InterviewSolution
Saved Bookmarks
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.
| 151. |
For the neutralisation of `1` mol of `H_(2)SO_(4)` with `2` mols of `NaOH` in dilute solutions the heat evolved isA. `=57.0 kJ`B. ` lt 57.0 kJ`C. ` gt 57.0 kJ`D. Unpredictable |
|
Answer» Correct Answer - C `1 "mol"H_(2)SO_(4)=2"mol"H^(+)` `2"mol"NaOH=2"mol"OH^(-)` `2H^(+)(aq)+2OH^(-)(aq)to2H_(2)O(l)` `DeltaH=57kJmol^(-1)xx2"mol"` `=114kJ gt 57kJ` |
|
| 152. |
Calculate the resonance enegry of `N_(2)O` form the following data `Delta_(f)H^(Theta) of N_(2)O = 82 kJ mol^(-1)` Bond enegry of `N-=N, N=N, O=O,` and `N=O` bond is `946, 418, 498`, and `607 kJ mol^(-1)`, respectively.A. `-88KJmol^(-1)`B. `-44KJmol^(-1)`C. `-22KJmol^(-1)`D. None of these |
|
Answer» Correct Answer - A `N_(2)(g)+(1)/(2)O_(2)(g)rarrN_(2)O` `N-=N+(1)/(0)0=0rarrN=N=0` Calculated `DeltaH_(f)^(0)(N_(2)O)=[B.E._((N-=N))+(1)/(2)B.E.(0=0)]` `-[B.E._(N=N))+B.E._(N=0)]` `=[946+(498)/(2)]-[418+607]=+170KJ//mol` Resonance energy=observed `DeltaH_(f)^(0)` -calculated `DeltaH^(0)_(f)=82-170=-88KJmol^(-1)` |
|
| 153. |
Thermodynamic equilibrium involvesA. Chemical equilibriumB. Mechanical equilibriumC. Thermal equilibriumD. All of the above simultaneously |
|
Answer» Correct Answer - D A thermodynamic equilibrium involves all the three equilibrium. |
|
| 154. |
The heat of combustion of yellow phoshphorus and red phosphorus are `-9.91KJmol^(-1)` and `-8.78` KJ/mol respectivaly. The heat of transition from yellow phosphrous to red phosphorus isA. `-18.69 kJ`B. `+1.13 kJ`C. `+18.69 kJ`D. `-1.13 kJ` |
|
Answer» Correct Answer - D Subtract `(ii)` from `(i)`. |
|
| 155. |
The polymerisation of ethylene to linear polyethylene is represented by the reaction `nCH_(2)=CH_(2)rarr(-CH_(2)-CH_(2)-)_(n)` When `n` has a large integral value. Given theat the average enthalpies of bond dissociation for `C=C` and `C-C` at `298K`are `+590` and `+331kJ mol^(-1)` respectively, calculate the enthalpy of polymerisation per mole of ethylene at `298K`. |
|
Answer» `nCH_(2)=CH_(2)to(-CH_(2)-CH_(2)-)_(n)` there are equal number of C-H bonds on both sides but on reactant side there are nC=C bonds and on product side `(2n+1)C-C` bonds. Enthalpy of polymerisation, `=nDeltaH_((C=C))-(2n+1)DeltaH_((C-C))` `=590n-(2n+1)(331)` `=590n-662n` [2n+1`to2n` as n is very large] `=72nkJ` Enthalpy of polymerisation per mole `=(DeltaH)/(n)=(72n)/(n)=72kJ" "mol^(-1)`. |
|
| 156. |
The standard molar heats of formation of ethane, carbon dioxide, and liquid water are `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.A. `-372 kcal`B. `162 kcal`C. `-340 kcal`D. `183.5 kcal` |
|
Answer» Correct Answer - A Aim : `C_(2)H_(6)+7//2O_(2)to2CO_(2)+3H_(2)O` `Delta_(r)H^(@)=[2xxDelta_(f)H^(@)(CO_(2))+3Delta_(f)H^(@)(H_(2)O)]-[Delta_(f)H^(@)(C_(2)H_(6))+7//2Delta_(f)H^(@)(O_(2))]` `=2xx(-94.1)+3xx(-68.3)-(-21)` `=-188.2-204.9+21=372.1 kcal` |
|
| 157. |
The standard molar heats of formation of ethane, carbon dioxide, and liquid water are `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.A. `372 kcal`B. `-162 kcal`C. `-240 kcal`D. `-183.5 kcal` |
|
Answer» Correct Answer - A `C_(2)H_(6)+5O_(2) to 2CO_(2)+3H_(2)O` Substitute the values. |
|
| 158. |
If `C(s)+O_(2)(g)rarrCO_(2)(g),DeltaH=X` and `CO(g)+1//2O_(2)(g)rarrCO_(2)(g),DeltaH=Y`, then the heat of formation of `CO` isA. `X+Y`B. `X-Y`C. `Y-X`D. `XY` |
|
Answer» Correct Answer - B `C(s) + O_(2) (g) rarr CO_(2)(g), Delta H = X` ...(1) `CO_(g) + 1//20_(2)(g) rarr CO_(2)(g), DeltaH = Y` ....(2) Subtracting equaction (2) from equaction (1), we get `C(s) + (1)/(2) O_(2)(g) rarr CO_((g)) = 0` or `C(s) + (1)/(2) O_(2)(g) rarr CO (g), DeltaH = X - Y` `DeltaH = X - Y = Delta H_(f) (CO) - (0 + 0)` `Delta H_(4)(CO) = X - Y` |
|
| 159. |
Bond energies can be obtained by using the following relation: `DeltaH (reaction) = sum`Bond energy of bonds, broken in the reactants `-sum` Bond energy fo bonds, formed in the products Bond enegry depends on three factors: a. Greater is the bond length, lesser is the bond enegry. b. Bond energy increases with the bond multiplicity. c. Bond enegry increases with electronegativity difference between the bonding atoms. Bond enegry of differene halogen molecules will lie in the sequencesA. `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`B. `Cl_(2) gt Br_(2) F_(2) gt I_(2)`C. `I_(2)gt Cl_(2) gt Br_(2) gt F_(2)`D. `Br_(2) gt F_(2) gt I_(2) gt Cl_(2)`. |
|
Answer» Correct Answer - B Bond energy of `F_(2)` is surprisingle low due to strong repulsion between the lone pairs of two fluorine atoms. |
|
| 160. |
The enthalpy of vaporisation of liquid diethly ether `-(C_2H_5)_2O,` is `26.0kJmol(-1)` at its boiling point `(35.0^@C)`. Calculate `DeltaS` for conversion of : (a) liquid to vapour, and (b) vapour to liquid at `35^@C`. |
|
Answer» (a). `DeltaS_("vaporisation")=(DeltaH_("vaporisation"))/(T_(bp))=(26000)/(308)` `=84.42JK^(-1)mol^(-1)` |
|
| 161. |
Calculate free energy change for the reaction, `H_(2)(g)+Cl_(2)(g)to2H-Cl(g)` By using the bond energy and entropy data. Bond energies of `H-H,Cl-Cl and H-Cl ` bonds are 435kJ `mol^(-1),240jK" "mol^(-1) and 430jk" "mol^(-1)` respectively standard entropies of `H_(2),Cl_(2) and HCl` are 130.59,222.95 and 186.68 `JK^(-1)" "mol^(-1)` respectively. |
|
Answer» Correct Answer - 190.9kJ `DeltaG^(@)` can be calculated by using: `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `DeltaH^(@)=sum(BE)_("reactants")-sum(BE)_("products")` `=435+240-2xx430=-185kJ` `DeltaS^(@)=sumS_("products")^(@)-sumS_("reactants")^(@)` `=2xx186.68-130.59-222.95` `=19.82JK^(-1)=19.82xx10^(-3)kJ" "K^(-1)` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=-185-298xx19.82xx10^(-3)=-190.9kJ` |
|
| 162. |
How much heat is required to cahgen 15.6 g of benzene from liquid into vapour at its boiling point of `80^(@)C`? Entropy of vaporization of enzene is `87JK^(-1)" "mol^(-1)` |
| Answer» Correct Answer - 6142.2J | |
| 163. |
The standed free energy of fromation of NO(g) is 86.6 kj/ mol at 298 K what is the standed free energy of fromation of ` NO_(2) g` at 298 k? `K _(p)= 1.6 xx 10^(12)`A. `0.5[2xx86.600-R(298)ln(1.6xx10^(12))`B. `R(298)ln(1.6xx10^(12))-86.600`C. `86.600+r(298)ln(1.6xx10^(12))`D. `86.600-(ln(1.6xx10^(12)))/(R(298))` |
|
Answer» Correct Answer - A `DeltaG=DeltaG^@+RT" ln "K_(p)` . . (i) `DeltaG=0 and DeltaG^(@)=2DeltaG_(f)^(@)NO_(2)-2DeltaG_(f)^(@)NO-2DeltaG_(f)^(@)O_(2)` `2DeltaG_(f)^(@)NO_(2)=DeltaG^(@)+2DeltaG_(f)^(@)NO` (Since `DeltaG_(f)^(@)O_(2)=0`) `=[2xx86,600-RT" ln "1.6xx10^(12)]` ltbRgt `DeltaG_(f)^(@)NO_(2)=0.5[2xx86.600-RT" ln "1.6xx10^(-12)]` |
|
| 164. |
Ethanol boils at `78.4^@C` and the enthalpy of vaporisation of ethanol is `42.4kJ mol^(-1)`. Calculate the entropy of vaporisation of ethanol. |
| Answer» `120.9kJ^(-1)" "mol^(-1)` | |
| 165. |
For which of the following reactions, the entropy change will be positive?A. `H_(2)(g)+I_(2)(g)hArr2HI(g)`B. `HCl(g)+NH_(3)(g)hArrNH_(4)Cl(s)`C. `NH_(4)NO_(3)(s)hArrN_(2)O+2H_(2)O(g)`D. `MgO(s)+H_(2)(g)hArrMg(s)+H_(2)O(l)` |
|
Answer» Correct Answer - C `Deltan_(g)=3-0=3` Since `Deltan_(g)gt0`, there will be increase in entropy change. |
|
| 166. |
The free energy of formation of NO is `78 kJ mol^(-1)` at the temperature of an authomobile engine `(1000 K)`. What is the equilibrium constant for this reaction at `1000 K`? `1/2 N_(2)(g)+1/2 O_(2)(g) hArr NO(g)`A. `8.4xx10^(-5)`B. `7.1xx10^(-9)`C. `4.2xx10^(-10)`D. `1.7xx10^(-19)` |
|
Answer» Correct Answer - A `K="antilog"[(-DeltaG^(@))/(2.303RT)]` `="antilog"[(-78xx1000)/(2.303xx8.314xx1000)]` `=8.4xx10^(-5)` |
|
| 167. |
Equilibrium constant for the reaction: `H_(2)(g) +I_(2) (g) hArr 2HI(g) is K_(c) = 50 at 25^(@)C` The standard Gibbs free enegry change for the reaction will be:A. `-6.964kJ`B. `-9.694kJ`C. `-4.964kJ`D. `-6.496kJ` |
|
Answer» Correct Answer - B `DeltaG^(@)=-2.303RT" log "K_(C)` `=-2.303xx8.314xx298" log "50` `=-9694J=-9.694kJ` |
|
| 168. |
6 moles of an ideal gas expand isothermally and reversible from a volume of `1dm^(3)` to a volme of `10dm^(3)` at `27^(@)C`. What is the maximum work done? Express your answer in joule. |
| Answer» `-34464.8J`, work is done by the system. | |
| 169. |
A certain volume of dry air at NTP is expanded reversible to four times its volume (a) isothermally (b) adiabatically. Calculate the final pressure and temperature in each case, assuming ideal behaviour. (`(C_(P))/(C_(V))` for air=1.4) |
|
Answer» Let `V_(1)` be the intial volume of dry at NTP. (a) Isotermal expansion: During isothermal expansion, the temperature remains the same throughout. Hence, final temperature will be 273K. Since, `P_(1)V_(1)=P_(2)V_(2)` `P_(2)=(P_(1)V_(1))/(V_(2))=(1xxV_(1))/(4V_(1))=0.25atm` (b) Adiabatic expansion: `(T_(1))/(T_(2))=((V_(2))/(V_(1)))^(gamma-1)` `(273)/(T_(2))=((4V_(1))/(V_(1)))^(1.4-1)=4^(0.4)` `T_(2)=(273)/(4^(0.4))=156.79K` Final pressure: `(P_(1))/(P_(2))=((V_(2))/(V_(1)))^(gamma)` `(1)/(P_(2))=((4V_(1))/(V_(1)))^(1.4)=4^(1.4)` `P_(2)=(1)/(4^(1.4))=0.143`atm |
|
| 170. |
If 1.5 L of an ideal gas at a apressure of 20 atm expands isothermally and reversibly to a final volume of 15 L, the work done by the gas in L atm is:A. 69.09B. 34.55C. `-34.55`D. `-69.09` |
|
Answer» Correct Answer - D `w_("expansion")=-2.303nRT"log"_(10)((V_(2))/(V_(1)))` `=-2.303PV" "log_(10)((V_(2))/(V_(1)))` `=-2.303xx1.5xx20" log "((15)/(1.5))` `=69.09L" "atm` |
|
| 171. |
`5mol` of an ideal gas at `27^(@)C` expands isothermally and reversibly from a volume of `6L` to `60L`. The work done in `kJ` isA. `-14.7`B. `-28.72`C. `+28.72`D. `-56.72` |
|
Answer» Correct Answer - B `w=-2.303nRTlog((V_(2))/(V_(1)))` `=-2.303xx5xx8.314xx300log((60)/(6))` `=-28.72kJ` |
|
| 172. |
Calculate the change in entropy when 1 mole nitrogen gas expands isothermally and reversibly from an initial volume of 1 litre to a final volume of 10 litre at `27^(@)C` |
|
Answer» `DeltaS=2.303nR" "log((V_(2))/(V_(1)))` `=2.303xx1xx8.314log((10)/(1))` `=19.12JK^(-1)`. |
|
| 173. |
An ideal gas in thermally insulated vessel at internal `(pressure)=P_(1), (volume)=V_(1)` and absolute `temperature = T_(1)` expands irreversiby against zero external, pressure , as shown in the diagram, The final internal pressure, volume and absolute temperature of the gas are `p_(2), V_(2) and T_(2)`, respectively . For this expansion A. `q=0`B. `T_(2)=T_(1)`C. `P_(2)V_(2)=P_(1)V_(1)`D. `P_(1)V_(2)^(gamma)=P_(1)V_(1)^(gamma)` |
|
Answer» Correct Answer - A::B::C Since, the vessel is thermally insulated hence q=0 Work done (W=0) is zero for free expansion `DeltaU` is also zero for free expansion, so, `T_(1)=T_(2) and P_(1)V_(1)=P_(2)V_(2)` (Isothermal process). |
|
| 174. |
How much energy is absorbed by 10 moles of an ideal gas if it expands from an initial pressure of 8 atmosphere to 4 atmosphere at a constant temperature of `27^(@)C`? `(R=8.31J" "mol^(-1)" "K^(-1))` |
|
Answer» Correct Answer - `1.728xx10^(4)J` In isothermal process, `DeltaU=0` so, q(heat absorbed)=-w. Thus, apply the equation `q=2.303nRT" log "(P_(1))/(P_(2))` |
|
| 175. |
Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `93 kJ mol^(-1)`B. `-245 kJ mol^(-1)`C. `-93 kJ mol^(-1)`D. `+245 kJ mol^(-1)` |
|
Answer» Correct Answer - C `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`, `DeltaH^(@)=?` `Delta_(r )H^(@)=(1)/(2)BE(H_(2))+(1)/(2)BE(Cl_(2))-BE(HCl)` `=(1)(2)(434)+(1)/(2)(242)-431` `=217+121-431=-93 kJ mol^(-1)` |
|
| 176. |
The standard enthalpy of formation of `H_(2)(g)` and `Cl_(2)(g)` and `HCl(g)` are `218 kJ//mol`, `121.88 kJ//mol` and `-93.31 kJ//mol` respectively. Calculate standard enthalpy change in `kJ` for `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`A. `+431.99B. `-262.14`C. `-431.99`D. `+247.37` |
|
Answer» Correct Answer - B `Delta_(r )H=DeltaH_("products")-DeltaH_("reactants")` `=-92.3-[(218)/(2)+(121.68)/(2)]` `=-92.3-[109+60.84]` `=-262.14 kJ//mol` |
|
| 177. |
Given the following data : Determine at what temperature the following reaction is spontanous? `FeO(s)+C_("graphite")toFe(s)+CO(g)`A. 298 KB. 668 KC. 966KD. `DeltaG^(@)` is +ve, hence the reaction will never be spontaneous. |
|
Answer» Correct Answer - C `DeltaH_("reaction")=sum{DeltaH_(f" "Fe(s))^(@)+DeltaH_(f" "CO" "(g))^(@)}-{DeltaH_(f" "FeO(s))^(@)+DeltaH_(f" "C("graphite"))^(@)}` `=(0-110.5)-(-266.3+0)=155.8kJ" "mol^(-1)` `DeltaS_("reaction")=sum[S_(Fe(s))^(@)+S_(Co(g))^(@)]-[S_(FeO(s))^(@)+S_(C("graphite"))^(@)]` `=(27.28+197.6)-(57.49+5.74)` `=161.65JK^(-1)" "mol^(-1)` For spontaneous reaction, `T gt (DeltaH)/(DeltaS),T gt (155.8xx1000)/(161.65)` `thereforeT gt 966K` |
|
| 178. |
The emf of the cell reaction `Zn(s) +Cu^(2+) (aq) rarr Zn^(2+) (aQ) +Cu(s)` is `1.1V`. Calculate the free enegry change for the reaction. If the enthalpy of the reaction is `-216.7 kJ mol^(-1)`, calculate the entropy change for the reaction. |
|
Answer» `-DeltaG^(@)=nxxFxxE^(@)=2xx96500xx1.1=212.3kJ` `DeltaG^(@)=-212.3kJ" "mol^(-1)` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `DeltaS^(@)=(DeltaH^(@)-DeltaG^(@))/(T)=(-216.7-(-212.3))/(298)` `=-0.01476" kJ "K^(-1)mol^(-1)` `=-14.76" J "K^(-1)mol^(-1)`. |
|
| 179. |
The standard potential for the reaction, `Ag^(+)(aq.)+Fe^(2+)(aq.)toFe^(3+)(aq.)+Ag(s)` is 0.028V. What is the standard free energy change for this reaction? |
| Answer» `-2.702kJ" "mol^(-1)` | |
| 180. |
For the reaction `A(s) rarr B(s) +C(s)` Calculate the entropy change at `298K` and `1 atm` if absolute etropies (in `J K^(-1)mol^(-1))` are `A = 130, B = 203, C = 152` |
|
Answer» `DeltaS^(@)=sumS_("products"))^(@)-sumS_(("reactants"))^(@)` `=[S_(B)^(@)+S_(C)^(@)]-[S_(A)^(@)]` `=[203+152]-[130]` `=225JK^(-1)`. |
|
| 181. |
A gas mixture of `3.67L` of ethylene and methane on complete combustion at `25^(@)C` produces `6.11 L` of `CO_(2)`. Find out the heat evolved on buring `1L` of the gas mixture. The heats of combustion of ethylene and methane are `-1423` and `-891kJ mol^(-1)`, respectively, at `25^(@)C`. |
|
Answer» `underset("a litre")(C_(2)H_(4))+3O_(2)tounderset("2a litre")(2CO_(2))+2H_(2)O` `underset((3.67-a)litre)(CH_(4))+2O_(2)tounderet((3.67-a)litre)(CO_(2))+2H_(2)O` Given, `2a+3.67-a=6.11` a=2.44 litre Volume of ethylene in mixture =2.44 litre Volume of methane in mixture=1.23 litre Volume of ethylene in 1 litre mixture `=(2.44)/(3.67)=0.6649` litre Volume of methane in 1 litre mixture `=(1.23)/(3.67)=0.3351` litre 24.45 litre of a gas at `25^(@)C` correspond to 1 mole Thus, heat evolved by burning 0.6649 litre of ethylene thus, heat evolved by burning 0.6649 litre of ethylene `=-(1424)/(24.5)xx0.6649=-38.69kJ` and heat evolved by burning 0.3351 litre of methane `=-(891)/(24.45)xx0.3351=-12.21kJ` ltBrgt so, total heat evolved by buring 1 litre of mixture. `=-38.69-12.21` `=-50.90kJ` |
|
| 182. |
Standard heat of formation of `CH_(4),CO_(2)` and `H_(2)O_((g))` are `-76.2,-394.8` and `-241.6kJ mol^(-1)` respectively. Calculate the amount of heat evolved by burning `1m^(3)` of `CH_(4)` measured under normal conditions. |
|
Answer» The required equation for the combustion of methane is: `CH_(4)+2O_(2)toCO_(2)+2H_(2)O,DeltaH=?` `DeltaH=DeltaH_(f("products"))-DeltaH_(f("reactants"))` `=DeltaH_(f(CO_(2)))+2xxDeltaH_(f(H_(2)O)-DeltaH_(f(CH_(4)))-2DeltaH_(f(O_(2)))` `=-398.8-2xx241.6-(-76.2)-2xx0` `=-805kJ" "mol^(-1)` heat evolved by burning 22.4 litre (1 mole) methane `=-805.8kJ`. So, heat evolved by burning 1000 litre `(1m^(3))` methane `=-(805.8)/(22.4)xx1000=-35973.2kJ` |
|
| 183. |
For the reaction, `A(s)+3B(s)to4C(s)+D(l)` `DeltaH` and `DeltaU` are related as-A. `DeltaH=DeltaU`B. `DeltaH=DeltaU+3RT`C. `DeltaH=DeltaU+RT`D. `DeltaH=DeltaU-3RT` |
|
Answer» Correct Answer - D `DeltaH=DeltaV+Deltan_(g)RT` `A(s)+3B(g)to4C(s)+D(l)` `Deltan_(g)=0-3=-3` `DeltaH=DeltaU-3RT` |
|
| 184. |
Which one of the following is correct ?A. `-DeltaG=DeltaH-TDeltaS`B. `DeltaH=DeltaG-TDeltaS`C. `DeltaS=(1)//(T)[DeltaG-DeltaH]`D. `DeltaS=(1)//(T)[DeltaH-DeltaG]` |
|
Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` or `T Delta S=DeltaH-DeltaG` or `DeltaS=(1)/(T)(DeltaH-DeltaG)`. |
|
| 185. |
Determine the heat of the following reaction: `FeO(s)+Fe_(2)O_(3)(s)toFe_(3)O_(4)(s)` Given informations: `2Fe(s)+O_(2)(g)to2FeO(s)," "DeltaH^(@)=-544kJ` `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)," "DeltaH^(@)=-1648.4kJ` `Fe_(3)O_(4)(s)to3Fe(s)+2O_(2)(g),DeltaH^(@)=+1118.4kJ`A. `-1074kJ`B. `-22.2kJ`C. `+249.8kJ`D. `+2214.6kJ`b |
|
Answer» `2FeO(s)to2Fe(s)+O_(2)(g)," "DeltaH^(@)=+544kJ` `2Fe_(2)O_(3)(s)to4Fe(s)+3O_(2)(g),DeltaH^(@)=+1648.4kJ` `underline(6Fe(s)+4O_(2)(g)to2Fe_(2)O_(4)(s)," "DeltaH^(@)=-2xx1118.4kJ)` On adding, `underline(2FeO(s)+2Fe_(2)O_(3)(s)to2Fe_(3)O_(4)(s)," "DeltaH^(@)=44.4kJ)` `thereforeFeO(s)+Fe_(2)O_(3)(s)toFe_(3)O_(4)(s)," "DeltaH^(@)=-22.2kJ)`. |
|
| 186. |
Which thermodynamics property provides a measure of randomness in the system ?A. EnthalpyB. PV-workC. Free energyD. Entropy |
|
Answer» Correct Answer - D Entropy is a measure of randomness. |
|
| 187. |
Calculate `DeltaH` at `85^(@)C` for the reaction: `Fe_(2)O_(3)(s) +3H_(2)(g) rarr 2Fe(s) +3H_(2)O(l)` The data: `DeltaH_(298)^(Theta) =- 33.0 kJ mol^(-1)` and `{:("Substance",Fe_(2)O_(3)(s),Fe(s),H_(2)O(l),H_(2)(g)),(C_(P)^(@)(JK^(-1)mol),103.0,25.0,75.0,28.0):}` |
|
Answer» `Fe_(2)O_(3)(s)+3H_(2)to2Fe+3H_(2)O` `DeltaC_(P)=(C_(P))_(P)-(C_(P))_(R)` `=(2xx25.1+3xx75.3)-(103.8+3xx28.8)` `=276.1-190.2=85.9J` `(DeltaH_(T_(2))-DeltaH_(T_(1)))/(T_(2)-T_(1))=DeltaC_(P)` `(DeltaH_(358)-(-33.29))/(358-298)=85.9xx10^(-3)` `DeltaH_(358)=-28.14kJ//mol` |
|
| 188. |
Calculate heeat of formation of cane sugar following data: `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH=-68.4` kcal `C(g)+O_(2)(g)toCO_(2)(g),DeltaH=-94.4kcal` `C_(12)H_(22)O_(11)(s)+12O_(2)(g)to12CO_(2)(g)+11H_(2)O(l),DeltaH=-1350.0kcal` |
| Answer» Correct Answer - `-535.kcal` | |
| 189. |
`C_("diamond")+O_(2)(g)rarrCO_(2)(g),DeltaH=-395 kJ` ......`(i)` `C_("graphite")+O_(2)(g)rarrCO_(2)(g),DeltaH=-393.5KJ" "…(ii)` The `DeltaH` , when diamond is formed from graphite, isA. `-1.5KJ`B. `+1.5KJ`C. `+3.0KJ`D. `-3.0KJ` |
|
Answer» Correct Answer - B Equaction (ii) - Equaction (i) `C_(C) rarr C_(D): , Delta H = 1.5 kJ` |
|
| 190. |
According to third law of thermodynamics, the entropy at `0 K` is zero forA. elements in their stable formB. perfectly crystalline solidsC. substances at `1 "atm"` and `25^(@)C`D. `N_(2)O` |
|
Answer» Correct Answer - B For perfectly crystalline solids at `0 K`, all molecular motion ceases and particles are closely packed. Thus randomness and hence entropy is minimum. |
|
| 191. |
`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)` `DeltaH^(@)=+234.12KJ ` `C(s)+O_(2)(g)toCO_(2)(g) DeltaH^(@)=-393.5KJ` Use these equations and `DeltaH^(@)` value to calculate `DeltaH^(@)` for this reaction : `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`A. `-1648.7kJ`B. `-1255.3 kJ`C. `-1021.2kJ`D. `-129.4kJ` |
|
Answer» Correct Answer - A `2Fe(s)+(3)/(2)CO_(2)(g)toFe_(2)O_(3)(s)+(3)/(2)C(s)" "(DeltaH^(@)=-234.1kJ)` . . .(i) `C(s)+O_(2)(g)toCO_(2)(g)" "(DeltaH^(@)=-393.5kJ)` . . . (ii) Multiplying eq. (i) by 2 and eq. (ii) by 3 and on adding both equations, we get: `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)," "DeltaH^(@)=(-234.1xx2)+(-3xx393.5)` `=-1648.7kJ` |
|
| 192. |
For the following reaction, `C_("diamond")+O_(2)toCO_(2)(g), DeltaH=-94.3" "kcal` `C_("graphite")+O_(2)toCO_(2)(g), DeltaH=-97.6" "kcal` the heat require to change 1 g of `C_("diamond")toC_("graphite")` is: |
|
Answer» Subtracting eq. (ii) from (i), the reqruied equation is obtained `DeltaH_("transoformation")=-94.5-(94.0)` `=94.5+94.0=-0.5kcal` |
|
| 193. |
Given `C_(("graphite"))+O_(2)(g)toCO_(2)(g),` `Delta_(r)H^(0)=-393.5kJ" "mol^(-1)` `H_(2)(g)=+(1)/(2)O_(2)(g)toH_(2)O(1),` `Delta_(r)H^(0)=-285.8" kJ "mol^(-1)` `CO_(2)(g)+2H_(2)O(1)toCH_(4)(g)+2O_(2)(g)`, `Delta_(r)H^(0)=+890.3kJ" "mol^(-1)` Based on the above thermochemical equations, the value of `Delta_(r)H^(0)` at at 298 K for the reaction `C_(("graphite"))+2H_(2)(g)toCH_(4)(g)` will be:A. `+74.8kJ" "mol^(-1)`B. `+144.0kJ" "mol^(-1)`C. `-74.8kJ" "mol^(-1)`D. `-144.8kJ:" " mol^(-1)` |
|
Answer» Correct Answer - C From the given equations: `C_(("graphite"))+O_(2)(g)toCO_(2)(g),Delta_(r)H^(@)=-393.5Kj" "Mol^(-1)` `2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)H^(@)=-285.8xx2kJ" "mol^(-1)` `CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)H^(@)=+890.3kJ" "mol^(-1)` Adding above three equations we get `C_(("Graphite"))+2H_(2)(g)toCH_(4)(g)` `Delta_(r)H^(@)=-393.5-285.8xx2+890.3` `=74.8kJ" "mol^(-1)`. |
|
| 194. |
One gram-atom of graphite and one gram-atom of diamond were separately burnt to `CO_(2)` . The amount of heat liberated was `393.5KJ` and `305.4KJ` respectively. It is apparent thatA. graphite has greater affinity for oxygenB. diamond has greater affinity for oxygenC. graphite is stabler than diamondD. diamond is stabler than graphite. |
|
Answer» Correct Answer - C Since graphite on combustion give less heat, it means graphite has less heat content and is more stable. |
|
| 195. |
According to the first law of thermodynamics which of the following quantities represents change in a state function ?A. `q_(rev)`B. `q_(rev)-w_(rev)`C. `q_(rev)//w_(rev)`D. `q_(rev)+w_(rev)` |
|
Answer» Correct Answer - D `DeltaU=q+w`, `DeltaU` is a state function. |
|
| 196. |
Calculate the enthalpy of the reaction `SnO_(2)(s)+2H_(2)(g)toSn(s)+2H_(2)O(l)` given that , bond enthalpies of formation of `SnO_(2)(s)` and `H_(2)O(l)` are -580.7 kJ and -285.8kJ respectively. |
| Answer» Correct Answer - `+9.1kJ` | |
| 197. |
Calculate the heat of formation of acetic acid form the following date: a. `CH_(3)COOH(l) +2O_(2)(g) rarr 2CO_(2)(g) +2H_(2)O(l)DeltaH^(Theta) =- 200.0kcal` b. `C(s) +O_(2) (g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.0 kcal` c. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l), DeltaH^(Theta) =- 68.0 kcal` |
|
Answer» The required equation is `2C(s)+2H_(2)(g)+O_(2)(g)=CH_(3)COOH(l)," "DeltaH=`? this equation can be obtained by multiplying eq. (ii) by 2 and also eq. (iii) by 2 and adding both and finally subtracting eq (i) `[2C+2O_(2)+2H_(2)+O_(2)-CH_(3)COOH(l)-2O_(2)to2CO_(2)+2H_(2)O-2CO_(2)-2H_(2)O]` `DeltaH_(CH_(3)COOH(l))=2xx(-94.96)+2xx(-68.4)-(-207.9)` `=-188.96-138.8+207.9` `=-325.76+207.9=-117.86kcal` |
|
| 198. |
One gram mole of graphite and diamond were burnt to form `CO_(2)` gas: `C_(("graphite"))+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-399.5kJ` `C_(("diamond"))+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-395.4 kJ`A. graphite is more stable than diamondB. diamond is more stable than graphiteC. graphite has greater affinity with oxygenD. diamond has greater affinity with oxygen. |
|
Answer» Correct Answer - A Thermal stability of one isotope is directly proportional to the heat of combustion. |
|
| 199. |
In thermodynamics, a process is called reversible whenA. surrounding is no boundary between syatem and surroundingB. there is no boundary between system and surrouningsC. the surrounding are always in equilibrium with the systemD. the system change into the surroundings spontaneous |
|
Answer» Correct Answer - C In a reversible process, the driving and the opposite forces are nearly equal, hence the system and the surroundings always remain in equilibrium with each other. |
|
| 200. |
Enthalpy of `CH_(4)+(1)/(2)O_(2)rarrCH_(3)OH` is negative. If enthalpy of combustion of `CH_(4)` and `CH_(3)OH` are `x` and `y` respectively, then which relation is correct?A. `xgty`B. `xlty`C. `x=y`D. `xgey` |
|
Answer» Correct Answer - A `CH_(4)+O_(2)toCO_(2)+2H_(2)O`, `DeltaH_(1)=-xkJ……(i)` `CH_(3)OH+O_(2)+CO_(2)+2H_(2)O`, `DeltaH_(2)=-y kJ……..(ii)` Subtract `(ii)` from `(i)` We get `CH_(4)+(1)/(2)O_(2)toCH_(3)OH`, `DeltaH_(3)=-ve` or `-x-(-y)=-ve` or `y-x=-ve` Hence `x gt y` |
|