4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 60 InterviewSolution

2951.	Enthalpy of sublimation of a substance is equal toA. enthalpy of fusion + enthalpy of vapourisationB. enthalpy of fusionC. enthalpy of vapourisationD. twice the enthalpy of vapourisation.
Answer» Correct Answer - A `"Solid "overset("Sublimation")rarr "Vapour"` This process occurs in two steps, Step 1 : `"Solid" overset("Fusion")rarr"Liquid"` Step 2 : `"Liquid" overset("Vaporisation")rarr "Vapour"` Thus, Enthalpy of sublimation = Enthalpy of fusion `+` Enthalpy of vaporisation.

Discussion

2952.	Enthalpy of sublimation of a substance is equal toA. enthalpy of fusion `+` enthalpy of vaporisationB. enthalpy of fustionC. enthalpy of vaporisationD. twice the enthalpy of vaporisation
Answer» Correct Answer - a Sublimation is `:` Solid`rarr` Vapour . Writing in two steps , we have Solid`rarr` Liquid `rarr`Vapour . Solid `rarr` Liquid required enthalpy of fusion . Liquid`rarr`Vapour required enthalpy of vaporisation.

Discussion

2953.	The enthalpy of elements in their standard atates are taken as zero .The enthalpy of formation of a compound:A. is always negativeB. is always positiveC. may be positive or negativeD. is negative negative
Answer» Correct Answer - C Combustion of elements to form a compound can be exothermic endothermic e.g., `C + O_(2) rarr CO_(2)` is exothermic whereas, `C + 2S rarr CS_(2)` is endothermic Hence, enthalpy of formation can be positive or negative.

Discussion

2954.	The integral enthaiply of solution of one mole of `H_(2)SO_(4)` with n mole of water is given by the equation `DeltaH=(-75.6n)/(n+1.8)` which of the following option(s) is/are correct ?A. When 1 mole of `H_(2)SO_(4)` is dissolved in 2 mole of `H_(2)O.DeltaH_(sol)=-39.79 kJ`B. When 1 mole of `H_(2)So_(4)` is dissolved in 7 mole of `H_(2)O,DeltaH_(sol)=- 60.14 kJ`C. When 1 mole of `H_(2)So_(4)` is dissolved in 7 mole of `H_(2)O, DeltaH_(sol)=-23.5 kJ`D. When 1 mole of `H_(2)So_(4)` is dissolved in 7 mole of `H_(2)O, DeltaH_(sol)=-75.6 kJ`
Answer» Correct Answer - a,b,d

Discussion

2955.	The enthalpy of elements in their standard atates are taken as zero .The enthalpy of formation of a compound:A. Is always negativeB. is always positiveC. may be positive or negativeD. is never negative
Answer» Correct Answer - c

Discussion

2956.	The standard enthaply of fromation of `CO_(2)`will be given by:A. standard enthaply of combustion of diamondB. standard enthaply of combustion of graphiteC. standard enthaply of combustion of COD. sum of standard enthaply of formation ans enthalpy of combustion of CO.
Answer» Correct Answer - b,d

Discussion

2957.	The standard molar heats of formation of ethane, carbon dioxide, and liquid water ate `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.
Answer» The required chemicla equation for combustion of ethane is `2C_(2)H_(6)(g) +7O_(2)(g) = 4CO_(2)(g) +6H_(2)O(l), DeltaH^(Theta) = ?` The equation involves `2mol` of `C_(2)H_(6)`, heta of combustion of thene will be `=(DeltaH^(Theta))/(2)` `DeltaH^(Theta) = Delta_(f)H^(Theta) ("products") - Delta_(f)H^(Theta) ("reactants")` `[4 xx Delta_(f)H_((CO_(2)))^(Theta)+ 6Delta_(f)H_((H_(2)O))^(Theta)]` `-[2Delta_(f)H_((C_(2)H_(6)))^(Theta) + 7Delta_(f)H_((O_(2)))^(Theta)]` `= [4 xx (-94.1) + 6 xx (-68.3)] -[2xx (-2.11) +7 xx 0]` `=- 376.4 - 409.8 + 42.2` `= 744.0 kcal` `(DeltaH^(Theta))/(2)=` Heat of combustion of ethane ` = (744.0)/(2) = - 372.0 kcal`

Discussion

2958.	Can the temperature of an isolated system remain constant?
Answer» Yes, provided no physical or chemical changes take place in the system.

Discussion

2959.	The physical quantity that determins whether or not a given system A is in thermal equilibrium with another system B is calledA. PressureB. VolumeC. temperatureD. none of these
Answer» Correct Answer - C In thermal equilibrium, another system B , temperature

Discussion

2960.	An adiabatic cyclinder fitted with an adiabatic piston at the right end of cylinder, is dicided into equal halves with a monoatomic gas of the left side and diatomic gas on right side, using animpermeable movales adiabatic wall , . If the piston is pushed slowely to compress the diatomic gas to `(3)/(4)th` of its origanl volume. The new volume of the monoatomic gas would be.A. `V_("new")= V_("initial") xx[(4)/(3)]^((25)/(21))`B. `V_("new")= V_("initial") xx[(7)/(5)]^((3)/(4))`C. `V_("new")= V_("initial") xx[(3)/(4)]^((21)/(25))`D. `V_("new")= V_("initial") xx(3)/(4)`
Answer» Correct Answer - C `Pv^(gamma) = K` for diatomic gas `(P_(1))/(P_(2))=((V_(2))/(V_(1)))^(gamma)=((3)/(4))^(7//5)` for monoatomic gas `(P_(1))/(P_(2)) = (V_("new")/(V_("initial")))^(5//3) rArr V_("new") = V_("initial")xx ((3)/(4))^((21)/(25))`

Discussion

2961.	`(DeltaH-DeltaU)` for the formation of carbon monoxide (CO) from its elements at 298 K is : (R=8.314 `JK^(-1)mol^(-1)`)A. 1238.78 J `mol^(-1)`B. `-2477.57 J mol^(-1)`C. 2477.57 J `mol^(-1)`D. `-1238.78 mol^(-1)`
Answer» Correct Answer - A

Discussion

2962.	A substance is expanded in adiabatic process from 2 L to 5 L against constant pressure of 1 bar then internal energy change will be :A. 3 bar-LB. `-3 "bar"-L`C. 6 bar-LD. `-6 "bar"-L`
Answer» Correct Answer - B

Discussion

2963.	The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.A. `-22` kcalB. `+22` kcalC. `+184` kcalD. `-184` kcal
Answer» Correct Answer - A `(1)/(2)H_(2)+(1)/(2)Cl_(2) rarr HCl` `DeltaH=(1)/(2)xx104+(1)/(2)xx58-103=-"22 kcal"`

Discussion

2964.	What is the enthalpy change for the given reaction, if enthalpies of formation of `Al_(2)O_(3)` and `Fe_(2)O_(3)` are `-"1670 kJ mol"^(-1) and -"834 kJ mol"^(-1)` respectively? `Fe_(2)O_(3) +2Al rarr Al_(2)O_(3)+2Fe`A. `-"836 kJ mol"^(-1)`B. `+"836 kJ mol"^(-1)`C. `-"424 kJ mol"^(-1)`D. `+"424 kJ mol"^(-1)`
Answer» Correct Answer - A `DeltaH=SigmaDeltaH_(p)-SigmaDeltaH_(R)` `=-1670-(-834)=-"836 kJ mol"^(-1)`

Discussion

2965.	Assuming that water vapour is an ideal gas, the internal energy change `(DeltaU)` when 1 mole of water is vaporised at `1 ba r` pressure and` 100^(@)C, (` given`:` molar enthalpy of vaporization of water `41kJ mol^(-1)` at `1 ba r` and `373 K ` and `R=8.3 J mol^(-1)K^(-1))` will be `:`A. `41 kJ mol^(-1)`B. `4.1 kJ mol^(-1)`C. `3.7904 kJ mol^(-1)`D. `37.904 kJ mol^(-1)`
Answer» Correct Answer - D `DeltaH=DeltaU+Delta nRT`

Discussion

2966.	Define surroundings.
Answer» The rest of the universe which might be in a position to exchange energy and matter with the system is called its surroundings.

Discussion

2967.	Define enthalpy.
Answer» A system in thermodynamics refers to that part of the universe in which observations are made.

Discussion

2968.	Change in internal energy is a state function while work is not, why?
Answer» The change in internal energy during a process depends only upon the initial and final state of the system. Therefore it is a state function. But the wonk is related the path followed. Therefore, it is not a state function.

Discussion

2969.	Name the different types of the system.
Answer» There are three types of system – (i) Open system (ii) Closed system (iii) Isolated system.

2969.

Name the different types of the system.

Answer»

There are three types of system –

(i) Open system

(ii) Closed system

(iii) Isolated system.

Discussion

2970.	How may the state of thermodynamic system be defined?
Answer» The state of thermodynamic system may be defined by specifying values of state variables like temperature, pressure, volume.

Discussion

2971.	What will happen to internal energy if work is done by the system?
Answer» The internal energy of the system will decrease if work is done by the system.

Discussion

2972.	From thermodynamic point of view, to which system the animals and plants belong?
Answer» Open system.

Discussion

2973.	At 373 K, steam and water are in equilibrium and `DeltaH=40.98" kJ mol"^(-1)`. What will be `DeltaS` for conversion of water into system ? `H_(2)O_((l))rarrH_(2)O_((g))`A. `"109.8 J K"^(-1)"mol"^(-1)`B. `"31 J K"^(-1)"mol"^(-1)`C. `"21.98 J K"^(-1)"mol"^(-1)`D. `"326 J K"^(-1)"mol"^(-1)`
Answer» Correct Answer - A `DeltaS_("vap")=(DeltaH_("vap"))/(T_(b))=(40.98xx1000)/(373)` `="109.8 J K"^(-1)"mol"^(-1)`

Discussion

2974.	The difference between `C_(p) " and " C_(v)` can be derived using the empirical relation `H = U + pV`. Calculate the difference between `C_(p) " and " C_(v)` for 10 moles of an ideal gas.
Answer» Given that, `C_(v)` = heat capacity at constant volume, `C_(p) =` heat capacity at constant pressure Difference between `C_(p) " and C_(v)` is equal to gas constant (R). `C_(p) - C_(v) = nR` (where, `n =` no. of moles) `= 10 xx 4.184 J` `= 41.84 J`

Discussion

2975.	In which of the following processes involving ideal gas magnitude of heat exchange will be maximum for same change in temperture and same moles.[P is in atm and V in litre]:A. `PV^(3)=20`B. Isochoric proceesC. Isobaric processD. `PV^(1//2)=10`
Answer» Correct Answer - D

Discussion

2976.	Which of the following statements is incorrect?A. `DeltaH_(vap)^(2)H_(2)O(l)gt21xx373cal`B. `oint(dq)/(T)=0 "for all cyclic processes".`C. Adiabatic reversible process is isoentropic.D. Heat exchange at constant volume condition will be independent of path.
Answer» Correct Answer - B

Discussion

2977.	Which of the following decreases with increasing tempreature ?A. The volume of an ideal gas in an adiabatic processB. Internal energy of a systemC. The pressure of an ideal gas in a fixed volumeD. Entropy of a pure substance
Answer» Correct Answer - A

Discussion

2978.	Which of the following partical derivations are incorrect for one mole of an ideal gas?A. `(delS)/(delT))_(V)=(C_(v,m))/(T)`B. `(delS)/(delV))_(T)=(nR)/(T)`C. `(delG)/(delP))_(T)=V`D. `(delH)/(delP))_(T)=0`
Answer» Correct Answer - B

Discussion

2979.	a. A cylinder of gas is assumed to contain `11.2 kg` of butane. If a normal family needs `20000kJ` of energy per day for cooking, how long will the cylinder last? Given that the heat of combustion of butane is `2658 kJ mol^(-1)`. b. If the air supply of the burner is insufficient (i.e. you have a yellow instead of a blue flame), a portion of the gas escape without combustion. Assuming that `33%` of the gas is wasted due to this inefficiency, how long would the cylinder last?
Answer» a. `1mol` of butane, i.e., `C_(4)H_(10) (= 58g)` gives heat `= 2658 kJ` `:. 58g of C_(4)H_(10)` gives heat `=2658 kJ` `11.2 xx 1000 g` gives `rArr (2658 xx 11.2 xx 1000)/(58) rArr 513268.96 kJ` `:. 20000 kJ `of heat is required for `1` day. `:. 513268.96 kJ` of heat is required for `= (513268.96)/(20000) = 25.66 days` `~~ 26 days` b. `33%` of heat is wasted, therefore, `67%` of heat is utilised. `:.` Heat utilised `= (513268.96xx67)/(100) = 343890 kJ` Number of days `= (343890)/(20000) = 17.19 days` `~~ 17 days`.

Discussion

2980.	Calculate the standard free energy change for the formation of methane at `300K`: `C("graphite") +2H_(2) (g) rarr CH_(4)(g)` The following data are given: `Delta_(f)H^(Theta) (kJ mol^(-1)): CH_(4)(g) =- 74.81` `Delta_(f)S^(Theta)(kJ mol^(-1)): C("graphite") = 5.70, H_(2)(g) = 130.7 CH_(4)(g) = 186.3`
Answer» `Delta_(r)G^(Theta)` can be calculated form the relation: `Delta_(r)G^(Theta) = Delta_(r)H^(Theta) - TDelta_(r)S^(Theta)` `C("graphite") +2H_(2)(g) rarr CH_(4)(g)` `Delta_(f)H^(Theta) = Delta_(f)H^(Theta) (CH_(4)) -[Delta_(f)H^(Theta)(C ) +2Delta_(f)H^(Theta)(H_(2))]` `=- 74.81 -0 -0 =- 74.81 kJ mol^(-1)` `Delta_(r)S^(Theta) S_(m)^(Theta) (CH_(4)) -[S_(m)^(Theta) (C )+2S_(m)^(Theta)(H_(2))]` `= 186.3 -[5.70 +2 xx 130.7]` `= 180.3 - 267.1` `=- 80.8 J K^(-1) mol^(-1)` `:. Delta_(r)G^(Theta) = Delta_(r)H^(Theta) -TDelta_(r)S^(Theta)` `=- 74.81 -300 xx (-80.0 xx 10^(-3))` `=- 74.81 +24.4 =- 50.57 kJ mol^(-1)`

Discussion

2981.	When 2mole of `C_(2)H_(6)` are completely burnt `3129kJ` of heat is liberated. Calculate the heat of formation of `C_(2)H_(6).Delta_(f)H^(Theta)` for `CO_(2)` and `H_(2)O` are `-395` and `-286kJ`, respectively.
Answer» The equation for the combustion of `C_(2)H_(6)` is: `2C_(2)H_(6) + 7O_(2) rarr 4CO_(2) + 6H_(2)O, DeltaH =- 3129 kJ` `DeltaH^(Theta) = Delta_(f)H^(Theta) ("products") - Delta_(f)H^(Theta)("reactants")` `=[4xxDelta_(f)H^(Theta)underset((CO_(2))).+6xxDelta_(f)H^(Theta)underset((H_(2)O)).]` `-[2xxDelta_(f)H^(Theta)underset((CO_(2)H_(6))).+7xxDelta_(f)H^(Theta)underset((O_(2))).]` `-3129 = [4xx-395) + 6 xx (-286)]` `-[2xxDelta_(f)H^(Theta)underset((C_(2)H_(6))).+7xx0]` or `2 xx Delta_(f)H^(Theta)(C_(2)H_(6)) =- 167` So `Delta_(f)H^(Theta)(C_(2)H_(6)) =- (167)/(2) =- 83.5kJ`

Discussion

2982.	Calculate the heat of combustion of ethylene (gas) to form CO2 (gas) and H2O (gas) at 298 K and 1 atmospheric pressure. The heats of formation of CO2, H2O and C2H4 are -393.7 - 241.8 + 52.3 kJ per mole respectively.
Answer» C₂H₄ (g) + 3O₂ (g) → 2CO₂ (g) + 2H₂O (g) ΔH_f(CO₂) = -393.7 kJ ΔH_f(H₂O) = -241.8 kJ ΔH_f(C₂H₄) = +52.3 kJ ΔH_Reaction = (Sum of ΔH°_f values of products) - (Sum of ΔH°_f value of reactants) = [2 x ΔH°_f(CO₂) + 2 x ΔH°_f(H₂O) - [ΔH°_f(C₂H₄) + 3 x ΔH°_f(O₂)]] = 2 x (-393.7) + 2 x )-241.8) - [523.0 + 0] [∵ ΔH°_f for elementary substance = 0] = [-787.4 - 483.6] - 53.3 = -1323.3 kJ

2982.

Calculate the heat of combustion of ethylene (gas) to form CO2 (gas) and H2O (gas) at 298 K and 1 atmospheric pressure. The heats of formation of CO2, H2O and C2H4 are -393.7 - 241.8 + 52.3 kJ per mole respectively.

Answer»

C₂H₄ (g) + 3O₂ (g) → 2CO₂ (g) + 2H₂O (g)

ΔH_f(CO₂) = -393.7 kJ

ΔH_f(H₂O) = -241.8 kJ

ΔH_f(C₂H₄) = +52.3 kJ

ΔH_Reaction = (Sum of ΔH°_f values of products) - (Sum of ΔH°_f value of reactants)

= [2 x ΔH°_f(CO₂) + 2 x ΔH°_f(H₂O) - [ΔH°_f(C₂H₄) + 3 x ΔH°_f(O₂)]]

= 2 x (-393.7) + 2 x )-241.8) - [523.0 + 0]

[∵ ΔH°_f for elementary substance = 0]

= [-787.4 - 483.6] - 53.3

= -1323.3 kJ

Discussion

2983.	When 2mole of `C_(2)H_(6)` are completely burnt `-3129kJ` of heat is liberated. Calculate the heat of formation of `C_(2)H_(6).Delta_(f)H^(Theta)` for `CO_(2)` and `H_(2)O` are `-395` and `-286kJ`, respectively.
Answer» `2C_(2)H_(6)(g)+7O_(2)(g)rarr 4CO_(2)(g)+6H_(2)O(l) , Delta H=-3129 kJ` `Delta H_("reaction")=Sigma Delta H_("products")-Sigma Delta H_("reac tan ts")` `-3129 =[4xxDelta_(f)H_(CO_(2))+6xxDelta_(f)H_(H_(2)O)]-[2xxDelta_(f)H_(C_(2)H_(6)]` `2Delta_(f)H_(C_(2)H_(6))=-167` kJ [for elements `Delta H = 0`] `Delta_(f)H_(C_(2)H_(6))=-83.5 kJ`

Discussion

2984.	Calculate the ethalpy of combustion of ethylene (g) to form `CO_(2)` (gas) and `H_(2)O` (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation of `CO_(2),H_(2)O` and `C_(2)H_(4)` are `-393.7, -241.8+52.3` kJ per mole respectively.
Answer» Given : `{:((i),C(s)+O_(2)(g)rarr CO_(2)(g),,Delta H=-393.7 kJ mol^(-1)),((ii),H_(2)(g)+(1)/(2)O_(2)(g)rarr H_(2)O(g)",",,Delta H=-241.8 kJ mol^(-1)),((iii),2C(s)+2H_(2)(g)rarr C_(2)H_(4)(g)",",,Delta H=+52.3 kJ mol^(-1)):}` `"Objective reaction : " C_(2)H_(4)(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)," "Delta H = ?` `2xx` Equation `(i)+2xx` Equation (ii) - Equation (iii) gives `{:(2C(s)+2O_(2)(g)" "rarr " "2CO_(2)(g)),(+O_(2)(g)+2H_(2)(g)" "rarr " "2H_(2)O(g)),(ul(-2C(s)-2H_(2)(g)" "rarr " "-C_(2)H_(4)(g))),(" "3O_(2)(g)" "rarr " "2CO_(2)(g)+2H_(2)O(g)-C_(2)H_(4)(g)),(or C_(2)H_(4)(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)):}` Alternative Method : We aim t : `C_(2)H_()(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)` We are given : `Delta H_(f(CO_(2))=-393.7 kJ mol^(-1), Delta H_(f(H_(1)O))=-21.8 kJ mol^(-1), Delta H_(f(C_(2)H_(4))=+52.3 kJ mol^(-1)` `Delta H_("Reaction")` = (Sum of `Delta H_(f)^(@)` values of Products) - (Sum of `Delta H_(f)^(@)` values of Reactants) `=[2xxDelta H_(f(CO_(2)))^(@)+2xxDelta H_(f(H_(2)O))^(@)]-[Delta H_(f(C_(2)H_(4)))^(@)+3xxDelta H_(f(CO_(2)))^(@)]` `=[2xx(-393.7)+2xx(-241.8)]-[(-52.3)+0]` `(because Delta H_(f)^(@) "for elementary substance" = 0)` `=[-787.4-483.6]-52.3=-1323.3 kJ mol^(-1)`

Discussion

2985.	A cylinder of cooking gas is assumed to contain 11.2 kg of butane. The thermo chemical equation for the combustion of butane is : `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)rarr 4CO_(2)(g)+5H_(2)O(l) , Delta H = -2658 kJ` If a family needs 15000 kJ of energy per day for cooking, how would the cylinder last ?
Answer» Molecular mass of butane is = `58 g mol^(-1)` 58 g of butane on combustion gives 2658 kJ of heat `therefore` 11.2 g of butane on combustion gives `= (2658xx11.2xx1000)/(58)` kJ of heat The daily required of energy = 15000 kJ `therefore` The cylinder would last `=(2658xx11.2xx1000)/(58xx15000)` days `~~34` days

Discussion

2986.	all natural processes areA. SpontaneousB. non-spontaneousC. ExothermicD. Endothermic
Answer» Correct Answer - A

Discussion

2987.	The statement "Heat cannot flow from colder body to hotter body" is known asA. 1 st law of ThermodynamicsB. Zeroth lawC. 2nd law of ThermodynamicsD. Law of conservation of energy
Answer» Correct Answer - C

Discussion

2988.	418.4 J of heat is added to a `4xx10^(-3)m^(3)` rigid container containing a diatomic gas atm and 273 K. Calculate the pressure of the gas assuming ideal behaviour. The vibrational contributions may be neglected : (R=8.314 J `mol^(-1) K^(-1)`)A. `3.48xx10^(5) N//m^(2)`B. `7.27xx10^(7) N//m^(2)`C. `1.43xx10^(5) N//m^(2)`D. `9.2xx10^(7) N//m^(2)`
Answer» Correct Answer - C

Discussion

2989.	The heat of combustion of C, S and `CS_(2)` are `-393.3kJ, -293.7kJ and -1108.76kJ.` What will be the heat of formation of `CS_(2)` ?A. `-128.02kJ`B. `+970kJ`C. `+1108.7kJ`D. `+12kJ`
Answer» Correct Answer - A `C+2S rarr CS_(2)` `DeltaH=SigmaDeltaH_(P)-SigmaDeltaH_(R)=-1108.76-[-393.3+2xx(-293.7)]` `=-1108.76+393.3+587.4=-"128.02 kJ"`

Discussion

2990.	If the enthalpy of combustion of `C`(graphite) is `-393.3kJ mol^(-1)`, then for producing `39.3kJ` of heat the amount of carbon required isA. `1.5 mol`B. `0.5mol`C. `1.2 g`D. `12 mg`
Answer» `393.3 kJ` enegry produced by `12 gC` Therefore, `39.3 kJ` enegry produced by `(12 xx39.3)/(393.3) = 1.2 g of C` (graphite)

Discussion

2991.	Identify different steps in the following cyclic process:
Answer» a.`A rarr B:`Temperature and pressure are constant. Therefore, it is an isothermal and isobaric process. b. `B rarrC:` It is adiabatic expansion in which temperature falls form `T_(1)` to `T_(2)`. c. `C rarrD:` Temperature and volume are constant. therefore, this process is isothermal and isochroic. d. `DrarrE:` Temperature and pressure are constant. therefore, it is an isothernal and isobaric contraction. e. `E rarrF:`It is adiabatic compression in which temperature increases form `T_(2)` to `T_(1)`. f. `F rarr A:` Temperature and volume are constant. therefore, it is an isothermal and isochoric process.

Discussion

2992.	Calculate the work done when done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.
Answer» The volume occupied by water is very small and thus the volume change is equal to the volume occupied by `1g` mol of water vapour. `V = (nRT)/(P) = (1.0 xx 0.0821xx 373)/(1.0) = 31.0L` `W =- P_(ext) xx DeltaV =- (1.0) xx (3.10) L-atm` `=- (3.10) xx 101.3J =- 3140.3J`

Discussion

2993.	Given `Delta_(i)H^(Theta)(HCN) = 45.2 kJ mol^(-1)` and `Delta_(i)H^(Theta)(CH_(3)COOH) = 2.1 kJ mol^(-1)`. Which one of the following facts is true?A. `pK_(a)(HCN) = pK_(a)(CH_(3)COOH)`B. `pK_(a)(HCN) gt pK_(a)(CH_(3)COOH)`C. `pK_(a)(HCN) lt pK_(a)(CH_(3)COOH)`D. `pK_(a)(HCN) = (45.17//2.07)pK_(a)(CH_(3)COOH)`
Answer» `pK_(a)(HCN) gtpK_(a)(CH_(3)COOH)`

Discussion

2994.	Calculate the enthalpy of formation of water, given that the bond energies of `H-H, O=O` and `O-H` bond are `433 kJ mol^(-1), 492 kJ mol^(-1)`, and `464 kJ mol^(-1)`, respectively.
Answer» `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g)` `Delta_(f)H = BE` of product `-BE` of reactant `= (BE of H_(2) +(1)/(2)BE of O_(2)) - (2BE of O-H)` `= 433 +(1)/(2) xx492 -2 xx 464` `= 433 +246 - 928 =- 249 kJ`

Discussion

2995.	Calculate the work donw when`1.0 mol` of water at `100^(@)C` vaporises against an atmospheric pressure of `1atm`. Assume ideal behaviour and volume of liquid water to be negligible.
Answer» Volume of water vapour `=(nRT)/(P) = (1xx0.082 xx 373)/(1.0)` `= 30.592L` `DeltaV = 30.59 - 0 = 30.59L` `w = 1atm xx 30.59L =- 30.59Latm` `w =- 30.59 xx 101.3J =- 3100J =- 3.1kJ`

Discussion

2996.	For the reaction 2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS?
Answer» ΔH: Negative (-ve) because energy is released in bond formation. ΔS: Negative (-ve) because entropy decreases when atom combine to form molecules.

2996.

For the reaction 2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS?

Answer»

ΔH: Negative (-ve) because energy is released in bond formation.

ΔS: Negative (-ve) because entropy decreases when atom combine to form molecules.

Discussion

2997.	Calculate the enthalpy change for the process `C Cl_(4)(g) rarrC(g) +4Cl(g)` and calculate bond enthalpyof`C-Cl` in `C Cl_(4)(g)` Given `: V_(vap) H^(@) (C C l_(4)) =30.5 kJ mol^(-1), Delta_(f)H^(@) (C C l_94))= -135.5 kJ mol^(-1)` `Delta_(a)H^(@) ( C ) = 715.0 kJ mol^(-1) ` where`Delta_(a)H^(@)`is enthalpy of atomisation`Delta_(a) H^(@)( Cl_(2)) = 242 kJ mol^(-1)`
Answer» The given data imply as under `: (i)C Cl_(4) (l) rarr C C l_(4)(g) , DeltaH= 30.5 kJ mol^(-1)` (ii)`C(s) + 2Cl_(2)(g) rarrC C l_(4)(l) , DeltaH =-135.5 kJ mol^(-1)` (iii) `C(s)rarr C(g) , DeltaH= 715 .0 kJ mol^(-1)` (iv) `Cl_(2)(g) rarr2Cl(g), DeltaH = 242 kJ mol^(-1)` Aim `:C C l_(4)(g) rarr C(g) + 4Cl(g) , DeltaH = ?` Eqn. (iii)`+2 xx` Eqn. (iv) - Eqn. (i). Eqn. (ii) gives the required equation with `DeltaH=715.0 + 2 ( 242)-30.5 - ( - 135.5) kJ mol^(-1) = 1304 kJ mol^(-1)` Bond enthalpy ofC-Cl in `C C l_(4)` ( average value)`= ( 1304)/( 4) = 326 kJ mol^(-1)`

Discussion

2998.	For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1. At what temperature will the reaction become spontaneous? Considering ΔH and ΔS to be constant over the temperature range.
Answer» ΔG = ΔH - TΔS For ΔG = 0, ΔH = TΔS or T = ΔH/ΔS T = {400 kJ mol^-1}/{2 kJ K^-1 mol^-1} = 200 K Thus, reaction will be in a state of equilibrium at 200 K and will be spontaneous above this temperature.

2998.

For the following reaction at 298 K2A + B → CΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1. At what temperature will the reaction become spontaneous? Considering ΔH and ΔS to be constant over the temperature range.

Answer»

ΔG = ΔH - TΔS

For ΔG = 0, ΔH = TΔS or T = ΔH/ΔS

T = {400 kJ mol^-1}/{2 kJ K^-1 mol^-1} = 200 K

Thus, reaction will be in a state of equilibrium at 200 K and will be spontaneous above this temperature.

Discussion

2999.	For an isolated system, ΔV = 0, what will be ΔS?
Answer» ΔS should be greater than zero.

Discussion

3000.	For the equilibrium `H_(2) O (1) hArr H_(2) O (g)` at 1 atm `298 K`A. standard free energy change is equal to zero `(Delta G^(@) = 0)`B. free energy change is less than zero `(Delta G lt 0)`C. standard free energy change is less than zero `(Delta G^(@) lt 0)`D. standard free energy change is more than zero `(Delta G^(@) gt 0)`
Answer» Correct Answer - A For a reversible process at at equilibrium, under standard conditions (1 atm and `298 K`), `Delta G^(@) = 0`.

Discussion

Explore topic-wise InterviewSolutions in .

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