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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
The enthalpy changes at 298 K in successive breaking of `O-H` bonds of water, are `H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1)` `OH(g) rarr H(g)+O(g), DeltaH =428 kJ mol^(-1)` The bond energy of the O-H bond isA. `498 "KJ mol"^(-1)`B. `428 "KJ mol"^(-1)`C. `70 "KJ mol"^(-1)`D. `463 "KJ mol"^(-1)` |
| Answer» Correct Answer - D | |
| 102. |
Liquid water freezes at 273K under external pressure of 1atm . The process is at eruilibrium `H_(2)O(l)toH_(2)O(S)at 273 Kand 1atm` however , it required to calculate the thermodynamic parameters of the fusion process occurring at same pressure and different temperature . Using the following dat ,answer the questions thaat follow: `d_(ice)=0.9gm //cc,d_(H_(2)O(l))=1gm //c"c",1L-atm =101.3 J ` `C_(p)[H_(2)O(s)]=36.4JK^(-1)mol^(-1)` `C_(p)[H_(2)O(l)]=75.3Jk^(-1)mol^(-1)` `DeltaH_("fusion")=6008.2mol^(-1).Alldataat 273K.` `DeltaS_("fusion")`at 263 K and 1atm will be :A. `22.01JK^(-1)mol^(-1)`B. `22.84JK^(-1)mol^(-1)`C. `21.36JK^(-1)mol^(-1)`D. `20.557JK^(-1) mol_(-1)` |
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Answer» Correct Answer - d |
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| 103. |
Liquid water freezes at 273 K under external pressure of 1 atm. The process is at equilibrium `H_(2)O (I) hArr H_(2)O (s)` at 273 K & 1 atm However it was required to calculate the thermodyanamic parameters of the fusion prcess occuring at same pressure & different temperature. Using the following data, answer the question that follow. `d_("ice") = 0.9 gm//c c, d_(H_(2)O(I)) = 1 gm//c c, C_(P) [H_(2)O(s)] = 36.4 JK^(-1) mol^(-1), C_(P) [H_(2)O (I) ] = 75.3 JK^(-1) mol^(-1), Delta H_("fusion") = 6008.2 J mol^(-1)` `' Delta S_("fusion")'` at 263 K & 1 atm will be:A. `22.01 JK^(-1) mol^(-1)`B. `22.84 JK^(-1) mol^(-1)`C. `21.36 JK^(-1) mol^(-1)`D. `20.557 JK^(-1) mol^(-1)` |
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Answer» Correct Answer - D `Delta_("fussion") S = (Delta_("fussion")H)/(T_(f))` |
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| 104. |
Liquid water freezes at 273 K under external pressure of 1 atm. The process is at equilibrium `H_(2)O (I) hArr H_(2)O (s)` at 273 K & 1 atm However it was required to calculate the thermodyanamic parameters of the fusion prcess occuring at same pressure & different temperature. Using the following data, answer the question that follow. `d_("ice") = 0.9 gm//c c, d_(H_(2)O(I)) = 1 gm//c c, C_(P) [H_(2)O(s)] = 36.4 JK^(-1) mol^(-1), C_(P) [H_(2)O (I) ] = 75.3 JK^(-1) mol^(-1), Delta H_("fusion") = 6008.2 J mol^(-1)` The value of `'DeltaH_("fusion")'` at 263 K & 1 atm will be :A. `+6008.0 J "mole"^(-1)`B. `5619.2 J "mole"^(-1)`C. `-5619.2 J "mole"^(-1)`D. `6619.2 J K^(-1) mol^(-1)` |
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Answer» Correct Answer - B `H_(2)O (l) hArr H_(2)O (s)` `DeltaC_(p) = 36.4 - 75.3 = - 38.9 J.K^(-1) mol^(-1)` `Delta H_(273) - DeltaH_(263) = Delta C_(p) [T_(2) - T_(1)]` |
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| 105. |
The enthalpy change at 298 K for decompostion is Given in following steps:`Step-1 H_(2)O(g)toH(g)+OH(g)DeltaH=498Kj//mol^(-1)``Steps-2 Oh(g)to H(g)+O(g)DeltaH=428KJ//mol^(-1)` then value of mean bond enthalpy of O-H bond will be :A. 498KJ/molB. 463KJ/molC. 428KJ/molD. 70KJ/mol |
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Answer» Correct Answer - b |
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| 106. |
Liquid water freezes at 273K under external pressure of 1atm . The process is at eruilibrium `H_(2)O(l)toH_(2)O(S)at 273 Kand 1atm` however , it required to calculate the thermodynamic parameters of the fusion process occurring at same pressure and different temperature . Using the following dat ,answer the questions thaat follow: `d_(ice)=0.9gm //cc,d_(H_(2)O(l))=1gm //c"c",1L-atm =101.3 J ` `C_(p)[H_(2)O(s)]=36.4JK^(-1)mol^(-1)` `C_(p)[H_(2)O(l)]=75.3Jk^(-1)mol^(-1)` `DeltaH_("fusion")=6008.2mol^(-1).Alldataat 273K.` the value of `DeltaH_("fusion")` at 263 K and 1 atm will be :A. `+6008.0K"mole"^(-1)`B. `5619.2J "mole"^(-1)`C. `-5619.2J "mole"^(-1)`D. `6619.2 "mole"^(-1)` |
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Answer» Correct Answer - b |
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| 107. |
Calculate resonance enthalpy of `CO_(2)(g)` from following data: `DeltaH_("combustion")^(@)[C_("graphite")]=-390Kj//mol` `DeltaH_("Sublimation")^(@)[C_("graphite")]=-715Kj//mol` `DeltaH_(B.E.)[O=O]=500KJ//mol` `DeltaH_(B.E.)[C=O]=875KJ//mol`A. `-40KJ//mol`B. `-145KJ//mol`C. `-72.5KJ//mol`D. `-290KJ/mol` |
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Answer» Correct Answer - b |
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| 108. |
White phosphorus is a tetra-atomic solid `P_(4)`(s)at room temperature. find bond enthalpy (P-P) in kJ//mol. Given : `DeltaH_("sublimation")` of `P_(4)` (s)=`61` kL//mol DeltaH_("atomisation") of `P_(4)`(s)=`1321` kJ//mol] |
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Answer» Correct Answer - 210 |
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| 109. |
One mole of an ideal diatomic gas `(C_(v) =5 cal)` was transformed from initial `25^(@)` and `1 L` to the state when temperature is `100^(@)` C and volume `10` L. The entropy change of the process can be expressed as (`R=2` calories /mol/K)A. `3 "In" (298)/(373) + 2 "In" 10`B. `5 "In" (373)/(298) + 2 "In"10`C. `7 "In" (373)/(298) + 2 "In" (1)/(10)`D. `5 "In" (373)/(298) + 2 "In"( 1)/(10)` |
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Answer» Correct Answer - B `DeltaS = nC_(v) "In" ((T_(f))/(T_(i))) + nR "In" ((V_(f))/(V_(i))) = (5)/(2) xx 2 "In" (373)/(298) + 2 "In"10` |
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| 110. |
The bond energy of C=O if `DeltaH_(f)^(@)(CO_(2))`=`390 kJ`, `DeltaH_("sublimation")^(@)`(Graphite)=720 kJ, O=O bond energy = 490 kJ and renonce energy `(CO_2)`=36 kJ :A. 782 kJB. 1564 kJC. 500 kJD. 626 kJ |
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Answer» Correct Answer - A |
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| 111. |
What will be the maximum amount of heat realeased when 321 g of a mixuture of `fe_(2)O_(3)` and Al is subjected to sparkling in absence of air? `DeltaH_(f)(Fe_(2)O_(3)=-199kJ//`mole `DeltaH_(f)(Al_(2)O_(3)=-399kJ//`moleA. 200 kJB. 300 kJC. 400 kJD. 100 kJ |
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Answer» Correct Answer - B |
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| 112. |
By how much does the entropy of `3` mole of an ideal gas change in going from a pressure of `2` bar to a pressure of `1` bar without any change in Temperature . If the surrounding is at `1` bar and `300` K (Expansion is again of the cosntant extenal pressure of surrounding).A. `+7.29 J-K^(-1)`B. `+4.82 J-K^(-1)`C. `-5.29 J-K^(-1)`D. `-8.35J-K^(-1)` |
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Answer» Correct Answer - B `DeltaS (gas) = "nR In" (P_(1))/(P_(2)) = 3 xx 8.314 xx "In"2 = 17.29` J/K Since `DeltaT =0 , "therefore" " " DeltaU =0.` So, `q=-W` `="nRT"[1 + (P_(1))/(P_(2))]` `q=T xx 12.47` J/K `DeltaS_("surr") =-(q)/(T) =- 12.47` J/K Change in `DeltaS =+ 4.82` J/K |
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| 113. |
For a perfectly crystalline solid `C_("p,m") =T^(3)+bT`, where a & b are constant . If `C_("p,m")` is `0.04 J//k-"mole"` at 10 K and io `0.92 J//K-"mole"` at 20 k, Then molar entropy at 20 k is.A. `0. 92 J//K -mol`B. `8.66 J// K-mol`C. `0.813 J//K-mol`D. None |
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Answer» Correct Answer - C `.04 = aT_(1)^(3)+ bT_(1) " "rArr " "0.40 = a xx (1000) + b xx 10` `" "rArr " "0.4 = 1000a + 10b" ".......(i)` `0.92 =aT_(2)^(3) + bT_(2) " " rArr " "0.92 = a xx 8000 + 20b" "......(ii)` On solving `0.12 = 6000 a " "," "a= xx 10^(-5)` `0.40= 2 xx 10^(-5) xx 1000+b+10` `therefore b = 0.038` `S_(m)= int (aT^(3) + bT)/(T)dT " "rArr " " (a[T_(2)^(3)- T_(1)^(3)])/(3) + b[T_(2) - T_(1)]` `" "rArr (2xx 10 ^(-5) xx (8000-0))/(3) +b(20) rArr (2xx 10^(-5) xx 8000)/(3) +0.038xx(20)` `" "0.053 + 0.76 rArr 0.813 J//K-mol` |
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| 114. |
The hydrolysis of Adienosine triphisphate [ATP] to give adienosine diphosphate [ADP] is represented by ATP hArr ADP. This reaction is expthermic zero[K] . The entropy change for the reaction is `982` J/K at `310` K . The free energy for the reaction is:A. `31.01` KJB. `9.188`KJC. `-9.188` KJD. `31.012`KJ |
| Answer» Correct Answer - D | |
| 115. |
For a perfectly crystalline solid `C_(p.m.)=aT^(3)`, where ais constant. If `C_(p.m.)` is 0.42.J//K-"mol" at 10K, molar entropy at 10K is:A. `0.42J//K-"mol"`B. `0.14J//K-"mol"`C. `4.2 J//K-"mol"`D. zero |
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Answer» Correct Answer - B |
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| 116. |
For a perfectly crystalline solid `C_(p.m.)=aT^(3)+bT`,where a and b constant . If `C_(p.m.)is 0.40J//Kmol` at 10K and `0.92J//K "mol" at 20K` then molar entropy at 20k is:A. `0.92 J//K mol`B. `8.66 `J/k molC. `0.813 J//K mol`D. None of these |
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Answer» Correct Answer - D `0.40 = aT_(1)^(3) + bT_(1) " " implies " " 0.40 = a xx (1000) + b xx 10 implies 0.4 =1000a + 10b " ""…."(1)` `0.92 = aT_(2)^(3) + bT_(2) " "implies " "0.92 = a xx 9000 + 20b" "….(2)` From Eqs. (1) and (2) `" "a=2xx10^(-5) , b = 0.038` `" "S_(n) = int(aT^(3) + bT)/(T).dT " "=(a[T_(2)^(3) - T_(1)^(3)])/(3) + b[T_(2) - T_(1)] = 0.427` J/K- mol |
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| 117. |
Energy hidden in a definite quantity of substanceA. EnthalpyB. Internal energy increasesC. Free energyD. Entropy |
| Answer» Correct Answer - B | |
| 118. |
For a perfectly crystalline solid `C_(p.m.)=aT^(3)`, where ais constant. If `C_(p.m.)` is 0.42.J//K-"mol" at 10K, molar entropy at 10K is:A. 0.42 J/k molB. 0.14 J/K molC. 1.12 J/K molD. zero |
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Answer» Correct Answer - C `0.42=a(10)^(3)impliesa=0.42xx10^(-3)` `S_(m)=int_(0)^(20)(C_(p,m))/(T)dT` `=int_(0)^(20)aT^(2)dT=(a)/(3)[20^(3)-0]` `=1.12 J//K-"mol"` |
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| 119. |
Heat of a reaction (Q) at constant pressure is equal toA. `E_(P)-E_(R)`B. `E_(R)-E_(P)`C. `H_(P)-H_(R)`D. `H_(R)-H_(P)` |
| Answer» Correct Answer - C | |
| 120. |
The enthalpies of the elements in their standard states are arbitrarily assumed to beA. unityB. zeroC. `lt0`D. different for each element |
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Answer» the enthalpy of alll elements in their standrad state is zero. Therefore, alternative (ii) is correct. |
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| 121. |
The standard enthalpy is zero for the substanceA. `C("graphite")`B. `C ("diamond")`C. `CO_(2)`D. `O_(3)` |
| Answer» Correct Answer - A | |
| 122. |
The enthalpies of the elements in their standard states are arbitrarily assumed to beA. Zero at 298 K and 1 atmB. Unity at 298 K and 1 atmC. Zero at all temperaturesD. Zero at 273 K and 1 atm |
| Answer» Correct Answer - A | |
| 123. |
Regarding a thermochemical equation a wrong statement isA. It tells about the physical states of reactants and productB. It tells whether the reaction is exothermic or endothermicC. It tells about the allotropic from (if any) of the reactantD. It tells whether the reaction is possible or not |
| Answer» Correct Answer - D | |
| 124. |
The heat change associated with reactions at constant volume is due to the difference in which property of the reactants and the productsA. Internal energyB. EnthyalpyC. Heat capacityD. Free energy |
| Answer» Correct Answer - A | |
| 125. |
Assume each reaction is carried out in an open container. For which reaction will `DeltaH=DeltaU` ?A. `2CO(g)+O_(2) (g) rarr 2CO_(2) (g)`B. `H_(2)(g)+Br_(2)(g) rarr 2HBr(g)`C. `C(s)+2H_(2)O (g) rarr 2H_(2)(g)+CO_(2) (g)`D. `PCl_(5)(g) rarr PCl_(3)(g)+Cl_(2)(g)` |
| Answer» Correct Answer - B | |
| 126. |
If `(1)/(2)X_(2)O_((s)) rarr X_((s))+(1)/(4)O_(2(g)),DeltaH=90kJ` then heat change during reaction of metal `X` with `1 mol e O_(2)` to form oxide to maximum extent is `:`A. `-360 kJ`B. `-180 kJ`C. `+360 kJ`D. `+180 kJ` |
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Answer» Correct Answer - A `1/4` mole of `O_(2) rarr 90 kJ` 1 mole `rarr` ? |
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| 127. |
Determine the enthalpy change for the reaction of 5.00 g `Fe_(2)O_(3)`with aluminium metal according to the equation : `Fe_(2)O_(3)(s)+2Al(s)toAl_(2)O_(3)(s)Fe(l)` A. `-26.6KJ`B. `-28.2Kj`C. `-52.4KJ`D. `-77.9KJ` |
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Answer» Correct Answer - a |
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| 128. |
use the thermodymanmic information : `(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)DeltaH^(@)=90.4Kj//mol` `(1)/(2)N_(2)(g)+O_(2)(g) DeltaH^(@)=33.8KJ//mol` `2NO_(2)(g)toN_(2)O_(4)(g) DeltaH^(@)=-58.0KJ//mol` to Calcualate `DeltaH^(@)` Kj /mol for the reaction : `2NO_(2)(g)+O_(2)(g)toN_(2)O_(4)`A. `-171.2`B. `-114.6`C. `114.6`D. `171.2` |
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Answer» Correct Answer - a |
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| 129. |
Dissociation of `NH_(3)(g)` over solid platinum follows zero order kinetics. `2NH_(3)(g) to N_(2)(g)+3H_(2)(g)` The rate of reaction is `2xx10^(-3)Msec^(-1)`. Also at 300K, thermodynamic data are: `Delta H_(f)^(@)NH_(2)`=-45kJ/mole `S_(N_(2))^(@)`=190J/K mole `S_(NH_(3))^(@)`=200J/K mole, `S_(H_(2))^(@)`=130J/K mole From the above data and the assumption that `DeltaH_("Rxn")^(@)` are independent of temperature, anwer the question that follows. [Take `Rxx300`kJ] What is the rate at which heat is absorbed at time t=50 sec. if volume of vessel is kept at 1 litre?A. 0.18kJ`sec^(-1)`B. `0.36kJsec^(-1)`C. 0.09kD. 0.17kJ `sec^(-1)` |
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Answer» Correct Answer - d |
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| 130. |
The value for `DeltaU` for the reversible isothermal evaporation of 90 g water at `100^(@)C` will be `(DeltaH_("evap")" of water "=40.8" kJ mol"^(-1), R=8.314"J K"^(-1)"mol"^(-1))`A. 4800 kJB. `188.494 kJ`C. `40.8 kJ`D. `125.03 kJ` |
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Answer» Correct Answer - B `DeltaH" for 18 g water = 40.8 kJ"` `"For 90 g water "=(40.8)/(18)xx90=204kJ` `"n for 90 g water "=90//18=5` `Deltan_(g)=5-0=5` `DeltaH=DeltaU+Deltan_(g)RT` `DeltaU204000-(5xx8.314xx373)=188494 or 188.494 kJ` |
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| 131. |
Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as `(DeltaG)_(PT) = DeltaH - T DeltaS` The magnitude of `DeltaH` does not change much with the change in temperature but the enrgy factor `T DeltaS` changes appreciably. Thus, spontaneity of a process depends very much on temperature. When `CaCO_(3)` is heated to a high temperature, it undergoes decomposition into `CaO` and `CO_(2)` whereas it is quite stable at room temperature. The most likely explanation of its isA. The enthalpy of reaction `(DeltaH)` overweighs the term `T DeltaS` at high temperature.B. The term `T DeltaS` overweights the enthalpy of reaction at high temperatureC. At high temperature, both enthalpy of reaction and entropy change becomes negative.D. None of these. |
| Answer» `DeltaG = DeltaH - T DeltaS`, at high temperature `T DeltaS` factor dominates of `DeltaH` and hence `DeltaG` becomes negative and reaction occurs spontaneously. | |
| 132. |
Gibbs-Helmoholtz equation relates the free energy change to the enthalpy and entropy changes of the process as `(DeltaG)_(PT) = DeltaH - T DeltaS` The magnitude of `DeltaH` does not change much with the change in temperature but the enrgy factor `T DeltaS` changes appreciably. Thus, spontaneity of a process depends very much on temperature. For the reaction at `298K, 2A +B rarr C` `Deltah = 100 kcal` and `DeltaS = 0.020 kcla K^(-1)`. If `DeltaH` and `DeltaS` are assumed to be constant over the temperature range, at what temperature will the reaction become spontaneous?A. `1000 K`B. `3500 K`C. `5000 K`D. `1500K` |
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Answer» `DeltaG = DeltaH - T DeltaS` For spontaneous process, `DeltaG =- ve` or `DeltaH lt T DeltaS` `:. T gt (DeltaH)/(DeltaS) = (100K cal)/(0.02 Kcal K^(-1)) = 5000K` |
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| 133. |
The enthalpy of solution , sodium and sodium oxide in large volume of water , are -18KJ/mole and -238KJ/mol, respectively . If the enthalpy of formation of water is -286 KJ/mol, then what is the enthalpy of formation of sodium oxide ? All the enthalpies are at 298K and 1 bar pressure. [Given : reaction involved are `2Na(s)+2H_(2)O(l)to 2NaOH(aq)+H_(2)(g)`` Na_(2)O(s)+H_(2)O(l)to2NaOH(aq)`A. `+54KJ//mol`B. `-130KJ//mol`C. `-416KJ//mol`D. `+156KJ//Mol` |
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Answer» Correct Answer - c |
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| 134. |
Given, `H_(2)(g)+Br_(2)(g)to2HBr(g),DeltaH_(1)^(@)` and standard enthalpy of condensation of bromine is `DeltaH_(2)^(@)`, standard enthalpy of formation of HBr at `25^(@)` C is :A. `(DeltaH_(1)^(@))/(2)`B. `(DeltaH_(1)^(@))/(2)+DeltaH_(2)^(@)`C. `(DeltaH_(1)^(@))/(2)-DeltaH_(2)^(@)`D. `((DeltaH_(1)^(@)-DeltaH_(2)^(@)))/(2)` |
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Answer» Correct Answer - D |
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| 135. |
Thermodynamic is concerned withA. Total enegry of a systemB. Enegry chnages in a systemC. Rate of a chemicle changeD. Mass changes in nuclear reactions |
| Answer» Enegrgy changes in a system. | |
| 136. |
When a student mixed `50mL` of `1M HCI` and `150mL` of `1M NaOH` in a coffee cup calorimeter, the temperature of the resultant solution increases from `21^(@)C` to `27.5^(@)C`. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the total volume of solution is `100mL`, its density `1gmL^(-1)` and that its specific heat is `4.18 J g^(-1)`. calculate: a. The heat change during mixing. b. The enthalpy change for the reaction `HCI(aq) +NaOH(aq) rarr NaCI(aq) +H_(2)O(aq)` |
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Answer» Number of moles of `HCI` and `NaOH` added `= (MV)/(1000) = (1xx150)/(1000) = 0.15` Miss of mixture `= V xx d = 1000 xx 1 = 100 g` Heat evolved, `q = ms DeltaT = 100 xx 4.18 xx (27.5 - 21.0)` `= 100 xx 4.18 xx 6.5J = 2717J = 2.717 kJ` The involved reaction is `HCI(aq) +NaOH(aq)rarr NaCI(aq) +H_(2)O` `DeltaH^(Theta) =` Heat evolved per mol `= (-2.717)/(0.15) =- 18.11 kJ` |
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| 137. |
`{:(,"Column I( Solutions mixed)",,"Column II ( Heat evolved)"),((A),500 mL of 0.1 "M HCl acid"+200 mL of 0.2 M NaOH "solution",(p),4568J),((B),200 mLof 0.2MH_(2)SO_(4) + 400 mL of 0.5 M KOH "solution",(q),2284J),((C ),500 mL of 0.1M HCl + 500 mL of 0.1 M NH_(4)OH,(r),2760 J),((D),500 mL of 0.1 M "acetic acid " + 500 mL of 0.1 M NaOH ,(s),2575J):}` (a) `A-r,B-p,C-s,D-q` (b) `A-p, B-r, C-q,D-s` (c ) ` A-q,B-p, C-s, D-r` (d) `A-,B-s,C-p,D-r` |
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Answer» Correct Answer - c A. Refer to Solved Problem B. Refer to Solved Problem C. Enthalpy of neutralization`NH_(4)OH` with HCl`= 51.5 kJ mol^(-1)=51500 J mol^(-1)` 500mLof 0.1 M `NH_94)OH =500 xx 0.1` millimoles`=50` millimole `= 0.05` mole `:. `Enthalpy of neutralisation. `=51500 xx 0.05 J = 2575J`. D. Enthalpy of neutralisation of acetic acid with NaOH `=55.2 kJ mol^(-1) = 55200 J mol^(-1)` 500mL of 0.1 M NaOh` =500 xx 0.1 ` millimole `= 50` millimole `= 0.05 ` mole `:. `Enthalpy ofneutralisation `=55 200 xx 0.05 J = 2760 J`. |
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| 138. |
Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`A. `14 kcal`B. `35 cal`C. `10 cal`D. `7.5 cal` |
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Answer» Number of moles of `HCI = (MV)/(1000) = (0.01xx25)/(1000)` `= 25 xx 10^(-5)` `HCI rarr H^(o+) + CI^(Theta)` `n_(H)^(o+) = 25 xx 10^(-5)` Number of moles of `Ca(OH)_(2) = (MV)/(1000) = (0.01xx50)/(1000) = 50 xx 10^(-5)` `Ca(OH)_(2) hArr Ca^(2+) +2 overset(Theta)OH` `n_(OH)^(Theta) = 2 xc 50 xx 10^(-5) = 10^(-3)` In the process of neutralisation, `25 xx 10^(-5) "mole" H^(o+)` will be completely neutralised. `:. DeltaH^(Theta) = 140 xx 25 xx 10^(-5) kcal = 0.035 kcal = 35 cal` |
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| 139. |
For the change C (diamond) `rarr` C(graphite) , `Delta H=-1.89` KJ, if 6 g of diamond and 6g of graphite are seperately burnt to yield `CO_(2)` the heat liberated in first case is :A. Less than in the second case by 1.89 KJB. Less than in the second case by 11.34 KJC. Less than in the second case by 14.34 KJD. More than in the second case by 0.945 KJ |
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Answer» Correct Answer - D `C_("Diamond")rarr C_("graphite") , Delta H =-1.89 KJ mol^(-1)` For 6g conversion of diamond into graphite `Delta H=(-1.89)/(2)=-0.945 KJ` `(1)/(2)C_("Diamond")rarr (1)/(2)C_("Graphite"), Delta H=-0.945 KJ` |
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| 140. |
If `H_(2)+1//2O_(2)rarr H_(2)O , Delta =-68.39` Kcal `K+H_(2)O+ "water" rarr KOH(aq)+1//2H_(2) , Delta H=-48.0` Kcal `KOH+"water" rarr KOH(aq)Delta H=-14.0` Kcal the heat of formation of KOH is -A. `-68.39+48-14.0`B. `-68.39-48.0+14.0`C. `+68.39-48.0+14.0`D. `+68.39+48.0-14.0` |
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Answer» Correct Answer - B `K_((s))+(1)/(2)O_(2(g))+(1)/(2)H_(2(g))rarr KOH_((s)) , Delta H_(f) = ?` Desired equation = eq(i) + (ii) - eq (iii) `Delta H_(f)=(-68.39)+(-48)-(-14)` |
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| 141. |
From the following data, the heat of formation of `Ca(OH)_(2)(s)` at `18^(@)C` is …. Kcal. (i) `CaO(s)+H_(2)O(l)=Ca(OH)_(2)(s) , Delta H18^(@)C=-15.26` Kcal….. (ii) `H_(2)O(l)=H_(2)(g)+1//2O_(2)(g) , Delta H18^(@)C=68.37` Kcal…. (iii) `Ca(s)+1//2O_(2)(g)=CaO(s) , Delta H 18^(@)C=-151.80` Kcal....A. `-98.69`B. `-235.43`C. `194.91`D. `98.69` |
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Answer» Correct Answer - B `Ca_((s))+O_(2(g))+H_(2(g))rarr Ca(OH)_(2) , Delta H_(f) = ?` Desired equation = eq (iii) + eq(i) - eq (ii) `Delta H_(f)=(-151.80)+(-15.26)-(-68.37)` `Delta H_(f)=-235.43 KCal mol^(-1)` |
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| 142. |
If `S +O_(2) rarr SO_(2), DeltaH =- 298.2 kJ` `SO_(2)+1//2O_(2) rarr SO_(3), DeltaH =- 98.7 kJ` `SO_(3)+H_(2)O rarr H_(2)SO_(4), DeltaH =- 130.2 kJ` `H_(2) +1//2O_(2) rarr H_(2)O, DeltaH =- 287.3 kJ` Then the enthalpy of formation of `H_(2)SO_(4)` at `298K` isA. `-814.4` KJB. `-650.3` KJC. `-320.5` KJD. `-433.5` KJ |
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Answer» Correct Answer - A `H_(2(g))+S_((s))+2O_(2(g))rarr H_(2)SO_(4) , Delta H_(f) = ?` Desired equation = eq(i) + eq(iii) + eq(iv) + eq(ii) `Delta H_(f)=(-298.2)+(-130.2)+(-287.3)+(-98.7)` `Delta H_(f)=-814.4 KJ` |
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| 143. |
Given : `S_((s))+(3)/(2)O_(2(g))rarrSO_(3(g)+2X Kcal` `SO_(2(s))+(1)/(2)O_(2(g))rarrSO_(3(g)+Y Kcal` The heat of formation of `SO_(2)` is `:-`A. (2x + y)B. `-(2x - y)`C. x + yD. 2x / y |
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Answer» Correct Answer - B (i) `S_((s))+(3)/(2)O_(2(g))rarr SO_(3(g)) , Delta H=-2x KCal` (ii) `SO_(2(g))+(1)/(2)O_(2)(g)rarr SO_(2(g)) , Delta H = -y KCal` `S_((s))+O_(2(g))rarr SO_(2(g)) , Delta H_(f) = ?` Desired equation = eq(i)-eq(ii) `Delta H_(f)=-2x-(-y)=(-2x-y)` |
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| 144. |
Which plot represents for an exothermic reaction ?A. B. C. D. |
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Answer» Correct Answer - A For exothermic reaction, enthalpy of reactants is more than enthalpy of products. |
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| 145. |
An ideal gas filled at pressure `2 atm` and temp of `300 K`, in a balloon is kept in vaccum with in a large container wall of balloon is punchtured then container temperature :A. DecreasesB. IncreasesC. Remain constantD. Unpredictable |
| Answer» Correct Answer - C | |
| 146. |
`N_(2)(g)+20_(2)(g)rarr2NO_(2)(g)DeltaH_("rxn")gt0` Under what temperature conditions is this reaction spontaneous at standard preassure?A. At low temperatures onlyB. At high temperatures onlyC. At all temperaturesD. At no temperature |
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Answer» Correct Answer - d |
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| 147. |
`N_(2)(g)+20_(2)(g)rarr2NO_(2)(g)DeltaH_("rxn")gt0` Which relationship is correat for this reaction at a pressure of 1 atm ?A. `DeltaE_(rxn)gtDeltaH_(rxn)`B. `DeltaE_(rxn)ltDeltaH_(rxn)`C. `DeltaE_(rxn)=DeltaH_(rxn)+DeltaS_(rax)`D. `DeltaE_(rxn)=DeltaH_(rxn)-DeltaS_(rax)` |
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Answer» Correct Answer - a |
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| 148. |
Calculate `Delta_(r)G^(Theta)` at `298K` for the following reaction if the reaction mixtur econsists of `1atm` of `N_(2),3` atm of `H_(2)`, and `1atm` of `NH_(3)`. `N_(2)(g) +3H_(2)(g) W hArr 2NH_(3)(g), Delta_(r)G^(Theta) =- 33.32 kJ` |
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Answer» `N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g), Delta_(r)G^(Theta) =- 33.32 kJ` Using: `Delta_(r)G = Delta_(r)G^(Theta) +RT In Q` where `Q = Q=(p^(3)underset(NH_(3)).)/(p_(N_(2)p_(H_(2)))) = (1^(2))/(1xx3^(2)) = (1)/(27)`, `T = 298 K: R = 8.314 J mol^(-1)K^(-1)` `rArr Delta_(r)G =- 33.2 +(8.314 xx 10^(-5)) xx 298 xx 2.303 log_(10).(1)/(27)` `rArr -33.2 - 8.16 = - 41.36 kJ mol^(-1)` |
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| 149. |
Determine whether or not is possible for sodium to reduce aluminium oxide to aluminium at `298K`. Also, calculate equilibrium constant for this reaction at `298K`. `Delta_(f)G^(Theta)AI_(2)O_(3)(s) =- 1582 kJ mol^(-1)` `Delta_(f)g^(Theta)Na_(2)O(s) =- 377.7 kJ mol^(-1)` |
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Answer» The reaction is: `AI_(2)O_(3)(s) +6Na(s) rarr 3Na_(2)O(s) +2AI(s)` `Delta_(r)G^(Theta) = Delta_(f)G^(Theta) (Na_(2)O) - Delta_(f)G^(Theta)(AI_(2)O_(3))` `=3 xx (-377.07) -(-1582.0) =+ 451.21 kJ mol^(-1)` Hence, the reaction cannot occur, since `DeltaG^(Theta)` is positive. Also, `Delta_(r)G^(Theta) = 2.303 RT log K` `rArr 451.21 =- 2.303 xx 8.314 xx 10^(-3) xx 298 log K_(eq)` `rArr log K_(eq) =- 79.0 rArr K_(eq) = 1.0 xx 10^(-79)` |
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| 150. |
Calculate the heta of formation of anhydrous `AI_(2)CI_(6)` from the following data:A. `2AI(s) +6HCI(aq) rarr AI_(2)CI_(6)(aq) +3H_(2)+1004.2 kJ`B. `H_(2)(g) +CI_(2)(g) rarr 2HCI(g) +184.1 kJ`C. `HCI(g) +aq rarr HCI (aq) +73.2 kJ`D. `AI_(2)CI_(6)(s) +aq rarr AI_(2)CI_(6)(aq) +643.1 kJ` |
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Answer» To calculate `DeltaH` of the following reaction: `2AI(s) +3CI_(2)(g) rarr AI_(2)CI_(6)(s)` `DeltaH = DeltaH_(1) +3DeltaH_(2) +6DeltaH_(3) - 4DeltaH_(4)` `=- 1004.2 +3 (-184.1) +6 (-73.2) -(-643.1)` `=- 1004.2 - 552.3 - 439.2 + 643.1` `=- 1995.57 +643.1` `=- 1352.6 kJ mol^(-1)` |
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