COMEDK 2024 Chemistry: Top 30 Thermodynamics Questions | COMEDK Chemistry Revision

hello students welcome back to our Channel dsha Karnataka so your Comet K 2024 exam is approaching so we have started a series where we are discussing the top 30 questions from all the important chapters from chemistry right so in this video we'll be discussing the top 30 questions from one of the most important topics from physical chemistry that is thermodynamics right so now if you uh see like it's a moderate weighted chapters where you can expect two to three questions from this chapter but since the chap like syllabus is deleted so so we can explore little higher questions so the formulas are very basic the concepts are very basic just the point is this chapter needs little uh like more focus on calculation okay so if you are able to solve these questions we have already covered like almost all the concepts from this chapter so you can assure that this chapter is ready for your comment K 2024 examination right so let us see the first question now the first question is what is the work done when one mole of a gas so what is the work done when one mole of a gas expands isothermally from 25 L to 250 L at a constant pressure of 1 atm and the temperature is 300 and kin so this question is direct formula based question from the work done so we have this formula work done is equal to - 2.303 okay n value is 1 then r value is 8.314 temperature is 300 and Kelvin log of we have log of uh V2 by V1 right so 2250 L by 25 5 that is log 10 so 1 once you solve this you'll be getting the answer around 5744 JW okay so the correct answer for this question will be option number B now let us see the next question when 1 mole of h204 is completely neutralized by sodium hydroxide the heat liberated is 11 14.64% right this is called as the enthalpy of neutralization and the value here is heat is liberated so it is -4.6 4 now if you see consider one mole of h204 will give you 2 mol of H+ so um sorry not this one this value we have to calculate okay now according to the question in one mole of H2S we have 2 mol of H+ and it will react with 2 moles of O minus to give you 2 mol of water right sorry we'll get uh yeah so now if you see the Delta H for this value is given to you as -4.6 but then for 1 moles it will be divided by 2 so we can just divide by this one with two so we will get - 57.3 2 correct answer is option number c okay now let us see the next question the entropy change for a non-spontaneous reaction is this much the reaction is okay so now if the uh we have this formula Delta G from gives fre energy Delta G is equal to Delta H minus t Delta s right right now Delta g non-spontaneous means this value is positive and this value is also positive now to get this one as a positive value see if the reversible and irreversible we cannot say on this basis but we can say about exothermic and endothermic so if it is exoic Delta H will be negative if Delta H is negative then both the negative terms will give you Delta G as a negative and the reaction has to be spontaneous but according to the question this value is positive and Delta H minus t Delta s because Delta s is positive now if you see so generally it will depend on temperature but Endo exothermic cannot be the answer because once it is negative the whole equation will become negative answer but if you take endothermic then it depends on the temperature whether temperature is high or low so this reaction the Delta H has to be endothermic okay that is Delta H value has to be positive so the correct answer for this question will be option number D now coming to next question which of the following statements regarding the gives free energy is correct so we know Delta G should be negative then the reaction is spontaneous so Delta G is negative then the process is spontaneous so correct answer is option number c now let us see the next question what will be the amount of heat released by burning 10 lit of methane under the standard conditions given the mass given the heat of formation of CH4 CO2 and H2O are- 76.2 398.36 okay so now if you calculate here uh we have been given with the heat of formation of these thing and the question is asked about the combustion of methane right so this we have to use H's law summation so when you see the heat we have this reaction CH4 + 22 gives CO2 + H2O right so that is reaction first should be balanced now if you see we have to calculate we have been given with the heat of formation of all now I want to calculate the enthalpy of this reaction okay so now if you calculate the enthalpy of this reaction so that will be the enthalpy of formation of CO2 product minus reactant so that is enthalpy of water so 1 second for this let us write the reaction formula in detail so if you see this one we will have the enthalpy of reaction first we will calculate the enthalpy of reaction is Delta H not formation of products minus Delta H formation of reactants right so according to this question we will get Delta H formation of product is CO2 right plus 2 * Delta H formation of water okay then minus products minus reactants so O2 value will be zero because it is in the native state so that will be equal to Delta H formation of the reactant is CH4 okay now you can can see the values are already given to us for CH4 it is- 76.2 for so you can write down the values separately for avoiding confusion so Delta H formation of CO2 is - 39 8.8 Now Delta H formation of CH4 is equal to - 76.2 and then Delta H formation of um we have the this one is H2 okay so H2O is - 241.7 okay so now what we can do is we can substitute the values here so now let us substitute the values and calculate so one second okay so now if you calculate the values here what you will get so for CO2 it is - 3 98.8 so I'll substitute here - 3 98.8 plus 2 * of water like that is 2 * of - 241.7 okay then uh let us check the values once again for CO2 it is - 398.62 41.6 then minus enthalpy of formation of CH4 is - 76.2 okay so once you solve this value okay if you solve this value we will get the Delta H formation okay of um sorry Delta h of this reaction will be equal to - 85.8 okay but the question here is you have to see the question is asked for 10 L okay it is so now if this is under standard conditions so this methane is 1 mole so 1 mole under standard conditions is 22.4 L volume so I can say that value whatever you have got- 85.8 is for 22.4 L but we have to calculate for 10 L so now if I want to calculate here I can say for 20 2.4 L the value is - 85.8 for 1 lit it will be - 85.8 by 22.4 and then for 10 L it is equal to - 85.8 by 22.4 into 10 right so now if you solve this the correct answer for this question we will be getting is option number c that is 359.06 here you can be little confusing once you get 8005 can directly put this value but you have to be careful that it has to be 39.7 you will get minus 359.06 this is your H's law of heat summation so correct answer for this question is option number c now the equilibrium constant for the reaction is 0.8 at 298 Kelvin the standard free energy change of the reaction at the same temperature is so you know Delta G formula is equal to minus 2.303 okay nrt nrt log of K right so we have been given with uh for this one so we can take it as one mole since it is not mentioned so we can take - 2.303 into 8.314 into the temperature is 298 K into log of 8 into 10^ ofus 3 now you can remember this value okay so because most of the time this value will be coming so it will be around 5.7 - 5.7 K okay so - 5.7 K and then we have this side log 8 okay log 8 is 3 * of log 2 so that is equal to .9 0.9 - 3 so we will get around 2.1 okay and this 5.7 I can round it of something around 6 so we'll get 6 into 2 that is somewhere around -2 okay so no sorry this will get - 2.1 so we'll get + 12 this is minus this is minus we'll get + 12 so the correct answer for this question will be option number a okay so this value you can remember for 298 Kelvin because most of the time the temperature is around 298 kin that will be around 5.7 K now coming to the next question 1 mole of a perfect gas expands isothermally and reversibly from 10 decim Cube to 20 decim Cube at 300 kin Delta U value you have to calculate q and W so isothermal reversible process first we can calculate work done okay this question you can solve in another way by seeing the options also so now if you see here the gas is expanding right the work is done by the gas so it will have uh like uh negative work done will be there so you can see in all the other options the work done are given positive right so these two options cannot be there none of these there's like less chance for any answer to be none of this but so we can like most probably the answer should come out to be a but we can check with calculation of work done - 2.303 n is again 1 Mo R is 8.314 then temperature is 300 and Kelvin then log of v2- V1 log 20 by log 10 so now if you solve this we'll get - 1726 K okay so the correct answer for this question so if you got now for ismal process Delta U is equal to Z right for isothermal process Delta U and Delta H will be always equal to zero now if you see this one is Z then I can say Q is equal to we have this formula Q + W so Q will be equal to - W so minus of work done is - 1.76 726 so we'll get + 1. 726 right so the correct answer for this question will be option number a now next question based on the first law of thermodynamics which of the following is correct for isocoric process we have this formula Delta U is equal to Q + W for isocoric process work done will be zero then Delta U should be equal to plus Q This one is wrong for adiabetic process Q is equal to Z so Delta U should be equal to plus W not minus W so that is incorrect for an isothermal process Delta U is equal to Z so Q should be equal to minus W so this relation is also incorrect now for a cyclic process also Delta U is equal to 0 for cyclic process also Delta U is equal to 0 so Q should be equal to minus W so correct answer is option number D okay next question when two moles of ethane are completely burnt 3129 K of heat is liberated if Delta H for CO2 h and H2 are this much given values per mole respectively then the heat of formation of C2 H6 is so again the question is from hes law first understand the question 2 moles of ethane are completely combusted the heat liberated is given so now we can first write down the ethane reaction so C2 H6 + O2 gives um 2 CO2 sorry 2 CO2 and we have three Water 2 CO2 2 + 3 water and then it will be if you balance this one we'll get oxygen will be 7x2 right now the question is for 2 moles so we can write for 2 moles the value is 3129 then for 1 mole it will be minus 3129 by 2 okay so because this is for two moles I took it for one Mo now the question is asked another values are given Delta H formation of CO2 so if I want to form CO2 the reaction will be equal to the Delta H value should be equal to - 395 and another reaction is formation of H2 sorry water so H2 +/ O2 is also equal to H2O so this value is - 286 okay now if you see here per mole respectively then heat of formation of C2 H6 so we need to calculate the heat of formation of c26 so what is our required reaction so C2 H6 we have to form right so we need 2 carbon and 3 H2 okay then we can get this one so you can see how we can arrange this equation from here so what you can do is multiply the equation 1 with two so we will multiply equation 1 with two so what I will get 2 C + 22 is giving me 2 CO2 and the Delta H value for this one will be - 395 into 2 next one is we need 3 H2 so multiply the second equation with 3 so I'll get 3 H2 + 3x 2 O2 okay is giving you uh we'll get around uh 3 H2 okay so this value will be minus 286 into 3 okay now you see if you solve this one we need C2 H6 on the right hand side of the equation so we can reverse this equation and write down so I will write the third equation is so I have multiplied second equation with three okay and this is my third equation okay this is my first equation second equation so I multipli the first equation with two to get two carbon I multiplied the second equation with three so to get three hydrogen then I need C2 H6 on the product side that is in this side but c26 here is on the left hand side so we can reverse this equation so if I reverse this equation third equation I have revers so if I add that one what I'll get so 2 CO2 right 2 CO2 + 3 water gives C2 H6 + 7x 22 now here the Delta sign will also change so it will become + 3129 by 2 now you just add these reactions and check whether we are getting the required reaction or not so if you check here uh 2 CO2 2 CO2 is getting cancel three water three water will get cancel now what is remaining 2 C is remaining okay and then we have see 2 22 + 3x 22 so that is 7x 22 so this 7x 22 and this 7x 22 will also get cancelled so what is remaining now for the equation we have 2 c 2 C + 3 H2 okay so that will give you C6 C2 H6 okay so that is what our required equation is so we have to just add the same values for the enthalpy so what calculations we be requiring so our answer will be - 395 into 2 okay plus - 2863 into 3 then we have + 3129 by 2 okay so the after solving this one the value you'll be getting around is - 83.5 K so correct answer for this question is option number B okay so now let us see the next question when two moles of uh Ethan are okay so this is the same question so next question let us move question number 11 the heat of formation of ncl3 in terms of H1 H2 and H3 is okay so let us first write down the reactions again we have ncl3 formation so I want to form N2 I want to form ncl3 so whenever you write heat of formation equation the elements must be in their native state right so N2 should be there and then chlorine should be there now for balancing this one I can write 3x2 cl2 right so from these three equations we have to get this equation and then we have to calculate the Delta H value so now let us write all the three equations that is given to us so I'll write the second equation first so second equation is N2 + 3 H2 gives 2 NH3 okay and the Delta H value here is minus H2 now third reaction I'll write third reaction is H2 + cl2 okay here also I need to multiply half right this is 1 n so H2 + cl2 gives 2 HCL and this equation is plus H3 okay now you see we need half N2 so multiply equation 2 with half so what I'll get equation 2 into half so if I do then I will get half n + 3x2 H2 gives NH3 okay this is the reaction now multiply we need 3x2 cl2 so multiply this equation with so third I'm multiplying with 3x2 okay so if I multiply the third equation with 3x2 so I'll get 3x2 H2 okay and then 3x2 cl2 so 3x2 H2 + 3x2 cl2 and we'll get 3 by 2 into 2 so we'll get 3 HCL 3 HCL okay so now what I have done so I have multiplied the the equation 2 with 1X 2 and equation 3 with 3x2 okay so now I can write down these two equations separately so I'll write here the equations so what I have done I have multiplied 1 by 2 N2 + 3x 2 H2 gives n H3 right so what will be the Delta H value now you're multiplying 1 by 2 so I'll get - 1 by 2 Delta H2 now equation multiplying here is 3x2 so 3x 2 H2 plus uh then again we have 3x2 cl2 okay so we have just multiplied the same thing here 3x2 H2 + 3x2 will give us 3 HCL right 3x2 2 two will get canceled so we will get 3x 2 Delta H3 okay now what you can do is you can subtract these two equations okay so now if you subtract these two equations so what I will get I'll subtract equation 2 minus equation 3 okay 2 - 3 so what I'll get 1X 2 N2 okay then 3x2 H2 3x2 H2 will get cancelled - 3x2 cl2 will give here if you see NH3 - 3 HCL okay so this is my equation number four that is equal to equation number four four now what you can do is add equation number one with equation number four so what is there so what is my equation number four after subtracting this one I got half N2 okay and we got - 3x 2 cl2 and we got here NH3 okay now why we are doing this one because we need N2 and cl2 on the product on the reactant side right now we need to add the third equation that is NH3 Plus that is your first equation sorry NH3 + 3 cl2 gives ncl3 + 3 HCL okay we have this one also minus HCL right minus 3 HC now if you add these two equations so just check the equations properly now if you add these two equations we can see this NH3 and this NH3 we can cancel okay now next what we have this one plus h C and minus HCL we can cancel and now if I have this one 3 - 3x2 so that is equal to 3x2 right so this is 3x2 HCl so I can write it as 3x2 chlorine so yeah so now you see half N2 okay and then this is 3x 2 chlorine and here we are left with ncl3 so if you see in the first slide I have written this was our required equation right so that is what we have done here so what is my Delta H enthalpy value so for the fourth reaction the Delta H value was equal to - half H2 and then plus 3x 2 H3 okay and this Value First equation Delta H value is minus Delta H1 so if you see here minus Delta H1 so we have to add all the three equations now so minus Delta H1 - 1X 2 Delta H2 + 3x2 Delta H3 so now if you see the options here we have minus Delta H1 okay - 1 by2 of Delta H2 and then - 3x2 of Delta H3 okay sorry yeah- 3x2 of Delta H3 so correct answer for this question will be option number a little lengthy calculation you have to be very careful and focused while doing such questions okay now coming to next question for the gas phase reaction pcl5 giving you pcl3 + cl2 which of the following options are correct so now if you see when a substance is breaking we have to provide energy endothermic so Delta H has to be positive okay so these two equations are there and now if you see there was one reactant gaseous reactant and now it is forming two gaseous products so entropy is increasing or the randomness is increasing so Delta s is also positive correct answer will be option number D now next question for the five moles of a gas are put so you have to just tell the process about AB BC and CA so volume versus temperature graph is there so now AB is at constant volume constant volume processes are called as isocoric right so we can eliminate these two options now if you see AC right so AC is at constant temperature so that is isothermal so we can eliminate this option also and when volume and temperature varying pressure is kept constant so BC is is isobar correct option is option number a next the process is spontaneous at a given temperature is so we have this formula again Delta G is equal to Delta H minus t Delta s right spontaneous means Delta G has to be negative now if this one has to be negative then Delta H should be negative and Delta s should be positive so I'll get minus Delta H minus t Delta s so addition of two negative numbers will give you negative only so the correct answer should be Delta H should be less than zero and Delta s should be positive so correct answer for this question is option number c now coming to next question the standard enthalpy of the decomposition of n204 to N2 NO2 is 58.4 K and the standard enthropy of the reaction is 16 76.7 JW per Kelvin therefore the standard free energy change in this reaction is so very simple question we have been given with Delta H value Delta H not value is 50 8.04 KJ and entropy value is also given standard entropy value is 1767 into 10^ ofus 3 Jew so this one we have to be very careful this is in Jew so we have to make it KJ because both has to be in the same units and the temperature is 298 K 25° C 25 + 273 so just now we have to substitute the formula Delta G is equal to Delta H - t Delta s so enthalpy is 58.4 minus temperature is 298 and Delta s is 1.76 sorry 176 we'll have uh 176 1 second okay so we will get this one as 176 into 10 minus 0.176 okay so now if you solve this value we'll be getting the value we'll be getting around is 5.39 okay so we'll be getting around 5.39 positive value we'll get so the correct answer for this question will be option number c okay now let us see the next question a reaction is spontaneous at a temperature when so just now we have discussed this one right so Delta H should be negative and Delta s should be positive so that at every temperature the addition of two negative numbers will be coming to be negative so the correct answer for this question is option number a okay next question is from your bone Haber cycle so for one mole of NAC the latis enthalpy is okay so now if you see from First Step okay very important question na+ to na gas is getting converted that is your 11 18.4 so I can write that so I can write first of all let us um if I write the equation so first is your Delta h of the reaction okay is equal to Delta h of ioniz so first this one is sublimation okay then what is done here na+ gas is converted to na gas uh na+ Na gas is converted to na+ gas so that is what ionization enthalpy so Delta h of ionization enthalpy now if you see chlorine now cl2 gas is converted to CL gas so bond is broken that is bond dissociation enthalpy so Delta h of bond dissociation enthalpy next you see chlorine is getting one ion CL minus so that is your electron gain enthalpy so plus Delta h of electron gain enthalpy now if you see na+ gases formed NaCl formed is next one is this one conversion is the ltis enthalpy so plus Delta h of ltis enthalpy now this one this Na and cl2 this reaction in this reaction if you see NaCl is formed from the reactants in their native state so when one mole of a solid is formed from the reactants in their native state that value is called as your enthalpy of formation so enthalpy of formation here is how much - 41.2 so now we can substitute all the values clearly so if I substitute this one we'll get uh - 41.2 okay is equal to sublimation enthalpy is 108.4 plus ionization enthalpy is 495.00 okay and then we have next is your ltis enthalpy okay so 1 2 3 4 values we have already substituted and then lce enthalpy will solve so now once you solve this we'll be getting around - 788 K per mole Okay so so the ltis enthalpy will be getting around is - 7 788 K per mole so correct answer for this question is option number a okay so this reaction you should know but it is like addition of the intermediate steps now next question an endothermic reaction is found to have positive entropy change the reaction will be okay so endothermic reaction we have this formula on the sign of Delta G we have already seen three questions endothermic reaction is positive mean Delta H has to be positive right then uh the entropy is also positive right so that means the equation is in this form right now when this value will come out to be negative we need the reaction will be like at what temperature the reaction will take place so for taking place it has to be spontaneous so the temperature must be very high so that this second term is a very big term so the addition of one positive term and one negative term comes out will negative if only one term is much more higher right so then the temperature value should be very high so the reaction is possible at high temperature correct answer is option number a now next question for an adiabetic change in the system the condition which is applicable is for for adiabetic condition we have q is equal to Z so the correct answer for this question is option number D next standard M enthalpies of formation of C3 Co and CO2 are given the Delta HR for the decomposition or reaction of the C3 so if you see we have this reaction C a CO3 decomposes to give you CAO plus CO2 right now the the M enthalpies of formation is given so I can write again the Delta h of the reaction is equal to the Delta h of the products minus the Delta h of formation of the reactants right formation of products minus reactants so I can write here Delta H formation of CAO plus Delta H formation of CO2 right these to our products clear minus Delta H formation of uh we have ca3 okay so now just substitute the values so first we can write the values here correctly without for C CO3 the value is - 120. 692 okay then for CAO the value is um 6359 okay and then for CO2 CO2 the value is - 393.130 n okay and then for CO2 we have - 3 93.5 1 then minus for C A3 we have 120. 692 right so now if you uh solve this okay so you'll be getting uh the value you'll be getting around is 178.6 okay the correct answer for this question will be option number a okay so little calculation is involved in this question so now let us see the next question this is very important question okay to understand and how to solve it so it is not a very direct question little trick based question is there when 5 L of a gas mixture of methane and propane is perfectly combusted at 0° c and 1 atmosphere pressure so that is the standard conditions right 16 L of oxygen at the same temperature and pressure is consumed so what it is telling we have a mixture of gas so first let us understand the question we have a mixture of gas where it you have CH4 and C3 h8 this total volume is 5 L now this was combusted under uh standard conditions because 0° c and 1 atm pressure is given so this one will react with O2 so this volume of o2 that required total was 16 L okay now if you see here so the calculation you have to calculate the amount of heat released so when CH4 will react with O2 for that some amount of heat will be released for C3 h8 will react with O2 for that some amount of heat will released so you can see combustion of methane is given combustion of propane is given to you so now what is basically you have to calculate is you have to calculate or you have to find out the number of moles of CH4 here and the number of moles of C3 h8 here okay so let us see this question okay so first first uh first of all we will calculate the number of moles of total number of moles of gases okay which gas this gas is your both C3 H like methane and propane so the total volume you know 5 lit so given volume by by m volume 22.4 L is your total number of moles of gas so this value will'll be getting around 0.22 moles okay now sorry uh yeah so now we'll get around 0 22 moles now we will also calculate the number of moles of oxygen required so that was your 16 L right 16 divided by 22 16 lit is the given volume and then divided by 22.4 so we will get around 0.71 Mo okay now let us write the reaction for both okay so if you know the reaction here for methane it is CH4 + 2 O2 will give you CO2 plus 2 water just check whether it is balanced or not four hydrogen there is four hydrogen and there is four oxygen now for ethane uh sorry propane C3 h8 plus O2 will give you we have 3 CO2 and we will get four water okay so how many oxygen are there uh 6 + 10 so we will get 52 here right now let us assume that the number of moles of CH4 is X right the total number of gaseous moles is 0.22 then what will be the number of moles of C3 h8 0.22 - x right now if you see one mole so how much oxygen will be required here so here if you see 1 mole of CH4 requires 2 moles of oxygen so X moles of CH4 will require 2x moles of oxygen so moles of oxygen required here will be 2x right now even if you see this equation Now 1 mole of propane requires 5 mol of oxygen 1 mole of propane requires 5 moles of oxygen so this much moles 22 - x moles of propane will require five times of this one oxygen required here is 5 * of 0.22 - x so if you did not understand that I'll repeat here this one see 1 according to the balanced equation 1 mole of CH4 requires 2 moles of oxygen right now we have assumed that let the number of moles of uh CH4 is X then X moles of CH4 will require 2 x mol of o 2 like normal unitary method now the total number of gaseous moles was 0.22 right now if x if CH4 is X then what will be the remaining amount of propane total number of moles minus X that is 22 - x so I can say now according to again balanced equation 1 mole of C3 h8 okay reacts with 5 moles of oxygen right then we have how many moles of C3 H22 - x moles of C3 h8 will react with one mole reacts with five so these many moles will react with 5 into 22 - x right so that is what we have written so the number of moles of oxygen is 2X and 5 into this one now we have already calculated the total number of moles of oxygen right so now here we will calculate the moles of oxygen so I I'll erase this part okay so if you now see the first number of moles of oxygen for methane was 2X and for this one for propane we got 5 * of 22 - x so according to the equation that we have got total number of moles of oxygen is 0.71 so I can write 2x okay + 5 into 22- X this is the number of moles of oxygen required for the combustion of methane and this is the number of moles of oxygen required for the combustion of propane is equal to how much that is your 0.71 okay so now you can solve for X okay so just it is a linear equation okay so once you solve for x you'll be getting around 0.13 okay so now what I can say now we have the number of so what is the number of moles of methane number of moles of methane is 0.13 and number of moles of propane C3 h8 will be 0.22 minus 0.13 right so we will get U 0.09 right so we will getting the number of moles of propane is 0.09 and number of moles of methane is 0.13 the question is still not not over the question has asked you to calculate the enthalpy of combustion so now we'll go to this one and we'll solve here so till now what we have done so if you see we have explained the question here so if you see here now we have the number of moles of CH4 I got already have soled it is 0.13 and number of moles of C3 h8 which I got was 0.09 now this value combustion of CH4 is 890 K per mole means if you come if you burn 1 mole of methane the energy releases 890 now if we are releasing .13 so the total so I can say the heat liberated from CH4 will be equal to13 into 890 and the heat liberated from C3 h8 will be C3 h8 is what from 1 mole we are getting 2 220 so we have these many moles so this into this so that is 09 okay into 2220 now once you solve this and you add it you'll be getting around so now we have to add both of them we'll be getting around 3177 K so the correct answer for this question will be option number B okay the question is little lengthy but more than that you have to know how to solve it okay the approach should be known otherwise we'll waste a lot of time in this question okay so now let us see the next question here okay three thermodynamical equations are given Below based on the above equations find out the relationship which is correct okay so we have these three re uh reactions C + O2 is giving you CO2 then C +/ O2 will produce Co and then we have Co +/ O2 is giving you CO2 now just see if we add these two reaction the second and the third reaction what will happen yeah if we add this one and this one this this two will get cancelled and I'll get C + O2 is giving you CO2 so the addition of second and third reaction is the uh we'll get the first reaction so I can say x is equal to y + z right this value was X this value was y this value was Z so x equal to y + z correct answer is option number B okay now let us see the next question the heat of formation of water liquid is - 286 K the heat of formation of H2O gas is likely to be okay so this question is little I'll say this is a tricky question you have to know some analysis you have to do so heat of formation of water liquid right so what is it formed of H2 +2 okay H2 +/ O2 will give you H2 and this value is given to you as - 286 now H2 liquid to gas H2O liquid to H2O gas right so this value we have to calculate the heat of formation of H2O gas we have to find this value okay now if you see the reaction what is the enthalpy of this reaction so the enthalpy of this reaction is H2 enthalpy of formation of H2O gas minus the enthalpy of formation of H2O liquid right so we have to find this value now if you see in this reaction we have to provide energy liquid to gas is moving so energy should be provided so that is why what happens this value has to come come out to be positive this is an endothermic process now in that case H2 formation of gas that is that we don't know we have to find this value okay then we have for liquid it is minus so we have this one so minus is minus is here and then again minus 286 so we'll get + 286 okay now you have to see this options we can substitute here and check in which case we can get a positive value uh we can get a positive value because this reaction is an endothermic process now you have to be little careful and think you have to just know that already liquid gas is formed and then we have to give some amount of energy to convert it into gas so I can say the enthalpy of gas that is water is slightly greater than slightly greater than okay not very much difference will be there the enthalpy of water liquid water right clear this one now if this is slightly gler now we can substitute the values and check if I put Delta H gas is - 286 this will become Zer - 286 + 286 this one cannot be the answer now if you substitute here 286 so it is almost double right once you it is added two times so I told you it is just slightly greater not too much difference will be there so this option we can eliminate based on the common logic okay now if you see here - 341 so if I put - 341 + 286 the answer will come negative negative answer but we want a positive answer for this one so this option can also be the is incorrect now if you substitute - 242 + 286 so we'll be getting around 4S right which will be a positive value so the correct answer for this question is option number D okay now let us see the next question the temperature of 1 mole of an ideal gas increases from 298 Kelvin when it absorbs 200 je of F at constant volume the change in internal energy is so we have this formula Delta U is equal Q + W constant volume is given so work done will be equal to zero because work done formula is pdv constant volume DV 0 work done Z Now Q is given to you as 200 J so Delta U will also be 200 J correct answer is option number a okay now next question the enthalpy of neutralization of HCL and no is x k the heat evolved when 500 mL of 2 normal HCL is mixed with 250 ml of 4 normal NaOH okay so the question here is we have this one h++ o minus the value given to you is x k now from this one 500 mL of two normal H so we know for H the N factor is one so two normal H will be equal to 2 m HCL so how many number of millimoles are there 1,000 moles okay 1,000 Mill of H+ are there and here if you multiply we'll get 1,000 moles of O minus right so that means 1,000 milles means we can say 1 M okay and here we have um also 1 M sorry we have 1 mole and here also we have one mole so in this from this so what I'm saying is from 500 mL of 2 normal HCL okay we are multiplying this two 500 into 2 I'm getting 1 mole of H+ okay and then similarly from 250 ml of 4 normal NaOH okay I'll again get 1 mole of h o minus right so when 1 mole of H+ and 1 mole of O minus will react the value is already given as x k so the correct answer for this question will be option number a okay easy question this one intensive properties intensive properties are those properties which do not depend on the amount of substance present right so the correct answer for this question will be option number a now let us see the next question the enthalpy of formation of two compounds X and Y so x value is given to you as - 36 and Y value is given to you as -72 then which one of the following statement is correct so you know once the more energy is released more stable the product will be so X so in case of Y more energy is released 72 KJ of heat is given out so y will be more stable so if you see compound Y is more stable than compound X so correct answer for this question is option number B okay very easy question this one next the enthalpy change for the reaction is given H2 + cl2 gives 2 HCL the heat of formation of HCL is okay so if you see here H2 + cl2 gives you 2 HCL the Delta H for this reaction is given to you as minus 194 now what is the definition of heat of formation heat of formation is the amount of energy released when one mole of a substance is formed from the reactants in its native state here how many moles of the product is formed 2 moles of HC so heat of formation we have to calculate for 1 mole so for 2 moles of H if the value is- 19.4 then for 1 mole it will be 1 so Delta H for 1 mole will be equal to - 194 by 2 so that will be equal to minus we'll have 97 okay so the correct answer for this question will be option number c now let us see the next question for the reaction C plus O2 CO2 Delta okay so this formula we know Delta H is equal to Delta U + Delta NG RT now if you see for the given Delta NG number of moles of products minus number of moles of reactants so we have here 1 minus so this one we cannot consider right because this is solid so here 1 mole 1 mole so 1 - 1 is 0 so this value becomes Z so Delta NG is equal to 0 so I can see Delta H is equal to Delta U so that is your option number c okay now the last question if the work done on an adiabetic wall then which of the following is true so if the work done is on the adiabetic wall then Q will be equal to Z right then Delta U will be equal to W so the correct answer for this question will be option number B okay these were the top 30 questions from the thermodynamics chapter most of the questions were taken from previous year papers and all almost all the concepts of this chapters are covered little lengthy calculations are there in this chapter so you have to invest time in this like you have to save time from other questions and have to invest in this question so you can easily expect two to three questions from this chapter okay so please prepare all these questions practice similar questions also if you have any doubt in any question you can post in the comment section so we'll definitely address it and one more thing the mock test 3 uh is going to be released today we have started the comet K mock test so mock test 3 is going to be released so please attempt it the link of the WhatsApp channel is there in the description section so join the WhatsApp Channel download the question paper and solve it okay so many more videos will be coming for your comment care preparation if you are liking these videos you founding these videos helpful so please like share and subscribe our Channel thank you and all the very best for your exam

Channel: Deeksha Karnataka

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Length: 49min 35sec (2975 seconds)

Published: Wed May 08 2024