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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.
| 1301. |
Explain the concept of bond energy and bond dissociation energy in diatomic molecules. |
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Answer» For a diatomic molecules, the bond energy is equal to the bond dissociation energy. The bond dissociation energy of a diatomic molecule is equal to the heat energy (ΔH) needed to break the bonds in one mole of molecules so that free gaseous atoms are produced. For example (a) The bond energy of H-H. is equal to the enthalpy change for breaking one mole of H-H bonds is H2 molecules and to produce free gaseous H atoms. H-H (g) → 2H (g) ΔH°d (H2, g) = ΔH° (H-H) = 436 kJ mol-1 (b) The bond energy of O = O bond is equal to the enthalpy change for breaking one mole O = O bonds in O2 molecules and to produce free gaseous O atoms. O = O (g) → 2O (g) ΔH°d (O2, g) = ΔH° (O - O) = 498 kJ mol-1 (c) The bond energy of H-Cl bond is equal to the enthalpy change in breaking one mole H-Cl bonds in HCl molecules so that 1 mole H atoms and 1 mole Cl atoms are produced. H-Cl (g) → H (g) + Cl (g) ΔH°d (HCl) = Δ° (H-Cl) = 432 kJ mol-1 |
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| 1302. |
(a) A 1000 watt hour is kept on in a room whose dimensions are `5m xx 5m xx 3m`. Howmuch temperature of the room will rise after half an hour? Given the following data `:` Heat capacity ofair at room temperature and 1 atm`= 0.71 Jg^(-1)K^(-1)` Density ofair `= 1.22xx 10^(-3) g mL^(-1)` (b) How much temperature of the room willrise if `25%` heat is lost by radiation ? |
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Answer» (a) 1 Watt `= 1 J s^(-1)` `:. `Heat evolved by 1000 W heaterin 30min. `= 100 xx 30 xx 60 J = 18 xx 10^(5) J` Volume of the room ` 5 xx4 xx 3m^(3) = 60 m^(3) = 60 xx 10^(6) cm^(3) ` Density of air`=1.22 xx 10^(-3) g mL^(-1)` Mass of air in the room`= ( 60xx 10^(6)) xx ( 1.22 xx 10^(-3)) g = 73.2 xx 10^(3)g` Heat evolved by the heater `=` Heat absorbed by walls, roof and air inside. `18 xx10^(5) J =( 5 xx 10^(4) xx Delta T ) + ( 73.2 xx 10^(3)) xx 0.71 xx Delta T` `= ( 50000+ 51972 )Delta T` `= 101972 Delta T = 1.01 972 xx 10^(5) DeltaT ` or `Delta T = ( 1.8 xx 10^(5))/( 1.01972 xx 10^(5)) = 17.65^(@) ` (b) If `25%` heat is lost by radiation, net heat available in the room ` = (75)/( 100) xx 18 xx 10^(5) J = 13.5 xx 10^(5) J` ,brgt `13.5 xx 10^(5) J = ( 5 xx 10^(4) xx Delta T ) + ( 73.2 xx 10^(3)) xx 0.71xx DeltaT` or` Delta T =( 13.5 xx 10^(5))/( 1.01972xx 10^(5)) = 13.24^(@)` |
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| 1303. |
For the reversible process, the value of `DeltaS` is given by the expression:A. `DeltaH//DeltaT`B. `T//q(rev)`C. `q(rev)xxT`D. `q(rev)//T` |
| Answer» `DeltaS = (q_(rev))/(T)` | |
| 1304. |
`50` students sitting in the room of `5 xx 10 xx 3m^(3)` dimensions. The air inside the room is at `27^(@)C` and `1atm` pressure. If each student loses `100` watt heat per second assuming the walls, ceiling floor, and all the material present inside the room is perfectly insulated as well as neglecting loss of air to the outside as the temperature is raised, how much rise in temperature will be noticed in `10min`? Given `C_(P) = (7)/(2)R` for air.A. `15.90^(@)C`B. `16.90^(@)C`C. `14.90^(@)C`D. `17.90^(@)C` |
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Answer» Correct Answer - B Vol. of room `1=5xx10xx3=150 m^(3)` `=15xx10^(4) litre` Moles of air in the room at `27^(@)C` and `1 atm P` `n=(PV)/(RT)=(1xx15xx10^(4))/(0.0821xx300)=6.1xx10^(3)` Heat produced by `50` student `=100xx50` `5000` watt `sec^(-1)` Heat produced in `10` minutes `=500xx10xx60` `=3xx10^(6)` watt `sec^(-1)` or `Jsec^(-1)` `:.` change in enthalpy of air, `DeltaH=nC_(p)xxDeltaT` `3xx10^(6)=6.1xx10^(3)xx7/2xx8.314xxDeltaT` `:. DeltaT=16.90^(@)` |
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| 1305. |
For a process `H_(2)O(s) rarr H_(2)O(l) at 273 K`A. `G (ice) = G ("water") = 0`B. `G (ice) = G ("water") 1= 0`C. `G(ice) gt G (water)`D. `G(ice) lt G(water)` |
| Answer» `G_((ice)) = G_((water)) = 0` | |
| 1306. |
In which of the following process `DeltaH` and `DeltaU` are of same magnitudeA. Evaporation of `C CI_(4)(l)`B. `CaCO_(3)(s) rarr CaO(s) +CO_(2) (g)`C. `NH_(4)CI(s) rarr NH_(4)CI(g)`D. `2HI(g) rarr H_(2)(g) +I_(2)(g)` |
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Answer» `DeltaH = DeltaU +DeltanRT` `Deltan = 0 n_(P) -n_(R)` `,. DeltaH = DeltaU` |
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| 1307. |
Two moles of ideal gas at `27^(@)C` temperature is expanded reversibly from 2 litre to 20 liter. Find entropy change `(R=2cal//molK)` .A. zeroB. `9.2`C. `92.2`D. `5` |
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Answer» Correct Answer - B `DeltaS=2.303nR"log"(V_(2))/(V_(1))` `=2.303xx2xx2"log"(20)/(2)` `=9.2` |
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| 1308. |
Why does heat get released`//` absorbed during chemical reactions? |
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Answer» Because the reactant have a fixed enthalpy content before the reaction and when these are converted into the products which have a different enthalpy content. So, heat gets released or absorbs. Even if temperature of reactions remains constant yet due to change in bonding energies `Delta E != 0`. If `H_("Product") gt H_("reactants")` Reaction should be endothermic as we have to give extra heat to reactants to get these converted into products and if `H_("Product") lt H_("reactants")` Reaction will be exothermic as extra heat content of reactant will be released during the reaction. Enthalpy change of a reaction: `Delta H_("reaction") = H_("products")-H_("reactant")` `{:(DeltaH_("reactions"),H_("products"),-H_("reactants"),,),(,="positive",-,"endothermic",),(,="negative",-,"exothermic",):}` |
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| 1309. |
The enthalpy of tetramerization of X in gas phase `(4 X(g) to X_(4)(g))` is -100 kJ/mol at 300K. The enthalpy of vaporisation for liquid X and `X_(4)` are respectively `30kJ//mol` and `72kJ//mol` respectively . `DeltaS` for tetramerization of X in liquid phase is - 125J/K mol at 300K. What is the `DeltaG` at 300n K for tetramerization of X liquid phase ?A. `- 52 kJ//mol`B. `- 89. 5 kJ//mol`C. `-14.5 kJ//mol`D. None the these |
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Answer» Correct Answer - C `" "4X(l) " "to" "X_(4) (l)" " DeltaS = - 0.125 kJ` `" "4X(l) " "to" "X_(4) (l)" " DeltaH = ?` `(i) " "4X(l) " "to" "X(g)" " DeltaH = 120` `(ii) " "X_(4)(g) " "to" "X_(4)(l)" " DeltaH = -72` `(iii) " "4X(g) " "to" "X_(4)(g)" " DeltaH = -100` ` " "4X(l) " "to" "X_(4)(l)" " DeltaH = -52` ` " "DeltaG = -52 - 300 (-0.125) = -14.5kJ` |
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| 1310. |
At `25^(@)C`, when 1 mole of `MgSO_(4)` was dissolved in water, the heat evolved was found to be 91.2 kJ. One mole of `MgSO_(4) . 7H_(2)O` on dissolution gives a solution of the same composition accompanied by an absorption of 13.8 kJ. The enthalpy of hydration, i.e., `DeltaH_(h)` for the reaction `MgSO_(4)(s) + 7H_(2)O(l)rarrMgSO_(4) . 7H_(2)O(s)` :A. `-105 kJ//mol`B. `-77.4 kJ//mol`C. `105kJ//mol`D. `77.4 kJ//mol` |
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Answer» Correct Answer - A Given that : `Delta_(r)H_(1) = -91.2 kJ//mol` ....(i) `MgSO_(4) 7H_(2)O (s)+(n-7)H_(2)O rarr MgSO_(4)(nH_(2)O)` `Delta_(r)H_(2) = 13.8 kJ//mol` ....(ii) Equation (i)-(ii) or `Delta H_(hyd) = Delta_(r)H_(1)-Delta_(r)H_(2)` `= -91.2 kJ//mol-13.8 kJ//mol` `= -105 kJ//mol` |
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| 1311. |
A 0.05 L sample of 0.2 M aqueous hydrochloric acid is added to 0.05 L of 0.2 M aqueous ammonia in a calorimeter. Heat capacity of entire calorimeter system is 480 J/K. The temperature increase is 1.09 K. Calculate `Delta_(r )H^(@)` in kJ/mol for the following reaction : `HCl(aq.)+NH_(3)(aq.)rarrNH_(4)Cl(aq.)`A. `-52.32`B. `-61.1`C. `-55.8`D. `-58.2` |
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Answer» Correct Answer - A m mole of acid `=0.05 xx 0.2 =0.01` `Delta_(r )H^(@)=-(480xx1.09)/(0.01 xx 1000)=-52.32 " kJ"//"mol"` |
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| 1312. |
As area under P-V diagram represent work done by gas in a thermodynamic process, area under temperature (T) entropy (S) graph represents heat supplied to the thermodynamic system. Consider the following graph A. Process AB is isochoricB. Process BC is isothermalC. Process AB is adiabaticD. heat is absorbed by the gas during the process ACBA |
| Answer» Correct Answer - B::C::D | |
| 1313. |
Statement-1: Work and Heat are two equivalent form of energy. Statement-2: Work is transfer of mechanical energy irrespective of temperature-differene whereas heat is transfer of thermal energy because of temperature difference only.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - A | |
| 1314. |
Heat equivalent to `30J` is supplied to a thermodynamic system and `10 J` of work is done on the system. What is change in its internal energy? |
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Answer» Here, `dQ= +30 J, dW= -10 J`. `dU=dQ-dW=30 -(-10)= 40J` |
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| 1315. |
Heat of combustion of ethanol at constant pressure and at temperature TK is found to be `-q J" mol"^(-1)`. Hence, heat of combustion (in J `mol^(-1)`) of ethanol at the same temperature and at constant volume will be :A. RT - qB. `-(q+RT)`C. q - RTD. q + RT |
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Answer» Correct Answer - A `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)` `Deltan_(g)=2-3=-1` `DeltaH=DeltaU+Deltan_(g)RT` `DeltaU=DeltaH+RT=RT-q` |
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| 1316. |
For which process will `DeltaH and DeltaG^(@)` be expected to be most similar?A. `2Al(s)+Fe_(2)O_(3)(s) to2Fe(s)+Al_(2)O_(3)(s)`B. `2Na(s)+2H_(2)O(l) to 2NaOH(aq)+H_(2)O(g)`C. `2NO_(2)(g) to N_(2)O_(4)(g)`D. `2H_(2)(g) + O_(2)(g) to 2H_(2)O(g)` |
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Answer» Correct Answer - A |
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| 1317. |
From the following data at`25^(@)C` `{:(Reaction ,Delta_(r)H^(o)kJ//mol),((1)/(2)H_(2)(g)+(1)/(2)O_(2)(g)toOH(g),42),(H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(g),-242),(H_(2)(g)to 2H(g),436),(O_(2)(g)to2O(g),495):}` Which of the following Statement (s) is /are Correct: StatementA:`Delta_(r)H^(@) "for the reaction"` `H_(2)O(g)to2H(g)+O(g)"is 925.5kJ/mol"` Statement B:`Delta_(r)H^(@) "for the reaction"` `H_(2)O(g)toH(g)+O(g)"is 423.5kJ/mol"` Statement C:Enthalpy of formation of H(g) is-218 kJ/mol Statement D: Enthalpy of formation of OH(g) is 42 kJ/molA. Statement CB. Statement A,B,DC. Statement B,CD. Statement A,B only |
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Answer» Correct Answer - b |
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| 1318. |
Which of the following graphs given below show (s) adiabatic process? A. `II,III`B. `I,III`C. `II,IV`D. `I,IV` |
| Answer» Adiabatic slopes are move steeper than isothermal. Slope of adiabatic process `= gamma xx` Slope of isothermal process. | |
| 1319. |
Which of the following graphs given below show (s) adiabatic process? A. `I` and `III`B. `II` and `IV`C. `II` and `III`D. `I` and `IV` |
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Answer» Correct Answer - C Adiabatic slope are more steeper than isothermal. |
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| 1320. |
If the enthalpy of formation and enthalpy of solution of HCl (g) are-92.3kj /mol and -75.14kJ/mol respectively then find the enthalpy of `Cl^(-)`(aq):A. `-17.16` kJ/molB. `-167.44` kJ/molC. 17.16 kJ/molD. None of these |
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Answer» Correct Answer - B `HCl(g)+aqrarrH^(+)(aq.)+Cl^(-)(aq.),` `Delta_(r )H=-75.14` `-75.14 = Delta_(f)H(H^(+),aq)+Delta_(f)H(Cl^(-),g)-Delta_(f)H(HCl, g)` `because " "Delta_(f)H(H^(+),aq)=0` `Delta_(f)H(Cl^(-),aq)=-75.14-92.3` `=-167.44 " kJ"//"mol"` |
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| 1321. |
For which of the following process `abs(DeltaH)ltabs(DeltaE)` :A. Vaporisation of liquid bromine at constant pressureB. Dissociation of `NH_(3)(g)` to give `N_(2)(g)` and `H_(2)(g)` at constant pressureC. Adiabatic free expansion of ideal gasD. Conversion of graphite to diamond occuring at constant pressure condition |
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Answer» Correct Answer - D |
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| 1322. |
Entropy is a state function and its depends on two or three variable temperature (T), pressure(P) and volume (V). Entropy change for an ideal gas having number of moles (n) can be determined by the following equation: `DeltaS=2.303nC_(v)" "log((T_(2))/(T_(1)))+2.303 nR" "log ((V_(2))/(V_(1)))` `DeltaS=2.303nC_(p)" "log((T_(2))/(T_(1)))+2.303 nR" "log ((P_(2))/(P_(1)))` Since free energy change for a process or a chemical equation is a deciding factor of spontaneity, which can be obtained by using entropy change (`DeltaS)` according to the expression, `Delta G=DeltaH-TDeltaS` at a temperature T. For a reaction `M_(2)O(s) to 2M(s)+(1)/(2)O_(2)(g),DeltaH=30` kJ/mol and `DeltaS`=0.07kJ/K-mol at 1 atm. Calculate upto which temperature the reaction would not be spontaneous.A. `Tgt428.6K`B. `Tgt300.8K`C. `Tlt300.8K`D. `Tlt428.6K` |
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Answer» Correct Answer - d |
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| 1323. |
Standard Gibbs energy of reaction `(Delta_(r)G^(@))` at a certain temperature can be completed as `Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@)` and the change in the value of `Delta_(r)H^(@)` and `Delta_(r)S^(@)` for a reaction with temperature can be computed as follows: `Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1))` `Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2)))` `Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) ` `Delta_(r)^(@)G^(@)=-RTInK_(eq)` Consider the following reaction : `CO(g)+2H_(2)(g)toCH_(3)OH(g)` Given `Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol` `Delta_(r)H^(@)(CO,g)=-114KJ//mol`` s^(@)(CH_(3)OH,g)=240J//mol-k,` `S^(@)(H_(2)g)=198J//mol-K``C_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1)``C_(p.m(CO)=29.4J//mol-K` `C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K` `and In ((320)/(300))=0.06,"all data at"300K.` `Delta_(r)S^(@) ` at 300 K for the reaction is :A. 152.6J/K -molB. 181.6J?K-molC. `-16J/K--mol`D. none of these |
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Answer» Correct Answer - c |
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| 1324. |
The enthalpy of formation , `DeltaH_(f)^(@)` equals zero at `25^(@)`C for which of the following in their standard states?A. elementsB. compoundsC. gasesD. solids |
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Answer» Correct Answer - a |
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| 1325. |
Standard enthalpy of combustion of cyclopropane is `-2091 kJ//"mole"` at `25^(@)C` then calculated the enthalpy formation of cyclopropane. If `DeltaH_(f)^(o)(CO_(2))= -393.5 kJ//"mole"` and `Delta H_(f)^(o)(H_(2)O)= -285.8 kJ//"mole"`. |
| Answer» Correct Answer - 53 | |
| 1326. |
When freezing of a liquid takes place in a system it:A. may have `q gt 0` or `q lt 0` depending on the liquidB. is represented by `q gt 0`C. is represented by `q lt 0`D. has q = 0 |
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Answer» Correct Answer - C In freezing heat released so `q lt 0` |
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| 1327. |
Specific heat capacity of water isA. `2.09 J g^(-1) K^(-1)`B. `1 J K^(-1)`C. `4.18 J g^(-1) K^(-1)`D. `1.74 J g^(-1) K^(-1)` |
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Answer» Correct Answer - C Specific heat of liquid water is `4.18 J g^(-1) K^(-1)` or `1 cal g^(-1) K^(-1)` |
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| 1328. |
The standard enthalpies of combustion of `C_(6)H_(6) (l) ` ,C ( graphite) and`H_(2)(g)` are respectively `-3270 kJ mol^(-1), - 394kJ mol^(-1)` and`-286 kJmol^(-1)` . What is the standard enthalpy of formation of`C_(6)H_(6)(l)`in `kJ mol^(-1)` /A. `- 48`B. `+48`C. `-480`D. `+480` |
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Answer» Correct Answer - b (i) ` C_(6)H_(6)+ ( 15)/(2) O_(2) rarr 6CO_(2)+3H_(2) O, DeltaH= - 3270 kJ mol^(-1)` (ii) `C+ O_(2) rarr CO_(2), DeltaH = -394kJ mol^(-1)` (iii) `H_(2)+(1)/(2) O_(2) rarr H_(2)O,DetlaH=-286kJ mol^(-1)` Aim `: 6C+ 3H_(2) rarr C_(6)H_(6) , DeltaH=?` `6 xx`Eqn. (ii) `+ 3 xx ` Eqn. (iii) -Eqn. (i) ,gives required eqn.with `DeltaH =+ 48 kJ mol^(-1)` |
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| 1329. |
Given that `DeltaH = 0 ` for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not ? |
| Answer» It is a spontaneous process because although `DeltaH = 0 , i.e., ` energy factor has no role to play but randmness increases, i.e.,randomness factor favours the process. | |
| 1330. |
When the gas is ideal and process is isothermal, thenA. `P_(1)V_(1) = P_(2)V_(2)`B. `Delta U = 0`C. `Delta W = 0`D. `Delta H_(1) = Delta H_(2)` |
| Answer» Correct Answer - A::B::D | |
| 1331. |
Which is an irreversible procees ? (A) Mixing of two gases by diffusion (B) Evaporation of water at 373 K and 1 atm pressure (C) Dissolution of `NaCl` in water (D) Burning of coal the correct answerA. A, BB. B, CC. A, B, CD. A, C, D |
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Answer» Correct Answer - D All natural processes are thermodynamically irreversible. |
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| 1332. |
In which case of mixing of strong acid and strong base each of 1 N concentration temperature increase of solution will be same (assume heat evolved in neutralisation is only used up to increase the temperature of solution)?A. 20 ml acid and 30 ml alkaliB. 10 ml acid and 10 ml alkaliC. 35 ml acid and 15 ml alkaliD. 50 ml acid and 50 ml alkali |
| Answer» Correct Answer - B::D | |
| 1333. |
For a diatomic gas, which options is/are correct?A. `gamma = 1.40`B. `C_(P) = (7R)/(2)`C. `C_(v) = (5R)/(2)`D. `gamma = 1.67` |
| Answer» Correct Answer - A::B::C | |
| 1334. |
A mole of steam is condensed at `100^(@)` C, the water is cooled to `0^(@)`C and frozen to i.e . What is the difference in entropies of the stem and ice? The heat of vaporization and fusion are `540 cal" "gm^(-1)` and `80 cal" "gm ^(-1)` respectively . Use the average heat capacity of liquild water as `1 cal" "gm^(-1)` `degree^(-1)`. |
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Answer» Correct Answer - B `DeltaS_("condensation")= (-DeltaH_("vap"))/(T)=(-540 xx 18)/(373)` `DeltaS_("cooling")= + nC_(p)ln .((T_(2))/(T_(2))) = 18 ln. ((273)/(373))` `DeltaS_("fusion") = (-80 xx 18)/(273)` `= - [18((80)/(273)+ (540)/(373)) + (18ln.(273)/(373))]` `= - 18((80)/(273) +(540)/(373)+ ln.(373)/(273))" "rArr 36.95 cal//^(@)` |
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| 1335. |
A mole of steam is condensed at `100^(@)` C, the water is cooled to `0^(@)`C and frozen to i.e . What is the difference in entropies of the stem and ice? The heat of vaporization and fusion are `540 cal" "gm^(-1)` and `80 cal" "gm ^(-1)` respectively . Use the average heat capacity of liquild water as `1 cal" "gm^(-1)` `degree^(-1)`.A. Entropy change during the condensation of steam is `-26.06 cal//^(@)C`B. Entropy change during cooling of water from `100^(@)C "to" 0^(@)C "is" - 5.62 cal//^(@)C`C. Entropy change during freezing of water at `0^(@)C "is" -5.27 cal//^(@)C`D. Total entropy changwe is `-36.95 cal//^(@)C` |
| Answer» Correct Answer - A::B::C::D | |
| 1336. |
Calculate the magnitude of standard entropy change for reaction `X hArr Y` if `DeltaH^(@) = 25 KJ` and `K_("eq") "is" 10^(-7)` at `300` K. |
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Answer» Correct Answer - 51 Given : `DeltaH^(@) = 25KJ` `T=300 k` `K_("eq") = 10^(-7)` Asked : standard entropy change Formula :`DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)` Since `DeltaG^(@) =- 2.303 "RT log K"_("eq")` `therefore" "-2.303 "RT log K"_("eq") = DeltaH^(@)-TDeltaS^(@)` Explanation: `DeltaG^(@)`= Standard free energy change `DeltaH^(@) ` = Standard enthalpy change T=Temperature `DeltaS^(@)` = standard enthropy change R = Gas constant `K_("eq")` = equilibrium constant Substitution & calculation: `-2.303 xx (25//3) xx (300) "log"10^(-7)=25000-(300) DeltaS^(@)` `DeltaS^(@) = -51 J//K.` |
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| 1337. |
The vapour pressure of liquid Hg at `433` K is`5` mm Hg . Calculate the free energy change accompanying the expansion of one mole of Hg vapour in equilibrium with liquid at `433` K to a pressure of `750` mm Hg at the same temperature assuming the vapour bahaves like an ideal monoatomic gas in KJ `"mole"^(-1)` . (Approximate integer and `e^(5) = 150`) |
| Answer» Correct Answer - 18 | |
| 1338. |
For an endothermic process to be spontaneous, the factor `T Delta S` should be `"…................." Delta H `in magnitude ( equal to or less than or greater than ) . |
| Answer» Correct Answer - greater | |
| 1339. |
The entropies of `H_(2)(g)` and `H(g)` are `60 " and " 50 J"mole"^(-1)K^(-1)` respectively at `300` K. Using the data given below calcualate the bond enthalpy of `H_(2)(g)` in Kcal `"mole"^(-1)`. `" "H_(2)(g) rarr 2H(g)," "DeltaG^(@)=21.6 KJ "mole"^(-1)` |
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Answer» Correct Answer - 8 `DeltaG^(@) = DeltaH^(@) - tDeltaS^(@)` `21.6 = DeltaH_(H-H) - 300 (2 xx S_(H)^(@) - S_(H_(2))^(@))` `DeltaH_(H-H) = 33.6 KJ "mole"^(-1) =8Kcal "mole"^(-1)` |
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| 1340. |
For the vaporisation of water : `H_(2)O (I) hArr H_(2)O(g)` [1 atm. Pressure] Given : `Delta S = 120 JK^(-1)` and `Delta H =+45.0 kJ`. Evaluate the temperature at which liquid water and water vapour are in equilibrium at 1 atm. Pressure- |
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Answer» We know that : at equilibrium `Delta G = 0` `because " " Delta G = Delta H - T Delta S " " therefore " " T Delta S = Delta H rArr T=(Delta H)/(Delta S)=(45.0xx10^(3)J)/(120 JK^(-1))=375 K` |
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| 1341. |
Given these reactions : `Ato2B DeltaH=40KJ`` BtoCDeltaH=-50KJ`` 2CtoD DeltaH=-20KJ` Calculate `DeltaH` for the reaction : `D+Ato4C`A. `-100KJ`B. `-60KJ`C. `-40KJ`D. `100Kj` |
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Answer» Correct Answer - c |
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| 1342. |
Statement-1: For a process to be spontaneous, `Delta G` as well as `Delta S` has to be less than zero. Statement-2 : For spontaneous change, `Delta S_("total") gt 0`A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - D | |
| 1343. |
Which of the following process is/are always exothermic ?A. Enthalphy of combustionB. Enthalphy of neutralisationC. Enthalphy of atomisationD. Enthalphy of fromation |
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Answer» Correct Answer - a,b |
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| 1344. |
Enthalpy change equal internal energy change whenA. All the reactants and products are in solutionB. Reaction is carried out in a closed vesselC. Number of moles of gaseous reactants and that of products is equalD. Reaction is carried out at constant pressure |
| Answer» In (a), volume change in negligible while in (b) and (c ), there is no volume change. | |
| 1345. |
Choose the correct statements:A. The enthalpy of combustion of a fuel per gram (expressed without a negative sign) is called its specific enthalpyB. The enthalpy of combustion of a fuel per gram (expressed without a negative sign) is called its enthalpy densityC. The enthalpy of combustion of fuel per litre (expressed without a negative sign) is called its specific enthalpyD. The enthalpy of combustion of fuel per litre (expressed without a negative sign) is called its enthalpy density |
| Answer» Correct Answer - A::D | |
| 1346. |
In which of the following reactions the entropy change is positive ?A. `H_(2(g))+(1)/(2)O_(2(g)) rarr H_(2)O_((l))`B. `Na_((g))^(+) + Cl_((g))^(-) rarr NaCl_((s))`C. `NaCl_((g)) rarr NaCl_((s))`D. `H_(2)O_((l)) rarr H_(2)O_((g))` |
| Answer» Correct Answer - D | |
| 1347. |
A sample of `0.20` mole of a gas at `44^@C` and `1.5 atm` pressure is cooled to `27^@C` and compressed to `3.0atm`. Calculate `DeltaV`. Suppose the original sample of gas was heated at constant volume until its pressure was `3.0atm` and then cooled at `27^@C`, what would have been `DeltaV`? |
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Answer» `V_(1)=(nRT_(1))/(P_(1))=(0.2xx0.0821xx317)/1.5=3.470 litre` `V_(2)=(nRT_(2))/(P_(2))=(0.2xx0.821xx300)/3.0` `=1.642 litre` `:. DeltaV=V_(2)-V_(1)=1.642-3.470` `=-1.828 litre` In second case the initial and final states at constant volume are same and thus `DeltaV=0`, since volume is a state function. |
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| 1348. |
One mole of an ideal gas is heated at constant pressure from `0^(@)C` to `100^(@)C`. a. Calculate the work done. b. If the gas were expanded isothermally and reversibly at `0^(@)C` from `1 atm` to some othe pressure `P`, what must be the final pressure if the maixmum work is equal to the work in `(a)`? |
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Answer» (a) Work involved in heating of gas `W_(a)=-PDeltaV=-P(V_(2)-V_(1))` `=-P((nRT_(2))/P-(nRT_(1))/P)` `=-nR(T_(2)-T_(1))=-1xx1.987(373-273)` `=-198.7 cal` (b) Now the work equivalent to `198.7 cal` is used up in causing reversible isothermal expansion of gas at `0^(@)C`, then `W_(R)=-2.303 nRT "log"_(10)(P_(1))/(P_(2))` `:. -198.7=-2.303xx1xx1xx1.987xx273"log"_(10)(1)/(P_(t))` `:. P_(t)=0.694 atm` |
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| 1349. |
The enthalpies of combustion of carbon and carbon monoxide in excess of oxygen at `298K` and constant pressure are `-393.5` and `-283.0 kJ mol^(-1)`, respectively. Calculate the heat of formation of carbon monoxide at constant volume. |
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Answer» `C(s) +(1)/(2)O_(2)(s) rarr CO(s) DeltaH = ?` Given: `C(g) +O_(2)(g) rarr CO_(2)(g), DeltaH_(1) = - 393.5 ….(i)` `CO(g) +(1)/(2)O_(2)(g) rarr CO_(2)(s), DeltaH_(2) =- 283 …(ii)` `Delta_(f)H = DeltaH_(1) - DeltaH_(2)` `= 393.5 -(-283) =- 110.5 kJ` `Deltan_(g) = 1 - (1)/(2)=(1)/(2)` `DeltaU^(Theta) = DeltaH - Deltan_(g)RT` `= - 11.05 -(-(1)/(2)) xx 8.314 xx 10^(-3) xx 298` `=- 111.7 kJ mol^(-1)` |
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| 1350. |
In which of the following reaction does the heat change represent the heat of formation of water ?A. `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((l)),Delta H = 116 Kcal`B. `H_(2(g))+(1)/(2)O_(2(g)) rarr H_(2)O_((l)),Delta H = -6.8 Kcal`C. `H_((aq))^(+)+OH_((aq))^(-) rarr H_(2)O_((l)),Delta H = 13.7 Kcal`D. `C_(2)H_(2(g)) +2O_(2(g))+H_(2)O_((l)),Delta H = 310 Kcal` |
| Answer» Correct Answer - B | |