InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Which of the following has lowest fusion temperatureA. NaphthaleneB. DiamondC. NaClD. Mn |
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Answer» Correct Answer - A The fusion temperature of napthalene is minimum, because it is non-polar covalent compound and has less fusion temperature. |
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| 102. |
The lattice enthalpy and hydration enthalpy of four compounds are given below `{:("Compound","Lattice enthalpy","Hydration enthalpy"),(,("in kJ mol"^(-1)),("in kJ mol"^(-1))),(P,+780,-920),(Q,+1012,-812),(R,+828,-878),(S,+632,-600):}` The pair of compounds which is soluble in water isA. P and QB. Q and RC. R and SD. P and R |
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Answer» Correct Answer - D A compount is soluble in water when its hydration enthalpy is greater than its lattice enthalpy. |
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| 103. |
For which reaction change of entropy will be positiveA. `H_(2(g))+I_(2(g))iff2HI_((g))`B. `HCl_((g))+NH_(3(g))iffNH_(4)Cl_((s))`C. `NH_(4)NO_(3(s))iff N_(2)O_((g))+2H_(2)O_((g))`D. `MgO_((s))+H_(2(g))iffMg_((s))+H_(2)O_((l))` |
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Answer» Correct Answer - C Entropy of gas is greater than that of liquid and solid. In option (c ) solid is converted into gaseous phase. Entropy of gas `gt` liquid `gt` solid. |
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| 104. |
Which of the following is correctA. Evaporation of water causes an increase in disorder of the systemB. Melting of ice causes a decrease in randomness of the systemC. Condensation of steam causes an increase in disorder of the systemD. There is practically no change in the randomness of the system when water is evaporated |
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Answer» Correct Answer - A Randomness or entropy order is gas `gt` liquid `gt` solid `H_(2)O_((l))overset("Evaporation")rarrH_(2)O_((g))" "DeltaS " increases"` `H_(2)O_((s))overset("Melting")rarrH_(2)O_((l))" "DeltaS " increases"` `H_(2)O_((g))overset("Condensation")rarrH_(2)O_((l))" "DeltaS " increases"` |
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| 105. |
The second law of thermodynamics says that in a cyclic processA. Work cannot be converted into heatB. Heat cannot be converted into workC. Work cannot be completely converted into heatD. Heat cannot be completely into work |
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Answer» Correct Answer - D According to second law, work can always be completely converted into heat but heat cannot be completely converted into work. |
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| 106. |
The heat required to raise the temperature of a body by 1 K is calledA. Specific heatB. Thermal capacityC. Water equivalentD. None of these |
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Answer» Correct Answer - B Heat required to rise the temperature of a body by 1 K called thermal capacity of the body. |
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| 107. |
The incorrect statement among the following isA. The entropy of the universe remains constantB. Heat cannot be completely converted into workC. This absolute entropy of a perfectly crystalline solid at absolute zero temperature is zeroD. The total energy of an isolated system remains constant |
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Answer» Correct Answer - A The entropy if the universe is increasing `DeltaS_("universe")=DeltaS_("system")+DeltaS_("surrounding")gt0`. |
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| 108. |
Heat required to raise the temperature of 1 mol of a substance by `1^(@)C` is calledA. Specific heatB. Molar heat capacityC. Water equivalentD. Specific gravity |
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Answer» Correct Answer - B It is molar heat capacity. |
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| 109. |
When water is cooled to ice its entropyA. IncreasesB. DecreasesC. Remains sameD. Becomes zero |
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Answer» Correct Answer - B When water freezes to form ice, the randomness or disorderliness decreases i.e., entropy decreases. |
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| 110. |
Which one of the following process is non-spontaneousA. Dissolution of `CuSO_(4)` in waterB. Reaction between `H_(2)` and `O_(2)` to form waterC. Water flowing down hillD. Flow of electric current from low potential to high potential |
| Answer» Correct Answer - B | |
| 111. |
Which of the following statements is correct for the spontaneous adsorption of a gasA. `DeltaS` is positive and, therefore, `DeltaH` should be negativeB. `DeltaS` is positive and, therefore, `DeltaH` should also be highly positiveC. `DeltaS` is negative and, therefore, `DeltaH` should be highly positiveD. `DeltaS` is negative and therefore, `DeltaH` should be highly negative |
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Answer» Correct Answer - D For adsorption `DeltaS=-ve, DeltaH=-ve`. |
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| 112. |
Which is the following is the correct option for free expansion of an ideal gas under adiabatic condition.A. `q=0, DeltaT ne 0, W = 0`B. `q=0, DeltaT ne 0, W ne 0`C. `q=0, DeltaT = 0, W = 0`D. `q=0, DeltaT = 0, W ne 0` |
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Answer» Correct Answer - C For the free expansion of an ideal gas, the opposing force is zero, hence work done = 0. `DeltaU=q+w, w=0`, hence `DeltaU=q` But the process is adiabatic, hence q=0 `therefore q=0, DeltaT=0, W=0` |
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| 113. |
The work done to contract a gas in a cylinder, is 462 joules. 128 joule energy is evolved in the process. What will be the internal energy change in the processA. `+ 590` joulesB. `- 334` joulesC. `+ 334` joulesD. `- 590` joules |
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Answer» Correct Answer - C As the work is done on system, it will be positive i.e. W=+462 joule, q=-128 joule (heat is evolving) From the `I^(st)` law of thermodynamics `DeltaE=q+w=(-128)+(+462)=+334` Joules. |
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| 114. |
For which of the following `DeltaE = DeltaH`A. `N_(2)O_(4)(g) iff 2NO_(2)(g)`B. `2SO_(2)(g)+O_(2)(g)iff2SO_(3)(g)`C. `H_(2)(g)+Cl_(2)(g)iff2HCl(g)`D. `H_(2)(g)+(1)/(2)O_(2)(g)iffH_(2)O(l)` |
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Answer» Correct Answer - C `Deltan=0` for this reaction so, `DeltaE=DeltaH` |
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| 115. |
Point out the wrong statement in relation to enthalpyA. It is a state functionB. It is an intensive propertyC. It is independent of the path followed for the changeD. Its value depends upon the amount of substance in the system |
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Answer» Correct Answer - B Enthalpy is an extensive property. |
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| 116. |
A reaction occurs spontaneously ifA. `TDeltaS lt DeltaH` and both `DeltaH` and `DeltaS` are +veB. `TDeltaS gt DeltaH` and both `DeltaH` and `DeltaS` are +veC. `TDeltaS = DeltaH` and both `DeltaH` and `DeltaS` are +veD. `TDeltaS gt DeltaH` and `DeltaH` is +ve and `DeltaS` is -ve |
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Answer» Correct Answer - B For spontaneous reaction `DeltaG` should be negative `DeltaG=DeltaH-TDeltaS=(+ve)-T(+ve)` If `TDeltaS gtDeltaH` then `DeltaG` will be negative and reaction will be spontaneous at high temperature. |
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| 117. |
Which one of the following statement is falseA. Work is a state functionB. Temperature is a state functionC. Change in the state is completely defined when the initial and final states are specifiedD. Work appears at the boundary of the system |
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Answer» Correct Answer - A::B Work and temperature are not state functions as their value are not specific for a state. |
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| 118. |
At 300 K, the reactions which have following values of thermodynamic parameters occur spontaneouslyA. `DeltaG^(@)=-400 kJ mol^(-1)`B. `DeltaH^(@)=200 kJ mol^(-1), DeltaS^(@)=-4 JK^(-1) mol^(-1)`C. `DeltaH^(@)=-200 kJ mol^(-1), DeltaS^(@)=4 JK^(-1) mol^(-1)`D. `DeltaH^(@)=200 J mol^(-1), DeltaS^(@)=40 JK^(-1)mol^(-1)` |
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Answer» Correct Answer - A::C::D When `DeltaH=+ve` and `DeltaS=-ve` than the reaction is non-spontaneous. |
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| 119. |
In `omega` is the amount of work done by the system and q is the amount of heat supplied to the system. Identify the type of the systemA. Isolated systemB. Closed systemC. Open systemD. System with thermally conducting walls |
| Answer» Correct Answer - B | |
| 120. |
`DeltaU` equal toA. Isobaric workB. Adiabatic workC. Isothermal workD. Isochoric work |
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Answer» Correct Answer - B For Adiabatic process, Q = 0 Now, `DeltaU=Q+Wimplies DeltaU=W` |
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| 121. |
For an isolated system, `DeltaU = 0`, thenA. `DeltaS = 0`B. `DeltaS lt 0`C. `DeltaS gt 0`D. The value of `DeltaS` cannot be predicted |
| Answer» Correct Answer - C | |
| 122. |
For the reaction `CO_((g))+(1)/(2)O_(2(g))rarrCO_(2(g)),DeltaH and DeltaS " are "-283 kJ and -87 JK^(-1)`. Respectively. It was intended to carry out this reaction at 1000, 1500, 3000 and 3500 K. At which of these temperature would this reaction be thermodynamically spontaneousA. 1500 and 3500 KB. 3000 and 3500 KC. 1000, 1500 and 3000 KD. 1500, 3000 and 3500 K |
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Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS` For spontaneous reaction `DeltaG` should be negative When T=1000 K, `DeltaG`=(-283 -1000 `xx` 0.087) kJ =-370 kJ When `T=1500 K, DeltaG=(-283-1500xx0.087) kJ =-413.5 kJ` `T=3000 K, DeltaG=(-283-3000xx0.087) kJ=-544 kJ` When `T=3500 K, DeltaG=(-283-3500xx0.087) kJ=+21.5 kJ` Hence at 3500 K, reaction will be non-spontaneous. |
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| 123. |
A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. It a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last for … Days `(DeltaH_(c)" of "C_(4)H_(10)=-2658 "KJ per mole")`A. 15 daysB. 20 daysC. 50 daysD. 32 days |
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Answer» Correct Answer - D Calorific value of butan = `(DeltaH_(c))/("mol. wt.")=(2658)/(58)=45.8 KJ//gm` Cylinder consist 14 Kg of butane means 14000 gm of butane `because` 1 gm gives 45.8 kJ `therefore` 14000 gm gives `14000xx45.8 = 641200 kJ` So gas full fill the requirement for `(641200)/(20,000)=32.06` days. |
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| 124. |
In which of the following conditions a chemical reaction can not occurA. `DeltaH` and `DeltaS` increase and `TDeltaS gt DeltaH`B. `DeltaH` and `DeltaS` decrease and `DeltaH gt TDeltaS`C. `DeltaH` increase and `DeltaS` DecreasesD. `DeltaH` decrease and `DeltaS` increase |
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Answer» Correct Answer - C If `DeltaH=+ve` and `DeltaS=-ve` then the reaction is non-spontaneous i.e. can not occur. |
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| 125. |
The free energy change for a reversible reaction at equilibriumA. Large positiveB. Small negativeC. Small positiveD. 0 |
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Answer» Correct Answer - D `DeltaG` at equilibrium = 0. |
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| 126. |
A chemical reaction is spontaneous at 298 K but non-spontaneous at 350 K. Which one of the following is true for the reactionA. `{:(DeltaG,DeltaH,DeltaS),("_","_","+"):}`B. `{:(DeltaG,DeltaH,DeltaS),("+","+","+"):}`C. `{:(DeltaG,DeltaH,DeltaS),("-","+","-"):}`D. `{:(DeltaG,DeltaH,DeltaS),("-","-","-"):}` |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` For first condition, reaction will be spontaneous at all temperatures. For second option as `DeltaH=+ve` and `DeltaS` is +ve thus `DeltaG=underset(+ve)underbrace(DeltaH)-underset((-ve))underbrace(TDeltaS)` If temperature is high, `TDeltaS` will have higher negative value and reaction will be spontaneous Thus not true for given reaction. In case (c), `DeltaG`=ve, `DeltaS=+ve` implies spontaneous at all temperatures In case (e), `DeltaG=underset(-ve)underbrace(DeltaH)-underset(+ve)underbrace(TDeltaS)` At high temperature, second term will have high positive value and reaction will be non-spontaneous, thus case (e) is true for given example. |
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| 127. |
What is the weight of oxygen that is required for the complete combustion of 2.8 kg of ethyleneA. 9.6 kgB. 96.0 kgC. 6.4 kgD. 2.8 kg |
| Answer» Correct Answer - A | |
| 128. |
Which of the following salt on dissolution in water absorbs heatA. `NH_(4)Cl`B. `CaO`C. `Na_(2)SO_(4)`D. `Na_(2)CO_(3)` |
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Answer» Correct Answer - A Generally enthalpy of solution is positive but here in this question, the three other compounds have -ve enthalpy of solution. |
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| 129. |
Mass and energy are conserved is demonstrated byA. First law of thermodynamicsB. Law of conservation of energyC. Law of conservation of massD. Modified form of `I^(st)` law of thermodynamics |
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Answer» Correct Answer - D It is a modified form of Ist of thermodynamics. It is a conservation law of mass and energy. |
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| 130. |
For reaction `Ag_(2)O(s)rarr2Ag(s)+(1//2)O_(2)(g)` the value of `DeltaH=30.56 " kJ mol"^(-1)` and `DeltaS = 0.066 kJK^(-1)mol^(-1)`. Temperature at which free energy change for reaction will be zero, isA. 373 KB. 413 KC. 463 KD. 493 K |
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Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS` `DeltaG=0` at equilibrium `therefore DeltaH=TDeltaS` or `30.56=Txx0.066` T=463 K. |
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| 131. |
An engine operating between `150^(@)C` and `25^(@)C` takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine isA. 147.7 JB. 157.75 JC. 165.85 JD. 169.95 J |
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Answer» Correct Answer - A `T_(2)=150+273=423 K , T_(1)=25+273 = 298 K` Q=500 J `(W)/(Q)=(T_(2)-T_(1))/(T_(2)), W=500((423-298)/(423))=147.7 J`. |
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| 132. |
Conditions of standard state used in thermochemistry isA. `0^(@)C` and 1 atmB. `20^(@)C` and 1 atmC. `25^(@)C` and 1 atmD. 0 K and 1 atm |
| Answer» Correct Answer - C | |
| 133. |
Maximum entropy will be in which of the followingA. IceB. Liquid waterC. SnowD. Water vapours |
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Answer» Correct Answer - D Entropy is the measure of randomness in the molecules. Randomness is maximum in case of gases. Hence, entropy is maximum for water vapours. |
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| 134. |
In which reaction there will be increase in entropyA. `Na(s)+H_(2)O(l)rarrNaOH(l) + (1)/(2)H_(2)(g)`B. `Ag^(+)(aq)+Cl^(-)(aq)rarrAgCl(s)`C. `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l)`D. `Cu^(2+)(aq)+4NH_(3)(g) rarr[Cu(NH_(3))_(4)]^(2+)(aq)` |
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Answer» Correct Answer - A In reaction (a) Na (s) reacts with `H_(2)O(l)` to give NaOH(l) and `(1)/(2)H_(2)(g)`. Since, the entropy of a gas is higher than that of liquid and the entropy of liquid is higher than that of solid, therefore, the change in entropy of this reaction increases. |
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| 135. |
For a reaction `DeltaH=(+3kJ), DeltaS=(+10J//K)` beyond which temperature this reaction will be spontaneousA. 300 KB. 200 KC. 273 KD. 373 K |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS`. For the spontaneous reaction the `DeltaG` must be negative. `DeltaH=3kJ=+3000 J` `DeltaS=+10 J//K` If `T=300 K implies DeltaG=3000-300xx10=0` If `T=200 K implies DeltaG=3000-200xx10=1000 J` If `T=273 K implies DeltaG=3000-273xx10=270 J` If T = 373 K `DeltaG=3000-373xx10=-730 J` Hence, beyond 373 temperature the reaction will be spontaneous. |
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| 136. |
`2C+O_(2)rarr2CO, Delta H= - 220 kJ` which of the following statements is correct for this reactionA. Heat of combustion of carbon is 110 kJB. Reaction is exothermicC. Reaction needs no initiationD. All of these are correct |
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Answer» Correct Answer - B `2C+O_(2)rarr2CO, DeltaH=-220 KJ` This reaction does not represent complete combustion of carbon, hence heat of combustion of carbon will not be equal to 110 kJ. The negative sign of `DeltaH` indicates that this reaction is exothermic. Also, despite being spontaneous reaction, it requires initiation. |
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| 137. |
For the process `H_(2)O(l)` (1 bar, 373 K) `rarr H_(2)O` (1 bar, 373 K), the correct set of thermodynamic parameters isA. `DeltaG=0, DeltaS=+ve`B. `DeltaG=0, DeltaS=-ve`C. `DeltaG=+ve, DeltaS=0`D. `DeltaG=-ve, DeltaS=+ve` |
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Answer» Correct Answer - A Since, liquid is passing in to gaseous phase so entropy will increase and at 373 K the phase transformation remains at equilibrium. So `DeltaG=0`. |
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| 138. |
Assertion : Enthalpy and entropy of any elementary substance in the standard state are taken as zero. Reason : At zero degree completely motionless.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
| Answer» Correct Answer - D | |
| 139. |
For isothermal expansion of an ideal gas, the correct combination of the thermodynamic parameters will beA. `DeltaU=0, Q=0, wne0 and DeltaHne0`B. `DeltaUne0, Qne0, wne0 and DeltaHne0`C. `DeltaU=0, Qne0, w=0 and DeltaHne0`D. `DeltaU=0, Qne0, wne0 and DeltaHne0` |
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Answer» Correct Answer - D `{:("For isothermal process,"DeltaT=0,"From first law of"),(,"thermodynamics"),(,DeltaU=Q+W),(thereforeDeltaU=nC_(v)DeltaT=0,"As " DeltaU=0),(DeltaH=nC_(p)DeltaT=0,thereforeQ=Wne0):}` |
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| 140. |
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where E is internal energy of the system)A. `-DeltaP//P`B. ZeroC. `-V DeltaP`D. `-DeltaE` |
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Answer» Correct Answer - B From 1 st law of thermodynamic `DeltaE=q+w. "Now w = "PDeltaV. "for "DeltaV=0` w=0 |
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| 141. |
Which of the following is zero for an isochoric processA. dPB. dVC. dTD. dE |
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Answer» Correct Answer - B dV=0 for an isochoric process. |
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| 142. |
Internal energy of an ideal gas depends onA. VolumeB. TemperatureC. PressureD. None of these |
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Answer» Correct Answer - B Internal energy of an ideal gas is a function of temperature only. |
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| 143. |
Assertion : We feel cold on touching the ice. Reason : Ice is a solid form of water.A. If both assertion and reason are true and the reaction is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If the assertion and reason both are false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B It is correct that on touching the ice we feel cold because ice absorb heat from our hand. |
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| 144. |
Heat produced in calories by the combustion of one gram of carbon is calledA. Heat of combustion of carbonB. Heat of formation of carbonC. Calorific value of carbonD. Heat of production of carbon. |
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Answer» Correct Answer - C It is the definition of calorific value. |
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| 145. |
The total internal energy change for a reversible isothermal cycles isA. Always 100 calories per degreeB. Always negativeC. 0D. Always positive |
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Answer» Correct Answer - C `DeltaE=0` for isothermal reversible cycle. |
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| 146. |
For chemical reactions, the calculation of change in entropy is normally doneA. At constant pressureB. At constant temperatureC. At constant temperature and pressure bothD. At constant volume |
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Answer» Correct Answer - C Calculation of change in entropy is done at constant temperature and pressure both. |
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| 147. |
In the reaction `C+2SrarrCS_(2)+DeltaH, DeltaH` is theA. Heat of combustionB. Heat of neutralizationC. Heat of solutionD. None of these |
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Answer» Correct Answer - D `CS_(2)` is formed from its initial components carbon and sulphur so, `DeltaH` is heat of formation of `CS_(2)`. |
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| 148. |
Which of the following is an example of endothermic reactionA. `C_(2)H_(2)+2H_(2)rarrC_(2)H_(6),DeltaE=-314.0 kJ`B. `C+O_(2)rarrCO_(2),DeltaE=-393.5 kJ`C. `N_(2)+O_(2)rarr2NO,DeltaE-180.5 kJ`D. `2H_(2)+O_(2)rarr2H_(2)O, DeltaE+571.8 kJ` |
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Answer» Correct Answer - C Heats of combustions are always exothermic except oxidation of N as. `N_(2)+(1)/(2)O_(2)rarrN_(2)O DeltaH=+ve` `N_(2)+O_(2)rarr2NO DeltaH=+ve` |
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| 149. |
The following thermochemical reactions are given `M+(1)/(2)O_(2)rarrMO+351.4 kJ` `X+(1)/(2)O_(2)rarrXO+90.8 kJ` It follows that the heat of reaction for the following process `M+XOiffMO+X` is given byA. 422.2 kJB. 268.7 kJC. `-442.2 kJ`D. 260.6 kJ |
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Answer» Correct Answer - D By (i) - (ii) and find required result. |
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| 150. |
The volume of a gas decreases from `500 c c` to `300 c c` when a sample of gas is compressed by an average pressure of 0.6 atm. During this process 10 J of heat is liberated. The change in internal energy isA. `-2.16 J`B. `12.156 J`C. `2.16 J`D. `101.3 J` |
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Answer» Correct Answer - C Here: Change in Volume (V) = 500 - 300 = 200cc = 0.2 litre, Pressure (P) = 0.6 atm and heat liberated (q) = 10 J Work done (W) = `PDeltaV=(0.2xx0.6)=0.12` litre-atm But 1 litre - atm = 101.3 J. hence `W=0.12xx101.3=12.156J`. We also know that heat is liberated, therefore it would be negative. Thus change in `DeltaE=q+W=-10+12.16=2.16J`. |
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