Ch.6 Chemical Kinetics PDF

# Chapter 20: Chemical Kinetics ## Contents - Chemical Kinetics - Reaction Rate - Units of Rate - Rate Laws - Order of a Reaction - Zero Order Reaction - Molecularity of a Reaction - Molecularity versus Order of Reaction - Pseudo-Order Reactions - Integrated rate equations - Zero Order Reactio...

# Chapter 20: Chemical Kinetics ## Contents - Chemical Kinetics - Reaction Rate - Units of Rate - Rate Laws - Order of a Reaction - Zero Order Reaction - Molecularity of a Reaction - Molecularity versus Order of Reaction - Pseudo-Order Reactions - Integrated rate equations - Zero Order Reactions - First Order Reactions - Second Order Reactions - Third Order Reactions - Units of Rate Constant - Half-life of a Reaction - How to determine the order of a Reaction - Collision Theory of Reaction Rates - Effect of increase of temperature on reaction rate - Limitations of the collision theory - Transition State Theory - Activation Energy and Catalysis ## Chapter Summary So far, we have studied equilibrium reactions. In these reactions, the rates of the two opposing reactions are equal and the concentrations of reactants or products do not change with lapse of time. But most chemical reactions are spontaneous reactions that occur from left to right till all reactants are converted to products. For example, the reactions between aqueous sodium chloride and silver nitrate is a fast reaction. The precipitate of AgCl is formed as fast as AgNO3 solution is added to NaCl solution. On the contrary, the rusting of iron is a slow reaction that occurs over the years. The branch of physical chemistry which deals with the rate of reactions is called chemical kinetics. The study of chemical kinetics includes: 1. The rate of the reactions and rate laws. 2. The factors as temperature, pressure, concentration and catalyst, that influence the rate of a reaction. 3. The mechanism or the sequence of steps by which a reaction occurs. The knowledge of the rate of reactions is very valuable to understand the chemical of reactions. It is also of great importance in selecting optimum conditions for an industrial process so that it proceeds at a rate to give maximum yield. ### Reaction Rate The rate of a reaction tells as to what speed the reaction occurs. For example, consider the simple reaction ``` A --> B ``` The concentration of the reactant A decreases and that of B increases as time passes. The rate of reactions is defined as the change in concentration of any of reactant or products per unit time. The rate of reaction may be equal to the rate of disappearance of A which is equal to the rate of appearance of B. ``` rate of reaction= rate of disappearance of A = rate of appearance of B ``` or ``` rate = -d[A]/dt = +d[B]/dt ``` where [] represents the concentration in moles per litre whereas 'd' represents infinitesimally small change in concentration. Negative sign shows the concentration of the reactant A decreases whereas the positive sign indicates the increase in concentration of the product B. ### Units of Rate Reaction rates have the units of concentration divided by time. We express concentrations in moles per litre (mol/litre or mol/l or mol 1-¹) but time may be given in any convenient unit second (s), minutes (min), hours (h), days (d) or possible years. Therefore, the units of reaction rates may be - mole/litre sec or mol 1-¹ s - mole/litre min or mol 1-1 min-1 - mole/litre hour or mol 1-1 h-¹ and, so on ### Average Rate of Reaction is a Function of Time Consider the reaction between carbon monoxide (CO) and nitrogen dioxide. ``` CO(g) +NO₂(g) -->CO₂(g) + NO(g) ``` The average rate of reaction may be expressed as ``` -Δ[CO]/Δt = d[CO]/dt ``` The concentration of CO was found experimentally every 10 seconds. | Conc. of CO | Time (sec) | |---|---| | 0.100 | 0 | | 0.067 | 10 | | 0.050 | 20 | | 0.040 | 30 | | 0.033 | 40 | As the reaction proceeds the concentration of CO decreases rapidly in the initial stages of the reaction. Then the concentration of CO decreases more and more slowly. Obviously the rate of reaction is a function of time. Over the first 10 seconds, the average rate is ``` -d[CO]/dt = (0.067 -0.100)/(10 - 0) = 0.033/10 = 0.0033 moll-¹ g-1 ``` ### Instantaneous Rate of Reaction The average rates obtained by finding the slope of the curve are not always useful. They cover a large time interval during which the rate of reaction changes significantly. A better way to estimate the rate of a reaction is to make the time interval as small as possible. If the interval be infinitesimally small (that is as At approaches zero), the rate is referred to as the instantaneous rate and is written in calculus as ``` rate = d[ ]/dt ``` where [ ], is the concentration at time t. In the present case, ``` instantaneous rate = d [CO]/dt ``` Thus at any time the instantaneous rate is equal to the slope of a straight line drawn tangent to the curve at that time. For example, the instantaneous rate at 10 seconds is found to be 0.0022 mol 1-1 s-1. ### Rate Laws At a fixed temperature the rate of a given reaction depends on concentration of reactants. The exact relation between concentration and rate is determined by measuring the reaction rate with different initial reactant concentrations. By a study of numerous reactions it is shown that: the rate of a reaction is directly proportional to the reactant concentrations, each concentration being raised to some power. ``` rate = k[A]^n ``` For a reaction ``` 2A +B -->products ``` The reaction rate with respect to A or B is determined by varying the concentration of one reactant, keeping that of the other constant. Thus the rate of reaction may be expressed as ``` rate = k[A]^m[B]^n ``` An expression which shows how the reaction rate is related to concentrations is called the rate law or rate equation. The power (exponent) of concentration n or m in the rate law is usually a small whole number integer (1, 2, 3) or fractional. The proportionality constant k is called the rate constant for the reaction. ### Examples of rate law | Reactions | Rate Law | |---|---| | 2N2O5 --> 4NO2 + O2 | rate = k[N2O5] | | H2+I₂ --> 2HI | rate = k[H2][I2] | | 2NO2 --> 2NO + O2 | rate = k[NO₂]2 | | 2NO +2H₂ --> N2+2H2O | rate = k[H2][NO]2 | In these rate laws where the quotient or concentration is not shown, it is understood to be 1. That is [H2] = [H2]. It is apparent that the rate law for a reaction must be determined by experiment. It cannot be written by merely looking at the equation with a background of our knowledge of Law of Mass Action. However, for some elementary reactions the powers in the rate law may correspond to coefficients in the chemical equation. But usually the powers of concentration in the rate law are different from coefficients. Thus for the reaction (4) above, the rate is found to be proportional to [H2] although the quotient of H₂ in the equation is 2. For NO the rate is proportional to [NO]² and power '2' corresponds to the coefficient. ### Order of a Reaction The order of a reaction is defined as the sum of the powers of concentrations in the rate law. Let us consider the example of a reaction which has the rate law ``` rate = k [A]^m[B]^n ``` The order of such a reaction is (m + n). The order of a reaction can also be defined with respect to a single reactant. Thus the reaction order with respect to A is m and with respect to B it is n. The overall order of reaction (m + n) may range from 1 to 3 and can be fractional. ### Examples of reaction order | Rate Law | Reaction Order | |---|---| | rate = k [N₂O₃] | 1 | | rate = k [H2] [12] | 1+1=2 | | rate = k [NO₂]2 | 2 | | rate = k [[H2] [NO]2 | 1+2=3 | | rate = k [CHC13] [C12]1/2 | 1 +1/2 = 1 1/2 | Reactions may be classified according to the order. If in the rate law (1) above - m+n = 1, it is first order reaction - m+ n = 2, it is second order reaction - m+n = 3, it is third order reaction ### Zero Order Reaction A reactant whose concentration does not affect the reaction rate is not included in the rate law. In effect, the concentration of such a reactant has the power 0. Thus [A] = 1. A zero order reaction is one whose rate is independent of concentration. For example, the rate law for the reaction ``` NO₂+CO -->NO+CO2 ``` at 200° C is ``` rate = k [NO₂]2 ``` Here the rate does not depend on [CO], so this is not included in the rate law and the power of [CO] is understood to be zero. The reaction is zeroth order with respect to CO. The reaction is second order with respect to [NO2]. The overall reaction order is 2 + 0 = 2. ### Molecularity of a Reaction Chemical reactions may be classed into two types : (a) Elementary reactions (b) Complex reactions An elementary reaction is a simple reaction which occurs in a single step. A complex reaction is that which occurs in two or more steps. #### Molecularity of an Elementary Reaction The molecularity of an elementary reaction is defined as: the number of reactant molecules involved in a reaction. - **Unimolecular** - **Bimolecular** - **Termolecular** Thus the molecularity of an elementary reaction is 1, 2, 3, etc., according as one, two or three reactant molecules are participating in the reaction. The elementary reactions having molecularity 1, 2 and 3 are called unimolecular, bimolecular and termolecular respectively. #### Unimolecular reactions - A --> product - Br₂ --> 2Br - H-C-COOH --> COOH-C-H #### Bimolecular reactions - A+B -->products - A+A -->products - CH3COOC2H5 + H₂O --> CH3COOH + C2H5OH #### Termolecular reactions - A+B+C --> products - 2NO +0₂ --> 2NO₂ - 2NO + Cl2 --> 2NOCI Most of the reactions involve one, two or at the most three molecules. The reactions involving four or more molecules are very rare. The rarity of reactions with high molecularity can be explained on the basis of the kinetic molecular theory. According to this theory, the rate of a chemical reaction is proportional to the number of collisions taking place between the reacting molecules. The chances of simultaneous collision of reacting molecules will go on decreasing with increase in number of molecules. Thus the possibility of three molecules colliding together is much less than in case of bimolecular collision. For a reaction of molecularity 4, the four molecules must come closed and collide with one another at the same time. The possibility of their doing so is much less than even in the case of termolecular reaction. Hence the reactions involving many molecules proceed through a series of steps, each involving two or three or less number of molecules. Such a reaction is called a complex reaction and the slowest step determines the overall rate of the reactions. #### Molecularity of a Complex Reaction Most chemical reactions are complex reactions. These occur in a series of steps. Each step is an elementary reaction. The stepwise sequence of elementary reactions that convert reactions to products is called the mechanism of the reaction. In any mechanism, some of the steps will be fast, others will be slow. A reaction can proceed no faster than its slowest step. Thus the slowest step is the rate-determining step of the reaction. For example, the decomposition of N2O5 ``` 2 N₂O₅ --> 4NO₂ + O₂ ``` is an example of a complex reaction. It occurs by the following steps: - **Step 1** 2N2O5 --> 2NO2 + 2NO3 (slow) - **Step 2** NO2 + NO3 --> NO + NO2 + O2 (slow) - **Step 3** NO + NO3 --> 2NO2 (fast) Each elementary reaction has its own molecularity equal to the number of molecules or atoms participating in it. It is meaningless to give the molecularity of the overall reaction because it is made of several elementary reactions, each, perhaps with a different molecularity. At best could be thought of as: the number of molecules or atoms taking part in the rate-determining step. Thus step 2 in the above mechanism is rate-determining and has molecularity '2' which could be considered as the molecularity of the decomposition reaction of N2O5. ### Molecularity versus Order of Reaction The term molecularity is often confused with order of a reaction. The total number of molecules or atoms which take part in a reaction as represented by the chemical equation, is known as the molecularity of reaction. The sum of the powers to which the concentrations are raised in the rate law is known as the order of reaction. #### Molecularity and Order are Identical for Elementary Reactions or Steps The rate of an elementary reaction is proportional to the number of collisions between molecules (or atoms) of reactions. The number of collisions in turn is proportional to the concentration of each reactant molecule (or atom). Thus for a reaction. ``` 2A+B -->products rate = k [A]² [B] ``` Two molecules of A and one molecule of B are participating in the reaction and, therefore, molecularity of the reaction is 2 + 1 = 3. The sum of powers in the rate law is 2 + 1 and hence the reaction order is also 3. Thus the molecularity and order for an elementary reaction are equal. | Reactions | Molecularity | Rate law | Order | |---|---|---|---| | A --> products | 1 | rate = k[A] | 1 | | A+A --> products | 2 | rate = k[A]2 | 2 | | A+B --> products | 2 | rate = k[A][B] | 2 | | A+2B --> products | 3 | rate = k[A][B]2 | 3 | | A+B+C --> products | 3 | rate = k[A][B][C] | 3 | ### Pseudo-Order Reactions A reaction in which one of the reactants is present in a large excess shows an order different from the actual order. The experimental order which is not the actual one is referred to as the pseudo order. Since for elementary reactions molecularity and order are identical, pseudo-order reactions may also be called pseudo molecular reactions. Let us consider a reaction ``` A + B -->products ``` in which the reactant B is present in a large excess. Since it is an elementary reaction, its rate law can be written as ``` rate = k [A] [B] ``` As B is present in large excess, its concentration remains practically constant in the course of reaction. Thus the rate law can be written as ``` rate = k' [A] ``` where the new rate constant k' = k [B]. Thus the actual order of the reaction is second-order but in practice it will be first-order. Therefore, the reaction is said to have a pseudo-first order. ### Examples of Pseudo-order Reactions 1. **Hydrolysis of an ester.** For example, ethyl acetate upon hydrolysis in aqueous solution using a mineral acid as catalyst forms acetic acid and ethyl alcohol. ``` CH3COOC2H5 + H₂O --> CH3COOH + C2H5OH ``` Here a large excess of water is used and the rate law can be written as ``` rate = k[CH3COOH] [H₂O] = K [CH3COOH] ``` The reaction is actually second-order but in practice it is found to be first-order. Thus it is a pseudo-first order reaction. 2. **Hydrolysis of sucrose.** Sucrose upon hydrolysis in the presence of a dilute mineral acid gives glucose and fructose. ``` C12H22O11 + H2O --> C6H12O6 + C6H12O6 ``` If a large excess of water is present, [H₂O] is practically constant and the rate law may be written as ``` rate = k [C12H22O11] [H₂O] = k [C12H22O11] ``` The reaction though of second-order is experimentally found to be first-order. Thus it is as pseudo-first-order reaction. ### Zero Order Reactions In a zero order reaction, rate is independent of the concentration of the reactions. Consider a zero-order reaction of the type | Initial Conc. | A --> Products | |---|---| | a | 0 | | a-x | x | The rate of reaction can be written as: ``` Rate of reaction = d[A]/dt = ko [A]° ``` or ``` dx/dt = -d (a - x)/dt = ko (a - x)° = ko ``` On integrating we get ``` x = kot ``` where k is the rate constant of a zero-order reaction, the unit of which is concentration per unit time. In zero order reaction, the rate constant is equal to the rate of reaction at all concentrations. ### First Order Reactions Consider a first order reaction ``` A --> products ``` Suppose that at the beginning of the reaction (t = 0), the concentration of A is a moles litre-¹. If after time t, x moles of A have changed, the concentration of A is a - x. We know that for a first order reaction, the rate of reaction, dx/dt, is directly proportional to the concentration of the reactant. Thus, ``` dx/dt = k (a - x) ``` or ``` dx/(a-x) = k dt ``` Integration of the expression (1) gives ``` ∫ dx/(a-x) = ∫ k dt ``` or ``` - 1n (a-x) = kt + I ``` where I is the constant of integration. The constant k may be evaluated by putting t = 0 and x = 0. Thus, ``` I = - In a ``` Substituting for I in equation (2) ``` In(a/(a-x)) = kt ``` or ``` k =-1/t ln(a/(a-x)) ``` Changing into common logarithms ``` k = 2.303/t log (a/(a-x)) ``` The value of k can be found by substituting the values of a and (a-x) determined experimentally at time interval t during the course of the reaction. Sometimes the integrated rate law in the following form is also used: ``` k = 2.303/(t₂-t₁) log ((a - x₁)/(a - x₂)) ``` where x₁ and x₂ are the amounts decomposed at time intervals t₁ and t₂ respectively from the start. ### Examples of First order Reactions - Decomposition of N2O5 in CCl4 solution: Nitrogen pentoxide in carbon tetrachloride solution decomposes to form oxygen gas. ``` N₂O₅ --> 2NO₂ + O₂ ``` The reaction is carried in an apparatus shown in . The progress of the reaction is monitored by measuring the volume of oxygen evolved from time to time. - Decomposition of H2O2 in aqueous solution: The decomposition of H2O2 in the presence of Pt as catalyst is a first order reaction. ``` H2O2 --> H2O + O2 ``` The progress of the reaction is followed by titrating equal volumes of the reaction mixture against standard KMnO4 solution at different time intervals. A solution of H₂O₂ when titrated against KMnO4 solution at different time intervals gave the following results: | t (minutes) | Vol KMnO₁ used for 10 ml H2SO4 | |---|---| | 0 | 23.8 ml | | 10 | 14.7 ml | | 20 | 9.1 ml | Show that the decomposition of H₂O₂ is a first order reaction. ### Solved Problem From the following data for the decomposition of N₂O₃ in CCl4 solution at 48°C, show that the reaction is of the first order | t (mts) | Vol of O₂ evolved | |---|---| | 10 | 6.30 | | 15 | 8.95 | | 20 | 11.40 | | 34.75 | | For a first order reaction the integrated rate equation is ``` 1/t log(V/(V-Vt)) = k ``` In this example, V = 34.75 | t | V-Vt | 1/t log(V/(V-Vt)) | |---|---|---| | 10 | 28.45 | 0.00868 | | 15 | 25.80 | 0.00862 | | 20 | 23.35 | 0.00863 | Since the value of k is fairly constant, it is a first order reaction. ### Second Order Reactions Let us take a second order reaction of the type ``` 2A --> products ``` Suppose the initial concentration of A is a moles litre-¹. If after time t, x moles of A have reacted, the concentration of A is (a – x). We know that for such a second order reaction, rate of reaction is proportional to the square of the concentration of the reactant. Thus, ``` dx/dt = k (a – x)² ``` where k is the rate constant, Rearranging equation (1), we have ``` dx/(a-x)² = k dt ``` On integration, it gives ``` 1/(a-x) = kt + I ``` where I is is integration constant. I can be evaluated by putting x = 0 and t = 0. Thus, ``` 1/a = I ``` Substituting for I in equation (3) ``` 1/(a-x) = kt + 1/a ``` ``` kt = 1/(a-x) - 1/a ``` ``` k = 1/(ta(a - x)) ``` This is the integrated rate equation for a second order reaction. ### Examples of Second order Reaction Hydrolysis of an Ester by NaOH. This is typical second order reaction. ``` CH3COOC2H5 + NaOH --> CH3COONa + C2H5OH ``` The reaction is carried in a vessel at a constant temperature by taking equimolar amounts of ethyl acetate and NaOH. Measured volumes of the reaction mixture (say, 25 ml) are withdrawn at various times and titrated against a standard acid. The volume of the acid used is a measure of the concentration of NaOH or ester. Thus the volume of the acid used when t = 0, gives the initial concentration (a) of the reactants. The volume of acid consumed at any other time t gives (a – x). The value of x can be calculated. The rate constant k can be determined by substituting values in the second order integrated rate equation. ### Solved Problem Hydrolysis of ethyl acetate by NaOH using equal concentration of the reactants, was studied by titrating 25ml of the reaction mixture at different time intervals against standard acid. From the data given below, establish that this is a second order reaction. | 1 (mts) | ml acid used | |---|---| | 0 | 16.00 | | 5 | 10.24 | | 15 | 6.13 | | 25 | 4.32 | The second order integrated rate equation is ``` k = 1/(at a(a - x)) ``` The volume of acid used at any time is a measure of concentration of the unreacted substances at that time. Therefore, a, initial concentration = 16.00 after 5 mts (a-x) = 10.24 and x = 5.76 after 15 mts (a-x) = 6.13 and x = 9.85 after 25 mts (a-x) = 4.32 and x = 11.68 Substituting values in the rate equation (1), we have - k = 1/(16 x 5 x 10.24) = 0.0070 - k = 1/(16 x 15 x 6.13) = 0.0067 - k = 1/(16 x 25 x 4.32) = 0.00675 The values of are fairly constant, this reaction is of the second order. ### Third Order Reactions Let us consider a simple third order reaction of the type ``` 3A --> products ``` Let the initial concentration of A be a moles litre-¹ and after time t, x, moles have reacted. Therefore, the concentration of A becomes (a - x). The rate law may be written as : ``` dx/dt = k(a – x)³ ``` Rearranging equation (1), we have ``` dx/(a - x)³ = k dt ``` On integration, it gives ``` 1/2(a - x)² = kt + I ``` where I is the integration constant. I can be evaluated by putting x = 0 and t = 0. Thus, ``` I = 1/2a² ``` By substituting the value of I in (3), we can write ``` kt = 1/2(a - x)² - 1/2a² ``` Therefore, ``` k = 1/t 2a²(a-x)² × (2a - x) ``` This is the integrated rate equation for a third order reaction. ### Examples of Third order Reactions There are not many reactions showing third order kinetics. A few of the known examples are: - 2FeCl3(aq) + SnCl2(aq) → 2FeCl2 + SnCl4 - 2NO(g)+O2(g) — 2NO2(g) - 2NO(g)+Cl2(g) → 2NOCl(g) ### Units of Rate Constant The units of rate constant for different orders of reactions are different. #### Units of Zero order Rate constant For a zero order reaction, the rate constant k is given by the expression ``` k = d[A]/dt = mol/litre × 1/time ``` Thus the units of k are ``` mol 1-¹ time-1 ``` Time may be given in seconds, minutes, days or years. #### Units of First order Rate constant The rate constant of a first order reaction is given by ``` k = 2.303/t log [A]o/[A]t ``` Thus the rate constant for the first order reaction is independent of the concentration. It has the unit ``` time-1 ``` #### Units of Second order Rate constant The rate constant for a second order reaction is expressed as ``` k = 1/t × 1/( [A]o([A]t - x) ``` or ``` k = 1/ (concentration x concentration x time) = 1/ (concentration x time) ``` or ``` k = 1/(mole/litre x time) = mol-¹ 1 time-1 ``` Thus the units for k for a second order reactions are ``` mol-¹ 1 time-1 ``` #### Units of Third order Rate constant The rate constant for a third order reaction is ``` k = 1/t 2a²(a-x)² × (2a - x) ``` or ``` k = 1/ (concentration x concentration x (concentration)² x time) = 1/ (concentration)² x time ``` or ``` k = 1/(mol/litre)² x time = mol-2 12 time-1 ``` Thus the units of k for third order reaction are ``` mol-2 12 time-1 ``` ### Half-life of a Reaction Reaction rates can also be expressed in terms of half-life or half-life period. It is defined as: the time required for the concentration of a reactant to decrease to half its initial value. In other words, half-life is the time required for one-half of the reaction to be completed. It is represented by the symbol 11/2 or 10.5 ### Calculation of Half-life of a First order Reaction The integrated rate equation (4) for a first order reaction can be stated as: ``` k = 2.303/t log [A]o/[A] ``` where [A]o is initial concentration and [A] is concentration at any time t. Half-life, 11/2, is time when initial concentration reduces to 1/2 i.e., Substituting values in the integrated rate equation, we have ``` k = 2.303/t1/2 log [A]o/(1/2[A]o) = 2.303/t1/2 log 2 ``` or ``` t 1/2 = 2.303/k log 2 = 0.3010/k ``` or ``` t1/2 = 0.693/k ``` It is clear from this relation that: 1. half-life for a first order reaction is independent of the initial concentration. 2. it is inversely proportional ot k, the rate-constant. ### Calculation of Time for Completing any Fraction of the Reaction As for half-change, we can calculate the time required for completion of any fraction of the reaction. For illustration, let us calculate the time in which two-third of the reaction is completed. First order integrated rate equation is ``` t = 2.303/k log [A]o/[A] ``` Here, the initial concentration has 2/3 reacted reducing it to 1/3. Thus, ``` [A] = [A]o/3 ``` Substituting values in the rate equation ``` t1/3 = 2.303/k log [A]o / (1/3[A]o) = 2.303/k log 3 ``` ``` t1/3 = 2.303/k × 0.4771 ``` ### Solved Problem Compound A decomposes to form B and C the reaction is first order. At 25°C the rate constant for the reaction is 0.450 s¯¹. What is the half-life of A at 25°C ? SOLUTION We know that for a first order reaction, half-life t1/2, is given by the expression Substituting the value of k = 0.450 s¯¹, we have ``` t1/2 = 0.693/0.450s-1 = 1.54 s ``` Thus half-life of the reaction A→B + C is 1.54 seconds. ### Solved problem The half-life of a substance in a first order reaction is 15 minutes. Calculate the rate constant. SOLUTION For a first order reaction Putting t1/2 = 15 min in the expression and solving for k, we have ``` k = 0.693/t1/2 = 0.693/15 min = 4.62 × 10-2 min-1 ``` ### Solved problem For the reaction 2N2O5 → 4NO2 + O2, the rate is directly proportional to [N2O5]. At 45°C, 90% of the N2O5 reacts in 3600 seconds. Find the value of the rate constant k. SOLUTION Since rate is < [N2O5] it is first order reaction. The integrated rate equation is ``` k = 2.303/t log [N2O5]o/[N2O5]t ``` When 90% of N₂O, has reacted, the initial concentration is reduced to 1/10. That is, ``` [N2O5] = 1/10 [N2O5]o ``` Substituting values in the rate equation, ``` k = 2.303/3600 log [N2O5]o / (1/10 [N2O5]o) ``` ``` k = 2.303/3600 log 10 = 2.303/3600 × 1 ``` ``` k = 6.40 × 10 s-1 ``` ### Solved Problem The rate law for the decomposition of N2O5 (1) is: rate = k [N₂O₃] where k = 6.22 × 104 sec-1. Calculate half-life of N₂O₃ (1) and the number of seconds it will take for an initial concentration of N₂O₃ (1) of 0.100 M to drop to 0.0100 M. SOLUTION Calculation of half-life ``` t1/2 = 0.693/k = 0.693/6.22 x 104 sec-1 = 1.11 x 103 sec ``` Calculation of time in seconds for drop of [N₂O₃] from 0.100 M to 0.0100 M From first order integrated rate equation, ``` t = 2.303/k log [N2O5]o/[N2O5]t ``` or Substituting values ``` t = 2.303/6.22 x 104 sec-1 log 0.100/0.0100 ``` ``` t= 2.303/6.22 x 104 sec-1 x1 ``` ``` t= 3.70 x 103 sec ``` ### Solved Problem For a certain first order reaction t0.5 is 100 sec. How long will it take for the reaction to be completed 75%? SOLUTION Calculation of k For a first order reaction ``` t 1/2 = 0.693/k ``` or ``` 100 = 0.693/k ``` ``` k = 0.693/100 = 0.00693 sec-1 ``` Calculation of time for 75% completion of reaction The integrated rate equation for a first order reaction is ``` t = 2.303/k log [A]o / [A] ``` or ``` t = 2.303/k log [A]o/ [A] ``` When 3/4 initial concentration has reacted, it is reduced to 1/4. Substituting values in the rate equation ``` t3/4 = 2.303/ 0.00693 log [A]o / (1/4 [A]o) ``` ``` t3/4 = 2.