COMEDK 2024 Chemistry | Top 30 Chemical Kinetics Questions for #COMEDK

hello students welcome back to our Channel dsha Karnataka so as you all know your Comet K exam is approaching and we have started a series where we are discussing the top 30 questions from all the high weighted chapters okay so in this video we'll be discussing the top 30 questions from one of the most high weightage topic that is chemical kinetics so you can easily expect three to five questions from this chapter and most of these questions are from your previous year paper so if you are able to solve these questions you can have a assurance that this chapter is done for the exam okay so let us start so what you have to do now is just try solving so just see the questions and try solving it by your own and then you have to see the solutions if any question you are stuck in so just Mark those questions for the revision okay so let us start with the first question so the first question is the temperature dependence of the rate constant K is expressed as k equal a into e ^ of minus E A by RT when a plot between log K and 1 BYT is plotted we get the graph as shown what is the value of slope in the graph okay so this is very important question from which topic that is your aranan equation okay so if you see here we have this equation K is = a into e^ of minus E A by RT right so if I take log on both sides we'll get log K isal to log a - e a by 2. 303 r 1 by T right now if you see here the slope this is a equation where 1 by T is X this is your Y and this is your M so what is the slope value minus E A by 2.303 R so the correct answer for this question will be uh yes EA by 2.33 R so what is the value of the slope okay so value means you have to just tell the magnitude so EA by 2.303 R so the correct answer for this question will be option number B okay so next question consider the following reaction P2 Q the concentration of both the reactants and the products vary exponentially with time which of the following figures correctly describe the change in concentration of reactants and products with time so this is very basic question you know as the reactants Pro as the reaction proceed the reactant concentration decreases and product concentration increases right so p is the reactant here so concentration of P will decrease and concentration of Q will increase so let us see all the graphs one by one here P concentration is decreasing Q concentration is increasing so this can be an answer now coming to next one uh option number B here what is happening P concentration is decreasing Q concentration is increasing yeah this is also correct now if you see here the graph for Q is not exponential right but in the question it was given it is varying exponentially with time okay now if you see here Q concentration is decreasing that cannot be there because Q concentration has to increase now coming to these two options right so now as the uh graph is growing exponentially if you have seen the graph for this one will be in this form okay exponential graph so for reactant it is in this form for product it is in this form okay it is not in this form the a option is in this form okay so that is incorrect so the correct answer for this question is option number B next question uh for a reaction 2 n + H2 gives N2 + 2 H2O the possible mechanism is given what is the rate Law and Order of the reaction so if you know the rate law is always dependent on the slowest step of the reaction right so in this three you can see this is the slowest step so I can write rate is equal to K into n2o2 into H2 right and the Order of the reaction here will be because this is one and this is one so we can see order two here order two here so this will be the correct answer that is your option number B very simple question next the energies of activation of forward and reverse reactions of A2 + B2 gives 2 AB are 180 K and 200 KJ per mole respectively okay so now let us draw this diagram first very important question this one so this question is important here uh first is your A2 B2 A2 + B2 is there and we have 2 AB okay so this is your reactance and I will have here 2ab okay the product site now the activation energy of the forward reaction is 180 K so this this is your activation energy value okay this is how much given 180 K and next uh and 200 k for the reverse reaction so for reverse reaction the activation energy will be 200 Kil right now the question is asked if the presence of catalyst the activation energy of both the reactions uh by reduces by 500 sorry reduces by 100 K right so the activation energy is decreased so we will draw another graph here okay the same one but the activation energy will decrease here so now suppose it was earlier 180 now it is decreased to 100 times means now it is 80 so we will draw a reaction like this so I'll use a different pen here so this activation energy this value okay is 80 and the another one which was 200 now this one will become 100 right clear now if you see the question is asked what is the uh um enthalpy change of the reaction so the enthalpy change of the reaction will be - 20 okay 80 by 80 - 100 that is your minus 20 so the correct answer for this question will be option number a okay so that is correct now let us go to the next question the plots are shown below between concentration and time T which of the given orders are shown by the graph respectively so if you see here the graph is versus A Min - x and t okay so a - x is nothing but the final concentration so in zero order reaction we have a is not - A is equal to KT now if I want to draw a graph for a I'll take this side then I'll get A- K into T right so this the negative slope value is minus K so this is for which order zero order reaction right so this two options we can eliminate now if you see 1 by a so for this second order reaction if you see the value is 1 by a - 1 by a is equal to KT so if I want to calculate here 1 by a that value will be equal to k t + 1 by a so here this is a positive slope okay with the Y intercept so this is your second order reaction the correct answer for this question is option number c okay so let us see the next question in a reaction 2x to Y the concentration of X decreases from 0.5 to 0.38 in 10 minutes what is the rate of the reaction so very easy question this one so we know rate is equal to change in concentration change in concentration by change in time okay so now what is the concentration change that has taken place 0.50 - 0.38 and you can see the unit is in seconds so 10 minutes we will convert it into seconds so now we will get uh 0.12 / by 600 or we can write it as 12 by 600 into 10 ^ of 2 so I'll get uh 1 by 5 that is 0.2 into 10 ^ ofus 3 or I can write it as 2 into 10 ^ of - 4 so the correct answer for this question will be option number a okay so easy one next for the reaction this one then the correct relation between K1 K2 and K3 okay very important question this one first of all let us write the uh differential rate equation so 1X 2 of D NH3 by DT okay will be equal to dn2 by DT will be equal to 1x3 of so I'll write in this one 1 second we'll write it the full equation we can write on the other side of this one I'll get - 1X 2 D NH3 by DT okay is equal to D N2 by DT is equal to 1x 3 D H2 by DT right now you see what is the value of D NH3 by DT minus D NH3 by DT is k1 NH3 right so I can write here 1 by 2 and minus D NH3 by DT is k1 to the K1 into NH3 okay and then dn2 by DT is K2 into NH3 so just substitute this one and then D H2 by DT is K3 NH3 so 1X 3 K3 NH3 okay so now you can cancel all NH3 is from the equation so I'll get 1X 2 of K1 isal K2 isal 1X 3 of K3 okay so now you can like um take the LCM of 2 and three and just that is six so multiply the whole equation with this one so I will get 3 K1 is equal to 6 K2 so I'm multiplying the whole equation with 6 okay just to see the options are not in the fraction form so to get the fraction form we'll get this one 3 K1 6 K2 and then we will get 2 K K3 now you see none of the option is matching this one so but you can see if we divide this equation the whole equation with two so I'll get 1.5 K1 is = 3 K2 isal to K3 right so that is this option so the correct answer for this question is option number c okay so little tricky question but a good question from Kinetics okay next during the kinetics study of the reaction 2 A + B gives C plus d the following results were obtained based on the above data which one is the following is correct so you have to find the rate law so you did not solve it just you have to understand here so we want to find first let us first find the order with respect to B so when I want to find the order with respect to B the a con uh the concentration of a should be constant so in equation 2 and 3 you see a is constant so now what has been done to the concentration of b b concentration is doubled when B concentration is doubled if you see the rate has become four times right the rate has become here four times that means when we are doubling the concentration and the rate is becoming four times so that is a second order reaction so second order reaction means with respect to B it is a second order so we can eliminate these two options now if we see the First with respect to a b should be constant so in these two reactions in these two reactions 1 and four the concentration of B is constant now concentration of a is changed so what is done this concentration of 1 and four has been four times like from0 1 to point 4 it has been like four times and what has happened to the reaction if you see the reaction has also become four times right so if the concentration is four times reaction is also four times then it is a first order reaction so here this with respect to a it is a so I can write with respect to a it is first order with respect to B it is second order so k a b squ so correct answer for this question is option number D okay next okay what will be the rate equation for the reaction 2x + y is equal to Z if the order of the reaction is zero so if order of the reaction is zero so here the order is 2 here the order is one here the order is 1 but here the order is zero so the correct answer for this question will be option number B next if the rate equation for the reaction is k a b half the order of the reaction is so we know order of the reaction is what the sum of the power of the co concentration terms so 1 + 1 by 2 that is your 3x2 that is 1.5 so the order will be here 1.5 right next question an endothermic reaction with high activation energy for the forward reaction is given by the diagram okay so if you know this one in endothermic the product energy will be higher right this will be the diagram so the diagrams where product energies are less can be eliminated now the question has also mentioned High activation energy so here the activation energy is very high so the correct answer for this question will be option number c okay very easy one next a reaction is given as 2 p + Q gives product if the concentration of Q is kept constant and the concentration of p is doubled then the rate of the reaction is so if I want to write the rate law for this one I'll write k p² q see every time the coefficient will not be equal to order but for elementary reactions we can take where the coefficient of the reaction or the reactance is equal to its um order of the reaction so now what is told p is doubled so 2 p q right so that will be equal to four times okay so the reaction is quadruple so the correct answer for this question is option number c okay next 3 a + 2 B the rate of the reaction d by DT is equal to so we can write first the differential rate law - 1 by3 sorry uh - 1X 3 da by DT okay will be equal to 1x 2 DB by DT okay so now what will be equal to DB by DT is equal to - 2 by3 Da a by DT okay so that is your opt option number B so correct answer for this question will be option number B very simple and easy question next one 90% of a first order reaction is completed in 70 minutes the velocity constant of the reaction is velocity constant is nothing but your rate constant okay so first order reaction we know the integrated rate law for first order K is equal to 2.303 by t log of a by a right so now if you see 2.303 * 70 log of 10 90% is completed okay then how much will be left 10% so log of 100 by 10 that is log 10 which is 1 so we need to calculate for 2.33 by 70 okay so the correct answer for this question will be option number a that is 0.329 okay you can just see 70 70 into 3 will be 210 so something related to so that will be 03 okay next uh this is your hydrolysis of acid hydrolysis of eser so we all know this is an example of pseudo unimolecular reaction so correct answer for this is option number a direct fact based question next how enzymes increases the rate of the reaction so enzymes provide an alternative pathway which reduces the activation energy of a reaction so that is why they do so by in by lowering the activation energy okay so the correct answer for this question is again option option number a very easy one next question is okay this is one important question for a reaction X to Y the graph of the product concentration versus time okay the graph of product concentration versus time is a straight line passing through the origin okay okay so now if I see we have this equation um if it is a straight line passing through the origin product concentration divided by time okay now if you see I can write rate is equal to for a uh zeroth order reaction if I take is equal to d p by DT is equal to K right so now this one is a slope is K and this is a straight line passing through the origin now the question is asked hence the graph of minus DX by DT and the time would beus DX by DT is your reactant right so for a zero order reaction it doesn't depend on the concentration of the reactant right that is what is the meaning of zero order reaction so how the graph will come with reactant concentration and time if you take it will be a straight line it doesn't depend on the initial concentration of the reactant so it will be a straight line parallel to x-axis so the correct answer for this question is option number c okay next in a chemical reaction a + 2 B gives product so we have a + 2 B gives product the concentration of a is double the rate of the reaction increases four times so when you are doubling it it is becoming four times so with respect to a it is a second order reaction and when the concentration of B alone is double the rate continues to be the same that means with respect to B it is a zeroth order reaction right so what will be the overall order of the reaction 2 + 0 that is 2 so the correct answer is option number B next the rate of a chemical reaction doubles for every 10° rise in the temperature if the temperature increases by 60° C the rate of the reaction increases okay so this is from your temperature coefficient part right so for every 10° like for every 10° the rate doubles right so just you have to calculate from 0° to 60° okay so from 10° 20° 30 40 and then 50 right so 1 2 3 4 5 and 6 six times we have done so that means 2 ^ of 6 that is 64 so the correct answer for this question is option number c that is 64 next an endothermic reaction A to B has an activation energy of x k per mole okay so let us draw this we have an endothermic reaction so in endothermic reaction the graph will be like this this is a this is B the activation energy is x k per mole okay next if the energy change of the reaction is y so this is the energy change energy change value is what energy of the reactants energy of the products minus the energy of the reactants so this one will be y okay next the activation energy of the reverse reaction so we have to calculate this value okay so this total value is X this value is y this is the activation energy of the reverse reaction so this value is your what x - y so the correct answer for this question will be option number B okay very easy one total value is X this one is y so what will be this value this value is nothing but x - y next A Plus BQ product so there's a reaction when p is doubled keeping Q constant the rate increases by two times so you're doubling p and the rate is also becoming double so with respect to P it is a first order reaction then when p is constant and Q is double the rate increases four times so when you are doubling q but rate is becoming four times then it is a with respect to Q it is a second order reaction then what is the overall overall order of the reaction that is 2 + 1 3 so the correct answer for this question is option number c next for a reaction between A and B the order with respect to a is 2 okay and the other with respect to B is three okay so this is the rate now what you have done the concentration of both A and B is doubled so we will have our 2 a and then 2 B okay so we have here so I can write 4 a² and 8 B Cub okay so we'll get 32 * K a² and B CU okay so the correct answer for this question will be 32 the rate has increased Now by 32 times okay so easy question next a reaction is 50% complete in 2 hours and 75% completed in 4 hours the order of the reaction is so if you see here 100 was there okay initially 100% was there then 50% was completed in 2 hours okay the next half that is 50 + next half will be 75 means from 50 to 25 that will again converted into 2 hours so now 100 to this one is your 75% so now if you see the half life is not depending on the initial concentration when the initial concentration was 100 the half life was 2 hours when the initial concentration is 50 then also the half life is 2 hours so in this occurs in your first order reaction right so the order of the reaction is first order correct answer is option number a okay next the time required for 100% completion of a zero order reaction is so for zero order reaction what what is the value a minus a is equal to KT right now if you see that the final concentration this in 100% completion of the reaction so I can write a not this one will be zero is equal to KT right now T value will be equal to a by k or a by K so the correct answer for this question is option number a next decomposition of NH3 on the surface of platinum has a rate constant of 2.4 into 10 2.5 into 10 ^ ofus 4 mole decim Cub second inverse okay the order of the reaction is so this is how we to the formula to calculate the rate uh order of the reaction using the rate constant so you should know the general formula of rate constant that is mole 1 - n l nus1 second inverse right so in this question what is the unit of moles moles unit is 1 so I can write mole 1 in place of liter 1 liter is 1 decim Cub okay so I can write decim cub and that is -1 2 inverse so we can equate these two so 1 - n is = 1 so n value will be equal to 1 - 1 that is 0 okay so the correct answer for this question will be option number a that is it is a zeroth order reaction next the rate equation of the gaseous reaction is given by K AB if the volume of the reaction is suddenly reduced to half so if you know what is the formula of concentration or marity is number of moles by volume so if volume is reduced to half concentration will increase so if it is half it will be doubled so now when the volume is double what has happened the concentration will become two times now okay so that will be equal to 4 * of K A and B okay so what will be the the reaction rate relating to the original rate will be four times okay so correct answer for this question will be option number B next the 6.25% of radioactive substance is left after 480 minutes the half lifee of uh half lifee period is okay so we know all the radioactive reactions are your first order reactions so I can write here k equal to 2.303 by t log of a by a okay now if I solve this what we have to put 6.25% is left so we can write 2.303 by 480 minutes okay and log of 100 by 6.25 right so I can write it as 2.303 by 480 log of 100 by 6 25 okay so we can solve this value so whatever value we get here is K now T half value is what 0.693 by K so what you have to do just substitute the K value that we get here and in this and then you have to calculate the th value so if you calculate it like if you substitute it correctly the value you will be getting around is around 120 Minutes okay so the correct answer for this question will be option number D okay next in case of reversible gaseous reaction a positive catalyst so what does a positive Catalyst do it increases the rate of both forward and backward reaction by increasing the activation energy so activation energy will decrease whenever the rate of reaction has to increase so increases the rate of both forward and Rea uh backward reaction by decreasing the activation energy so option b is the correct answer next with respect to the velocity constant means that is your rate constant K of a reaction which one of the following statements is not correct this is very important when questions is asking incorrect correct so be very careful there so it is a measure of velocity of the chemical reaction yes the statement is correct it is about the rate or the how fast a reaction is second it is dependent on the initial concentration of the reactants no rate constant depends only on the temperature it depends on the temperature this is correct statement during the progress of the reaction even though the velocity decreases the velocity constant remains constant yes rate constant value is always same for a particular reaction so the incorrect statement for this is option number B so the answer for this question will be option number B next which one of the following is an example of pseudo first order reaction so we have two examples if you remember alkaly hydrolysis is incorrect acid hydrolysis of Esther is an example of pseudo first order reaction and another is your inversion of sucrose that is acid hydrolysis so correct answer for this question will be option number B okay yeah so this were the top 30 questions or these were the previous year questions that were asked in your Comet K exam so you can see mostly the questions are asked from the topic order of the reaction and then to calculate the rate law very few questions are like from the rnes equation is there so just practice such questions and if you have any doubt in any question please comment down we'll be addressing it in the next video okay so practice hard prepare hard for your exam all the very best if you have found this video helpful so please uh share like And subscribe our channel for such informative videos for your preparation thank you

Channel: Deeksha Karnataka

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Id: mGBcU81WTw8

Length: 28min 4sec (1684 seconds)

Published: Thu May 02 2024