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Chemical Kinetics, EXERCISE [PAGES 135 - 137], , Solution:, The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles, if [A] is doubled and [B] is kept constant, Exercise | Q 1.02 | Page 135, Choose the most correct option., The order of the reaction for which the units of the rate constant are mol dm-3 s-1 is, _______., 1. 1, 2. 3, 3. 0, 4. 2, Solution:, The order of the reaction for which the units of the rate constant are mol dm-3 s-1 is 0.
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Solution:, , 1., 2., 3., 4., , Exercise | Q 1.04 | Page 135, Choose the most correct option., The time required for 90 % completion of a certain first-order reaction is t. The time, required for 99.9 % completion will be _________., t, 2t, t/2, 3t, Solution:, The time required for 90 % completion of a certain first-order reaction is t. The time, required for 99.9 % completion will be 3t ., , Solution:, The slope of the graph ln[A]t versus t for the first-order reaction is -k., Exercise | Q 1.06 | Page 135, Choose the most correct option.
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1., 2., 3., 4., , What is the half-life of a first-order reaction if the time required to decrease the, concentration of reactant from 0.8 M to 0.2 M is 12 h?, 12 h, 3h, 1.5 h, 6h, Solution: 6 h, , Solution:, , Exercise | Q 1.08 | Page 135, Choose the most correct option., 1., 2., 3., 4., , unimolecular and second order, bimolecular and first order, bimolecular and second order, unimolecular and first order, Solution:, Exercise | Q 1.09 | Page 135, Choose the most correct option.
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1., 2., 3., 4., , 1., 2., 3., 4., , increase in temperature, increase of concentration of NO, increase of concentration of Cl2, increase of concentrations of both Cl2 and NO, Solution:, , Exercise | Q 1.1 | Page 135, Choose the most correct option., For an endothermic reaction, X ⇌ Y. If Ef is the activation energy of the forward reaction, and Er that for the reverse reaction, which of the following is correct?, Ef = E r, Ef < E r, Ef > E r, ∆H = Ef - Er is negative, Solution: Ef > Er, , Solution:
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Solution:, , Exercise | Q 2.03 | Page 136, Answer the following in one or two sentences., What is the relationship between coefficients of reactants in a balanced equation for an, overall reaction and exponents in the rate law? In what case the coefficients are the, exponents?, Solution:, Coefficients of reactants in a balanced chemical equation may or may not be the same, as the exponents in rate law for the same reaction. For elementary reaction, coefficients, in a balanced chemical equation are the same as the exponents in the rate law., Exercise | Q 2.04 | Page 136, Answer the following in one or two sentences., Why all collisions between reactant molecules do not lead to a chemical reaction?, Solution:, All collisions of reactant molecules do not lead to a chemical reaction because the, colliding molecules need to possess certain energy which is greater than the activation, energy Ea and proper orientation., Exercise | Q 2.05 | Page 136, Answer the following in one or two sentences, What is the activation energy of a reaction?, Solution:, For the reaction to occur the colliding reactant molecules must possess the minimum, kinetic energy. This minimum kinetic energy is the activation energy., Exercise | Q 2.06 | Page 136, Answer the following in one or two sentences.
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What are the units for rate constants for zero-order and second-order reactions if time is, expressed in seconds and concentration of reactants in mol/L?, Solution:, Units of rate constant, i. Zero-order reaction is mol L-1 s-1, ii. The second-order reaction is mol-1 L s-1, Exercise | Q 2.07 | Page 136, Answer the following in one or two sentences., Write the Arrhenius equation and explain the terms involved in it., Solution:, , where k is the rate constant, Ea is the activation energy, R molar gas constant, T, temperature in kelvin and A is the pre-exponential factor., Exercise | Q 2.08 | Page 136, Answer the following in one or two sentences., What is the rate-determining step?, Solution:, When a chemical reaction occurs in a series of steps, one of the steps is slower than all, other steps. Such a slowest step in the reaction is called a rate-determining step., Exercise | Q 2.09 | Page 136, Answer the following in one or two sentences., Write the relationships between rate constant and half-life of the first order and zerothorder reactions., Solution:, , Exercise | Q 2.1 | Page 136, Answer the following in one or two sentences.
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How do the half-lives of the first order and zero-order reactions change with the initial, concentration of reactants?, Solution:, i. The half-life of the first-order reaction is independent of the initial concentration of the, reactant., ii. The half-life of zero-order reactions is proportional to the initial concentration of the, reactant., Exercise | Q 3.01 | Page 136, Answer the following in brief., How instantaneous rate of reaction is determined?, Solution:, i. To determine the instantaneous rate of a reaction the progress of a reaction is, followed by measuring the concentrations of a reactant or product for different time, intervals., ii. The concentration of a reactant or a product is plotted against time., iii. A tangent drawn to the curve at time t1 gives the rate of the reaction. The slope thus, obtained gives the instantaneous rate of the reaction at time t 1., , Determination of instantaneous rate
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iv. Mathematically the instantaneous rate is expressed by replacing Δ by derivative dc/dt, in the expression of the average rate., , Exercise | Q 3.02 | Page 136, Answer the following in brief., Distinguish between order and molecularity of a reaction., Solution:, Order, , Molecularity, , i., , It is an experimentally determined property., , It is a theoretical entity., , ii., , It is the sum of powers of the concentration, terms of reactants that appear in the rate, equation., It may be an integer, fraction, or zero., , It is the number of reactant, molecules taking part in an, elementary reaction., It is an integer., , iii., iv., , It is not based on a balanced chemical, equation., , It is based on a balanced, chemical equation.
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Solution:, , ii. Since, NOCl2 is formed in the first step and consumed in the second step. Hence, it is, the reaction intermediate., iii. The molecularity of each step is 2 because two reactants are involved in each of the, steps., Exercise | Q 3.04 | Page 136, Answer the following in brief., Obtain the relationship between the rate constant and half-life of a first-order reaction., Solution:, , Exercise | Q 3.05 | Page 136
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Answer the following in brief., How will you represent the zeroth-order reaction graphically?, Solution:, , Exercise | Q 3.06 | Page 136, Answer the following in brief., What are pseudo-first-order reactions? Give one example and explain why it is pseudofirst-order., Solution:, i. Certain reactions that are expected to be of higher order follow the first-order kinetics., Such reactions are said to be pseudo-first-order reactions., ii. Example: Consider the hydrolysis of methyl acetate., , The rate law is rate = k' [CH3COOCH3][H2O]
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iii. Explanation: The reaction was expected to follow the second-order kinetics,, however, obeys the first order. The reason is that solvent water is present in such large, excess that the change in its concentration is negligible compared to the initial one or its, concentration remains constant. Thus [H2O] = constant = k''. The rate law becomes rate, = k' [CH3COOCH3]k'' = k [CH3COOCH3], where, k = k'k'', The reaction is thus of first order., Exercise | Q 3.07 | Page 136, Answer the following in brief., What are the requirements for the colliding reactant molecules to lead to products?, Solution:, The requirements for colliding molecules to form the product are as follows:, i. Activation energy: For the reaction to occur the colliding reactant molecules must, possess the minimum kinetic energy. This minimum kinetic energy is the activation, energy. The reaction would occur only if colliding molecules possess kinetic energies, equal to or greater than the activation energy., ii. The orientation of reactant molecules:, a. For reactions involving complex molecules or ions, reactant molecules must collide, with proper orientation. The molecules need to be so oriented relative to each other that, the reacting groups approach closely., b. Consider, A + C - B → A - B + C, The collision of A with C approaching toward A would not lead to a reaction., , The reactant molecules would collide and separate owing to the improper orientation of, C – B., c. The reaction is successful as a result of proper orientation of C – B. A fraction of such, collisions brings forth the conversion of reactants to products., , iii. Potential energy barrier:, Consider the reaction A + C – B → A – B + C
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a. During a course of the collision, new bond A – B develops, and at the same time, bond C – B breaks. An activated complex is formed in which all three atoms are weakly, connected together. A + C – B → A-----B-----C → A – B + C, b. To attain the configuration A-----B-----C atoms need to gain energy, which comes, from the kinetic energy of colliding molecules. The energy barrier between reactants, and products is as shown in the diagram., , Potential energy barrier, c. To form an activated complex, the reactant molecules need to climb up and, overcome the energy barrier before they get converted to products., d. The height of the barrier is called activation energy (Ea). Thus, the reactant molecules, transform into products only if they possess energy equal to or greater than such, activation energy. As a result, only a few collisions lead to products., Exercise | Q 3.08 | Page 136, Answer the following in brief., How a catalyst increases the rate of reaction? Explain with the help of a potential, energy diagram for catalyzed and uncatalyzed reactions., Solution:, i. A catalyst provides an alternative pathway associated with lower activation energy.
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Potential energy barriers for catalyzed and uncatalyzed reactions, ii. The potential energy diagram compares the potential energy barriers for the, catalysed and uncatalysed reactions. The barrier for uncatalysed reaction (Ea) is larger, than that for the same reaction in the presence of a catalyst E a., , At room temperature, the rate of reaction is slower in the absence of a catalyst with its, activation energy being 76 kJ mol-1. In the presence of catalyst, iodide ion (I- ), the, reaction is faster since the activation energy decreases to 57 kJ mol-1., Exercise | Q 3.09 | Page 136, Answer the following in brief., Explain with the help of the Arrhenius equation, how do the rate of reaction changes, with (a) temperature and (b) activation energy., Solution:, , Exercise | Q 3.1 | Page 137, Answer the following in brief., Derive the integrated rate law for the first-order reaction., Solution:
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Exercise | Q 3.11 | Page 137, Answer the following in brief., How will you represent first-order reactions graphically?, Solution:, , Variation of rate with [A]
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Exercise | Q 3.12 | Page 137, Answer the following in brief., Derive the integrated rate law for the first-order reaction,, , Solution:, i. For the gas phase reaction,, AX(g)⟶BX(g)+CX(g), Let initial pressure of A be Pi that decreases by x within time t., ii. Pressure of reactant A at time t, PA = Pi - x ....(1), The pressures of the products B and C at time t, PB = P C = x, iii. The total pressure at time t is then, P = Pi - x + x + x = Pi + x
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Hence, x = P - Pi …(2), The pressure of A at time t is obtained by substitution of Eq. (1) into Eq. (2)., Thus, PA = Pi - (P - Pi) = Pi - P + Pi = 2Pi - P, iv. The integrated rate law turns out to be, , Exercise | Q 3.13 | Page 137, Answer the following in brief., What is a zeroth-order reaction? Derive its integrated rate law. What are the units of the, rate constant?, Solution:, i. Reactions in which the rate is independent of the reactant concentration are zerothorder reaction., ii. The integrated rate law for zero-order reactions:, For zero-order reaction, A → P, the differential rate law is given by
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or [A]t - [A]0 = -kt, Hence, kt = [A]0 - [A]t ....(2), iii. The unit of rate constant for the zero order reaction is mol dm -3 s-1 ., Exercise | Q 3.14 | Page 137, Answer the following in brief., How will you determine activation energy:, (a) graphically using Arrhenius equation, (b) from rate constants at two different temperatures?, Solution:
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Variation of log10k with 1/T, ii., Determination of activation energy from rate constants at two different, temperatures.
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Exercise | Q 3.15 | Page 137, Answer the following in brief., Explain graphically the effect of temperature on the rate of reaction., Solution:, i. The average kinetic energy of molecules is proportional to temperature., ii. At a given temperature, the fraction of molecules with their kinetic energy equal to or, greater than Ea may lead to the product., iii. With an increase in temperature, the fraction of molecules having their energies (E a), would increases. The rate of the reaction thus would increase., iv. This is depicted by plotting a fraction of molecules with given kinetic energy versus, kinetic energy for two different temperatures T1 and T2 (T2 > T1). The area between the, curve and the horizontal axis is proportional to the total number of molecules. The total, area is the same at T1 and T2., , Comparison of fraction of molecules activated at T1 and T2, v. The entire shaded area which represents a fraction of molecules with kinetic energy, exceeding Ea is larger at T2 than at T1 (since T2 > T1)., vi. This indicates that a fraction of molecules possessing energies larger than, Ea increase with temperature. The rate of reaction increases accordingly., Exercise | Q 3.16 | Page 137, Answer the following in brief., Explain graphically the effect of a catalyst on the rate of reaction., Solution:
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i. Diagram:, , Comparison of fraction of molecules for catalyzed and uncatalyzed reactions, ii. A catalyst lowers the threshold energy., iii. Consequently, more molecules acquire the minimum amount of energy and tend to, cross the energy barrier., iv. A fraction of activated molecules is greater for the catalyzed reaction. The rate of, catalysed reaction thus is larger than the reaction with no catalyst., Exercise | Q 3.17 | Page 137, Answer the following in brief., For the reaction 2A + B → products, find the rate law from the following data., [A]/M, [B]/M, rate/M s-1, 0.3, , 0.05, , 0.15, , 0.6, , 0.05, , 0.30, , 0.6, , 0.2, , 1.20, , Solution:, From above observations (i) and (ii), 0.15 = (0.3)x (0.05)y ......(i), 0.30 = (0.6)x (0.05)y .......(ii), Dividing (2) by (1)
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Exercise | Q 4.1 | Page 137, Solve, In a first-order reaction, the concentration of the reactant decreases from 20 mmol dm 3, , to 8 mmol dm-3 in 38 minutes. What is the half-life of reaction?, , Solution:, Given:, [A]0 = 20 mmol dm-3 , [A]t = 8 mmol dm-3 , t = 38 min, To find:, Half life of reaction t1/2
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Exercise | Q 4.2 | Page 137, Solve, The half-life of a first-order reaction is 1.7 hours. How long will it take for 20% of the, reactant to react?, Solution:, Given:, Half-life t1/2 = 1.7 hours, [A]0 = 100%, [A]t = 100 - 20 = 80%, To find:, Time for 20% of reactant to react = t, Formulae:
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Exercise | Q 4.3 | Page 137, Solve, The energy of activation for a first-order reaction is 104 kJ/mol. The rate constant at, 25°C is 3.7 × 10–5 s –1. What is the rate constant at 30°C? (R = 8.314 J/K mol), Solution:
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Exercise | Q 4.4 | Page 137, Solve, What is the energy of activation of a reaction whose rate constant doubles when the, temperature changes from 303 K to 313 K?, Solution:, Given:, Rate constants; k2 = 2k1,, Temperatures: T1 = 303 K, T2 = 313 K, To find:, Activation energy of the reaction (Ea), Formula:
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Exercise | Q 4.5 | Page 137, Solve, The rate constant of a reaction at 500°C is 1.6 × 103 M–1 s –1. What is the frequency, factor of the reaction if its activation energy is 56 kJ/mol?, Solution:
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Exercise | Q 4.6 | Page 137, Solve, Show that time required for 99.9% completion of a first-order reaction is three times the, time required for 90% completion., Solution:, For a first-order reaction,
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ii., Time is taken for 90% completion:, Let the time taken for 90% completion of the reaction be t90%., Let initial concentration, [A]0 = a, Then, final concentration, [A]t = a - 90% of a
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Therefore, for a first-order reaction, the time required for 99.9% completion is 3 times, that required for 90% completion., Exercise | Q 4.7 | Page 137, Solve, A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life., Solution:, Given:, [A]0 = 100%, [A]t = 100 - 30 = 70%, t = 40 min, To find:, Half life of reaction (t1/2), Formula:
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Exercise | Q 4.8 | Page 137, Solve, The rate constant for the first-order reaction is given by log10 k = 14.34 – 1.25 × 104 T., Calculate activation energy of the reaction., Solution:, , ∴ Ea = 1.25 × 104 × 2.303 × 8.314 = 239339 K mol-1 = 239.3 kJ mol-1, The energy of activation of the reaction is 239.3 kJ mol-1, Exercise | Q 4.9 | Page 137, Solve, What fraction of molecules in a gas at 300 K collide with an energy equal to the, activation energy of 50 kJ/mol?
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Solution:, Given:, Activation energy (Ea) = 50 kJ mol-1 = 50 × 103 J mol-1, Temperature (T) = 300 K, To find:, Fraction of molecule (f) with energy equal to Ea, Formula:
