540 + Interview Questions in CHEMICAL-THERMODYNAMICS IN GENERAL AWARENESS Page 8 InterviewSolution

351.	If an endothermic reaction occurs spontaneously at constant `T` and `P`, then which of the following is trueA. `DeltaGgt0`B. `DeltaHgt0`C. `DeltaSgt0`D. `DeltaSlt0`
Answer» Correct Answer - C Since reaction is endothermic, `DeltaH=+ve`. For the reaction to spontaneous, `DeltaG=-ve`. But `DeltaG=DeltaH-TDeltaS`. For `DeltaG` to be `-ve`, `DeltaS` mus be `+ve` i.e. `DeltaS gt 0`

Discussion

352.	Assertion: An exothermic reaction in principle cannot have zero activetion energy. Reason: In exothermic reaction `SigmaH` (Products) `ltSigmaH` (Reactants).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.
Answer» Correct Answer - D For exothermic reaction `(E_(a))_("minimum")ge0` .

Discussion

353.	Identify the correct statement regarding entropyA. At absolute zero of temperature, the entropy of perfectly crystalline substance is `+ve`B. At absolute zero of temperature entropy of perfectly crystalline substance is taken to be zeroC. At `0^(@)C` the entropy of a perfectly crystalline substance is taken to be zeroD. At absolute zero of temperature, the entropy of all crystalline substance is taken to be zero
Answer» Correct Answer - B Third law of thermodynamics.

Discussion

354.	The enthalpies of formation of `N_(2)O` and `NO` are respectively `82` and `90 kJ mol^(-1)`. The enthalpy of reaction `2N_(2)O(g)+O_(2)(g)to4NO(g)` isA. `8 kJ`B. `88 kJ`C. `-16 kJ`D. `196 kJ`
Answer» Correct Answer - D `(i) N_(2)+(1)/(2)O_(2)toN_(2)O` `(ii) (1)/(2)N_(2)+(1)/(2)O_(2)toNO` Multiply `(i)` with `2` and `(ii)` with `4` and subtract `(i)` from `(ii)` `DeltaH_(f)(NO_(2))=90xx4-82xx2` `=360-164=196 kJ`

Discussion

355.	In the reaction `CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g),` `DeltaH=2.8 kJ` `DeltaH` representsA. Heat of reactionB. Heat of combustionC. Heat of formationD. Heat of solution
Answer» Correct Answer - C Heat of reaction `(Delta H ) sum n H_(("products")) - sumn H_(("rectants"))`

Discussion

356.	For the reaction `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?`A. `DeltaE+2RT`B. `DeltaE-2RT`C. `DeltaE+RT`D. `DeltaE-RT`
Answer» Correct Answer - D `Delta H = Delta E + Delta n_(g) RT` `Delta n_(g) = 2 - u = -2` So, `Delta H = Delta E + (-2RT)` `= Delta E - 2RT`

Discussion

357.	In a closed insulated container, a liquid is stirred with a paddle to increase the temperature, which of the following is true?A. `DeltaU-W= 0,q=0`B. `DeltaU+W=qne 0`C. `DeltaU+Q,W=qne 0`D. `W=0,DeltaU=qne 0`
Answer» Correct Answer - A For insulated container `q=0` .

Discussion

358.	In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?A. `DeltaE=W ne 0`, `q=0`B. `DeltaE=W= qne0`C. `DeltaE=0`, `W=qne0`D. `W=0`, `DeltaE=qne0`
Answer» Correct Answer - A As the system is closed and insulated no heat enter or leave the system, i.e., `q=0` `:. DeltaE=q+W=W`.

Discussion

359.	for the reaction ` C+ O_(2) to CO_(2)`A. `DeltaH gt DeltaE`B. `DeltaH lt DeltaE`C. `DeltaH = DeltaE`D. none of these
Answer» Correct Answer - c for the reaction `Deltan=1-1=0 ` `DeltaH=DeltaE`

Discussion

360.	`DeltaH_(comb)^(@)` of carbon is `-x kJ mol^(-1)`. The standard formation of enthalpy of `CO_(2)(g)` will beA. `-xKJmol^(-1)`B. `+xKJmol^(-1)`C. `-x//3KJmol^(-1)`D. Data is insufficient to predict it
Answer» Correct Answer - B `C(s) + O_(2) (g) rarr CO_(2)(g)` `Delta H_("comb")^(@) = -x kJ//mol` `-x = DeltaH^(@) (O_(2)) - (0 + 0)` `DeltaH^(@) (O_(2)) = -x kJ//mol`

Discussion

361.	Given that `DeltaH_("comb")of C(s) , H_(2) (g) and CH_(4) (g) are , -394 , -294 and -829 kJ//mol` respectively. The heat of formation fo `CH_(4)` isA. `70 kJ//mol`B. ` -71. 8 kJ // mol`C. `-244 kJ//mol`D. `-748 kJ // mol `
Answer» Correct Answer - a the formation of `CH_(4)` is represented as : `C(s)+2H_(2)(g)to CH_(4)(g),` ` DeltaH_(f)=(DeltaH_("comb"))_(R)-(DeltaH("comb")_(P)` ` =[-394+2 (-284)-(-892)kJ` `-962 + 892 =-70 kJ mol^(-1)`

Discussion

362.	Calculate the maximum efficiency of an engine operating between `110^(@)C` and `25^(@)C` .A. `11.1%`B. `22.2%`C. `33.3%`D. `44.4%`
Answer» Correct Answer - A Maximum efficiency of an engine working between temperature `T_(2)` and `T_(1)` is given by `etn=(T_(2)-T_(1))/(T_(2))=(383.293)/(383)=0.222eta=22.2%`

Discussion

363.	Calcualte the maximum efficiency of an engine operating between`110^(@)C`and`25^(@)C`.A. `11.1%`B. `22.2%`C. `33.3%`D. `44.4%`
Answer» Correct Answer - B Maximum efficiency of an engine working between temperature `T_(2)` and `T_(1)` is given by `eta=(T_(2)-T_(1))/(T_(2))=(383.293)/(383)=0.222eta=22.2%`

Discussion

364.	Assertion (A) : There is a natural asymmetry between converting work to heat and converting heat of work. Reason (R ) : No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.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.
Answer» Correct Answer - A Statement represents second law of thermodynamics which is an argument for statement.

Discussion

365.	For the conversion of 1 mole of `SO_(2)(g)` into `SO_(3)(g)` the enthalpy of reaction at constant volume. `DeltaU`, at 298 K is `-97.027kJ`. Calculate the enthalpy of reaction, `DeltaH`, at constant pressure.
Answer» Correct Answer - `DeltaH=-98.267J` Use the rection, `SO_(2)(g)+(1)/(2)O_(2)(g)toSO_(3)(g),Deltan=1-(3)/(2)=-(1)/(2)`.

Discussion

366.	What is the value of internal energy change `(DeltaU)` at `27^(@)C` a gaseous reaction `2A_(2)(g)+5B_(2)(g)to2A_(2)B_(5)(g)` (whose heat chhange at constant pressure is -50700J)? `(R=8.314JK^(-1)" "mol^(-1))`A. `-50700J`B. `-63171J`C. `-38229J`D. `+38299J`
Answer» Correct Answer - C `DeltaH=DeltaU+Deltan_(g)RT` `-50700=DeltaU+(-5)xx8.314xx300` `DeltaU=-38229J`

Discussion

367.	Work is the mode of transference of energy. If the system involves gaseous substance and there is difference of pressure between system and surroundings, such a work is referred to as pressure - volume work `(W_(PV)= -P_(ext)DeltaV)`. It has been observed that reversible work done by the system is the maximum obtainable work. `w_(rev) gt w_(ir r)` The works of isothermal and adiabatic processes are different from each other. for isothermal reversible proces, `W_("isothermal reversible")=2.303 nRT log_(10) (V_(2)/V_(1))` `W_("adiabatic reversible")=C_(V) (T_(1)-T_(2))` Calculate work done when 1 mole of an ideal gas is expanded reversibly from 20 L to 40 L at a constant temperature of 300 K.A. 7.78kJB. `-1.73kJ`C. `11.73kJ`D. `-4.78kJ`
Answer» Correct Answer - B `w=-2.303nRT" "log((P_(1))/(P_(2)))` `=-2.303xx1xx8.314xx273" "log((1)/(0.1))` `=-5.227kJ` `q=-w=+5.227kJ` for isothermal process.

Discussion

368.	Which of the following thermodynamic relation is correct?A. `dG=VdP-SdT`B. `dU=PdV+TdS`C. `dH=-VdP+TdS`D. `dG=VdP+SdT`
Answer» Correct Answer - A `dG=dH-TdS-SdT` `(G=H-TS)` `dH=dU+PdV+VdP` `(H=U+PV)` and `dU=TdS-PdV` `thereforedG=(TdS-PdV)+PdV+VdP-TdS-SdT` `dG=VdP-SdT`

Discussion

369.	The pressure-volume of various thermodynamic process is shown in graphs: Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. `w_(rev) gt w_(irr)` The works of isothermal and adiabatic processes are different from each other. `w_("isothermal reversible") = 2.303 nRT log_(10) ((V_(2))/(V_(1)))` `= 2.303 nRT log_(10)((P_(2))/(P_(1)))` `w_("adiabatic reversible") = C_(V) (T_(1)-T_(2))` A thermodynamic system goes in a cyclic process as represented in the following `P -V` diagram: The net work done during the complete cycle is given by the areaA. cycle ACBDAB. `A A_(1),B_(1)BDA`C.D. `A A_(2)B_(2)B`
Answer» Correct Answer - A w=work done=Area under curve.

Discussion

370.	A reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively. The correct option (s) is (are)A. `q_(AC)=DeltaU_(BC) and w_(AB)=P_(2)(V_(2)-V_(1))`B. `w_(BC)=P_(2)(V_(2)-V_(1)) and q_(BC)=DeltaH_(AC)`C. `DeltaH_(CA) lt DeltaU_(CA) and q_(AC)=DeltaU_(BC)`D. `q_(BC)=DeltaH_(AC) and DeltaH_(CA) gt DeltaU_(CA)`.
Answer» Correct Answer - B::C In the given cyclic process: `AC to` Isochoric process `AB to` Isothermal process `BC to ` isobaric process (b). `w_(BC)=-P_(2)(V_(2)-V_(1))` work of isobaric expansion (correct) `=P_2(V_(1)-V_(2))` `q_(BC)=DeltaH_(AC)=nC_(p)(T_(2)-T_(1))` Enthalpy change in isobaric process (correct) (c) `nC_(p)(T_(1)-T_(2)) lt nC_(V)(T_(1)-T_(2))` `DeltaH_(CA) lt DeltaU_(CA)` (correct) `q_(AC)=DeltaU_(BC)`.

Discussion

371.	The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of `10 dm^(3)` to a volume of `100dm^(3)` at `27^(@)C` isA. `42.3J" "mol^(-1)K^(-1)`B. `38.3J" "mol^(-1)K^(-1)`C. `35.8J" "mol^(-1)K^(-1)`D. `32.3J" "mol^(-1)K^(-1)`
Answer» Correct Answer - B `DeltaS=2.303nR" "log_(10)(V_(2)//V_(1))` `=2.303xx2xx8.314log_(10)(100//10)` `=38.3JK^(-1)" "mol^(-1)`

Discussion

372.	Assertion: The heat absorbed during the isothermal expansion of an ideal gas againts vacuum is zero. Reason: The volume occupied by the molecules of an ideal gas is zero.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.
Answer» Correct Answer - B Isothermal expansion of an ideal gas against vacuum is zero because expansion is isothermal. The reason that volume occupied by the molecules of an ideal gas is zero is fals.

Discussion

373.	Work done during isothermal expansion of one mole of an ideal gas form 10atm to 1atm at 300K is (Gas constant=`2`)A. `938.8cal`B. `1138.8cal` .C. `1381.8cal` .D. `1581.8cal` .
Answer» Correct Answer - A `-W=+2.303nRT "log" (P_(1))/(P_(2))`

Discussion

374.	The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` againts a constant external presuure of `3atm` is `(1atm-L=101.32J)`A. `-6J`B. `-608J`C. `+304J`D. `-304J`
Answer» Correct Answer - B Work done `(W)=-P_(ext.)(V_(2))-V_(1))` `=-3xx6(6-4)=-6L-atm` `=-6xx101.32J ( :. 1L atm =101.32J)` `=-607.92~~608J.`

Discussion

375.	A gas expands againts a constant external pressure of `2.00` atm, increasing its volume by `3.40L` . Simultaneously, the system absorbs `400J` of heat from its surroungs. What is `DeltaE` , in joules, for this gas?A. `-689`B. `-289`C. `+400`D. `+289`
Answer» Correct Answer - C Given, `P=2.00` atm, `DeltaV=3.4L` `q=400J` Work done by das `w=-PDeltaV` `=-2xx3.4` `=-6.8L` atm `=-6.8xx101.32J` `=-688.976J` `DeltaE=q+(-w)` `=400+(-688.976)` `=-288.976~~-289J`

Discussion

376.	A gas absorbs `250 J` of heat and expands from `1` litre to `10` litres against the pressure `0.5` atmosphere at constant temperature. The values of `q`, `w` and `DeltaE` are respectivelyA. `+250`, `-455 J`, `-205 J`B. `250J`, `455 J`, `205 J`C. `-250J`, `205 J`, `455 J`D. `-205J`, `-250 J`, `-455 J`
Answer» Correct Answer - A `q=250 J`, `W=-P(V_(2)-V_(1))` `=-0.5(10-1)=4.5 L-"atm"` `=-4.5xx101J=-455J` `DeltaE=q+w=250+(-455)` `=-205 J`

Discussion

377.	Which of the following value of `DeltaH_(f)^(@)` represent that the product is least stable ?A. `-94.0 kcal mol^(-1)`B. `-231.6 kcal mol^(-1)`C. `+21.4 kcal mol^(-1)`D. `+64.8 kcal mol^(-1)`
Answer» Correct Answer - D More `+ve` is `DeltaH_(f)` less is the stability.

Discussion

378.	A swimmer coming out from a pool is covered with a film of water weighing about `80g`. How much heat must be supplied to evaporate this water ? If latent heat of evaporation for `H_(2)O` is `40.79 kJ mol^(-1)` at `100^(@)C`.
Answer» `q=mL` `=(80)/(18)xx40.79` `=181.28kJ`

Discussion

379.	A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at `100^(@)`, `[Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]`A. `35.67 kJmol ^(-1)`B. `35.67 kJmol ^(-1)`C. `36.57kJmol ^(-1)`D. `38.75kJmol ^(-1)`
Answer» Correct Answer - b we can repesent the process of evaporation as `18gH_(2)O(l)overset("veporistion")to 18 g H_(2)O(g)` number of moles in ` 18 g H_(2)O(l)` `(18g)/(18g mol^(-1))=1 mol` `Delta_(vap)U=Delta_(vap)H^(-)-pDeltaV` `Delta_(vap)H^(-)-Deltan_(g)RT` assume steam behave as an ideal ga `Delta_(vap)U=(40.66)-(1)(8.314xx10^(-3))(373) ` 40.66 - 3.10 ` 37.56 KJ mol^(-1)`

Discussion

380.	A gas absobs `100J` of heat and is simultaneously compressed by a constant external pressure of `1.50` atmosphere fron a volume of `8L` to `2L` . `DeltaE` will beA. `-812J`B. `812`C. `1011J`D. `911J`
Answer» Correct Answer - B `DeltaH=100J` Work done `PDeltaV=1.5(-6)` . Now, this work done is in litre atmosphere. To obtain the same in joules, multiply by `8.314` and divide by `0.0821` `:. DeltaH=DeltaE+PDeltaV` or, `100=DeltaE+(1.5(-6))/(0.0821)xx8.314` or, `DeltaE=1011.4J`

Discussion

381.	Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` isA. `-17Kcal`B. `111Kcal`C. `-170Kcal`D. `-85Kcal`
Answer» Correct Answer - D For reaction, `C(s)+2H_(2)(g)rarrCH_(4)(g),DeltaH^(@)=`? `DeltaH^(@)=-[(Delta_(c)^(0)` of `CH_(2)-(Delta_(c)^(0)` of `C+2xxDeltaH_(c)^(0)` oof `H_(2))]` `C+O_(2)rarrCO_(2),DeltaH=-94Kcal` (i) `2H_(2)+O_(2)rarr2H_(2)O,DeltaH=-68xx2Kcal` (ii) `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O,DeltaH=-213Kcal` (iii) `"eq". (i)+(ii)-Eq(iii)` `=-[(-213)-(-94+2xx-68)]Kcal//mol` `=-[-213+230]=-17Kcal//mol`

Discussion

382.	Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` isA. `-17 kcal`B. `-111 kcal`C. `-170 kcal`D. `85 kcal`
Answer» Correct Answer - A `(i) C(s)+2H_(2)(g)toCH_(4)(g)` `(ii) C(s)+O_(2)(g)toCO_(2)(g)` , `DeltaH=-94 kcal mol^(-1)` `(iii) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)`, `DeltaH=-68kcal mol^(-1)` `(iv) CH_(4)+3//2O_(2)toCO_(2)+2H_(2)O`, `DeltaH=-213 kcal mol^(-1)` To obtain eqn. `(i)` operate `(ii)+2xx(iii)-(iv)` `:. DeltaH=-94+2(-68)-(-213)` `=-94-136+213` `=-230+213=-17 kcal`

Discussion

383.	Five mole of a gas put through a series of change as shown below graphically in a cyclic process. The processes `A rarr B, B rarr C` and `C rarr A`, respectively, are A. isochoric, isobaric, isothermalB. isobaric, isocharic, isothermalC. isothermal,isobaric, isochoricD. isochoric,isothermal,isobaric
Answer» Correct Answer - A Process `A to B` is isochoric i.e., volume remains constant Process `B to C` is isobaric i.e., pressure remains constant. Process `C to A` is isothermal i.e., temperature remains constant

Discussion

384.	The enthalpy of vaporisation of a liquid is `30 kJ mol^(-1)` and entropy of vaporisation is `75 J mol^(-1) K^(-1)`. The boiling point of the liquid at `1atm` is :A. `250 K`B. `400 K`C. `450 K`D. `600 K`
Answer» Correct Answer - B `DeltaH=30 kJ mol^(-1)=30,000 J mol^(-1)` `DeltaS=75 J mol^(-1)K^(-1)` At boiling point, the reversible process `"liquid" hArr "vapour"` is in equilibrium at one atmospheric pressure `:. DeltaG=0` `DeltaG=DeltaH-TDeltaS` `:. T=(DeltaH)/(DeltaS)=(30,000 Jmol^(-1))/(75 J mol^(-1)K^(-1))=400 K`

Discussion

385.	Which of the following reactions will define `DeltaH_(f)^(@)` ?A. `C("diamond")+O_(2)(g)toCO_(2)(g)`B. `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g)toHF(g)`C. `N_(2)(g)+3H_(2)(g)to2NH_(3)`D. `CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)`
Answer» Correct Answer - B Standard enthalpy of formation is equal to enthalpy change for the preparation of `1` mole of a compound from its standard state. Therefore, for the reaction `(1)/(2)H_(2)(g)+(1)/(2)H_(2)(g)toHF(g)` `DeltaH_(f)^(@)=DeltaH^(@)`

Discussion

386.	What is entropy change for the conversion of one gram of ice to water at 273 K and one atmospheric pressure? `(DeltaH_("fusion")=6.025kJ" "mol^(-1))`
Answer» `DeltaH_("fusion")=6.025xx1000J" "mol^(-1)` `=(6025)/(18)J" "g^(-1)=334.72Jg^(-1)` `DeltaS_("fusion")=(DeltaH_("fusion"))/(T_(f))` `=(334.72)/(273)=1.226JK^(-1)g^(-1)`

Discussion

387.	The heats of formation of `CH_(4)(g),C_(2)H_(6)(g) and C_(4)H_(10)(g)` are `-74.8, -84.7 and -126.1 kJ" "mol^(-1)` respectively. Arrange them in order of their efficiency as fuel per gram. Heats of formation of `CO_(2)(g) and H_(2)O(l)` are `-393.5` and `-285.8`kJ `mol^(-1)` respectively.
Answer» Correct Answer - `CH_(4) gt C_(2)H_(6) gt C_(4)H_(10)` First determine heat of combustion in each case and then find the calorific value.

Discussion

388.	the following data (s) are given as the standed enthalpies of combustion of `C(s),H_(2)(g) and CH_(4)(g)and -393.5kJ mol ^(-1) 285.8 kJmol^(-1) and -890.4 kJ mol^(-1)` respectively at 298 K . The standed enthalpy of fromation of methation of methane `[CH_(4)(g)]` isA. `+ 724 . 42 kJ mol^(-1)`B. `+74.7 kJ mol^(-1)`C. `-114.82 kJ mol^(-1)`D. `-194.62 kJ mol ^(-1)`
Answer» Correct Answer - b `underset(-393.5)(C(s))+ underset(-285.8)(2H_(2)(g))tounderset(-890.4)(CH_(4)(g)` `Delta_(r)H(CH_(4)(g))=Delta_(c)H^(-)(C)-Delta_(C)H^(-)(H_(2)(g))xx2` ` -890.4 -(-393.5)-(-285xx2)` `+74.7 kJ//mol^(-1)`

Discussion

389.	The heat of combustion of carbon to `CO_(2)`is`-395.5kJ//mol`.The heat released upon the formation of`35.2g` of`CO_(2)`from carbon and oxygen gas is
Answer» Correct Answer - `-315kJ`

Discussion

390.	The heat of combustion of carbon to `CO_(2)`is`-395.5kJ//mol`.The heat released upon the formation of`35.2g` of`CO_(2)`from carbon and oxygen gas isA. `-630KJ`B. `-3.15KJ`C. `-315KJ`D. `+315KJ`
Answer» Correct Answer - D Formation of `CO_(2)` from carbon and dioxygen gas can be represented as `C(s)+O_(2)(g)rarrCO_(2(g)),Delta_(f)H=-393.5KJmol^(-1)` `(1"mole"=44g)` Heat released on the formation of `44gCO_(2)` `=-393.5KJmol^(-1)` `=(-393.5KJmol^(-1))/(44g)xx35.2g=-315KJ`

Discussion

391.	the heat of combustion of carbon to `CO_(2) is - 393 . 5 kJ // mol` . The heat released upon formation of 35.2g of `CO_(2)` from carbon and oxygen gas isA. `+ 315 kJ`B. `-31 . 5kJ`C. `-315 kJ`D. `+ 31.5KJ`
Answer» Correct Answer - c `C+O_(2)toCO_(2), DeltaH=-393.5kJ//mol` heat releaded during the formation of 35.2 g(given) of `CO_(2)=-(393.5xx35.2)/44=-315 kJ`

Discussion

392.	When `20mL` of a strong acid is added to `20mL` of an alkali, the temperature rises by `5^(@)C` . If `200mL` of each liquid is mixed, the temperature rise will beA. `5^(@)C`B. `50^(@)C`C. `20^(@)C`D. `0.5^(@)C`
Answer» Correct Answer - A Heat evolved in the first case `=40xx5=200cal` Heat evolved in the second case `=10` times to the first, that is `10xx200=2000cal` `:. 2000=400xxDeltaT` or `DeltaT=5^(@)C` .

Discussion

393.	The surface of copper gets tarnished by the formation of copper oxide. `N_(2)` gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the `N_(2)` gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: `2Cu(s) + H_(2)O(g) rarr Cu_(2)O(s) + H_(2)(g)` is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln is ____. (Given: total pressure = 1 bar, R (universal gas constant) = `8 J K−1 mol^(−1), ln(10) = 2.3. Cu(s) and Cu_(2)O(s)` are mutually immiscible. At 1250 `K: 2Cu(s) + 1//2 O_(2)(g) rarr Cu_(2)O(s)` `triangle H^(theta) = − 78,000 J mol^(−1)` `H_(2)(g) + 1//2 O_(2)(g) rarr H_(2)O(g), triangle G^(theta) = − 1,78,000 J mol^(−1)`, G is the Gibbs energy
Answer» Given, `2Cu(s)+(1)/(2)O_(2)(g)toCu_(2)O,DeltaG^(@)=-78kJ" "mol^(-1)` `H_(2)O(g)toH_(2)(g)+(1)/(2)O_(2)(g)," "DeltaG^(@)=+178kJ" "mol^(-1)` `overline(2Cu(s)+H_(2)(g)hArrCu_(2)O(s)+H_(2)(g),DeltaG^(@)=+100kJ" "mol^(-1)` `overline(p_(H_(2)O)=(1)/(100)xx1=0.01`bar we know, `DeltaG=DeltaG^(@)+RT" ln "Q` . . .(i) At equilibrium `DeltaG=0` `Q-K=(p_(H_(2)))/(p_(H_(2)O))=(pH_(2))/(0.01)` `R=8JK^(-1)mol^(-1)` `T=1250K` Putting values in (i) `0=+100kJ+(8)/(1000)xx1250" log"_(e)(p_(H_(2)))/(0.01)` `log_(e)pH_(2)=-14.6`

Discussion

394.	Equal volumes of monoatomic and diatomic gases a same initial temperature and pressure are mixed.The ratio of specific heats of the mixture `(C_(p)//C_(v))`will beA. `1`B. `2`C. `1.67`D. `1.5`
Answer» Correct Answer - A `C_(v)=(3)/(2)RT,C_(P)=(5)/(2)RT` for monoatomic gas `C_(v)=(5)/(2)RT,C_(P)=(7)/(2)RT` for monoatomic gas Thus for mixture of 1 mole each, `C_(v)=((3)/(2)RT(5)/(2)RT)/(2)` and `C_(p)=((5)/(2)RT(7)/(2)RT)/(2)` Therefore, `C_(p)//C_(v)=(3RT)/(2RT)=1.5` .

Discussion

395.	Equal volumes of two monmoatomic gases,`A,B,`at the same temperature and pressure are mixed.The ratio of specific heats`(C_(p)//C_(v))`of the mixture will beA. `0.83`B. `1.50`C. `3.3`D. `1.67`
Answer» Correct Answer - C For monoatomic gases `C_(p):C_(v)` is `1.67` . Note that equal volumes are mixed.

Discussion

396.	Equal volumes of two monoatomic gases,`A,B,`at the same temperature and pressure are mixed.The ratio of specific heats`(C_(p)//C_(v))`of the mixture will beA. `0.83`B. `1.50`C. `3.3`D. `1.67`
Answer» Correct Answer - D `c_(P)` for a mixture of monoatomic gas having equal volume `=5//2 R` , `c_(P)=3//2R` `c_(P)//c_(v)=(5//2R)/(3//2)=5//3=1.67`

Discussion

397.	A pressure cooker reduces cooking time becauseA. food particles are efffectively mashedB. water boils at higher temperture inside the pressure cookerC. food is cooked at constant volumeD. loss of heat due to radiation is minimum
Answer» Correct Answer - B Food is cooked faster at higher temperature in a pressure cooker as water boils at a higher temperature inside the pressure cooker.

Discussion

398.	The enthaplpy changes state for the following processes are listed below: `Cl_(2)(g)=2Cl(g)` : `242.3KJmol^(-1)` `I_(2)(g)=2I(g)` , `151.0KJ mol^(-1)` `ICl(g)=I(g)+Cl(g)` : `211.3KJ mol^(-1)` `I_(2)(s)=l_(2)(g)` , `62.76KJ mol^(-1)` Given that the standard states for iodine chlorine are `I_(2)(s)` and `Cl_(2)(g)` , the standard enthalpy of formation for `ICl(g)` is:A. `+244.8KJmol^(-1)`B. `-14.6KJmol^(-1)`C. `-16.8KJmol^(-1)`D. `+16.8KJmol^(-1)`
Answer» Correct Answer - C `I_(2)+Cl_(2)rarr2ICl` `DeltaH=e_(I_(2)srarrg)+e_(I-I)+e_(Cl-Cl)-2xxe_(I-Cl)` `=62.76+151.0+242.3-2xx211.3` `=33.46KJ` for 2 moles of `ICl` :. `DeltaH//mol=(33.46)/(2)=16.73KJmol^(-1)`

Discussion

399.	A reaction has both `DeltaH and DeltaS` negative. The rate of reactionA. increses with increase of temperatureB. increases with decrease of temperatureC. ramains unaffected by change of temperatureD. cannot be predicated for change in tempeature
Answer» Correct Answer - b `DeltaG=DeltaH-TDeltaS` Negative `DeltaH` favours the process. Negative `DeltaS` opposes the process. If temperature is decreased, opposing factor decreases. Hence , rate of reaction increases.

Discussion

400.	Consider the following processes :- `{:(,DeltaH(kJ//mol)),((1)/(2)A rarr B,+150),(3B rarr2C+D,-125),(E+A rarr 2D,+350),("For "B+D rarr E+2C",",Delta H" will be"):}`A. `525 kJ//mol`B. `-175 kJ//mol`C. `-325 kJ//mol`D. `325 kJ//mol`
Answer» Correct Answer - B Desired equation in `B+DtoE+2C`, `DeltaH=?` `2xx "Equation "(i)+"Equation"(ii)-"Equation"(iii)` `=` Desired equations `:. DeltaH=2xx150+(125)+(-125)-350` `=-175 kJ// mol`

Discussion

Explore topic-wise InterviewSolutions in .

If an endothermic reaction occurs spontaneously at constant `T` and `P`, then which of the following is trueA. `DeltaGgt0`B. `DeltaHgt0`C. `DeltaSgt0`D. `DeltaSlt0`

The enthalpies of formation of `N_(2)O` and `NO` are respectively `82` and `90 kJ mol^(-1)`. The enthalpy of reaction `2N_(2)O(g)+O_(2)(g)to4NO(g)` isA. `8 kJ`B. `88 kJ`C. `-16 kJ`D. `196 kJ`

In the reaction `CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g),` `DeltaH=2.8 kJ` `DeltaH` representsA. Heat of reactionB. Heat of combustionC. Heat of formationD. Heat of solution

For the reaction `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?`A. `DeltaE+2RT`B. `DeltaE-2RT`C. `DeltaE+RT`D. `DeltaE-RT`

In a closed insulated container, a liquid is stirred with a paddle to increase the temperature, which of the following is true?A. `DeltaU-W= 0,q=0`B. `DeltaU+W=qne 0`C. `DeltaU+Q,W=qne 0`D. `W=0,DeltaU=qne 0`

In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?A. `DeltaE=W ne 0`, `q=0`B. `DeltaE=W= qne0`C. `DeltaE=0`, `W=qne0`D. `W=0`, `DeltaE=qne0`

for the reaction ` C+ O_(2) to CO_(2)`A. `DeltaH gt DeltaE`B. `DeltaH lt DeltaE`C. `DeltaH = DeltaE`D. none of these

`DeltaH_(comb)^(@)` of carbon is `-x kJ mol^(-1)`. The standard formation of enthalpy of `CO_(2)(g)` will beA. `-xKJmol^(-1)`B. `+xKJmol^(-1)`C. `-x//3KJmol^(-1)`D. Data is insufficient to predict it

Given that `DeltaH_("comb")of C(s) , H_(2) (g) and CH_(4) (g) are , -394 , -294 and -829 kJ//mol` respectively. The heat of formation fo `CH_(4)` isA. `70 kJ//mol`B. ` -71. 8 kJ // mol`C. `-244 kJ//mol`D. `-748 kJ // mol `

Calculate the maximum efficiency of an engine operating between `110^(@)C` and `25^(@)C` .A. `11.1%`B. `22.2%`C. `33.3%`D. `44.4%`

Calcualte the maximum efficiency of an engine operating between`110^(@)C`and`25^(@)C`.A. `11.1%`B. `22.2%`C. `33.3%`D. `44.4%`

For the conversion of 1 mole of `SO_(2)(g)` into `SO_(3)(g)` the enthalpy of reaction at constant volume. `DeltaU`, at 298 K is `-97.027kJ`. Calculate the enthalpy of reaction, `DeltaH`, at constant pressure.

What is the value of internal energy change `(DeltaU)` at `27^(@)C` a gaseous reaction `2A_(2)(g)+5B_(2)(g)to2A_(2)B_(5)(g)` (whose heat chhange at constant pressure is -50700J)? `(R=8.314JK^(-1)" "mol^(-1))`A. `-50700J`B. `-63171J`C. `-38229J`D. `+38299J`

Which of the following thermodynamic relation is correct?A. `dG=VdP-SdT`B. `dU=PdV+TdS`C. `dH=-VdP+TdS`D. `dG=VdP+SdT`

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of `10 dm^(3)` to a volume of `100dm^(3)` at `27^(@)C` isA. `42.3J" "mol^(-1)K^(-1)`B. `38.3J" "mol^(-1)K^(-1)`C. `35.8J" "mol^(-1)K^(-1)`D. `32.3J" "mol^(-1)K^(-1)`

Work done during isothermal expansion of one mole of an ideal gas form 10atm to 1atm at 300K is (Gas constant=`2`)A. `938.8cal`B. `1138.8cal` .C. `1381.8cal` .D. `1581.8cal` .

The work done during the expansion of a gas from a volume of `4dm^(3)` to `6dm^(3)` againts a constant external presuure of `3atm` is `(1atm-L=101.32J)`A. `-6J`B. `-608J`C. `+304J`D. `-304J`

A gas expands againts a constant external pressure of `2.00` atm, increasing its volume by `3.40L` . Simultaneously, the system absorbs `400J` of heat from its surroungs. What is `DeltaE` , in joules, for this gas?A. `-689`B. `-289`C. `+400`D. `+289`

Which of the following value of `DeltaH_(f)^(@)` represent that the product is least stable ?A. `-94.0 kcal mol^(-1)`B. `-231.6 kcal mol^(-1)`C. `+21.4 kcal mol^(-1)`D. `+64.8 kcal mol^(-1)`

A swimmer coming out from a pool is covered with a film of water weighing about `80g`. How much heat must be supplied to evaporate this water ? If latent heat of evaporation for `H_(2)O` is `40.79 kJ mol^(-1)` at `100^(@)C`.

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at `100^(@)`, `[Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]`A. `35.67 kJmol ^(-1)`B. `35.67 kJmol ^(-1)`C. `36.57kJmol ^(-1)`D. `38.75kJmol ^(-1)`

A gas absobs `100J` of heat and is simultaneously compressed by a constant external pressure of `1.50` atmosphere fron a volume of `8L` to `2L` . `DeltaE` will beA. `-812J`B. `812`C. `1011J`D. `911J`

Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` isA. `-17Kcal`B. `111Kcal`C. `-170Kcal`D. `-85Kcal`

Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` isA. `-17 kcal`B. `-111 kcal`C. `-170 kcal`D. `85 kcal`

The enthalpy of vaporisation of a liquid is `30 kJ mol^(-1)` and entropy of vaporisation is `75 J mol^(-1) K^(-1)`. The boiling point of the liquid at `1atm` is :A. `250 K`B. `400 K`C. `450 K`D. `600 K`

Which of the following reactions will define `DeltaH_(f)^(@)` ?A. `C("diamond")+O_(2)(g)toCO_(2)(g)`B. `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g)toHF(g)`C. `N_(2)(g)+3H_(2)(g)to2NH_(3)`D. `CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)`

What is entropy change for the conversion of one gram of ice to water at 273 K and one atmospheric pressure? `(DeltaH_("fusion")=6.025kJ" "mol^(-1))`

The heats of formation of `CH_(4)(g),C_(2)H_(6)(g) and C_(4)H_(10)(g)` are `-74.8, -84.7 and -126.1 kJ" "mol^(-1)` respectively. Arrange them in order of their efficiency as fuel per gram. Heats of formation of `CO_(2)(g) and H_(2)O(l)` are `-393.5` and `-285.8`kJ `mol^(-1)` respectively.

The heat of combustion of carbon to `CO_(2)`is`-395.5kJ//mol`.The heat released upon the formation of`35.2g` of`CO_(2)`from carbon and oxygen gas is

The heat of combustion of carbon to `CO_(2)`is`-395.5kJ//mol`.The heat released upon the formation of`35.2g` of`CO_(2)`from carbon and oxygen gas isA. `-630KJ`B. `-3.15KJ`C. `-315KJ`D. `+315KJ`

the heat of combustion of carbon to `CO_(2) is - 393 . 5 kJ // mol` . The heat released upon formation of 35.2g of `CO_(2)` from carbon and oxygen gas isA. `+ 315 kJ`B. `-31 . 5kJ`C. `-315 kJ`D. `+ 31.5KJ`

When `20mL` of a strong acid is added to `20mL` of an alkali, the temperature rises by `5^(@)C` . If `200mL` of each liquid is mixed, the temperature rise will beA. `5^(@)C`B. `50^(@)C`C. `20^(@)C`D. `0.5^(@)C`

Equal volumes of monoatomic and diatomic gases a same initial temperature and pressure are mixed.The ratio of specific heats of the mixture `(C_(p)//C_(v))`will beA. `1`B. `2`C. `1.67`D. `1.5`

Equal volumes of two monmoatomic gases,`A,B,`at the same temperature and pressure are mixed.The ratio of specific heats`(C_(p)//C_(v))`of the mixture will beA. `0.83`B. `1.50`C. `3.3`D. `1.67`

Equal volumes of two monoatomic gases,`A,B,`at the same temperature and pressure are mixed.The ratio of specific heats`(C_(p)//C_(v))`of the mixture will beA. `0.83`B. `1.50`C. `3.3`D. `1.67`

A pressure cooker reduces cooking time becauseA. food particles are efffectively mashedB. water boils at higher temperture inside the pressure cookerC. food is cooked at constant volumeD. loss of heat due to radiation is minimum

A reaction has both `DeltaH and DeltaS` negative. The rate of reactionA. increses with increase of temperatureB. increases with decrease of temperatureC. ramains unaffected by change of temperatureD. cannot be predicated for change in tempeature

Consider the following processes :- `{:(,DeltaH(kJ//mol)),((1)/(2)A rarr B,+150),(3B rarr2C+D,-125),(E+A rarr 2D,+350),("For "B+D rarr E+2C",",Delta H" will be"):}`A. `525 kJ//mol`B. `-175 kJ//mol`C. `-325 kJ//mol`D. `325 kJ//mol`