Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Enthalpy of atomiation of C_(2)H_(6)(g)and C_(3)h_(8)(g) are 620 and 880 KJ mol^(-1) respectivelty. The C-C and C-H bond energies are respectively: |
|
Answer» `80and60KJMol^(-1)` |
|
| 2. |
Enthalpy of ammonia and water are -46.19KJ.mol^(-1) and -285.9KJ.mol^(-1). Calculate AH for the reaction, 4NH_(3)(g)+3O_(2)(g)to 2N_(2)(g)+6H_(2)O(l). |
|
Answer» |
|
| 3. |
Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A to B along one route is Delta_(r) H and Delta_(r) H_(1) , Delta_(r) H_(2), Delta_(r)H_(3)….. represent enthalpies of intermediate reactions leading to product B. What will be the relation between Delta_(r)H for overall reaction and Delta_(r) H_(1), Delta_(r) H_(2) ....... etc. for intermediate reactions. |
|
Answer» Solution :If enthalpy of an overall reaction `A to B` along ONE route is `Delta_(r) H and Delta_(r) H_(1), Delta_(r) H_(2), Delta_(r) H_(3) ….` REPRESENTING ENTHALPIES of reaction leading to same product B along another route, then we have `Delta_(r) H = Delta_(r) H_(1) + Delta_(r) H_(2) + Delta_(r) H_(3) + ......` |
|
| 4. |
Enthalpy is equal to |
|
Answer» <P>`T^(2) [(del(G//T))/(delT)]_(P)` `[ ( del(DeltaG//T))/(delT)] = - ( DeltaH)/(T^(2)) ` ( R efertopage) or `DeltaH= - T^(2)[(del(DeltaG//T))/(delT)]_(P)` This is ONE of theforms of Gibbs-Helmholtz EQUATION. |
|
| 5. |
Enthalpy is an extensive property. In general, if enthalpyof an overall reaction A rarr B alongone routeisDelta_(r) H and Delta_(r) H_(2),Delta_(r) H_(3)"…....." represent enthalpies of intermediate reactions leading to product B, what will be the relation between Delta_(r)H for overall reaction Delta_(r)H_(1), Delta_(r)H_(2)"......." etc. for intermediate reactions ? |
| Answer» Solution :By Hess's law, `Delta_(r)H = Delta_(r)H_(1)+ Delta_(r) H_(2)+Delta_(r)H_(3)+"…......"` | |
| 7. |
Enthalpy for the reaction C+ O_(2) to CO_(2)is…. |
|
Answer» `+ve` |
|
| 8. |
Enthalpy ''H'' can be given as |
|
Answer» `H= E- PV` |
|
| 9. |
Enthalpy for the oxidation of graphite to CO_2 and CO to CO_2 can easily be measured. For these conversions, the heat of combustion values are -393.5 kJ and - 283.5 kJrespectively.find the enthalpy of combustion of graphite to CO |
|
Answer» Solution :From these data the enthalpy of combustion of graphite to CO can be calculated HESS.s law. The reactions involved in this process can be expressed as follows  ACCORDING to Hess law , `DeltaH_1=DeltaH_2+DeltaH_3` -393.5 kJ =X- 283.5 kJ X=-110.5 kJ |
|
| 10. |
Enthalpy diagram for a particular reaction is given in figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain. |
| Answer» SOLUTION :No, enthalpy is one of the contributing factors in DECIDING spontaneity but it is not the only factor. ANOTHER CONTRIBUTORY factor, entropy factor has also to be taken into CONSIDERATION | |
| 11. |
Enthalpy change of CH_(4)+(1)/(2) O_(2) rarr CH_(3) OH is negative . If enthalpy of combustion ofCH_(4) andCH_(3)OH are x and y respectively, then which relation is correct ? |
|
Answer» `x gt y` (ii) `CH_(3) OH + (3)/(2) O_(2) rarr CO_(2)+ 2H_(2)O, DeltaH =y` (i) - (ii) GIVES `DeltaH ` for the given reaction, i.e., `DeltaH = x-y `.As it is -ve , `x lty` |
|
| 12. |
Enthalpy change when a solution is diluted from 4 M to 2 Mis -1.6 KJ/mol . Enthalpy change when 5 litre of such a solution is diluted, is: |
|
Answer» `-1.6KJ` |
|
| 13. |
Enthalpy change for the reaction, 4H(g) to 2H_(2)(g) is -869.6 kJ. The dissociation energy of H - H bond is |
|
Answer» `434.8 KJ` `2H_(2)(g) to 4H(g) DeltaH = 869.6 kJ` `H_(2)(g) to 2H(g) DeltaH = 44.8 kJ` |
|
| 14. |
Enthalpy change for the reaction , 4H(g) rarr 2H_(2)(g)is-869.6 kJ. The dissociation energy of H-H bondis |
|
Answer» `+217 .4kJ` or `H_(2)(g)rarr 2H(g), DeltaH= + ( 869.6)/( 2) Kj` `=434.8 kJ` |
|
| 15. |
The enthalpy change for the reaction C_(2)H_(6(g)) rarr 2C_((g)) + 6H_((g)) is x kJ. The bond energy of C-H bond is: |
|
Answer» `-434.8kJ` |
|
| 16. |
Enthalpy change ( Delta_(vap)H ) for the transition of liquid water to steam at 100^(@)C is40.8 kJ mol^(-1). Calculate the entropy change for the process. |
|
Answer» |
|
| 17. |
Enthalpy and entropy changes of a reaction are 40.63kJmol^(-1) and 108 J K^(-1) mol^(-1) respectively . Predict the feasibility of the reaction at 27^(@)C |
|
Answer» Solution :Here, we are given `: Delta H =40.63 kJ mol^(-1)= 40630 J mol^(-1) , Delta S = 108 J K^(-1) mol^(-1)` `T = 27^(@) C = 27 +273 =300K` `:. Delta G =Delta H = T Delta S = 40630 J mol^(-1) - 300 K xx 108.8 JK^(-1) mol^(-1) = 7990 J mol^(-1)` Since `Delta G` comes out to be positive ( i.e., ` Delta Ggt 0)` , the reaction is not feasible in the FORWARD DIRECTION. |
|
| 18. |
Enthalpy change and entropy change of a chemical reaction are -10.5 kJ mol^(-1) and -31.5 JK^(-1) mol^(-1) respectively. Predict whether the reaction is spontaneous or not at 300 K. |
|
Answer» Solution :`Delta G = Delta H - T Delta S` Given, `Delta S = -315 JK^(-1) MOL^(-1)` and `Delta H = -10.5 xx 10^(3) J mol^(-1)` `:. Delta G = -10500 - 300(-31.5) = -1050 J mol^(-1)` SINCE `Delta G` is negative, the reaction would be spontaneous. |
|
| 19. |
Enthalpies of formation of CO(g), N_(2)O(g) and N_(2)O_(4)(g) are-110, -393, 81 and 9.7 kJ mol^(-1) respectively . Find the value of Delta_(r)H for the reaction: N_(2)O_(4)(g)+3CO(g) rarr N_(2)O(g) +3CO_(2)(g) |
|
Answer» SOLUTION :`Delta_(r)H = Sigma Delta_(F)H(`Products) `-Sigma Delta_(f)H(` Reactants) `= [ DELTA _(f) H( N_(2)O) +3Delta_(f)H(CO_(2))]- [ Delta_(f)H(N_(2)O_(4)) +3Delta_(f) H(CO)]` `= [8] +3 ( -393)] - [9.7 +3( -110) ] kJ= -777.7 kJ` |
|
| 20. |
Enthalpies of formationof K^(+)(aq), Cl^(-)(aq) and KCl(s) are - 60.0 ,- 40.0and - 104.0 kcal mol^(-1) .If one mole of KCl(s) is dissolved in large excess of water at 25^(@)C, the enthalpy of solution will be |
|
Answer» `-4.0 kcal MOL^(-1)` |
|
| 21. |
Enthalpies of formation of CO_((g)) ,CO_(2(g)), N_(2) O_((g)) and N_(2) O_(4(g)) are -100, -393, 81 and 9.7 "kJ mol"^(-1) respectively. Find the value of Delta_(r) H for the reaction : N_(2) O_(4(g)) , 3CO_((g)) toN_(2)O_(4(g)) + 3CO_(2(g)) |
|
Answer» SOLUTION :`Delta_(r) H= sum Delta_(r) H_("(products)") - sum Delta_(f) H_( "(REACTANTS)")` `= [Delta_(f) H(N_(2) O) + 3 Delta_(f) H (CO_(2) ) ]- [Delta_(f) H (N_(2) O_(4) ) + 3 Delta _(f) H (CO) ]` `= [81+3 (-393 ) ] - [9.7 + 3(-110) ]` `=-777.7 "kJ"` |
|
| 22. |
Enlist the simularities between lithium and magnesium. |
Answer» SOLUTION :
|
|
| 23. |
Enlist the postulates of Bohr's atom model. |
|
Answer» Solution :Bohr.s atom is based on the following assumptions : `implies` The energies of electrons are quantised . `implies` The electron is revolving around the nucleus in a certain fixed CIRCULAR path called stationary ORBIT. `implies` Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron MUST be equal to an integral multiple of `h//2pi`. i.e., `mvr = nh//2pi` Where `n = 1,2,3...` etc. `implies` As long as an electron revolves in the fixed stationary orbit, it doesn.t LOSE its energy. However, when an electron jumps from higher energy state `(E_2)` to a lower energy state `(E_1)` the excess energy is emitted as radiation. The frequency of the emitted radiation is `E_2 - E_1 = hv` and `v = ((E_2 - E_1))/(h)` `implies ` Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit. |
|
| 24. |
Enlist the advantage using standard solutions. |
|
Answer» Solution :(i) The error DUE to WEIGHING the SOLUTE can be minimised by using concentrateed stock solution that requirs large quantity of solute (ii)We can prepare working standard of different concentrations by diluting the stock solutions which is more efficient since consistency is maintained. (III) Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments. |
|
| 25. |
Enilist the favourable conditions for orbital overlap. |
|
Answer» |
|
| 26. |
Write down the postulates of Bohr atom model. |
|
Answer» The electron is REVOLVING around the nucleus in a certain fixed circular PATH called stationary orbit. Electron can revolve only in those ORBITS in which the angular momentum (mvr) ofthe electron must be equal to an integral multipleof ` h//2pi` . i.e.mvr` = nh//2pi` Where n = 1, 2, 3, . . . . etc. As LONG as an electron revolves in the fixed stationary orbit, it doesn't lose its energy. However, when an electron jumps from higher energy state `(E_(2))` to a lower energy state `(E_(1))` the excess energy is emitted as radiation . The frequency of the emitted radiation is `E_(2)-E_(1)=hv`and `v= ((E_(2)-E_(1)))/(h)` Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit . |
|
| 27. |
Energy required to stop the ejection of electrons from Cu-plate 0.27 eV. Calculate the work function (in eV) When radiation of lambda= 235nm strikes the plate. |
|
Answer» =work function +`underset("STOPPING POTENTIAL")underset(to)(eV_0)"` ….(i) `E_("Photon") = (hc)/(lambda) = (6.625xx10^(-34) XX 3xx10^8)/(235 xx 10^(-9)) = 5.27 eV` from equation (i) `E_("photon") = W_0 + 0.27` `impliesW_0 = E_("Photon") - 0.27 = 5eV` |
|
| 28. |
Energy released when an electronis addedto a neutralgaseousatom iscalled ………oftheatom |
| Answer» SOLUTION :ELECTRON gainenthalpy | |
| 29. |
Energy profile for the following reaction will be NH_(3) + CH_(3)-I to CH_(3) - ""^(+)NH_(3) |
|
Answer»
|
|
| 30. |
Energy of the electron in Hydrogen atom is given by |
|
Answer» `E_n=-(131.28)/n^2 "KJ MOL"^(-1)` |
|
| 31. |
Energy of the state S_(1) in units of the hydrogen atom ground state energy is |
|
Answer» 0.75 `THEREFORE" "` Energy of the state `S_(1)` in the units of hydrogen atom ground state energy is : `E=E_(H)xx(Z^(2))/(n^(2))=E_(H)xx(3^(2))/(2^(2))=9/4E_(H)=2.25xxE_(H)` |
|
| 32. |
Energy of electron of hydrogen atom in second Bohr orbit is |
|
Answer» `-5.44 XX 10^(-19) J` `=-5.44xx10^(-19)` J |
|
| 33. |
Energy of C, C pi bond in k.cal is |
|
Answer» 96 |
|
| 34. |
Energy of an electron is given by Wavelength of light required to excite an electron in a hydrogen atom from level n = 1 to n = 2 will be… |
|
Answer» `1.214xx10^(-7)` m `=(6.26xx10^(-34)xx3xx10^(8))/(2.178xx10^(-18)(0.75))` `=12.169xx10^(-8)` `=12.169xx10^(-7)` m `=1.214 xx10^(-7)` m |
|
| 35. |
Energy of atomic orbitals in a particular shell is in the order |
|
Answer» s lt p lt dlt F |
|
| 36. |
Energy of an electron is given by E = -2.178 xx 10^(-18) J ((Z^(2))/(n^(2))). Wavelength of light required to excite an electron in a hydrogen atom from level 1 to level 2 will be (h = 6.62 xx 10^(-34) Js and c = 3 xx 10^(8) ms^(-1)) |
|
Answer» `8.500 XX 10^(-7) m` `= 2.178 xx 10^(-18) xx (3)/(4) xx 1 (Z = 1 " for " H)` or `lamda = (hc)/(Delta E) = ((6.62 xx 10^(-34)) (3 xx 10^(8)) xx 4)/(2.178 xx 10^(-18) xx 3)` `= 1.214 xx 10^(-7) m` |
|
| 37. |
Energy of an electron in the ground state of the hydrogen atomis -2.18xx10xx10^(-18)J. Calculate the ionisation enthalpy of atomic hydrogen in terms of kJ"mol"^(-1), |
|
Answer» Solution :Energy of an electron is the GROUND state of the hydrogen atom `=-2.18xx10^(-18)s` `HtoH^(+)+e^(-)` energy REQUIRED to ionize 1 mole of hydrogen atoms, we multiply by the Avagardo constant. `E=2.18xx10^(-18)xx6.023xx10^(23)` `=13.123xx10^(5)"J mol"^(-1)` `J.E=+1312"KJ mol"^(-1)` |
|
| 38. |
Energy of an electron in the ground state of the hydrogen atom is -2.18 xx 10^(-18) J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^(-1). |
|
Answer» Solution :Ionization enthalpy is the amount of energy REQUIRED to removed the ELECTRON from ground state to infinity. Energy of electron ground state ` = -2.18 xx 10^(-18) J` Energy of electron at infinity = 0 Energy required to remove electron ` = - (2.18 ) xx 10^(-18) J = 2.18 xx 10^(-18) J` Amount of energy required to remove 1 mol of electrons from 1 mol of H atom ` = 2.18 xx 10^(-18) J xx 6.022 xx 10^(23) `per mol ` :. `Ionization enthalpy of hydrogen ` = 13.130 xx 10^(5) J " mol "^(-1)` ` = 13.13 xx 10^(5) J " mol "^(-1)` NOTE: Ionisation enthalpy is always POSITIVE. |
|
| 39. |
Energy of an electron in the ground state of the hydrogen atom is-2.18 xx 10^(-18) J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^(-1). |
|
Answer» Solution :Energy of an electron in the ground STATE of the HYDROGEN atom=`2.18xx10^(-18)J` `HtoH^(+)+e^(-)` Energy required to ionize 1 mol e of hydrogen atoms, we MULTIPLY by the Avogadro constant. `E = 2.18 xx 10^(-18) xx 6.023 xx 10^(23)` = 13.123` xx 10^(5) J mol^(-1)` I.E = +1312 KJ `mol^(-1)` |
|
| 40. |
Energyof anelectron in the groundstate of thehydrogenatom is -2 . 18 xx 10^(-18)J. Calculatetheionizationenthalpyof atomichydrogenin termsof kJ mol^(-1) |
|
Answer» SOLUTION :Ionizationenergyis theamount ofenergyrequired toremovethe electronfrom thegroundstate to infinity. Now, energy ofthe electronin the groundstate = -`2.18 xx 10^(-15) J` ENERGYOF the electronat affinity=0 The energyrequiredto removedand electronin theground state ofhydrogen atom =0 - (its energyin thegroundstate ) = - `(-2 .18 xx 10^(-18)J )= 2. 18 xx 10^(-18) J` `:.` Ionizationenthalpy per MOLEOF HYDROGENATOMS `= (2.18 xx 10^(18)xx 6.02 xx 10^(23))/(1000) kJ` `= 1312 . 36 kJmol^(-1)= 1312 .36xx 10^(3)J mol^(-1)` |
|
| 41. |
Energy of an electron in n^(th) Bohr orbit is given as |
|
Answer» `-(n^(2)H^(2))/(4pi^(2)m ZE^(2))` |
|
| 42. |
Energy of an electron in the ground state of hydrogen atom is -2.18xx10^(-18)J. Calculate the ionization enthalpy of atomic hydrogen in terms of "Jmol"^(-1). |
|
Answer» |
|
| 43. |
Energy of an electron in the ground state of hydrogen atom is -2.18 xx 10^(-18) J. Calculate the inoisation enethalpy of atomic hydrogen in J mol^(-1). |
|
Answer» Solution :Energy of an electron at INFINITY is taken as zero. therefore The energy needed to REMOVE an electron from ground STATE to infinity = energy at infinity- Energy at ground state = `0 - (-2.18 xx 10^(-18)J)= 2.18 xx 10^(-18)J` IONISATION energy per mol of hydrogen = `2.18 xx 10^(-18) xx 6.02 xx 10^23/1000 = 1312.35 KJ mol^(-1)` |
|
| 44. |
Energy of an electron in hydrogen atom in ground state is -13.6eV.what is the energy of the electron in the second exited state? |
|
Answer» SOLUTION :Energy of an electron in ground state=-13.6eV. `:.` energy of an the second EXCITED state`=E_2`. N=2 `E_2=(-13.6eV)/(n^(2))=-13.6/(2^(2))=(-13.6)/(4)` |
|
| 45. |
Energy of an electron in hydrogen atom in ground state is –13.6 eV. What is the energy of the electron in the third excited state? |
|
Answer» SOLUTION :`E_1 =- 13.6=EV ` ` E_3= ( -13.6 )/( n^2)" where " n=3` `E_3 = ( -13.6 )/(9)= 1.511 eV ` ENERGYOF THEELECTRON in thethirdexcitedstate= 1.511eV . |
|
| 46. |
Energy levels of molecular orbitals have been experimentally determined by ________ studies . |
|
Answer» X- RAYS |
|
| 47. |
Energy of a photon with a wave length of 450 nm is |
|
Answer» `4.36xx10^(-12)` ERGS |
|
| 48. |
Energy is released in the process of |
|
Answer» `Na_((G)) to Na_((g))^(+) + e` |
|
| 49. |
Energy if an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the third excited state? |
|
Answer» SOLUTION :`E_1=-13.6eV` `E_3=(-13.6)/(n^(2))` where n=3 `E_3=(-13.6)/(9)=1.511eV` Energy of the electron IIN the THIRD excited state=1.511 eV. |
|
