KCET 2024 Chemistry | Chemical Kinetics Top 30 Questions for #kcet2024

hello students Welcome to our Channel dsha karnatak as we have already know that we have started two Series where we are discussing the important topics in a chapter and then we are solving 30 important questions from this chapter right so we have already done that for Solutions chapter the next turn is of another High weightage chapter that is chemical kinetics so in my previous video I have discussed what what all Concepts you have to revise from the chemical kinetics chapter so that in this last time you can effectively cover this chapter so in today's video I'll discuss top 30 questions of chemical kinetics so these questions are very much similar with your k at previous year questions from chemical kinetics so if you are able to do these questions you can uh assure yourself that you are very thorough with the chapter and you can get a good you can crack all the questions from this chapter coming in your exam right okay so let us see this okay and before this in my last video I have discussed top 30 questions from Solutions chapter I went through the comments that you have posted and I saw some doubts you have so first let me clear those doubts then we'll go with the kinetics chapter right okay so one question that was that you people have told is doubt is when the pure solvent diffuses out of the solution through the semi-permeable membrane then the process is called yeah so I think the answer for this I told osmosis actually they have to read the statement little carefully pure solvent diffuses out of of the solution okay so from the solution the solvent is going out so the correct answer for this will be reverse osmosis okay so that is one next is 5.5 mg of nitrogen gas dissolve in 180 g of water at 273 kin and 1 atm pressure due to nitrogen gas the mole fraction of nitrogen in 180 g of water at 5 ATM nitrogen pressure is approximately so this one I have told you this is question from Henry's law so what they're telling is we can use this formula P1 is = KH into X1 and P2 is = KH into X2 right so now if I take the ratio P1 by P2 is equal to X1 by X2 now let us little understand the question what is telling when the pressure is 1 atm how much gas was how much nitrogen gas was dissolved 5.5 mg now the question is asked if the pressure is increased to 5 ATM then how much mole fraction of nitrogen will dissolve okay that is what we have to calculate so now let us calculate X1 value X1 is equal to number of moles of nitrogen divided by number of moles of nitrogen plus number of moles of water so as you know this number of moles of nitrogen is very less as compared to the number of moles of water so we can ignore this part then if I calculate the value for number of moles of nitrogen it is 5.5 into 10 ^ of - 3 G / 28 whole divided by number of moles of water you can see 180 g 180 by 18 so that is your 10 so this value if you solve it that will give you the number of moles of nitrogen the mole fraction of nitrogen so I'll get 0.196 into 10 ^ of - 4 okay so this one value we'll get for X2 now we can substitute here 1 by 5 is equal to X1 is your1 9 6 into 10^ - 4 and we have to calculate the x value so now if you solve it we will get X is equal to this 196 we can approximate it to 2 so 5 into 2 is 1 into 10 ^ of - 4 okay so the correct answer for this question will be option D okay uh so like this you please solve it any doubts you are getting comment down I'll surely address them in my next video so now we are done with the solutions we'll go to the chemical kinetics important questions so first one is the question is from rate law I have also divided these questions based on the topic so that we can know which topic has to be focused more so first topic is your rate law the rate of gaseous reaction in the expression k a v square if the volume of the vessel is reduced to 1/2 the volume of the reaction rate as original rate is Dash so first of all we have to know what is the relation between volume and concentration so we know concentration is is equal to number of moles by volume so when volume decreases concentration will increase okay and vice versa so what they have done here the volume is reduced to 1/2 so the concentration will double now we can easily write down the rate law R is equal to K into a b² right now since the volume is double con volume is reduced to half concentration will be doubled so I can write all the values as 2 a and this one as 2 B whole to the^ of 2 so now if you solve this we'll get 8 * of K A and B to the^ of 2 so you can see suppose this is my final rate so I can write rate Dash is equal to 8 * of initial rate so the correct answer for this question is option C okay so this is very direct and easy question you will get similar questions from rate law now next question for the elementary Reaction 2 a + 3 B gives 4 C + D the rate of appearance of C at time T is 2 .8 into 10 ^ ofus 3 mole L second inverse rate of disappearance of B you this is the first topic from chemical kinetics where we learn about differential rate law right so very easy question you have to be little careful while writing the equation and one tip here you do not write the whole equation okay you write down the equation for those quantities which is asked in the question so the question is asking us about B and uh C so let us write the relation between B and C - 1 by 3 DB by DT will be equal to 1x 4 of DC by DT right so now let us calculate we have to calculate the DB by DT so DB by DT will be equal to - 3x4 DC by DT and the DC by DT value is given to you 2.8 into 10 ^ ofus 3 so the correct answer for this question will be option A now be thinking here minus is there but in option it is not given so when they have already told that rate of disappearance is there that means the concentration is decreasing so the rate of disappearance is 3x4 of DB by DT right so this is also easy question you'll get many questions of this type okay now again rate law question 1 L of 2 mol acetic acid is mixed with 1 L of three mol ethy alcohol to form an ester the rate of the reaction with respect to the initial rate when each of the solution is diluted with an equal volume of one water will be okay so first what is this 1 L 2 m means 2 moles are there in 1 lit so can I write the initial rate law as K into concentration of acetic acid concentration of acetic acid as 2 by 1 where 2 moles are present in 1 lit and for concentration of ethy alcohol I can write 3x 1 okay so this is the initial rate now what has happened the new rate is suppose let us add equal amount of water is I let let us add one lit to both the solutions okay you can do and you can take any quantity now if I add both the solutions by 1 so this will become 2x 2 and this will become 3x2 right the volume will become twice now what we can do just take the ratio of these two so if I take R by r d we can get uh 2 by 1 into 3x 1 iD 2x 2 into 3x2 right so if I solve this R by r d I will get 4 right so now what is r Dash value r d value is R by 4 or I can write 0.25 * of the initial rate so the correct answer for this question will be option A okay next again uh from the differential rate law question first of all write the reaction properly disappearance of o2 and the appearance of s SO3 so you just see what two reactants or two elements that have been related so I can write here minus D2 by DT is equal to uh s SO3 is 1x 2 of D SO3 by DT so we need to calculate this one so I can write -2 D2 by DT is equal to DS SO3 by DT okay so Min - 2 what is D O2 by DT that is 2 into 10 ^ of - 4 so I will get 4 into 10 ^ of - 4 see there are students of different categories this much you do not have to solve in the exam but for everyone to understand I'm solving in so much detail you can just Skip and you can write down the values and get the answer in quickly so the correct answer for this question will be option b okay next uh again our differential rate question for the reaction 1X 2 a gives 2 B the rate of disappearance of a is related to the rate of appearance of B by the expression here the question is little tricky because it is 1 by 2 okay so what I have to do is minus of 1 of 1x 2 D A by DT will be equal to 1x 2 of DB by DT okay so now if I make this one I'll get - 2 Da by DT is equal to 1 by 2 DB by DT right so we will get da by DT as 1x4 of DB by DT right so the correct answer for this question will be option A okay easy one next which of the following statement is incorrect okay molecularity is only applicable for element reaction this is a correct statement right we know molecularity is always applicable for elementary steps so statement a is correct now let us see the next statement the rate law for any reaction cannot be determined experimentally this is incorrect the rate of any the rate of any reaction is determined only and only experimentally okay so the incorrect statement is option b so the answer for this question is option b let us check the other options bimolecular reactions involve simultaneous Collis between two species that we know B molecular means it will include two species complex reactions can have fractional order it is also correct because it has many steps we can get fractional order for complex reaction so that is why statement B is the incorrect one now let us see quickly this one again rate law for a chemical reaction ma a gives XB the rate law is given by r k a squ if a is double the reaction rate will be very simple one R is equal K into 2 a² so so if we we'll get four * of K A squ so what is the new rate I can write this is my r d r d is equal to 4 * of R so the reaction has quadrupled the rate has quadrupled so the correct answer is option b easy one next uh DB by DT again our differential rate law question Min - 1x3 of Da by DT is equal to 1 by 2 of DB by DT right so DB by DT will be equal to - 2x3 of d by DT isal to DB by DT right so we'll get min - 2x 3 that is your option b correct answer is option b next okay this question is very similar to the one which we have already solved 100 ml 1 mol was mixed with 100 ml 2 m ethanol the change in the initial rate each solution is diluted with equal volume of water right okay so again we can do the same thing so we have the r is equal to K into concentration of acetic acid will be 1 mole in 100 ml into th000 okay divided by here 2 moles in 100 into 1,000 now what they have done is we have increased the volume so I can take it 200 and here also I can make it 200 okay so now if you take the ratio of these two we'll get R by r d as 4 okay so r d will be equal to R 1X 4 of r that is 0.25 time this question have already solved it so you can just check once again the correct answer for this option b okay next in which one of the following reaction the rate constant has the unit of this one okay so this question to solve such questions the thing is you should know the how to write the unit of rate constant for any nth order reaction so that is K is equal to mole 1 - n l nus1 2 inverse okay so what you have to do is you have to equate the power of mole given in the question to the power of mole to the general equation so what is the general equation power of mole 1 - n and here it is 1 so can I write 1 - n is = 1 then what will be n value n value is zero thir reaction so this is the first part of the question so what we can say is the question is directly asking us to find which of the following options is a zeroth order reaction so it is in an indirectly way it is asked so we have acid catalyze this is our first order chcl3 is first order no is also first order you know decomposition of hi on the surface of the gold is a zeroth order reaction so the correct answer will be option D now you'll be thinking how to know which is order which order these reactions are so go to the ncrt book you can see for every order there are some examples given and the questions will be from those examples only so revise those examples at least you should know two or three reactions for zeroth order and first order okay you have to remember those examples so that you can do such type of easy but Theory based questions next okay the rate of the reaction is given to us for any Ester and NaOH is giving you this one plus this one the rate of the reaction is given by the concentration of EST and NaOH if the concentration is expressed in mole lit inverse the unit of K is okay so first of all let us find the order so what's the definition of order order is the sum of the power of the concentration terms so can I say here the power is one and here the power is one so if you add them the order will be 2 so n is equal to 2 now if you find the unit of K it is I just told you the formula mole 1 - n so 1 - 2 liter N - 1 so 2 - 1 then second inverse so I'll get mole - 1 l 1 second inverse so the correct answer for this is option A okay this is also easy question next for a reaction a + 2 B products when concentration of B alone okay this question you have to pay little attention and you have to understand the meaning when the concentration of B alone is increased the half life remains same so what is a telling B concentration is increased and T half is same okay next let us read the next part concentration of a alone is doubled so a concentration is doubled and the re rate remains same okay now you should know some basic things that for first life first order half life it doesn't depend on the initial concentration whatever initial you concentration you check the half life will remain the same that is why if T half is remaining same for any concentration we can say that b is a first order reaction or with respect to B the reaction is first order so I can write any rate like this okay now what is told about a whatever concentration you increase the rate remains same so the rate will remain same only if it is a zeroth order reaction right so that is why it will be a to the^ of0 so what is the overall order of the reaction 1 for one what will be the unit again I told mole 1 - n so 1 - 1 liter N - 1 so 1 - 1 second inverse this will be 0 0 1 so we'll get second inverse and the correct answer for this question is option A okay next this one I'll tell little tricky question because we have to understand here something so the value of rate constant of sud of first order reaction okay listen first thing the rate constant always depends only on temperature okay rate can be uh rate can be dependent on concentration temperature and all but rate constant depends only temperature so the correct answer here will be depends only on temperature you need not think these things depends on the concentration of reactants present in small amount this is for rate but the question has asked for you rate constant which is only and only dependent on temperature so correct answer will be a okay so this one you have to know okay now we'll go to the another topic that is integrated rate law here you should remember or the formulas for zeroth order and first order reactions now if the rate constant of a first order reaction is K the time T required for the completion of 99% of the reaction is given by first let us write the integrated rate law K = to 2.303 by t log of a by a right now I know for 99% whenever percentage is given take the initial concentration as 100 so I can write T = 2.303 by K log of 100 by a see 99% is completed then how much reactant will be left 1% so I can write it as 1 so what I will get 2.303 by K log of 10 ^ of 2 we can write it or we'll get 4.66 by K the correct answer for this question will be option a easy direct based question just you see there's no calculation involved also okay you have to just write the expression next the time required for 60% completion of the first order reaction is 50 minutes the time required for 93.6% completion of this reaction will be okay so here you have to d uh again it is a first order reaction I can write t50 is equal to 2.303 sorry it is 60% T 60% T 60% will be 2.303 by K log of 6 60% is completed how much will be left then 40% so it will be 40 another T they have asked you is for 93.6 is equal to 2.303 by K log of 100 by 93.6 is completed then how much will be left 100us 93.6 that is your 6.4 okay now what you have to do this t60 is given to you is 50 minutes take the ratio of both so I can write 50 by T 93.6 is equal to log of 100 by 40 minus sorry divided by log of 1,000 by 64 okay so how you can solve this part we can take it as 0 canell Log of 10 - log of 4 divided by log of 1000us log of 64 so so you know log 10 value is 1 1 - log 4 is 602 okay and then here 10 log th000 will be 3 and log 64 means 64 can be written as 2 to the^ of 6 right so I can write log of 64 as 2 ^ of 6 that is 6 * of log 2 so that is equal to 6 * of 3 or we can take 1.8 okay so 1.8 and here also for easy calculation we'll take it 6 so how much I'll get here if we solve this part 1 - 6 will be 0.4 3 - 1.8 will be 1.2 so you see here I got the ratio 50 by T 93.6 is equal to4 by 1 that is 1x 3 so what I'll get X will be equal to 50 minutes okay so the correct answer for this question is option C not uh not a difficult question but little calculation is involved okay next so one more tip here you should remember log 22 log 9 these values keep it in your mind so that calculations will be faster okay and especially log 2 3 5 and 7 these values you must remember next half lifee period of a order first order reaction is 60 Minutes what percentage will be left after 240 minutes okay there are two ways of doing this question if it is a p question you have to go for uh substituting it in the first order equation and then do but when it is competitive you can easily solve it by one method that is 240 minutes are there and 60 Minutes is given so how many half life is there so total number of half life is four right so in each half life what will happen to the concentration it will decrease by half so we are starting with let us take we are starting with 100 then in the first half life what it will become 50 in the second half life it will become 20 25 in the third half life it will become 12.5 and in the fourth half life half of this one will be 6.25 so how much will be left 6.25% so the correct answer is option A okay next a gives PQ + r follows the first order kinetics with a half life of 69.3 seconds at 500° C starting from the gas a enclosed in a container at 500° C and a pressure of 4 ATM the total pressure of the system after 230 seconds will be okay so now this question is from your gas phase little not different uh but you have to understand how to write the equation so first see here a is giving us three gas p q and R now just imagine when you're starting the reaction just started you're taking a will there be any p q and R formed at that time no right so the initial pressure of a will be 0.4 p q and R will be zero okay this is when you have started the reaction now at 230 seconds uh 230 seconds after 230 seconds let X amount of a has reacted then how much a will be left 0.4 from that x amount has reacted so left will be 0.4 - x how much a is reacting same amount of P will form same amount of Q will form and same amount of R will form now we can write the equation for this A first order k equal to 2.303 by t log of a KN by a okay here we can see some calculation half life is given and we know the for first order T2 is equal to 0.693 by K so what will be K value 0.693 by 69.3 so this value will be 1 by 100 right so I'll put in case of K 1 by 100 and 2.30 I'll just take 2.30 so that my cancellation part will be easy and 2 30 seconds log of what is a not value4 and what is a value after 2 uh 2324 - x so if we solve this part I'll get this one as 1 by 100 again so 1 by 100 and 1 by 100 is cancel so 1 is = to log of 0.4 - 0.4 - x 1 can be written as log 10 yes or no 1 can be log 10 so log of 0.4 by 0.4 - x so this log can be canceled from both sides so now I have this equation I'll erase this side so if you have seen we have got already log 10 is equal to log of 0.4 4 / 0.4 - x so I can write 10 is = 0.4 by 0.4 - x if you solve you'll get x equal 36 but what is the point of calculating X here so the question has asked you what is the total pressure after 230 seconds so what is the total pressure after 230 seconds P total will be this addition of all these pressures so I'll get 4 x x will get cancelled 2x so we can subtit X is equal to so P total is equal to 0.4 + 2 * of 0.36 so we will get 2 6 are 12 we'll get this one so if you add this we'll get 1.12 so the correct answer for this question is option D 1.12 ATM so what mistake you could have done in this question generally after finding X if the option would have been 36 we'll directly mark it we forget what is the question asked here so the question is the total pressure okay next okay now now another very important topic half life our first order reaction is half completed in 45 minutes how long will it uh will take to complete 99% so for this we have a relation again from competitive point of view you should know this relation T 99.9 is equal to 10 * of T 50% T 50% is 450 uh 45 minutes so we'll get 450 minutes convert it into hours by dividing by 60 so we'll get 7.5 hours so the correct answer for this question is option D easy one okay next again from the general equation for the nth order of the reaction half life period is directly proportional so so you should know this T2 is directly proportional to 1 by a^ of n minus 1 okay so the correct answer for this question is option C from this formula you please note it down important one okay so we can use this uh some other numericals can be just solved by knowing this one formula yeah this formula this question uses that formula half life of the reaction is found to be inversely proportional to fifth power of its initial concentration so what it is telling half life is proportional to fifth power of the initial concentration what is the order of the reaction just now I told you T2 is proportional to 1 by a ^ of n minus 1 so I can equate these two okay so nus 1 is equal 5 so n will be equal to 6 so the order of the reaction is six and the correct answer for this question is option b I'll tell this question is little tricky okay not tricky means it is not this formula is not very commonly known so please note down this formula okay next uh the plot of T2 versus r0 for a reaction is a straight line parallel to xaxis what does this mean whenever you are doing a graph and we are having a graph of T2 versus R and the graph comes a straight line so what does this indicate it indicates that the T half is independent of R it doesn't depend on R not at all so we know this is true for which order reaction for first order reaction so if you know for first order reaction T2 is equal to 0.693 by K there is no term for a not here so what I can say the order of the reaction is is first order if it is first order reaction then what is the unit of k k unit will be again you can if you have not remembered you can use this formula later n minus 1 second inverse but it is my it is my suggestion that you should remember the zeroth order and first order the table everything their half life their integrated rate law their graph and then half life formulas also you should remember so if I substitute n is equal to 1 so in this case it will become zero this will become zero the unit will be second inverse correct answer will be option D okay next in a first order reaction the concentration of the reactant is reduced to 12.5% in 1 hour when was it half completed okay this is also very easy to solve if you know this trick so first of all what is telling the concentration is reduced to 12.5% so let us do first if we start with 100 after first half life it will become 50 after second half life 25 after third half life it will become 12.5 right according to question this whole thing has taken place in 1 hour okay that means each half life because it is a first order reaction it is also mentioned so each half life will be equal so 60 minutes divided by 3 so you'll get 20 20 and 20 so when it was 50% when did there occur 20 minutes the concentration become 50 so the correct answer for this question is option A okay next half lifee period of a first order reaction is 10 minutes starting with the initial concentration of 12 mol the rate after 20 minutes okay so first of all what is the formula of rate rate is K into concentration of a okay suppose you want to calculate the rate after 20 minutes you need the concentration of a after 20 minutes okay so first of all we'll calculate what is the concentration of a after 20 minutes how to do this again we'll use the half lifee table since it is a first order reaction 12 minute 12 mol was there in first 10 minutes it will become 6 molar and in the next 10 minutes it will become three M right so the concentration of a after 20 minutes is how much 3 so I can put here k a concentration is 3 MO okay now how to put the value of K we know for K is 0.693 by T2 right so I can write 0.693 by T2 is 10 minutes into 3 m so we'll get 0693 into 3 so you see you need not solve it also if you have solved only a by up to three molar you can easily see there is two options with three molar so we can eliminate these two options and we can just check with A and D so the correct answer for this question is option D now let us go to the next question arenus equation questions so we have got the rate constant K1 and K2 for two different reactions are given okay the temperature at which K1 is equal to K2 okay so first of all what is the K1 formula K1 is equal to K2 and K1 given to you is 10 ^ of 16 e the^ of - 2000 by T and that will be equal to 10 ^ of 15 into e ^ of -000 by T okay now take uh just we can cancel 10 here so I'll get 10 e^ of - 2000 by T is equal to e^ of -000 by T now take Ln on both sides so I'll take Ln of this part that is Ln of 10 plus Ln of e to the power of - 2000 by T is equal to Ln of e the ^ of- ,000 by T So Ln of 10 is 2.303 plus this one I can write -200 by T and this one can be written as - th000 by T now if I solve this equation I'll get 2.33 is equal to 1,000 by T or I can get T = 1,000 by 2.303 kelv so the correct option for this question is option number c okay so here little calculation is involved but not very difficult one easy to solve next question okay so this is some concept based question the rate of a reaction is given by this one under standard notation in order to speed up the reaction which of the following has to be decreased so it is a it is very well known that if activation energy decreases rate re rate of the reaction will increase so from there we can easily tell the correct answer is option b but we should also know what happens when T and Z okay so if you see K is equal to P into Z to 1 by e^ of e a by RT okay now you see it is telling the which factor has to be decreased so if T is decreased what will happen to this ratio denominator is increasing if T is decreased then this ratio will increase that means K value will decrease so T cannot be the answer now now what with zed so K and Zed are directly proportional if you decrease Zed K will also decrease so that also cannot be the answer now if you decrease EA if the only the activation energy is decreased then the this fraction will decrease so that means I can tell this whole term will decrease if this whole term decreases then K will increase inversely proportional so by that we can say EA or directly also we know whenever activation energy will decrease rate of the reaction will increase next question temperature coefficient of the reaction is two what is the meaning of temperature coefficient that means whenever you uh increase the uh temperature by 10° C the rate doubles okay that is the meaning of temperature coefficient is two when the temperature is from 30° to 90° the rate of the reaction is increased by okay very different question this one we are going from 30° C to 90° C okay so you have to tell here how many 10 how many times the temperature increased in terms of 10 means first we'll go 30 to 40 then 40 to 50 50 to 60 60 to 70 70 to 80 and then 80 to 90 so 1 2 3 4 5 and 6 so we have increased the temperature six times by 10° C so that is then what will be the rate so here you have to take 2 in 2 to the power of how many times we have increased that is six times so 2 to the^ of 6 is 64 times so the correct answer for this question is option D so this you have to let know what is done here next which of the following statement is in accordance with Aria's equation okay rate of a reaction increases with the increase in temperature we know this Formula K isal A into e^ ofus e a by RT so I can write this equation e^ of e a by RT now see if temperature increases this fraction will decrease the whole FRA the whole e to the power of that value will decrease and K will increase so the correct answer is a okay let us see the other options rate of a reaction does not change with the increase in activation energy this is wrong we know activation energy is a factor so it will definitely affect the rate rate constant decreases exponentially with the increase in temperature this is also incorrect whenever temperature increases rate constant will increase rate of a reaction decrease with the decrease in activation energy this is wrong when activation energy uh when uh activation energy increases rate of the reaction will decrease and vice versa okay so the correct statement is option A again a theory based question the activation energy of a chemical reaction can be determined by so we know this formula log of K2 by K1 is equal to EA by 2.303 r T1 1 by T1 - 1 by T2 okay so what does this equation is used for this equation is generally used to calculate the activation energy whenever you see numericals also this equation we always are asked to find the activation energy right so that is why this is done by evaluating the rate constants at two different temperatures so the correct answer is option A next which is a wrong statement again so on the basis of effect of catalyst I have told you in the effect of catalyst you should know that what does a catalyst do it increases the rate of the reaction by decreasing the activation energy so rate constant K is equal to arenus constant a if EA is equal to Zer okay this is also important statement I have this formula right uh k isal a into e^ of minus E A by RT okay if EA is zero then this whole term is zero then e to the^ of 0 is 1 so I can write K is equal to a right so that is why this statement is correct next e ^ of minus EA by RT gives the fraction of reactant molecules that are activated at the given temperature this is also correct statement from the B one curve we'll get the fraction of molecules that are activated is given by this fraction next lnk versus 1x2 plot is a straight line yes this graph is also there if you see if I take the uh Ln on both side I'll get l l n k is equal to Ln a minus E A by R 1 by T okay so if I draw a graph for Ln K and versus 1 by T okay so what will be the slope of this equation minus E A by R and the uh what is y intercept will be Ln a right so you can see it is a straight line Ln K versus this is a straight line okay next one presence of catalyst will not alter the value of EA so I have told you again and again whenever we use Catalyst the activation energy decreases as a result of which the rate of reaction increases so this statement is incorrect so the correct answer for the question is option D okay so all statements a c a b and c are correct and we have to tell the wrong statement so the correct answer will be option D next okay so we have reached the last question here higher order so this question is from Collision Theory higher order reactions are rare due to so we know Collision Theory speaks about two important points that is first of all the Collision should happen in proper orientation okay and second Factor the Collision should have sufficient energy right that two things you have to remember okay so higher order reactions are rare due to low probability of simultaneous collisions of all the reacting species so when two species are there they can react but when we increase the number of species so it is very rare chance that they all collide with proper orientation and also have the sufficient kinetic energy so that is why the correct answer for this question is option C okay so that's all for this video we have discussed all 30 important questions from Kinetics so give more importance to rate law and integrated equations and arous equation for the complete revision of this chapter okay and if you have found this uh video useful and informative so please like share and subscribe our channel so that you can boost up your preparation with such informative videos further thank you

Channel: Deeksha Karnataka

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Length: 41min 9sec (2469 seconds)

Published: Fri Mar 22 2024