4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 44 InterviewSolution

2151.	Which molecules, ice at 0°C, or water at 0°C have grater potential energy and why?
Answer» The potential energy of water molecules at 0°C is more, because heat spent in melting is used up in increasing the P.E.

Discussion

2152.	Five moles of an ideal gas are taken in a carnot engine working between `100^(@)C and 30^(@)C`. The useful work done in 1 cycle is `420J`. Calculate ratio of volume of the gas at the end and at the beginning of the isothermal expansion. Take `R=8.4J mol e^(-1)K^(-1)`
Answer» Correct Answer - `1.153` Here, `n=5, T_(1)= 100^(@)C= (100+273)K = 373K`, `T_(2)= 30^(@)C=(30+273)K=303K`, `W=420J, (V_(2))/(V_(1))=?, R= 8.4J mol e^(-1)K^(-1)` As `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))=(373)/(303) :. Q_(1)(373)/(303)Q_(2)`....(i) Now, `W=Q_(1)-Q_(2)=(373)/(303)Q_(2)-Q_(2)=(70)/(303)Q_(2)` `:. Q_(2)=(373)/(70)W= (303)/(70)xx420=1818J` From (i), `Q_(1)=(373)/(303)xx1818= 2238J` Work done during isothermal expansion `W_(1)=Q_(1)= 2.3026 n RT_(1)"log"_(10)(V_(2))/(V_(1))` `2238= 2.3026xx5xx8.4xx373 "log"_(10)(V_(2))/(V_(1))` `"log"_(10)(V_(2))/(V_(1))=(2238)/(2.3026xx5xx8.4xx373)=0.0620` `(V_(2))/(V_(1))= antilog0.0620= 1.153`

Discussion

2153.	When two moles of a gas is heated from `O^(0)` to `10^(0)C` at constant volume, its internal enernal changes by `420J` . The moles specifie heat of the gas at constant volumeA. `5.75 J K^(-1) "mole"^(-1)`B. `10.55 J K^(-1) "mole"^(-1)`C. `21 J K^(-1) "mole"^(-1)`D. `42 J K^(-1) "mole"^(-1)`
Answer» Correct Answer - C `dU=nC_(v)dT`

Discussion

2154.	The volume of steam produced by `1g` of water at `100^(@)C` is `1650 cm^(3)`. Calculate the change in internal energy during the change of state. Given `J= 4.2xx10^(7) erg`. `cal.^(-1), d= 981 cm^(s-2)`. Latent heat of stream `= 540 cal. G^(-1)`
Answer» Here, mass of water `= 1g` `:.` Initial volume of water, `V_(1)= 1cm^(3)` Volume of stream, `V_(2)= 1650 cm^(3)` change of internal energy, `dU=?` As the state of water is changing, `:. dQ=mL` `= 1xx540 cals= 540xx4.2xx10^(7) ergs` `22.68xx10^(9) ergs` Taking `P= 1` atmosphere `= 76xx13.6xx981 dyn e cm^(-2)` `dW=P dV= P(V_(2)-V_(1))` `76xx13.6xx981(1650-1)` `=76xx13.6xx981xx1649 erg` As `dQ=dU+dW` `:. dU=dQ-dW` `= 22.68xx10^(9)-1.67xx10^(9)` `= 21.01xx10^(9) ergs`

Discussion

2155.	The equilibrium constant of the reaction `2C_(3) H_(6)(g)iffC_(2)H_(4)(g) + C_(4)H_(8)(g)` is found to fit the expression `InK_(eq)= - 1.04-(1088)/(T//K)` (where T/k is temperature expressed in Kelvin scale ) Calculate the standard reaction enthalpy and entrophy at 400 K.
Answer» Correct Answer - `DeltaH^(@)= 9.04 kJ//mol; DeltaS^(@) = - 8.64J//mol^(-1)K^(-1)` `DeltaG^(@)= DeltaH ^(@)- TDeltaS^(@)` `DeltaG^(@) = -RT In K` `- RT In K = DeltaH^(@)- TDeltaS^(@)` `In K = - (DeltaH^(@))/(RT) + (T DeltaS^(@))/(RT)` `(1088)/(T)= (DeltaH^(@))/(RT)` `DeltaH^(@) = 9.045 KJ` `(DeltaS^(@))/(R)= - 1.04` `DeltaS^(@) = -8.64 J mol^(-1) K^(-1)`

Discussion

2156.	3 moles of ideal gas `X (C_(p,m) = (5)/(2))` and 2 moles of ideal gas `Y(C_(p,m) = (7)/(2)R)` are taken in vessa and compressed reversibly and adiabitically, during this process temperature of gaseous mixture increased from 300 K to 400 K. Calculate change in internal energy `(DeltaU)` in cal of gaseous mixture (Given `R = 2 cal//mol.K`)
Answer» Correct Answer - 1900 cal `DeltaU = DeltaU_(1) + DeltaU_(2)` `n_(1)C_(V_(1)m)DeltaT + n_(2)C_(V_(1)m)DeltaT` `=[3 xx(3)/(2)R xx 100 ]+[2 xx (5R)/(2)xx100]` `= 450 R + 500 R` `= 950 R` `= 1900 cal`

Discussion

2157.	One mole of an ideal monoatomic gas `(C_(V.M)= 1.5 R)` is subjected to the following sequence of steps : (a) The gas is heated reversibly at constant pressure of 1 atm from 298 K to 373 K. (b) Next, the gas is heated reversibly and isothermally to double its volume. (c) Finally , the gas is cooled reversibly and adiabatically to 308 K . Calcuated q, w, `DeltaH` for the overall porcess.
Answer» Correct Answer - (a) `q= DeltaH = 1558.88, DeltaU = 935.33 ; W= - P(DeltaU) = - 623.55 J mol^(-1)` (b) `W =- 2149.7 J//mol ; DeltaU & DeltaH = 0, q = - W` (c) `q = 0, W =- 810.62 , DeltaH = - 1351.03J mol^(-1)` for overall process `q = 3705.59 ; W = - 3583.88, DeltaU = 124.71 ; DeltaH = 207.85` (a) `" "W= intP_("ext").dV = - P_("ext). DeltaV` `= - nRDeltaT` `=- 1xx 8.14 xx (373 - 298)` ` = -623.55 J//mol` `DeltaH = q = nC_(P)DeltaT` `= - 2.5 xx 8.314 xx 75" " (C_(P)=2.5 R)` `= 1558.8 J//mol` `DeltaU = 1.5 xx 8.314 xx 75" "(C_(V) = 1.5 R)` ` =935.5 J//mol` (b) ` W =-2.303RT log ((V_(1))/(V_(2)))` ` = - 2.303 xx 8.314 xx 373 log 2` ` = - 2149. 7 J//mol` `DeltaU = 0 , DeltaH = 0 " "q =2149.7 J//mol^(-1)` (c) `W = Deltau = nC_(V)DeltaT" "("For adiabatic process "q = 0, DeltaU ne W)` `= 1 xx 1.5 xx 8.314 (308-373)` ` = - 810.62 J//mol` `q= 0` `DeltaH = nC_(P)DeltaT =- 1 xx 2.5 xx 8.314 (308-373)` `= - 1351 .03 J//mol` For overall process `q_(net) = 3708.59 J//mol` `W_("net") = - 3583. 88 J//mol" "DeltaU_("net")= 124.71 J//mol` `DeltaH_("net")= 207.85 J//mol`

Discussion

2158.	A closed bottle contains some liquid. We shake bottle vigoroulsy for some time. It is found that temperature of liquid is increased (neglect expansion on heating). If Q, `DeltaU and W ` represent heat, change in internal energy and work done, thenA. Q=0`B. `W` is negativeC. `DeltaU` is positiveD. We cannot apply `Q=DeltaU+W` on liquids
Answer» Correct Answer - A::B::C

Discussion

2159.	One mole of a monoatomic gas behving ideally is used as working substance in an engine working in the cycle as shown in the figure . The process AB, BC, CD and DA are respectively reversible isobaric, adiabatic isochoric and isothermal . The ratio of maximum to volume and temperature during the cycle is `8sqrt(2)` and 4 respectively. If the maximum T is 800 K and `gamma =5//3`. Calculate `DeltaE` (in kJ) for the process BC. Given `R= 8.3 J//K-mol`.
Answer» Correct Answer - `DeltaE = 2.49 KJ` Max. volume at `V_(A)` Min. volume at `V_(C)` & `V_(D)` `(V_(A))/(V_(C))= 8sqrt(2)` `V_(A)= 8sqrt(2) V_(C)` nin. Temp .at B max. temp at D or A `(T_(D))/(T_(B))= 4 or (T_(A))/(T_(B))= 4` `therefore T_(B) = 200 K` During isobaric process Ab `(V_(A))/(V_(B))= (T_(A))/(T_(B)) rArr V_(B) (T_(B))/(T_(A))= (8sqrt(2))/(4) V_(C) = 2 sqrt(2) V_(C)` During adiabatic proess BC `T_(C)= T_(B) ((V_(B))/(V_(C)))^(r-1)= 200 (2sqrt(2))^((5)/(3)-1)=400 K` `= 400 K` `DeltaE= (3)/(2)R[400-200]` `= 2494 .2 J = 2.494 kJ`

Discussion

2160.	A cyclic process ABCDA is shown in figure Which of the following statement is correct?A. Net work done by gas is negativeB. net work done by gas is positiveC. Above process represents a cycle usable for engineD. Above process represents a cycle usable for a refrigerator
Answer» Correct Answer - A::D

Discussion

2161.	4 moles of a monoatomic gas are filled in a rigid container. Temperature of the gas increases from `27^(@)C` to `127^(@)C` by heating. Let Q, `DeltaU` and W represent heat, change in internal energy and work done, thenA. Q=600RB. Q=1000RC. `DeltaU=600R`D. `DeltaU=800R`
Answer» Correct Answer - A::C

Discussion

2162.	Calcualte `q, w, DeltaU`, and `DeltaH` for the reversible isothermal expansion of one mole of an ideal gas at `127^(@)C` from a volume of `20dm^(3)` to `40dm^(3)`.
Answer» Since the process is isothermal, `DeltaU = DeltaH = 0` From first law of thermodynamics, `DeltaU = q +w = 0` `q =- w` `w =- 2.303 nRT log ((V_(2))/(V_(1)))` `=- 2.303 xx1xx 8.314 xx 400 "log" (40)/(20)` `=- 2.303 xx1xx 8.314 xx 400 xx 0.3010` `=- 2305.3J` (work is done by the system) `q =- w = 2305.3J` (Heat is abosrbed by the system)

Discussion

2163.	Assertion.The bond enthalpy of C-H bondin `CH_(4)` is nearly 416 k J `mol^(-1)` Reason . First , second, third and fourth C-H bonds in `CH_(4)` have same bond enthalpy.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false.
Answer» Correct Answer - c

Discussion

2164.	Given that (i) `O(g) + e^(-) rarr O^(-)(g) ,DeltaH= -142kJ mol^(-1)` (ii) `O(g) + 2e^(-)rarr O^(-2) (g), DeltaH = + 712 kJ mol^(-1)` Calculate `DeltaH` for the reaction `O^(-)(g) +e^(-) rarr O^(-2) (g)`.
Answer» Eqn. (ii) - Eqn. (i) gives `O^(-)(g) + e^(-) rarr O^(2-) (g), Delta H =712 -( - 142) kJ mol^(-1) = 854 kJ mol^(-1)`

Discussion

2165.	10 moles of an ideal gas expand isothermally and reversibly from a pressure of5 atm to 1 atm at300 K . What is the largest mass that can be lefted through a height of 1 metre by this expansion?
Answer» `w_(exp) = - 2.303 nRT log.(P_(1))/(P_(2))= -2.303 ( 10 mol) xx( 8.314 JK^(-1) mol^(-1)) ( 300 K ) log. ( 5)/(1) = - 40.15 xx 10^(3)J` If M is the mass that can be lifted by this work ,through a height of 1metre, then work done `=` Mgh ` 40.15xx 10^(3)J= Mxx9.81 m s^(-2) xx 1m` or ` M =( 40.15 xx10^(3) kg m^(2) s^(-2))/( 9.18 ms^(-2)xx 1 m ) (J = kg m^(2) s^(-2))` `= 4092 .76 kg`

Discussion

2166.	Is the bond energy of all thefour C-H bonds in `CH_(4)` molecule equal? If not then why? How is the C-H bond energy then reported ?
Answer» No because after breaking of C-H bonds one by one, the electronic environments change. The reported value is the average of the bond dissociation energies of the four C-H bonds.

Discussion

2167.	The enthalpy change for the reaction ,`Zn (s) + 2H^(+)(aq) rarr Zn^(2+) (aq)+ H_(2)(g)`, is `-154.0kJ mol^(-1)` . The formation of 2g of hydrogen expands the system by 22.4litresat1 atm pressure. What is the internal energy of the reaction ?
Answer» Taking the initial volume as negligible ( as no gaseous reactant is present ) , change in volume during expansion `(DeltaV)= 22.4 l`.External pressure `( P_(ext)) =1atm`. `DeltaH = DeltaU + P DeltaV`or`DeltaU = DeltaH- P DeltaV` `P DeltaV= 1 atm xx22.4 L =22.4L` atm` = 22.4 xx101.3 J =2307J =2.31kJ` `:. DeltaU= - 154 .4- 2.31 = - 156.71 kJ`

Discussion

2168.	Find the enthalpy of combustion of carbon ( graphite) to produce carbon monoxide( g)on the basis of data given below`:` C ( graphite ) `+ O_(2)(g) rarr CO_(2)(g) + 393. 4 kJ mol^(-1)` `CO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g) + 283.0 kJ mol^(-1)`
Answer» Correct Answer - `Delta H = -110 .4 kJ mol^(-1)` Aim `:C(s) + (1)/(2) O_(2)(g) rarr CO(g), DeltaH = ?` Eqn . (i) - Eqn. (ii) gives the required result.

Discussion

2169.	With the help of thermochemical equations given below, determine `Delta_(r )H^(Θ)` at `298 K` for the following reaction: `C("graphite")+2H_(2)(g) rarr CH_(4)(g),Delta_(r )H^(Θ) = ?` `C("graphite")+O_(2)(g) rarr CO_(2)(g), Delta_(r )H^(Θ) = -393.5 kJ mol^(-1)` ...(1) `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l)`, `Delta_(r )H^(Θ) = -285.8 kJ mol^(-1)` ...(2) `CO_2(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g)`, `Delta_(r )H^(Θ) = +890.3 kJ mol^(-1)` ...(3)A. `+ 78.8kJ mol^(-1)`B. `+144.0kJ mol^(-1)`C. `- 74.8kJ mol^(-1)`D. `-144.0kJ mol^(-1)`
Answer» Correct Answer - c

Discussion

2170.	What will be the enthalpy of combustion of carbon to produce carbon monoxide on the basis of data given below: `C_((s))+O_(2(g))rarrCO_(2(g))-393.4kJ` `CO_((g))+(1)/(2)O_(2(g))rarrCO_(2(g))-283.0kJ`A. `+676.4kJ`B. `-676.4kJ`C. `-110.4kJ`D. `+110.4kJ`
Answer» Correct Answer - C `C+(1)/(2)O_(2) rarr CO` is obtained by (i) - (ii). `-393.4-(-283.0)=-110.4kJ`

Discussion

2171.	The normal boiling point of a liquid X is 400 K. Which of the following statement is true about the process `X(l) to X(g)` ?A. At 400K and 1 atm pressure `Delta G = 0`B. At 400 K and 2 atm pressure `Delta G = +ve`C. at 400 K and 0.1 atm pressure `Delta G = -ve`D. at 410 K and 1atm pressure `Delta G = +ve`
Answer» Correct Answer - A::B::C `X(liq) overset(400K)underset(underset(1atm)(larr))rarr X (g), Delta G = 0`

Discussion

2172.	The correct statement among the following (i) heat of reaction depends on the temperature at which the reaction is carried (ii) heat of neutralisation depends on the temperature at which the experiment is carried. (iii) experimentally heat of combustion is `DeltaE`.A. i only correctB. ii only correctC. iii only correctD. all are correct
Answer» Correct Answer - D Standard definition

Discussion

2173.	One mole of an ideal gas is subjected to a two step reversible process (A-B and B-C). The pressure at A and C is same. Mark the correct statement(s) :A. Work involved in the path AB is zeroB. In the path AB work will be done on the gas by the surroundingC. Volume of gas at C = 3 `xx` volume of gas at AD. Volume of gas at B is 16.42 litres
Answer» Correct Answer - C::D At `V_(A)=(1xxRxx100)/(1)=100R` `V_(B)=(1xxRxx600)/(3)=200R` `becauseV_(B) gt V_(A)` so expansion of gas takes place `V_(B)=200xx0.0821=16.42 L`

Discussion

2174.	If one mole of an ideal gas with `C_(v) = (3)/(2) R` is heated at a constant pressure of 1 atm `25^(@)C` to `100^(@)C`. Which is correctA. `Delta U` during the process is 223.51 calB. `Delta H` during the process is 372.56 calC. entropy change during the process is `1.122 cal k^(-1) mol^(-1)`D. `Delta U, Delta H` are same for the process
Answer» Correct Answer - A::B::C `because C_(v) = (3)/(2)R :. C_(p)=C_(v) + R = (3)/(2)R+R = (5)/(2)R` `:.` Heat given at constant pressure ` m.C_(p).Delta T` ltbr. Or Now work done in the process `= -P Delta V` `Delta H` or `q_(p) = 1 xx (5)/(2) xx R xx (373-298)` or `Delta H = 1 xx (5)/(2) xx 1.987 xx 75 = 372.56 cal` `w = -p(V_(2)-V_(1)) = -p((nRT_(2))/(p)-(nRT_(1))/(p))` `(because pv = nRT)` `= -nRT(T_(2)-T_(1)) = -1 xx 1.987 xx (373 - 298)` `= -149.225 cal` `:.`from I law of thermodynamics `Delta U = q+Q = 372.56 - 149.05` `:. Delta U = 223.51 cal` Also, `dq_(rev) = nC_(p).dt` `ds = (dq_(rev))/(T) :. ds =(nC_(p).dt)/(T)` or `Delta S = int_(T_(1))^(T_(2))(nC_(p).dT)/(T) = nC_(p)log_(e).(T_(2))/(T_(1))` `:. Delta s = 2.303 nC_(p) log_(10).(T_(2))/(T_(1))` `= 2.303 xx 1 xx (5)/(2) xx (5)/(2) R xx log_(10).(373)/(298) =1.122 cal k^(-1) mol^(-1)`

Discussion

2175.	One mole of an ideal gas is subjected to a two step reversible process (A-B and B-C). The pressure at A and C is same. Mark the correct statement(s) :A. Woek involved in the path AB is zeroB. In the path AB work will be done on the gas by the surroundingC. Volume of gas at `C = 3 xx` volume of gas at AD. Volume of gas at B is 16.42 litres
Answer» Correct Answer - C::D At `V_(A) = (1 xx R xx 100)/(1) = 100 R V_(B) = (1 xx R xx 600)/(3) = 200 R` `because V_(B) gt V_(A)` so expansion of gas takes place `V_(B) = 200 xx 0.0821 = 16.42 L`

Discussion

2176.	`2H_(2(g))+2Cl_(2(g))rarr 4HCl(g), DeltaH^(@)=-92.3 kJ` (i) If the equation is reversed, the `DeltaH^(@)` value equal to `+92.3 kJ` (ii) The four `H-Cl` bonds are stronger than the four bonds in `2H_(2)` and `2Cl_(2)` (iii) The `DeltaH^(@)` value will be `-92.3` kJ if the the `HCl` is produced as a liquidA. All are correctB. i only correctC. i and ii are correctD. iii only correct
Answer» Correct Answer - C Enthalpy change is dependent on physical state

Discussion

2177.	Consider this equation and the associated value for `DeltaH^(@)`. `2H_(2)(g)+2Cl_(2)(g)to4HCl(g),DeltaH=-92.3Kj` which statement abount this information is incorrect?A. If the equation is reversed ,the `DeltaH^(@)` value equals +92.3KJ.B. The four HCl bonds are stronger then the four bonds in `H_(2) and Cl_(2)`.C. The `DeltaH^(@)` value will be -92.3 KJ if the HCl is produced as a liquid.D. 23.1 KJ of heat will be enolved when 1 mol of HCl(g)Is produced.
Answer» Correct Answer - c

Discussion

2178.	One mole Of an ideal triatomic gas undergoes compression process adiabatically . Then:A. Temperature change of gas will be more if process is carried out reversiblyB. Work involved during the process will be more if process is carreid out reversiblyC. Change in internal energy of gas will be less if process is carreid out reversiblyD. Change in enthalphy of gas will be less if process is carried out reversibly
Answer» Correct Answer - c,d

Discussion

2179.	For an ideal gas undergoing expansion compression process . The relationships which hold good are:A. `((delU)/(delP))_(T)=0`B. `((delH)/(delT))_(V)=C_(p)`C. `((delU)/(delT))_(P)=C_(V)`D. `C_(v)=C_(p)+R`
Answer» Correct Answer - a,b,c

Discussion

2180.	A reaction attains equilibrium state under standard conditions, then:A. Equilibrium constant `K = 0`B. Equilibrium constant `K =1`C. `DeltaG^(Theta) = 0` and `DeltaH^(Theta) = T DeltaS^(Theta)`D. `DeltaG = 0` and `DeltaH =T DeltaS`
Answer» `DeltaG^(Theta) = 0` at equilibrium under standard state. Also, at equilibrium, `DeltaG =0`. `:. DeltaH^(Theta) - T DeltaS^(Theta) = 0` Also, `DeltaG^(Theta) = 2.303 RT In K` `:. K = 1`

Discussion

2181.	Select the correct statements.A. The magnitude of work involved in an intermediate irreversible expansion is less than that involved in reversible expansion.B. Heat absorbed during intermediate irreversible expansion is more than that in intermediate reversible expansion.C. The magnitude of work involved in an intermediate reversible compression is more than that involved in intermdiate irreversible compression.D. Heat released during intermediate irreversible compression is more than that in intermediate reversible compression.
Answer» These are mathematically established facts.

Discussion

2182.	A gas `(C_(v.m) = (5)/(2)R)` behaving ideally is allowed to expand reversibly and adiabatically from `1` litre to `32` litre. Its initial temperature is `327^(@)C`. The molar enthalpy change (in `J//mol`) for the process is :A. `-1125 R`B. `-675`C. `-1575 R`D. 101 R
Answer» Correct Answer - C `(T_(2))/(T_(1)) = ((V_(1))/(V_(2)))^(gamma-1), T_(2) = T_(1)((1)/(32))^((7)/(5)-1) = 600((1)/(2^(5)))^(2//5)` `= 600(0.5)^(2)=150K, Delta H_(m) = (7)/(2)R xx(150-600) = -1575 R`

Discussion

2183.	For polytropic process `PV^(n)` = constant, molar heat capacity `(C_(m))` of an ideal gas is given by:A. `C_(v,m) +(R)/((n-1))`B. `C_(v,m) +(R)/((1-n))`C. `C_(v, m) +R`D. `C_(p, m) + (R)/((n-1))`
Answer» Correct Answer - B `dU = dp + dw, nC_(v,m).dT = nC_(m)dT -P.dV`, `C_(m) = C_(v, m) + (P.dV)/(n.dT)` ....(1) `PV^(n) = K` and `PV = nRT :. KV^(1-n) = nRT` `K(1-n)V^(-n).dV = nRdT` `(dV)/(dT) = (nR)/(K(1-n)V^(-n))` .....(2), From Eqs (1) and (2) `C_(m) = C_(v, m) + (R)/((1-n))`

Discussion

2184.	`CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l), DeltaH =- 890 kJ` what is the calorific or fuel value of `1kg` of `CH_(4)`?
Answer» Calorific value/kg ` =(890)/(16) xx 1000 = 55625 kJ g^(-1)`

Discussion

2185.	Given the enthalpy of formation of `CO_(2)(g)` is -94.0 KJ, of CaO(s) is -152 KJ, and the enthalpy of the reaction `CaCO_(3)(s) rarr CaO(s)+CO_(2)(g)` is 42KJ, the enthalpy of formation of `CaCO_(3) (s)` isA. `-268 KJ`B. `202 KJ`C. `-202 KJ`D. `-288 KJ`
Answer» Correct Answer - D `CaCO_(3) rarr CaO+CO_(2), Delta H = H_(P) - H_(R)`

Discussion

2186.	What is the heat of formation of `C_(6)H_(6)`, given that the heats of combustion of Benzene, carbon and Hydrogen are 782, 94 and 68K.Cal respectivelyA. `+14 K.Cal`B. `-14K.Cal`C. `+28 K.Cal`D. `-28K.Cal`
Answer» Correct Answer - A `6C+3H_(2) rarr C_(6)H_(6)`, `Delta H = H_(P) - H_(R)`

Discussion

2187.	Select the correct enthalpy at corresponding temperature using following datas (i) Heat capacity of solid from 0K to normal melting point 200K `C_(p.m)(s)=0.035T" "JK^(-1)"mol"^(-1)` `(ii)` Enthalpy of fusion `=7.5 KJ"mol"^(-1)` `(iii)` Enthalpy of vaporisation `=30KJ "mol"^(-1)` `(iv)` Heat capacity of liquid form 200K to normal boiling point 300K `C_(p.m)(l)=60+0.016T " "JK^(-1)"mol"^(-1)` `(v)` Heat capacity of gas from 300 K to 600 K at 1 atm `C_(p.m)(g)=50.0" " JK^(-1)"mol"^(-1)`A. `S_(200(s))`=7B. `S_(300(l))`=70.43C. `S_(300(g))=170.43`D. `S_(600(g)) = 205.09`
Answer» Correct Answer - A::B::C::D `S_(200(0)) = underset(0)overset(200)int (C_(p) dT)/(T) =7 J"mol"^(-1)K^(-1)` `S_(200(f))=7 + (7500)/(200) =44.5 J "mol"^(-1)K^(-1)` `S_(300(f)) = 44.5 + underset(200)overset(300)int (C_(p) dT)/(T)=44.5 + 60 "In" (300)/(200) + 1.6= 70.43 J "mol"^(-1)K^(-1)` `S_(300(g)) = 70.43 + (30000)/(300) = 170.43 J "mol"^(-1) K^(-1)` `S_(800(g)) = 170.43 "In" 50 "In" (600)/(300)=205.09 JK^(-1)"mol"^(-1)`

Discussion

2188.	Using `._(f)G^(Theta)(Hi) = 1.3 kJ mol^(-1)`, calculate the standard free enegry change for the following reaction: `H_(2)(g) +I_(2)(g) rarr 2HI(g)`
Answer» `DeltaG^(Theta) = sum_(f)G^(Theta) ("products") -sum_(f)G^(Theta)("reactants")` `= 2xx(1.3) -(0+0)` `= 2.6 -0 = 2.6 kJ mol^(-1)`

Discussion

2189.	The ΔH and ΔS for 2Ag2O(s) → 4Ag(s) + O2(g) are + 61.17 kJ mol-1 and +132 JK-1 mol-1 respectively. Above what temperature will the reaction be spontaneous?
Answer» The reaction 2Ag₂O(s) → 4Ag(s) + O₂(g) will be spontaneous when ΔG is negative. Since ΔH is +ve and ΔS is also +ve, the reaction ΔG = ΔH - TΔS suggests that ΔG would be negative when ΔH = TΔS < 0 or T > \(\frac{\Delta H}{\Delta S}\) = \(\frac{61170 mol^{-1}}{132 JK^{-1}mol^{-1}}\) = 463.4 K The given reaction will be spontaneous above a temperature of 463.4 K.

2189.

The ΔH and ΔS for 2Ag2O(s) → 4Ag(s) + O2(g) are + 61.17 kJ mol-1 and +132 JK-1 mol-1 respectively. Above what temperature will the reaction be spontaneous?

Answer»

The reaction 2Ag₂O(s) → 4Ag(s) + O₂(g) will be spontaneous when ΔG is negative. Since ΔH is +ve and ΔS is also +ve, the reaction ΔG = ΔH - TΔS suggests that ΔG would be negative when

ΔH = TΔS < 0

or T > \(\frac{\Delta H}{\Delta S}\) = \(\frac{61170 mol^{-1}}{132 JK^{-1}mol^{-1}}\) = 463.4 K

The given reaction will be spontaneous above a temperature of 463.4 K.

Discussion

2190.	For the reaction `2Ag_(2)O(s) rarr 4Ag(s) + O_(2)(g), DeltaH ` is 61.17 `KJ"mol"^(-1)` and `DeltaS` is 132 `JK^(-1) "mol"^(-1)` . Compute the temperature above which the given reaction will be spontaneous.A. `T gt 463.4 K`B. `T gt 190.25^(@) C`C. `T lt 190.25^(@) C`D. `T lt 463.4 K`
Answer» Correct Answer - A::B The given reaction will be spontaneous when `DeltaG` would be negative when `DeltaH-TDeltaS` is negative . That is , `DeltaH-TDeltaS lt 0` or `TDeltaS gt DeltaH` or `T gt (DeltaH)/(DeltaS)` or `T gt (61170 J "mol"^(-1))/(132 JK^(-1) "mol"^(-1))` or `T gt 463.4 K`

Discussion

2191.	Which always true for a specific system during a spontaneous reaction?A. `DeltaH lt 0`B. `DeltaH ge 0`C. `DeltaG lt 0`D. `DeltaS gt 0`
Answer» Correct Answer - C

Discussion

2192.	The standard Gibbd energies `(Delta_(f)S^(Theta))` for the formation of `SO_(2)(g)` and `SO_(3)(g)` are `-300.0` and `-371.0 kJ mol^(-1)` at `300K`, respectively. Calculate `DeltaG` and equilibrium constant for the following reaction at `300K`:
Answer» `2SO_(2)(g) +O_(2)(g) hArr 2SO_(2)(g)` `Delta_(r)G^(Theta) = 2Delta_(f)G^(Theta) (SO_(3)) - 2Delta_(f)G^(Theta) (SO_(2)) -Delta_(f)G^(Theta) (O_(2))` `= 2(-371) -2(-300)-0` `= -742 +600 =- 142 kJ mol^(-1)` Now `log K = (Delta_(r)G)/(2.303 RT)` `Delta_(r)G^(Theta) =- 142 kJ mol^(-1), R = 8.314 xx 10^(-3) kJ mol^(-1)K^(-1)`, `T = 300K` `log K = (-142)/(2.303 xx 8.314 xx 10^(-3) xx 300) = 24.72` `:. K = Antilog (24.72) = 5.248 xx 10^(24)`

Discussion

2193.	Calculate equilibrium constant for the reaction: `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g) at 25^(@)C` Given: `Delta_(f)G^(Theta) SO_(3)(g) = - 371.1 kJ mol^(-1)`, `Delta_(f)G^(Theta)SO_(2)(g) =- 300.2 kJ mol^(-1)` and `R = 8.31 J K^(-1) mol^(-1)`
Answer» Step I: Calculation of `DeltaG^(Theta)` for the reaction `DeltaG^(Theta) = sum Delta_(f)G^(Theta) (p) - sum Delta_(f)G^(Theta) (r )` `=[2mol xx Delta_(f)G^(Theta)SO_(3)(g)]` `-[2mol xx Delta_(f)G^(Theta) SO_(2),(g) +Delta_(f)G^(Theta) O_(2)(g)]` `=[2mol xx 371.1 kJ mol^(-1)]` `-[2mol xx (-300.2 kJ mol^(-1))]` `=- 742.2 +600.4 =- 141.8 kJ` Step II: Calculation of equilibrium constant `(K)` `DeltaG^(Theta) =- 2.303 RT log K` `DeltaG^(Theta) = - 141.8 kJ =- 141800 J, T = 25 +273 = 298K`, `R = 8.31 JK^(-1) mol^(-1)` `logK =- (DeltaG^(@))/(2.303 RT)` `=(-) ((-141800J))/(2.303xx(8.31JK^(-1)mol^(-1)xx298K)) = 24.864` `K = Antilog (24.864) = 7.31 xx 10^(24)`

Discussion

2194.	The equilibrium constant of the reactionCO2 (g) + H2 (g) ⇌ CO (g) + H2O (g) at 298 K is 73. Calculate the value of the standard free energy change, (Q = 8.314 J mol-1 K-1).
Answer» ΔG° = -2.303 RT log K_C = -2.303 x (8.314 mol^-1 K^-1) x 298 K x log 73 = -2.303 x 8.314 x 298 x 1.8633 = -10632 J = -10.632 kJ

2194.

The equilibrium constant of the reactionCO2 (g) + H2 (g) ⇌ CO (g) + H2O (g) at 298 K is 73. Calculate the value of the standard free energy change, (Q = 8.314 J mol-1 K-1).

Answer»

ΔG° = -2.303 RT log K_C

= -2.303 x (8.314 mol^-1 K^-1) x 298 K x log 73

= -2.303 x 8.314 x 298 x 1.8633

= -10632 J = -10.632 kJ

Discussion

2195.	In a constant pressure calorimeter, 224 mL of 0.1 M KOH (aq) solution is reacted with 50ml of 0.1 M `H_(2)SO_(4)(aq)` solution then increase in temperature of solution will be 9assume heat capacity of calorimeter is negligible): Given : Specific heat of solution =1cal/g-K Density of solution =1g/mLA. `0.5K`B. `1K`C. `2K`D. `4K`
Answer» Correct Answer - a

Discussion

2196.	When `1`pentyne `(A)` is trated with `4N` alcoholic `KOH` at `175^(@)C`, it is slowely converted into an equilibrium mixture of `1.3%` of `1`pentyne `(A), 95.2% 2`-pentyne `(B)` and `3.5%` of `1,2`-pentandiene `(C )`. The equilibrium was maintained at `175^(@)C`. calculate `DeltaG^(Theta)` for the following equilibria: `B hArr A, DeltaG^(Theta)underset(1) = ?` `B hArr C, DeltaG^(Theta)underset(2) =?` From the calculated value of `DeltaG^(Theta)underset(1)`and `DeltaG^(Theta)underset(2)`, indicate the order of stability of `A,B` and `C`. write a reasonable reaction meachanisum showing all intermediates leading to `A,B` and `C`.
Answer» `{:("Pentyne"-1hArr,"Pentyne"-2+,1.2-"pentadiene",,),((A),(B),(C),,),(t_(Eq)1.3,95.2,3.5,,):}` `K_(Eq) = ([B][C])/([A]) = (95.2xx3.5)/(1.3) = 256.31` `B hArr A` `K_(1) = ([A])/([B]) = ([C])/(K_(Eq)) = (3.5)/(256.31) = 0.013` `DeltaG^(Theta)underset(1).=-2.303 RT log_(10)K_(1)` `= -2.303 xx 8.314 xx 448 log 0.013` ` = 1617J = 16.178 kJ` for `B hArr C` `K_(2) =([C])/([B]) = (K_(Eq)[A])/([B]^(2)) = (256.31xx1.3)/((9.5.2)^(2)) = 0.037` `DeltaG^(Theta)underset(2). =- 2.303 Rt log_(10)K_(2)` `=- 2.303 xx 8.314 xx 448 log 0.037` ` = 12282 J = 12.282 kJ` Stability will lie in the order `B gt C gt A`

Discussion

2197.	Which of the following relationships is not correct for the relation between `DeltaH and DeltaU`?A. When `Deltan_(g)=0" then " DeltaH=DeltaU`B. When `Deltan_(g) gt 0 " then "DeltaH gt DeltaU`C. When `Deltan_(g) lt 0" then " DeltaU lt DeltaU`D. When `Deltan_(g)RT=0" then "DeltaH=DeltaU`
Answer» Correct Answer - D When `Deltan_(g)RT=0,` then `DeltaH=DeltaU`

Discussion

2198.	The boiling point of diethyl ether is `34.6^(@)C`. Which is true for the vaporization of diethyl ether at `25.0^(@)C`?A. `DeltaG_(vap)^(@) gt0`B. `DeltaH_(vap)^(@) lt0`C. `K_(vap)=1`D. `DeltaS_(vap)^(@) lt0`
Answer» Correct Answer - A

Discussion

2199.	The equilibrium constant for the reaction: `CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)` has been found to be equal to `4` at `25^(@)C`. Calculate the free energy change for the reaction.
Answer» `DeltaG^(Theta) =- 2.303 xx 8.314 xx 298 log 4` `=- 3.435 kJ mol^(-1)`

Discussion

2200.	The relation between `DeltaU` and `DeltaH`:A. `DeltaH=DeltaU-PDeltaV`B. `DeltaH=DeltaU+PDeltaV`C. `DeltaH=DeltaV+ DeltaH`D. `DeltaU= DeltaH+ PDeltaV`
Answer» Correct Answer - B

Discussion

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The ΔH and ΔS for 2Ag2O(s) → 4Ag(s) + O2(g) are + 61.17 kJ mol-1 and +132 JK-1 mol-1 respectively. Above what temperature will the reaction be spontaneous?

For the reaction `2Ag_(2)O(s) rarr 4Ag(s) + O_(2)(g), DeltaH ` is 61.17 `KJ"mol"^(-1)` and `DeltaS` is 132 `JK^(-1) "mol"^(-1)` . Compute the temperature above which the given reaction will be spontaneous.A. `T gt 463.4 K`B. `T gt 190.25^(@) C`C. `T lt 190.25^(@) C`D. `T lt 463.4 K`

Which always true for a specific system during a spontaneous reaction?A. `DeltaH lt 0`B. `DeltaH ge 0`C. `DeltaG lt 0`D. `DeltaS gt 0`

The standard Gibbd energies `(Delta_(f)S^(Theta))` for the formation of `SO_(2)(g)` and `SO_(3)(g)` are `-300.0` and `-371.0 kJ mol^(-1)` at `300K`, respectively. Calculate `DeltaG` and equilibrium constant for the following reaction at `300K`:

Calculate equilibrium constant for the reaction: `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g) at 25^(@)C` Given: `Delta_(f)G^(Theta) SO_(3)(g) = - 371.1 kJ mol^(-1)`, `Delta_(f)G^(Theta)SO_(2)(g) =- 300.2 kJ mol^(-1)` and `R = 8.31 J K^(-1) mol^(-1)`

The equilibrium constant of the reactionCO2 (g) + H2 (g) ⇌ CO (g) + H2O (g) at 298 K is 73. Calculate the value of the standard free energy change, (Q = 8.314 J mol-1 K-1).

Which of the following relationships is not correct for the relation between `DeltaH and DeltaU`?A. When `Deltan_(g)=0" then " DeltaH=DeltaU`B. When `Deltan_(g) gt 0 " then "DeltaH gt DeltaU`C. When `Deltan_(g) lt 0" then " DeltaU lt DeltaU`D. When `Deltan_(g)RT=0" then "DeltaH=DeltaU`

The boiling point of diethyl ether is `34.6^(@)C`. Which is true for the vaporization of diethyl ether at `25.0^(@)C`?A. `DeltaG_(vap)^(@) gt0`B. `DeltaH_(vap)^(@) lt0`C. `K_(vap)=1`D. `DeltaS_(vap)^(@) lt0`

The equilibrium constant for the reaction: `CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)` has been found to be equal to `4` at `25^(@)C`. Calculate the free energy change for the reaction.

The relation between `DeltaU` and `DeltaH`:A. `DeltaH=DeltaU-PDeltaV`B. `DeltaH=DeltaU+PDeltaV`C. `DeltaH=DeltaV+ DeltaH`D. `DeltaU= DeltaH+ PDeltaV`