5.26 % (ii) 4.62 × 10 5 seconds

5.

3.44 × 10– 3 M min– 1

6.

k = 5.21 × 10– 3 min– 1 = –

Chemistry / Chemical Kinetics

3.

(ii) 4.175 × 10– 5 mol L– 1 s– 1

(c)

( 200 ) 2.303 log (30 min) ( 233.8 )

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28

TRY YOURSELF - III 1.

(A), (D)

2.

very high temperature (or) zero activation energy

3.

(c)

4.

(b), (c)

6.

False

7.

False

8.

False (The rate of reaction increases with increasing temperature because the no. of molecules with Ea increases as the temperature increases. This has nothing to do with the reaction being exothermic/endothermic)

5.

(c)

9.

≈ 311 K

10.

k27° C = 0.0231 min– 1 Ea = 43.85 kJ/mol

11.

227 kJ/mol

12.

(a) Yes (b) No

13.

No

14.

reverse one

15.

5 times

16.

49.9 kJ/mol

17.

80.3 kJ/mol

18.

(b), (d)

19.

(b)

20.

283 K

Chemistry / Chemical Kinetics

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EXERCISE

[ ANSWERS KEY ] 1.

grow by 27 times.

2.

(a) No, (b) Yes

3.

3 × 10– 5, 7.2 × 10– 6

4.

(a) It will grow 27 times (b) It will grow 27 times (c) It will grow 9 times

5.

1, 0, 1.386 minutes

6.

(a) R0 = k[A0] (b) 0.5 sec– 1

7.

(a) (mol/L)– 1/2 s– 1 (b) 2.8

8.

(c)

9.

(a) one, zero (b) 1.386 minutes

10.

Hint: Use mean value theorem (differential calculus).

11.

0.25 M– 1 s– 1

12.

(a) first order

13.

9 seconds

14.

(A), (B), (D)

15.

(i) 1 (ii) 6.93 × 10– 3 min– 1 (iii) 200 min (iv) 950 mmHg

16.

first order related

17.

(a) 47.7 seconds (b) 0.180 atm.

18.

(b)

19.

20.4 minutes.

20.

Ea = 2.39 kJ/mol; T = 669 K

21.

A = 5.45 × 1010 s– 1; Ea = 22012.7 J/mol.

22.

100 kJ/mol

23.

34 kJ/mol.

24.

1.14 times

25. 26.

k Hint: (rate)100° C = (rate)50° C  100° C  k 50° C  Hint

(b) 6.9 × 10– 2 hr– 1

  

k 298 K use in two temperature 2.303 100 2.303 100 gives Ea = ? =t=  → → log log from of Arrhenius eqn to get E a k 298 K 90 k 308 K 75 k 308 K

k318K = ?

use k = k 318 K = 9.43 × 10 −4 s −1

E a = 76.5 kJ / mol Q. 27 76.83% B and 23.17% C Chemistry / Chemical Kinetics

A.e

− Ea / RT

↓ (pre − exp onential factor)