{"id":10244,"date":"2019-12-06T15:40:08","date_gmt":"2019-12-06T10:10:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10244"},"modified":"2021-07-02T15:43:52","modified_gmt":"2021-07-02T10:13:52","slug":"2nd-puc-chemistry-question-bank-chapter-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-4\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 4 Chemical Kinetics"},"content":{"rendered":"

You can Download Chapter 4 Chemical Kinetics Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 4 Chemical Kinetics<\/h2>\n
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2nd PUC Chemistry Chemical Kinetics NCERT Textbook Questions and Answers<\/h3>\n

Question 1.
\nFrom the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
\n(i) 3NO(g) \u2192 N2<\/sub>O (g) Rate = K[NO]2<\/sup>
\n(ii) H2<\/sub>O2<\/sub> (aq) + 3I–<\/sup> (aq) + 2H+<\/sup> \u2192 2H2<\/sub>O(l) +I3<\/sub>–<\/sup>
\nRate = K[H2<\/sub>O2<\/sub>][I–<\/sup>]
\n(iii) CH3<\/sub>CHO (g) \u2192 CH4<\/sub> (g) + CO(g)
\nRate = K [CH3<\/sub>CHO]3\/2<\/sup>
\n(iv) C2<\/sub>H5<\/sub>Cl (g) \u2192 C2<\/sub>H4<\/sub> (g) + HCl (g)
\nRate = K [C2<\/sub>H5<\/sub>Cl]
\nAns:
\n(i) 2
\n(ii) 2
\n(iii) 3\/2
\n(iv) 1<\/p>\n

Question 2.
\nFor the reaction:
\n2A + B \u2192 A2<\/sub>B
\nthe rate = k[A][B]2<\/sup> with k = 2.0 \u00d7 10-6<\/sup> mol-2<\/sup> L2<\/sup> s-1<\/sup>. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1<\/sup>, [B] = 0.2 mol L-1<\/sup>. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1<\/sup>.
\nAnswer:
\nInitial rate = K [A] [B]2<\/sup>
\n= 2.0 \u00d7 10-6 <\/sup>(mol-2<\/sup> L2<\/sup> s-1<\/sup>\u00a0\u00d7 0.1 (mol-1<\/sup>) \u00d7 (0.2)2<\/sup>mol2<\/sup>L-2<\/sup>
\n= 8 \u00d7 10-9<\/sup> mol L-1<\/sup> s-1<\/sup>
\nwe know that
\n\"2nd<\/p>\n

\\(\\frac { 1 }{ 2 }\\) [rate of disappearance of A] = rate of disappearance of B
\n\u2234 When [A] is reduced by 0.1 – 0.06 = 0.04 M
\n[B] is reduced by \\(\\frac { 1 }{ 2 }\\) \u00d7 0.04 = 0.02 M
\n\u2234 [B] remaining = 0.2 -0.02 = 0.18 M
\nRate of reaction = K [A] [B]2<\/sup>
\n= 2.0 \u00d7 10-6<\/sup> mol -2<\/sup> L21<\/sup>
\ns -1<\/sup> \u00d7 0.06 \u00d7 (0.18)2<\/sup> mol3<\/sup> L-3<\/sup>
\n= 3.89 \u00d7 10-9<\/sup> Ms-1<\/sup><\/p>\n

Question 3.
\nThe decomposition of NH3<\/sub>on platinum surface is zero order reaction. What are the rates of production of N2<\/sub> and H2<\/sub> if k = 2.5 \u00d7 10-4<\/sup>(H mol-1<\/sup>L s-1<\/sup> ?
\nAnswer:
\nFor zero order reaction rate of reaction = K
\n\"2nd
\n\u2234Rate of production of N2<\/sub>= k = 2.5 \u00d7 10-4<\/sup> Ms-1<\/sup>
\nRate of production of H2<\/sub>\u00a0= 3k
\n= 3 \u00d7 2.5 \u00d7 10-4<\/sup>
\n= 7.5 \u00d7 10-4<\/sup> Ms-1<\/sup><\/p>\n

Question 4.
\nThe decomposition of dimethyl ether leads to the formation of CH4<\/sub>, H2<\/sub> and CO and the reaction rate is given by
\nRate = K [CH3<\/sub>OCH3<\/sub>]3\/2<\/sup>
\nThe rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.
\nRate = k (PCH3<\/sub>OCH3<\/sub><\/sub>)3\/2<\/sup>
\nIf the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
\nAnswer:
\nUnits rate of reaction = bar min-1<\/sup>
\nunits of rate constant
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nMention the factors that affect the rate of a chemical reaction.
\nAnswer:
\n(i) Concentration of reactants
\n(ii) Temperature
\n(iii) Nature of reactants and products
\n(iv) Exposure to light (Radiation)
\n(v) presence of catalysts.<\/p>\n

Question 6.
\nA reaction is second order with respect to a reactant. How is the rate of reaction ‘ affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?
\nAnswer:
\n(i) Four times
\n[-rate1<\/sub> = K[CA<\/sub>]2<\/sup>
\nCA2<\/sub> = 2CA<\/sub>
\nrate2<\/sub> = K [2CA<\/sub>]2<\/sup> = 4K [CA<\/sub>]2<\/sup>] = 4 rate1<\/sub>]<\/p>\n

(ii) 1\/4 times
\n\"2nd<\/p>\n

Question 7.
\nWhat is the effect of temperature on the rate constant of a reaction ? How can this temperature effect on rate constant be represented quantitatively.
\nAnswer:
\nIncreasing the temperature on decreasing the activation energy will result in an increase in the rate of reaction and an exponential increase in the rate constant. On increasing the temperature the fraction of molecules which collide with energy greater than Ea increases and hence the rate constant (exponentially)
\nK = A -ea\/RT<\/sup> , quantitative representation of temperature effect on rate constant.<\/p>\n

Question 8.
\nIn a pseudo first order hydrolysis of ester in water, the following results were obtained:<\/p>\n\n\n\n\n
t\/s<\/strong><\/td>\n0<\/td>\n30<\/td>\n60<\/td>\n90<\/td>\n<\/tr>\n
[Ester]\/mol L-1<\/sup><\/strong><\/td>\n0.55<\/td>\n0.31<\/td>\n0.17<\/td>\n0.085<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
\n(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
\nAnswer:
\n(i)
\n\"2nd<\/p>\n

(ii) For pseudo first order reaction,
\nAverage rate = Rate constant \u00d7 Average concentration Average rate
\n\"2nd<\/p>\n

Question 9.
\nA reaction is first order in A and second order in B.
\n(i) Write the differential rate equation.
\n(ii) How is the rate affected on increasing the concentration of B three times?
\n(iii) How is the rate affected when the concentrations of both A and B are doubled?
\nAnswer:
\n(i)
\n\"2nd
\n(ii) When concentration of B is tripled, rate of reaction increases by 9 times.
\n(iii) Rate of reaction increases by 8 times.<\/p>\n

Question 10.
\nIn a reaction between A and B, the initial rate reaction (r0<\/sub>) was measured for different initial concentrations of A and B as given below.<\/p>\n\n\n\n\n\n
A\/ mol L-1<\/sup><\/td>\n0.20<\/td>\n0.20<\/td>\n0.40<\/td>\n<\/tr>\n
B\/ mol L-1<\/sup><\/td>\n0.30<\/td>\n0.10<\/td>\n0.05<\/td>\n<\/tr>\n
r\u03b8<\/sub>mol L-1<\/sup>S-5<\/sup><\/td>\n5.07×10-5<\/sup><\/td>\n5.07×10-5<\/sup><\/td>\n1.43 x 10-5<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

what is the order of the reaction with respect to A and B ?
\nAnswer:
\nLet the rate law be, r = K [A]x<\/sup> [B]y<\/sup>
\nr1<\/sub> = K[0.2]x<\/sup>[0.3]y<\/sup> = 5.07 \u00d7 10-5<\/sup> -(1) .
\nr2<\/sub> = K[0.2]x<\/sup> [0.1]y<\/sup> = 5.07 \u00d710-5<\/sup> -(2)
\nr3<\/sub> = K [0.4]x<\/sup> [0.05]y<\/sup> = 7.6 \u00d7 10-5<\/sup> -(3) ,
\n\"2nd
\n\u2234 The order of reaction with respect to A is 0.585 and with respect to B is 0.
\n\u2234 rate law = K[A]0.858<\/sup> [B]0<\/sup> = K [A]0.585<\/sup>.<\/p>\n

Question 11.
\nThe following results have been obtained during the kinetic studies of the reaction 2A + B \u2192 C + D<\/p>\n\n\n\n\n\n\n\n
Experiment<\/strong><\/td>\n[A]\/mol L-1<\/sup><\/strong><\/td>\n[B]\/mol L-1<\/sup><\/strong><\/td>\nInitial rate of formation of D\/mol L-1<\/sup>min-1<\/sup><\/strong><\/td>\n<\/tr>\n
I<\/td>\n0.1<\/td>\n0.1<\/td>\n6.0 x 10-3<\/sup><\/td>\n<\/tr>\n
II<\/td>\n0.3<\/td>\n0.2<\/td>\n7.2 x 10-2<\/sup><\/td>\n<\/tr>\n
III<\/td>\n0.3<\/td>\n0.4<\/td>\n2.88 x 10-1<\/sup><\/td>\n<\/tr>\n
IV<\/td>\n0.4<\/td>\n0.1<\/td>\n2.40 x 10-2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Determine the rate law and the rate constant for the reaction.
\nAnswer:
\nr = K [A]x<\/sup> [B]y<\/sup>
\nr, = K[0.1]x<\/sup> [0.1]y<\/sup> = 6.0 \u00d7 10-3<\/sup> -(i)
\nr2 = K [0.3]x<\/sup> [0.2]y<\/sup> = 7.2 \u00d7 10-2<\/sup> -(2)
\nr3 = K [0.3]x<\/sup> [0.4]y<\/sup> = 2.88 \u00d7 10-1<\/sup> -(3)
\nr4 = K [0.4]x<\/sup> [Q.l]y<\/sup> = 2.4 \u00d7 10-2<\/sup> -(4)
\n\"2nd
\n4x<\/sup> = 4
\n\u21d2 x =1
\n\"2nd
\n2y<\/sup> = 4 \u21d2 y = 2.<\/p>\n

\u2234 Rate law, r = K [A]1<\/sup> [B]2<\/sup>
\nRate constant,
\n\"2nd
\n= 6M-2<\/sup>min-1<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nThe reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:<\/p>\n\n\n\n\n\n\n\n
Experiment<\/strong><\/td>\n[A]\/mol L-1<\/sup><\/strong><\/td>\n[B]\/mol L-1<\/sup><\/strong><\/td>\nInitial rate \/mol L-1 <\/sup>min-1<\/sup><\/strong><\/td>\n<\/tr>\n
I<\/td>\n0.1<\/td>\n0.1<\/td>\n2.0 x 10-2<\/sup><\/td>\n<\/tr>\n
II<\/td>\nx<\/td>\n0.2<\/td>\n4.0 x 10-2<\/sup><\/td>\n<\/tr>\n
III<\/td>\n0.4<\/td>\n0.4<\/td>\ny<\/td>\n<\/tr>\n
IV<\/td>\nz<\/td>\n0.2<\/td>\n2.0 x 10-2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nrate law = K [A]
\nfrom I 2.0 \u00d7 10-2<\/sup> = K 0.1
\n\u21d2 K = 2.0 \u00d7 10-1<\/sup>min-1<\/sup>
\nfrom II 4 .0 \u00d7 10-2<\/sup> = 2 .0 \u00d7 10-31<\/sup> x
\n\u21d2 K = 2 .0 \u00d7 10-1<\/sup> mol L-1<\/sup>
\nfrom III y = 2.0 \u00d7 10-1<\/sup> \u00d7 0.4
\n\u21d2 y = 8 \u00d7 10-2<\/sup> mol 1L
\nfrom IV 2.0 \u00d7 10-2<\/sup> = 2.0 \u00d7 10-1<\/sup> \u00d7 z
\n\u21d2 z = 0.1 mol L-1<\/sup>.<\/p>\n

Question 13.
\nCalculate the half – life of a first order reaction from their rate constants given below.
\n(i) 200 s-1<\/sup>
\n(ii) 2 min-1<\/sup>
\n(iii) 4 years-1<\/sup>
\nAnswer:
\n\"2nd<\/p>\n

Question 14.
\nThe half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
\nAnswer:
\n[A]0<\/sub> = 100
\n[A] = 80
\n\"2nd
\n= 1845 years
\n\u2234The age of the sample is 1845 years.<\/p>\n

Question 15.
\nThe rate constant for a first order reaction is 60 s-1<\/sup>. How much time will it take to reduce the initial concentration of the reactant to its 1\/16 th value?
\nAnswer:
\nK = 60 s-1<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nDuring nuclear explosion, one of the products is MSr with half-life of 28.1 years. If mg of 90<\/sup>Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?
\nAnswer:
\nt1\/2<\/sup> = 28.1 years
\n\"2nd
\n[A]0<\/sub> = 2.47 \u00d7 1\u03bcg
\nFor 10 years
\n\"2nd
\n[A] =0.78 \u00d7 10-6<\/sup>g = 0.78\u03bcg
\nFor 60 years
\n\"2nd
\nThe amount of 90<\/sup>Sr that will remain in the child\u2019s body after 10 years is 0.78\u03bcg and after 60 years is 0.229 \u03bcg.<\/p>\n

Question 17.
\nFor a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
\nAnswer:
\n\"2nd
\nfor 99% completion,
\nif [A]0<\/sub> = 100
\n[A] = 100-99= 1
\nfor 90% completion
\n[A] = 100-99
\n\"2nd<\/p>\n

Question 18.
\nA first order reaction takes 40 min for 30% decomposition. Calculate t1\/2<\/sub>.
\nAnswer:
\n[A]0<\/sub> = 100.
\n[A] = 100 – 30 = 70
\nt = 40min
\n\"2nd<\/p>\n

Question 19.
\nThe rate constant for the decomposition of hydrocarbons is 2.418 \u00d7 10-5<\/sup>s-1<\/sup> at 546 K. If the energy of activation is 179.9 kJ\/mol, what will be the value of pre-exponential factor ?
\nAnswer:
\nPre – exponential factor = A = Arrhenius
\nfactor = frequency factor
\nK = 2.418 \u00d7 10-5<\/sup>s-1<\/sup>
\nT = 546 K.
\nEa= 179.9 \u00d7 10-3<\/sup>J mol-1<\/sup>
\n[By Arrhenius equation] K = \\(\\text { Ae }^{-\\mathrm{E}} \/ \\mathrm{kr}\\)
\n\"2nd<\/p>\n

Question 20.
\nonsider a certain reaction A \u2192 Products with k = 2.0 \u00d7 10-2<\/sup> s-1<\/sup>. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol Lr1. Ans: [A]0 = 1 mol L-1<\/sup>
\nAnswer:
\n[A]0<\/sub> = 1 mol L-1<\/sup>
\nt = 100s
\nk = 2.0 \u00d7 10-2<\/sup>s-1<\/sup>
\n\"2nd
\nlog [A] = -0.8684
\n\u21d2 [A] = 10 -0.8684<\/sup> = 0.354 mol L-1<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nSucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1\/2<\/sup> = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
\nAnswer:
\nt1\/2<\/sup> = 3 hrs
\n\"2nd<\/p>\n

Question 22.
\nThe decomposition of hydrocarbon follows the equation K = (1.5 \u00d7 1011<\/sup> s-1<\/sup>)
\n\\(e^{\\frac{-28000 K}{T}}\\) Calculate Ea<\/sub>
\nAnswer:
\nAccording to Arrehenius equation
\nK = Ae-Ea\/RT<\/sup>
\nGiven K = (4.5 \u00d7 1011<\/sup>s-1<\/sup>)\\(e^{\\frac{-28000 K}{T}}\\)
\ncomparing the two equations
\n\"2nd
\nEa<\/sub> = 28000 \u00d7 R
\n= 28000 \u00d7 8.314 J mol-1<\/sup>
\n= 232.79 KJ mol-1<\/sup>.<\/p>\n

Question 23.
\nThe rate constant for the first order decomposition of H2<\/sub>O2<\/sub> is given by the following equation:
\nlog k = 14.34 – \\(\\frac{1.25 \\times 10^{4} K}{\\mathbf{T}}\\)
\nCalculate Ea<\/sub> for this reaction and at what temperature will Its half-period be 256 minutes?
\nAnswer:
\nAccording to Arrehenius equation
\n\"2nd
\nEa<\/sub> = 2.303 R \u00d7 1.25 \u00d7 104<\/sup> K
\n= 2.303 \u00d7 8.3l4 JK-1<\/sup> mol-1<\/sup> \u00d7 1.25 \u00d7 104<\/sup> K
\n= 239.34 KJ mol-1<\/sup>
\nWhat t1\/2<\/sup> = 356 min
\n\"2nd
\nSubstituting K in given equation
\nlog(4.51 \u00d7 10-5<\/sup>) = 14.34 – \\(\\frac{1.25 \\times 10^{4}}{\\mathrm{T}}\\)
\nT = 669 K.<\/p>\n

Question 24.
\nThe decomposition of A into product has value of k as 4.5 \u00d7 103<\/sup> s-1<\/sup> at 10\u00b0C and energy of activation 60 kj mol-1<\/sup>. At what temperature would k be 1.5 \u00d7 104<\/sup>s-1<\/sup> ?
\nAnswer:
\nk1<\/sub> = 4.5 \u00d7 103<\/sup> s-1<\/sup>
\nk2<\/sub> = 1.5 \u00d7 104<\/sup>s-1<\/sup>
\nEa<\/sub> = 60 \u00d7 103<\/sup> J mol-1<\/sup>
\nR = 8.314 JK-1<\/sup> mol-1<\/sup>
\nT1<\/sub> = 283 k
\nT2<\/sub> = ?
\n\"2nd<\/p>\n

Question 25.
\nThe rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
\nAnswer:
\nT1<\/sub> = 283 k
\nT2<\/sub> = 313 k
\nk2<\/sub> = 4k
\n\"2nd<\/p>\n

2nd PUC Chemistry Chemical Kinetics Additional Questions and Answers<\/h3>\n

Question 1.
\nName the factors on which the rate of a particular reaction depends. (H. S .B 2001)
\nAnswer:
\nConcentration, temperature, presence of catalyst and light, surface area of reactants.<\/p>\n

Question 2.
\nWrite the expression for rate of reaction in terms of each reactant and product for the reaction.
\nN2<\/sub> + 3H2<\/sub> \u2192 2NH3<\/sub> (P.S.B 2011, HSB 2000)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nGive the example of a reaction having fractional order. (PSB 2000)
\nAnswer:
\nDecomposition of acetal dehyde (order =1.5)
\n\"2nd<\/p>\n

Question 4.
\nGive an example of a psueudo first order reaction. (AISB, DSB 2004, DSB 2006)
\nAnswer:
\nAcid catalysed hydrolysis of ethyl acetate
\n\"2nd
\nRate = K [CH3COOC2H5] [H20]\u00b0<\/p>\n

Question 5.
\nFor an elementary reaction,
\n2 A + B \u2192 3C
\nthe rate of appearance of C at time \u2018t\u2019 is 1.3 \u00d7 104<\/sup> mol-1<\/sup> S-1<\/sup>. Calculate at this time
\n(i) Rate of the reaction
\n(ii) Rate of disappearance of A. (CBSE sample paper II 2008)
\nAnswer:
\nRate
\n\"2nd
\n= 1\/3 \u00d7 1.3 \u00d7 10-4<\/sup>mol L-1<\/sup>S-1<\/sup>
\n= 4.33 \u00d7 10-5<\/sup> mol L-1<\/sup>S-1<\/sup><\/p>\n

Rate of disappearance of A
\n\"2nd
\n2\/3 \u00d7 1.3 \u00d7 10-4<\/sup> L-1<\/sup>S-1<\/sup>
\n= 8.66 \u00d7 10-5<\/sup> mol L-1<\/sup>S-1<\/sup><\/p>\n

Question 6.
\nThe rate law for a reaction is found to be
\nRate = K [NO–<\/sup>2<\/sub>] [I–<\/sup> ] [H+<\/sup>]2<\/sup> How would the rate of reaction change when
\n(i) Concentration of H+<\/sup> is doubled
\n(ii) Concentration of I–<\/sup> is halved
\n(iii) Concentration of each of NO2<\/sub>, I–<\/sup> and H+<\/sup> are tripled ?
\nAnswer:
\nSuppose initially the concentration are [NO–<\/sup>2<\/sub>] = a mol L-1<\/sup>, [I–<\/sup> ] = b mol L–<\/sup> and [H+<\/sup>] = c mol L-1<\/sup>
\n\u2234 Rate = K abc2<\/p>\n

(i) New [H+<\/sup>] = 2c
\n\u2234 New rate = 4ab (2c)2<\/sup> = 4abc2
\n= 4 times<\/p>\n

(ii) New [I–<\/sup> ] = \\(\\frac { b }{ 2 }\\)
\nNew Rate = Ka\\(\\frac { b }{ 2 }\\)c2<\/sup>= \\(\\frac { 1 }{ 2 }\\) Kabc2<\/sup>
\nie. rate of reaction is halved<\/p>\n

(iii) New [NO–<\/sup>2<\/sub>] = 3a, [I–<\/sup> ] = 3b, [H+<\/sup>] = 3c
\nNew rate = K (3a) (3b) (3c)2<\/sup>
\n= 81K abc2<\/sup> = 81 times.<\/p>\n

Question 7.
\nThe decomposition of NH3<\/sub> on platinum surface
\n\"2nd
\nis zero order with k = 2.5 \u00d7 10-4<\/sup> mol L-1<\/sup> S-1<\/sup>
\nWhat are the rates of production of N2<\/sub> and H2<\/sub>
\nAnswer:
\n2NH3<\/sub> -> N2<\/sub> + 3H2<\/sub>
\n\"2nd
\nFor zero order reaction, rate = K
\n\"2nd
\n= 2.5 \u00d7 10-4<\/sup> mol L-1<\/sup> S-1<\/sup>
\nRate of production of H2<\/sub> = \\(\\frac{\\mathrm{d}\\left[\\mathrm{H}_{2}\\right]}{\\mathrm{dt}}\\)
\n= 3 \u00d7 (2.5 \u00d7 10-4<\/sup> mol L-1<\/sup> S-1<\/sup>)
\n= 7.5 \u00d7 10-4<\/sup> mol -1<\/sup> S-1<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nWhy does coal not burn by itself in air, but once initiated by a flame, continues to bum ?
\nAnswer:
\nActivation energy for the combustion reaction is very high and is not available at room temperature. Oh applying flame, to a part of the coal and air in contact with the flame absorb heat which provides the necessary activation energy and combustion starts. The heat liberated further provides activation energy for the combustion to continue.<\/p>\n

Question 9.
\nFollowing reaction takes place in one step:
\n2NO(g) + O2<\/sub>(g) \u21cc 2NO2<\/sub>(g)
\n\"2nd
\n\\(\\frac{\\mathbf{r}_{2}}{\\mathbf{r}_{1}}\\) = 27 or r2<\/sub> = 27r1<\/sub> ie rate becomes 27
\ntimes
\nThere is no effect on the order of reaction.<\/p>\n

Question 10.
\nRate constant K of a reaction varies with temperature according to the equation
\nlog K = constant . \\(\\frac{\\mathbf{E} \\mathbf{a}}{2.303 \\mathbf{R}} \\frac{\\mathbf{1}}{\\mathbf{T}}\\)
\nAnswer:
\nwhere Ea<\/sub> is the energy of activation for the reaction. When a graph is plotted for
\nlog K versus\\(\\frac { 1 }{ T }\\), a straight line with a slope 6670 K is obtained. Calculate the energy of activation for this reaction. State units (R = 8.314JK-1<\/sup> mol-1<\/sup>)
\nAnswer:
\nSlope of the lin,e
\n\\(\\frac{-E a}{2.303 R}\\) =-6670K
\nEa = 2.303 \u00d7 8.314 (JK-1<\/sup> mol-1<\/sup>) \u00d7 6670K
\n= 127711.4J mol-1<\/sup>.<\/p>\n

\"KSEEB<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

You can Download Chapter 4 Chemical Kinetics Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 4 Chemical Kinetics 2nd PUC Chemistry Chemical Kinetics NCERT Textbook Questions and Answers Question …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10244"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10244"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10244\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10244"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10244"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10244"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}