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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.
| 301. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` is:A. `DeltaG=RT lnK_(c )`B. `-DeltaG=RT lnK_(c )`C. `DeltaG^(@)=RT lnK_(c )`D. `-DeltaG^(@)=RT lnK_(c )` |
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Answer» Correct Answer - D `DeltaG^(@)=-RT ln K_(c )` or `-DeltaG^(@)=RT ln K_(c)`. |
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| 302. |
Enthalpy change for the reaction `2H(2)(g)rarr4H(g)` is `-869.6kJ` The dissociation energy of `H--H`bond is:A. `+434.8 kJ`B. `+217.4 kJ`C. `434.8 kJ`D. `869.6 kJ` |
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Answer» Correct Answer - C `4H(g)to2H_(2)` , `DeltaH=-869.6 kJ` `4H_(2)to4(g)`, `DeltaH=869.6 kJ` `H_(2)to2H(g)` , `DeltaH=(869.6)/(2)=438.8 kJ` |
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| 303. |
Enthalpy change for the reaction `2H(2)(g)rarr4H(g)` is `-869.6kJ` The dissociation energy of `H--H`bond is:A. `+217.4 kJ`B. `-431.8 kJ`C. `-869.6kJ`D. `+434.8 kJ` |
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Answer» Correct Answer - B `2H_(2)(g)to4H(g)`, `DeltaH=869.6 kJ` or `H_(2)(g)to2H(g)`, `DeltaH=+(869.6)/(2)` `=434.8 kJ` |
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| 304. |
Enthalpy change for the reaction `4H(g)rarr2H_2(g)` is `-869.6kJ` The dissociation energy of `H-H` bond is:A. `=217.4KJ`B. `-434.8KJ`C. `-869.6KJ`D. `+434.8KJ` |
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Answer» Correct Answer - D `2H_(2)(g)rarr4H(g),DeltaH=+869.6KJ` `H_(2)(g)rarr2H(g),DeltaH=(869.6KJ)/(2)=434.8KJ` . |
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| 305. |
Which of the following statements is correct?A. For exothermic reactions, `DeltaH` is positiveB. For endothermic reactions, `DeltaH` is negativeC. The `DeltaH_(Neutralization)` for strong acid and strong base is always the sameD. The enthalpy of fusion, `DeltaH_(fus)` is negative |
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Answer» Correct Answer - C The enthalpy of neutralisation `(DeltaH_(necut.)` for a strong acid and strong base is always constant. |
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| 306. |
`C("diamond")toC("Graphite")`, `DeltaH=-ve`. Then shows thatA. Graphite is more stable than diamondB. Graphite has more energy than diamondC. Both are equally stableD. Stability cannot be predicted |
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Answer» Correct Answer - A Graphite has lower energy and therefore, more stable. |
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| 307. |
In view of the signs of `Delta_(r)G^(0)` for the following reactions `PbO_(2)+Pbrarr2PbO,Delta_(r)G^(0)lt0` `SnO_(2)+Snrarr2SnO,Delta_(r)G^(0)gt0` Which oxidation state are more characteristic for lead and tin?A. For lead `+4` and for tin `+2`B. For lead `+2` and for tin `+2`C. For lead `+4` and for tin `+4`D. For lead `+2` and for tin `+4` |
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Answer» Correct Answer - D `DeltaG^(@) lt 0` for `PbO_(2)+Pbtp2PbO` `:.` Reaction is spontaneous in the forward direction i.e., `Pb^(2+)` is more stable than `Pb^(4+)` `DeltaG^(@) gt 0` for the reaction, `Sn^(+4)O_(2)+Snto2Sn^(2+)O` `:.` Reaction is non-spontaneous in the forward direction but is spontaneous in the backward direction. Hence, `Sn^(4+)` will be more stable than `Sn^(2+)` |
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| 308. |
In view of the signs of `Delta_(r)G^(0)` for the following reactions `PbO_(2)+Pbrarr2PbO,Delta_(r)G^(0)lt0` `SnO_(2)+Snrarr2SnO,Delta_(r)G^(0)gt0` Which oxidation state are more characteristic for lead and tin?A. For lead `+4` , for tin `+2`B. For lead `+2` , for tin `+2`C. For lead `+4` , for tin `+4`D. For lead `+2` , for tin `+4` |
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Answer» Correct Answer - A Oxidation state `underset(+4)pbO_(2)+underset(0)Pbrarrunderset(+2)2PbO` Since `DeltaG^(@)lt0` , i.e., it is negative. Therefore, the reaction is spontaneous in the forward direction. This suggest `Pb^(2+)` is more stable than `Pb^(4+)` . Oxidation state `underset(+4)SnO_(2)+underset(0)Snrarrunderset(+2)2SnO` Since `DeltaG^(@)gt0` , i.e., it is positive, therefore, the reaction is non-spontaneous in the forward direction.But it will be spontaneous in the backward direction. This suggests that `Sn^(2+)` . These facts are also supported by the inert pair effect down the group. |
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| 309. |
In a fuel cell, methanol if used as fuel and oxygen gas is used as an oxidiser. The reaction is `CH_(3)OH(l) +(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)` Calculated standard Gibbs free enegry change for the reaction that can be converted into electircal work. If standard enthalpy of combustion for methanol is `-702 kJ mol^(-1)`, calculate the efficiency of converstion of Gibbs energy into useful work. `Delta_(f)G^(Theta)` for `CO_(2),H_(2)O, CH_(3)OH,O_(2)` is `-394.00, -237.00,-166.00`and `0 kJ mol^(-1)` respectively. |
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Answer» The reaction for combustion of methanol is : `CH_(3)OH(l)+(3)/(2)O_(2)(g)toCO_(2)(g)+2H_(2)O(l)` `DeltaG_("reaction")^(@)=[DeltaG_(f)^(@)CO_(2)(g)+2DeltaG_(f)^(@)H_(2)O(l)]-[DeltaG_(f)^(@)CH_(3)OH(l)+(3)/(2)DeltaG_(f)^(@)O_(2)(g)]` `=[-394.36+2(-237.13)]-[-166.27+0]` Efficiency of conversion of Gibbs free energy into useful work `=(DeltaG_("reaction")^(@)xx100)/(DeltaH_("reaction")^(@))` `=(-702.35xx100)/(-726)=96.7%` |
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| 310. |
Assertion:Phase transition involves change in internal energy only. Reason:Phase transition occurs at constant pressure.A. If both ther assertion and reason are true and the reason is the correct explanation of the assertion.B. If both the assertion and reason are true but reason is not the correct explantion of the assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Phase transition occurs at constant `P` and thus change is referred as `DeltaH` . |
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| 311. |
Assertion:The enthalpies of neutralised of strong acids and strong bases are always same. Reason:Neutralisation is heat of formation of water.A. If both ther assertion and reason are true and the reason is the correct explanation of the assertion.B. If both the assertion and reason are true but reason is not the correct explantion of the assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The enthalpies of neutralisation of strong acids and strong bases are the same they are `13.7Kcal` . The reason is that it is heat of formation of water from `H^(+)` and `OH^(-)` ions. `H^(+)+OH^(-)rarrH_(2)O,DeltaH=13.7Kcal` . Thus, both assertion and reason are true. |
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| 312. |
When water is added to quick lime, the reaction isA. explosiveB. endothermicC. exothermicD. photochemical |
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Answer» Correct Answer - C In this reaction, heat is released `(DeltaH=-ve)`. |
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| 313. |
Enthalpuy of solution of NaOH (s) in water is `-41 .6 KJ mol ^(-1)` .when NaOH is dissolved in water , the temperature of waterA. increasesB. decreasesC. does not changeD. fluctuates indefinitely |
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Answer» Correct Answer - a since the process is exothermic , heat is evolved due to this temperature of water increases. |
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| 314. |
Water is supercooled to `- 4^(@)C`. The enthalpy `(H)` isA. same as ice at `-4^(@)C`B. more than ice at `-4^(@)C`C. same as ice at `0^(@)C`D. less than ice at `-4^(@)C` |
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Answer» Correct Answer - D When ice is supercooled to `-4^(@)C`, heat is released, so enthalpy decreases. |
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| 315. |
In a flask, colourless `N_(2)O_(4)` is in equilibrium with brown-coloured `NO_(2)`. At equilibrium, when the flask is heated to `100^(@)C` the brown colour deepens and on cooling, the brown colour became less coloured. The change in enthalpy `DeltaH` for the ayatem isA. NegativeB. PositiveC. ZeroD. Undefined |
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Answer» Correct Answer - B The reaction is endothermic, so on getting energy, forward reaction is favoured producing more `NO_(2)`. This causes colour to deepen. |
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| 316. |
The bond energies of `C=C` and `C-C` at `298 K` are `590` and `331 kJ mol^(-1)`, respectively. The enthalpy of polymerisation per mole of ethaylene isA. `+259 kJ mol^(-1)`B. `+72 kJ mol^(-1)`C. `-259 kJ mol^(-1)`D. `-72 kJ mol^(-1)` |
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Answer» Correct Answer - D In the given polymerisation reaction, one mole of ethylene involves breaking of one `C=C` double bond a nd formation of three `C-C` single bonds. However, in the complete polymer chain, the number of single `C-C` bonds is two per `C=C` double bond broken. Energy required in breaking of one mole of `C=C` double bond `=590 kJ` Energy released in the formation of two moles of `C-C` single bonds `=2xx331=662 J` `:. "Net energy released per mole ethylene"` `=662-590=72 kJ` i.e., `DeltaH=-72kJ "mole"^(-1)` |
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| 317. |
A system absorbs `10 kJ` of heat at constant volume and its temperature rises from `27^(@)C` to `37^(@)C`. The `DeltaE` of reaction isA. `100 kJ`B. `10 kJ`C. `0`D. `1 kJ` |
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Answer» Correct Answer - B `DeltaE=q_(v)` (heat absorbed at constant volume) |
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| 318. |
what is `DeltaE` for a system that does 500 cal of work pm surrounding and 300 cal of heat is adsorbed by the system ?A. `-200 cal `B. `- 300 cal `C. `+ 200 cal `D. ` + 300 cal ` |
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Answer» Correct Answer - a `DeltaE =q+W=300+(-500)=-200 cal ` |
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| 319. |
themodynamic is not concerned about .........A. energy changes invoived in a chemical reactionB. the extent to which a chemical reaction proceedsC. the rate at which a reaction proceedsD. the feasibllilty of a chemical reaction |
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Answer» Correct Answer - c thermoynamics is not concerned with the rate at which a reaction proceeds. |
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| 320. |
The source of energy in a cellular reaction isA. chemical energyB. light energyC. heat energyD. solar energy |
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Answer» Correct Answer - A Cellylar reaction mainly refers to metabolism which is essentially bond making and bond breaking and chemical energy is the source that fuels such reactions. |
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| 321. |
Change in internal energy when `4 kJ` of work is done on the system and `1 kJ` of heat is given out of the system isA. `+1 kJ`B. `-5 kJ`C. `+5k J`D. `+3 kJ` |
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Answer» Correct Answer - D `DeltaU=q+w=(-1 kJ)+4kJ=+3kJ` |
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| 322. |
Which of the ions in the table below would have the largest value of enthalpy of hydration? Ionic radius in nm Charge of ionA. `{:("Ionic radii in nm","Charge of ion"),(0.0065, +2):}`B. `{:("Ionic radii in nm","Charge of ion"),(0.095, +1):}`C. `{:("Ionic radii in nm","Charge of ion"),(0.135, +2):}`D. `{:("Ionic radii in nm","Charge of ion"),(0.169, +1):}` |
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Answer» Correct Answer - A Smaller the size of the ions and greater the charge on the ion, greater is the enthalpy of hydration. |
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| 323. |
`Delta_(f)U^(-)` of formation of `CH_(4) (g)` at certain temperature is `-393 kJ mol ^(-1)` . The value of `Delta_(f)H^(-)`A. zeroB. `ltDelta_(f)U^(-)`C. `gtDelta_(f)U^(-)`D. equal to `Delta_(f)U^(-)` |
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Answer» Correct Answer - b `Delta_(f)H^(-) = Delta_(f)U+ pDeltaV. So Delta_(f)H^(-)ltDelta_(f)U^(-)` |
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| 324. |
the reaction which proceeds in the forwards direaction isA. `Fe_(2)O_(3) + 6HCl to 2FeCl_(3) + 3H_(2)O`B. `NH_(3)+ H_(2)O + NaCl to NH_(4)Cl + NaOH`C. `SnCl_(4)+ Hg_(2)Cl_(2) to SnCl_(2) + 2HgCl_(2)`D. `2"Cul"+l_(2)+4K^(+) to 2Cu^(+)+4Kl` |
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Answer» Correct Answer - d in (d) , ondine acts as oxidising agent , so this reaction proceeds in forward direaction. |
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| 325. |
The heat of combustion of butane is `2880kJ" "mol^(-1)`. What is the heat liberated by buring 1 kg of butane in excess of oxygen supply? |
| Answer» Correct Answer - 49655kJ | |
| 326. |
When 0.1 mol of a gas absorbs `41.75 J` of heat at constant volume, the rise in temperature occurs equal to `20^(@)C`. The gas must beA. triatomicB. diatomicC. polyatomicD. monoatomic |
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Answer» Correct Answer - D `C_(V)` (heat absorbed per deg.rise per mole) `=(41.75 J)/(0.1 mol xx20^(@))=20.875 JK^(-1)mol^(-1)` `C_(P)=C_(V)+R=20.875+8.314JK^(-1) mol^(-1)` `=29.189 JK^(-1)mol^(-1)` `(C_(P))/(C_(V))=(29.189)/(20.875)=1.40` `:.` The gas is monoatomic |
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| 327. |
One mole of non-ideal gas undergoes a change of state (`2.0 "atm"`, `3.0 L`, `95 K`) `to` (`4.0 "atm"`, `5.0 L`, `245 K`) with a change in internal energy, `DeltaU=30.0 L atm`. The change in enthalpy `(DeltaH)` of the process in `L "atm"` isA. `40.0`B. `42.3`C. `44.0`D. not defined because pressure is not constant |
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Answer» Correct Answer - C `H=U+PV` `DeltaH=DeltaU+DeltaPV` `DeltaH=DeltaU+(P_(2)V_(2)-P_(1)V_(1))` `=30+(5xx4-3xx2)` `=30+(20-6)=44 L` atm |
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| 328. |
From the standard enthalpies of formation values for some compounds (in `KJ mol^(-1)` ), predict which one of them will be most stable?A. Propane `(g)(DeltaH_(f)^(@)=-103.8)`B. `n`-Butane `(g)(DeltaH_(f)^(@)=-124.7)`C. `n`-Butane `(g)(DeltaH_(f)^(@)=-167.2)`D. `n`-Octane `(g)(DeltaH_(f)^(@)=-208.4)` |
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Answer» Correct Answer - B Greater the standard enthalpy of formation, the more stable will be the compound. |
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| 329. |
Which of the following statement is false ?A. Work is a state functionB. Temperature is a state functionC. Change of state is completely defind when initial and final states are specifiedD. Work appears at the boundary of the system. |
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Answer» Correct Answer - A These are the questions based on factual statements. |
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| 330. |
A well stoppered thermo flask containing some ice cubes is an example ofA. closed systemB. open systemC. isolated systemD. non-thermodynamic system. |
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Answer» Correct Answer - C Because there is no exchange of matter and energy. |
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| 331. |
Which among the following state functions is an extensive property of the system?A. TemperatureB. VolumeC. Refractive indexD. Viscosity |
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Answer» Correct Answer - B Volume is an extensive property because it depends upon the amount of the substance. |
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| 332. |
Heat evolved in the reaction `H_(2)+Cl_(2) rarr 2HCl` is 182 kJ Bond energies H- H = 430 kJ/mole, `Cl-Cl = 242kJ//"mole"`. The H-Cl bond energy isA. `245KJmol^(-1)`B. `427KJmol^(-1)`C. `336KJmol^(-1)`D. `154KJmol^(-1)` |
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Answer» Correct Answer - B `DeltaH=-2xxvarepsilon_(H-Cl)+varepsilon_(H-H)+varepsilon_(Cl-C)` `:. 182=-2xxa++430+242` `:. a=245KJmol^(-1)` |
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| 333. |
Change in entropy is negative forA. Bromine `(l)rarrBromine(g)`B. `C(s)+H_(2)O(g)rarrCO(g)+H_(2)(g)`C. `N_(2)(g, 10atm)rarrN_(2)(g, 1 atm)`D. `Fe(1 mol, 400K)rarrFe(1mol,300K)` |
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Answer» Correct Answer - B `DeltaS=SigmanDeltaS` (Products)`-SigmanDeltaS` (Reactants) In option `(d)` change in entropy is negative because at higher temperature degree of randomess is high. |
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| 334. |
Identify the correct statement from the following in a chemical reactionA. The entropy always increasesB. The change in entropy along the suitable change in enthalpy decides the fate of a reactionC. The enthalpy always decreasesD. Both the enthalpy and the entropy remains constant. |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS` For a reaction to be spontaneous, `DeltaG` should be negative. Therefore, resultant of `DeltaH` and `TDeltaS` decides the fate of a reaction. |
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| 335. |
Assuming that water vapour is an ideal gas, the internal energy change `(DeltaU)` when 1 mole of water is vaporised at 1 bar pressure and `100^(@)C` , (given: molar enthalpy of vaporisation of water `-41KJmol^(-1)` at 1 bar and `373K` and `R=8.3Jmol^(-1)Kmol^(-1)`) will be:A. `4.100KJmol^(-1)`B. `3.7904KJmol^(-1)`C. `37.904KJmol^(-1)`D. `41.00KJmol^(-1)` |
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Answer» Correct Answer - B `H_(2)O_((l))overset(vap o risation)(rarr)H_(2)O_((g))` `Deltan_(g)=1-0=1` `DeltaH=DeltaU+Deltah_(g)RT` `DeltaU=DeltaH-Deltan_(g)RT` `=41-8.3xx10^(-3)xx373=37.9KJmol^(-1)` |
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| 336. |
For spontaneous process:A. `DeltaS_("total")=0`B. `DeltaS_("total")gt0`C. `DeltaS_("total")lt0`D. none of these |
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Answer» Correct Answer - B `DeltaS_("total")=DeltaS_("system")+DeltaS_("surrounding")` `=+ve` for spontaneous process. |
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| 337. |
The a reaction to be spontaneous at all temperatureA. `DeltaG` and `DeltaH` should be negativeB. `DeltaG` for the reaction is zeroC. `DeltaG` and `DeltaH` should be positiveD. `DeltaH lt DeltaG` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` For spontaneous reaction at all temperature `DeltaH=-ve` `DeltaS=+ve` So `DeltaG=-ve` `DeltaG=DeltaH-TDeltaS` For spontaneous reaction at all temperature `DeltaH=-ve` `DeltaS=+ve` So `DeltaG=-ve` |
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| 338. |
Which of the following statements is true? The entropy of the universeA. Increases and moves towards maximumm valueB. Decreases and moves to zeroC. Remains constantD. Decreases and increases with a periodic rrate |
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Answer» Correct Answer - A Second law of thermodynamics states the entropy of the universe increases and moves towards maximum value. |
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| 339. |
A spontaneous reaction is impossible ifA. both `DeltaH` and `DeltaS` are negativeB. both `DeltaH` and `DeltaS` are positiveC. `DeltaH` is negative and `DeltaS` is positiveD. `DeltaH` is positive and `DeltaS` is negative |
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Answer» Correct Answer - D For a reaction to be non-spontaneous, `DeltaG gt 0`. It can happen when both `DeltaH` and `DeltaS` are unfavourable i.e., when `DeltaH` is `+ve` and `DeltaS` is `-ve` |
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| 340. |
For which one of the following equation is `DeltaH_(reaction)^(@)` equal to `DeltaH_(f)^(@)` for the product ?A. `N_(2)(g)+O_(3)(g)rarrN_(2)O_(3)(g)`B. `CH_(4)(g)+2Cl_(2)(g)rarrCH_(2)Cl_(2)(l)+2HCl(g)`C. `Xe(g)+2F_(2)(g)rarrXeF_(4)(g)`D. `2CO(g)+O_(2)(g)rarr2CO_(2)(g)` |
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Answer» Correct Answer - D Heat of formation is ginen when compound is formed from its components. |
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| 341. |
Assertion: The product of pressure and volume for a fixed amount of gas is equal to a constant represented by `RT` Reason: At constant temperature, for a fixed amount of a gas, pressure is inversely proportional to volume.A. If both the assertion and reason are true and the reason is the correct explanation of the assertion.B. If both the assertion and reason are true but reason is not the correct explantion of the assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Assertion is ture only for 1 mol of an ideal gas. |
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| 342. |
Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour. |
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Answer» The volume occupied by water is very small and thus the volume change is equal to the volume occupied by one gram mole of water vapour. `V=(nRT)/(P)=(1.0xx0.0821xx373)/(1.0)=31.0` litre `w=-P_(ext)xxDeltaV=-(1.0)xx(31.0)` litre-atm `=-(31.0)xx101.3J=-3140.3J` |
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| 343. |
Identify the correct statement regarding a spontaneous process:A. For a spontaneous process in an isolated system, the change in entropy is positiveB. Endothermic process are never spontaneousC. Exothermic processes are always spontaneousD. Lowering of energy in the reaction process is the only criterion for spontaneity |
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Answer» Correct Answer - B For a spontaneous process in an isolated system, the change in entropy is positive. |
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| 344. |
Standard entropt of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:A. `1250K`B. `500K`C. `750K`D. `1000K` |
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Answer» Correct Answer - A `(1)/(2)X_(2)+(3)/(2)Y_(2)rarr XY_(3)` `DeltaS_("reaction")=50-((3)/(2)xx40+(1)/(2)xx60)=-40Jmol^(-1)` `DeltaG=DeltaH-TDeltaS` At equilibrium `DeltaG=0` `DeltaH=TDeltaS` `30xx10^(3)=Txx40` `:. T=750` |
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| 345. |
In an endothermic reaction, the value of `DeltaH` isA. ZeroB. PositiveC. NegativeD. Constant |
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Answer» Correct Answer - B For exothermic reactions `sumH_("Products") gt sum H_("Reactants")` Hence, `(sumH_("Products")-sumH_("Reactants"))` is positive. |
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| 346. |
If liquids `A` and `B` from an ideal solution,A. The entropy of mixing is zeroB. The Gibbs free energy as well as the entropy of mixing are zeroC. The Gibbs free energy as well as the enthalpy of mixing are zeroD. The enthalpy of mixing is zero |
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Answer» Correct Answer - C For the ideal solution, `DeltaH=0,DeltaV=0` . |
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| 347. |
The correct statement regarding entropy is:A. At absolute zero temperature, entropy of a perfectly crystalline solid is zeroB. At absolute zero temperature, the entropy of a perfectly crystalline substance is +veC. At absolute zero temperature, the entropy of all crystalline substances is zeroD. At `0^(@)C` the entropy of a perfectly crystalline solid is zero. |
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Answer» Correct Answer - A Entropy of perfectly crystalline solids will be zero at absolute zero. It is predicted in the third law of thermodynamics. |
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| 348. |
Which reaction, with the following value of `DeltaH` and `DeltaS` at `350K` is spontaneous and endothermic? `DeltaHKJmol^(-1)DeltaSKJmol^(-1)DeltaHKJ mol^(-1)DeltaSKJ mol^(-1)`A. `-48,+135xx10^(-3)`B. `-48,-135`C. `+48,+135xx10^(-3)`D. `+48,-135xx10^(-3)` |
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Answer» Correct Answer - D For endothermic reaction `DeltaHgt0` Thus, `(c)` or `(d)` `DeltaG=DeltaH-TDeltaS` If `TDeltaSgtDeltaH` , then `DeltaGlt0` Thus, spontaneous `TdeltaS=135xx10^(-3)xx350=54KJgtDeltaH` Thus `(c)` is correct. |
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| 349. |
Which of the following is an endothermic reaction?A. `2H_(2)+O_(2)to2H_(2)O`B. `N_(2)+O_(2)to2NO`C. `2NaOH+H_(2)SO_(4)toNa_(2)SO_(4)+2H_(2)O`D. `C_(2)H_(5)OH+3O_(2)to2CO_(2)+3H_(2)O` |
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Answer» Correct Answer - C Factual questions. |
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| 350. |
The following is(are) endothermic reaction (s):A. combustion of methaneB. Decomposition of waterC. Dehydrogenation of ethane to ethyleneD. Conversion of graphite to diamond |
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Answer» Correct Answer - B::C::D All the processes represented in `(B)`, `(C )`, `(D)` are endothermic. |
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