4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 81 InterviewSolution

4001.	`{:("column"-1,"Column"-2),((A)(DeltaG_("system"))_(T.P) = 0,(p) "Process is in equilibrium"),((B)DeltaS_("system")+DeltaS_("surrounding") gt 0, (q)"Process is nonspontaneous"),((C)DeltaS_("system") +DeltaS_("surroumding") lt 0 , (r) "Process is spontaneous"),((D)(DeltaG_("system"))_(T.P) gt 0,(s)"System is unable to do useful work"):}`
Answer» Correct Answer - `(A) rarr (p,s); (B) rarr (r); (C) rarr(q,s); (D)rarr (q,s)` `(A)DeltaG_("sys") =0 implies` equilibrium and free energy is zero. So, no useful work. `(B) DeltaS_("universe") gt 0 implies` Spontaneous process and able to do usefull work. `(C) DeltaS_("universe") lt 0 implies` Nonspontaneous process and unable to do usefull work. `(D) DeltaG_("sys") gt 0 implies` Nonspontaneous process and unable to do useful work.

Discussion

4002.	For the hypothetical reaction `A_(2)(g)+B_(2)(g) to 2AB_(g)` `DeltaG_(r)^(@)" and " DeltaS_(r)^(@)" are " 20KJ//moland-20JK^(-1)mol^(-1) "respectively at" 200K` `Delta_(r)C_(r)" is "JK^(-1) " then " DeltaH_(r)^(@) "at" 400K is `A. `20KJ//mol`B. `7.98KJ//mol`C. `28kj//mol`D. None of these
Answer» Correct Answer - A

Discussion

4003.	For the hypothetical reaction `A_(2)(g)+B_(2)(g) to 2AB_(g)` `DeltaG_(r)^(@)" and " DeltaS_(r)^(@)" are " 20KJ//moland-20JK^(-1)mol^(-1) "respectively at" 200K` `Delta_(r)C_(p)" is "20 JK^(-1) " then " DeltaH_(r)^(@) "at" 400K is `A. 20 kJ/molB. `7.98` kJ/mosC. 28 kJ/molD. None of these
Answer» Correct Answer - A `DeltaG_(200)^(@)=DeltaH_(200)^(@)-TDeltaS_(200)^(@)` `DeltaH_(200)^(@)=20-4=16"kJ/mol"` `DeltaH_(T_(2))^(@)=DeltaH_(T_(1))^(@)+DeltaC_(P)[T_(1)-T_(2)]` `DeltaH_(400)^(@)=DeltaH_(200)^(@)+(20xx200)/(1000)"kJ/mol"` `=16+4=20"kJ/mol"`

Discussion

4004.	Two solids A and B having molar masses 200 and 300 react to form another solid C as shown `2A(s)+3B(s)toC(s)` if `DeltaH_("combustion")^(@)` of A, B and C are `-200,-300` and `-400` kJ respectively then maximum heat which can be released at constant pressure if total 2600 g of A and B is taken, is given by :A. 200 kJB. 1800 kJC. 900 kJD. 450 kJ
Answer» Correct Answer - B

Discussion

4005.	`{:(,"Column-I",,"Column-II"),((a),(DeltaG_(system))_(T.P^(=0)),(p),"Process in equilibrium"),((b),DeltaS_(system)+DeltaS_("sorrrounding")gt0,(q),"Process is non-spontaneous"),((c),DeltaS_(system)+DeltaS_("Surrounding")lt0,(r),"Process is spontaneous"),((d),(DeltaG_(system))+_(T,P)gt0,(s),"System is unable to do useful work"):}`
Answer» Correct Answer - `(a-p,s);(b-r);(c-q,s);(d-q,s)`

Discussion

4006.	One mole of an non linear triatomic ideal gas is expanded abiabatically at 300 K form 16 atm to 1 atm ltbr gt Find the value `DeltaS_(sys), DeltaS_(surr) & DeltaS_("total")` under the following conditions. (i) Expansion is carried out reversibly (ii) Expansion is carried out irreversibly (iii) Expansion is free.
Answer» Correct Answer - (i)` " "DeltaS_("sys")= nC_(v)ln .(T_(1))/(T_(2))+nRln .(V_(1))/(V_(2))= 0, q= 0,DeltaS_("surr")= - DeltaS_("sys")=DeltaS_("total")=0` (ii)` " "DeltaS_("sys")= 1.702R," "DeltaS_("surr")=0 , DeltaS_("total")=DeltaS_("sys")=1.702 R` (iii)` " "DeltaS_("sys")= nRln .(p_(1))/(p_(2))= 2.77 R, DeltaS_("surr")=(-q_("irr"))/(T) = 0, DeltaS_("total")= DeltaS_("sys") = 2.177 R`

Discussion

4007.	If one mole of an ideal gas `(C_(p.m.)=(5)/(2)R)` is expanded isothermally at 300K until it’s volume is tripled, then change in entropy of gas is:A. zeroB. infinityC. `(5)/(2)R In 3`D. `R In 3`
Answer» Correct Answer - D

Discussion

4008.	If one mole of an ideal gas `(C_(p.m.)=(5)/(2)R)` is expanded isothermally at 300K until it’s volume is tripled, then change in entropy of gas is:A. zeroB. infinityC. `(5)/(2)R ln 3`D. `R ln 3`
Answer» Correct Answer - D `DeltaS=nRln.(V_(2))/(V_(1))=1R ln.(3V)/(V)=Rln3`

Discussion

4009.	Match the column- I `{:(,"Column I",,"Column II"),((A),"Adiabtic expansion of an ideal gas against 1 atm",(P),DeltaS_("sys")gt0),((B),"Condensation of vapour at normal boiling point",(Q),DeltaU=0),((C),"Adiabatic reversible compression involving an ideal gas",(R),DeltaS_("uni")=0),((D),"Isothermal reversible expansion of an ideal gas",(S),"w"=0),(,,(T),"q"=0):}`
Answer» Correct Answer - (A) P,T (B) R (C) R,T (D) P,Q,R

Discussion

4010.	Match the column-1 `{:(,"Column I",,"Column II"),((A),DeltaS="n"C_(v)l"n"(T_(2))/(T_(1)),(P),DeltaS_("sys")" for an irreversible isochoric process"),((B),DeltaS=-(DeltaU_("sys"))/(T),(Q),\|DeltaS\|_("surr")" for a reversible isochroric process"),((C),DeltaS=nC_(v)l"n"(T_(2))/(T_(1))+nRl"n"(V_(2))/(V_(1)),(R),DeltaS_("surr")" for a reversible isochoric process"),(,,(S),DeltaS_("sys")" for a reversible process"):}`
Answer» Correct Answer - (A) P,Q (B) Q,R (C) P,Q,S

Discussion

4011.	One mole of an ideal gas is expanded isothermally at `300` K until its volume is tripled . Find the values of `DeltaS_("sys")` under the condition.
Answer» Correct Answer - `DeltaS_("total")=3 JK^(-1)`

Discussion

4012.	Standard molar entropies of `Mg_((s)), C_((s)), MgO_((s))` & `CO_((g))` respectively are `S_(1), S_(2), S_(3)` & `S_(4)` J/K/mol. Then `DeltaS_(sys)` for `MgO_((s))+C_((s)) rarr Mg_((s))+CO_((g))`A. `S_(1)+S_(2)+S_(3)+S_(4)`B. `(S_(1)+S_(4))-(S_(2)+S_(3))`C. `(S_(2)+S_(3))-(S_(1)+S_(4))`D. `(S_(4)-S_(3))`
Answer» Correct Answer - B `DeltaS=DeltaS_(P)-DeltaS_(R)`

Discussion

4013.	Calculate the entropy change `(J//mol K)` of the given reaction. The molar entropies `(J//K-mol)` are given in brackets after each substance: `2PbS(s)[19.2]+3O_(2)(g)[205.1]` ` to 2PbO(s)[66.5]+2O_(2)(g)[248.2]`A. `-113.5`B. `-168.3`C. `+72.5`D. `-149.2`
Answer» Correct Answer - B `Delta_(r )S=(2xx248.2)+(2xx66.5)` `-(2xx91.2)-3(205.1)` =-168.3 J

Discussion

4014.	An ideal gas `(gamma=1.4)` is used in a carnot cycle as a working substance . Substance. The efficency of the cycle, if as a result of an adiabatic expansion, the gas volume increases `2.75 times` is `[(1.5)^(2.5)=2.75]`A. `(100)/(3)%`B. `(200)/(3)%`C. `50%`D. `25%`
Answer» Correct Answer - A

Discussion

4015.	Caculate `DeltaG(kJ//"mole") "for the reaction at 300K"` `N_(2)(g)+O_(2)(g)Leftrightarrow 2NO(g)` at constant where partical pressure of `N_(2), O_(2)` and `NO are 10^(-1)"bar", 10^(-3)"bar"`. `DeltaH_(f)^(@)NO(g) at 300K =90.5kJ//"mole"and DeltaS_(f)^(@),NO(g) at 300K =12.5J//K"mole"and DeltaS_(f)^(@)NO(g)at 300k=12.5J//K"mole" "` `[2.303xxRxx300=5750J//"mole"]`A. `173.5kJ//"mole"`B. `185kJ//"mole"`C. `162 Kj//"mole"`D. `84.5kJ//"mole"`
Answer» Correct Answer - C

Discussion

4016.	Predict in which of the following, entropy increases or decreases.1. A liquid crystallizes into a solid2. Temperature of a crystallized solid is raised from 0 K to 115 K3. 2NaHCO3(s) → NaCO2CO3(s) + CO2(s) + H2O(s) 4. H2(s) → 2H(s)
Answer» 1. After freezing, the molecules attain an ordered state and therefore, entropy decreases. 2. At O K the constituent particles are in static form therefore, entropy is minimum. If the temperature is raised to 115 K particles begin to move and entropy increases. 3. Reactant, NaHCO₃ is solid. Thus, its entropy is less in comparison to product which has high entropy. 4. Here, one molecule gives two atoms. Thus, number of particles increases and this leads to more disordered form.

4016.

Predict in which of the following, entropy increases or decreases.1. A liquid crystallizes into a solid2. Temperature of a crystallized solid is raised from 0 K to 115 K3. 2NaHCO3(s) → NaCO2CO3(s) + CO2(s) + H2O(s) 4. H2(s) → 2H(s)

Answer»

1. After freezing, the molecules attain an ordered state and therefore, entropy decreases.

2. At O K the constituent particles are in static form therefore, entropy is minimum. If the temperature is raised to 115 K particles begin to move and entropy increases.

3. Reactant, NaHCO₃ is solid. Thus, its entropy is less in comparison to product which has high entropy.

4. Here, one molecule gives two atoms. Thus, number of particles increases and this leads to more disordered form.

Discussion

4017.	Predict in which of the following entropy of the system increases`//`decreases: (P) A liquid crystalizes into a solid (Q) Temperature of a crystalline solid is raised (R ) `2NaHCO_(3)(s)toNa_(2)CO_(3)(s)+H_(2)O(g)+CO_(2)(g)` (S) `H_(2)(g)to2H(g)`
Answer» (i) After freezing, the molecules attain an ordered state and therefore, entropy decreases. (ii) At 0 K, the contituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the lattice and system becomes more disordered, therefore entropy increases. (iii) Reactant, `NaHCO_(3)` is a solid and it has low entropy. Among products there are one solid and two gases. Therefore, the products represent a condition of higher entropy. (iv) Here one molecule gives two atoms i.e., number of particles increases leading to more disordered state. Two moles of H atoms have higher entropy than one mole of dihydrogen molecule

Discussion

4018.	Predict in which of the following entropy of the system increses`//`decreases: (P) A liquid crystalizes into a solid (Q) Temperature of a crystalline solid is raised (R ) `2NaHCO_(3)(s)toNa_(2)CO_(3)(s)+H_(2)O(g)+CO_(2)(g)` (S) `H_(2)(g)to2H(g)`A. increases in allB. decreases in (P) and (Q)C. Decreases in (P) onlyD. Increases in (P),(S) only
Answer» Correct Answer - C

Discussion

4019.	Predict in which of the following , entropy increases `//` decreases . (i) A liquid crystallizes into a solid (ii) Temperaure of a crystalline solid israised from 0Kto 115K (iii) `2 NaHCO_(3)(s) rarr Na_(2)CO_(3)(s) + CO_(2)(g) + H_(2)O(g)` (iv)` H_(2)(g) rarr 2H(g)`
Answer» (i) Decreases because a solid has ordered arrangement of constituent particles. (ii) Increases because at 0K,there is a perfect order of constituent particles. As temperature is increased , the particle start vibrating about their equilibrium position. Hence, a disorder sets in. (iii)Increases because the reactant is solid whereas products have gaseous substances. (iv) Increases because the products havelarger (double) number of particles.

Discussion

4020.	Predict the sign of entropy change in each of the following `:` (i) `H_(2) ( ` at 298 K, 1 atm ) `rarr H_(2)` at 298 K, 10 atm) (ii) `H_(2)O( ` at 298K, 1 atm ) ` rarr H_(2)O` ( at 330K, 1atm) (iii) `2NH_(4)NO_(3)(s) rarr 2N_(2)(g)+4H_(2)O(g) +O_(2)(g)` (iv) Crystallization of copper sulphate from its saturated solution. (v) `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)`
Answer» Entropy isa measure of randomness or disorder of a system. If randomness increases, entropy increases. (i) Gas at lower pressure has greater randomness than at high pressure ( compressed gas) at the same temperature. Hence, entropy decreases, i.e., `Delta S ` is -ve. (ii) Molecules at higher temperature have greater randomness at the same pressure . Hence, `Delta S` is `+` ve . (iii) Gaseous molecules have greater randomness than the solid . Hence, `Delta S` is `+ ` ve. (iv) `CuSO_(4)` crystals have ordered arrangement whereas ionsin the soluiton have greater randomness. Hence, `Delta S` is `-` ve. (v) There are 3 moles of two different gaseous reactants which have greater randomness than 2 moles of the only gaseous product, `SO_(3)` . Hence, randomness decrease, i.e., `Delta S` is `-` ve.

Discussion

4021.	Calculate the standard free energy change for the reaction, `H_(2)(g) + I_(2)(g) rarr 2HI(g), DeltaH^(@) = + 51.9 kJ` Given that the standard entropies `(S^(@))` of `H_(2), I_(2)` and HI are 130.6, 116.7 and 206.3 J `K^(-1) mol^(-1)` respectively. Predict whether the reaction is feasible at the standard state or not.
Answer» Correct Answer - `DeltaG^(@) = + 2640. 6 Jmol^(-1)` . Not feasible `Delta_(r)S^(@)= 2xxS_((HI))^(@) - [ S_((H_(2)))^(@) +S_((I_(2)))^(@)]= 2 xx 206.3 - ( 130.6+ 116.7) = 165.3 J mol^(-1)` `Delta_(r)G^(@) = Delta_(r)H^(@) - T Delta_(r)S^(@) = 51900 - 298 xx 165.3 = 2640.6 J mol^(-1)`

Discussion

4022.	Using the data `Delta_(f) H^(@) (Nf_(3), g) = - 114 kH mol^(-1)`, `Delta_(N -= N) H^(@) = 946 kJ mol^(-1)`, and `Delta_(f - f) H^(@) = 158 kJ mol^(-1)`, calculate the average bond enthalpy of `N - F` bond in `NF_(3)`. Strategy : First write the thermochemical equation corresponding to `Delta_(f) H^(@) (NF_(3), g)`: `(1)/(2) N_(2) (g) + (3)/(2) F_(2) (g) rarr NF_(3) (g)` Now define `Delta_(r) H^(@)` of this reaction in terms of bonds made and bonds broken. Notice the `1//2` mol of `N -= N` bonds and `3//2` mol of `F - F` bond are broken, while 3 mole of `N - F` bonds are formed each `NF_(3)` has three `N - F` bonds: we are given the blood enthalpies of `N -= N` and `F - F` bonds, while the average bond enthalpy of `N -F` bond is not known. Applying Eq. for the reaction, we can calculate this unknown.
Answer» According to Eq. we can write `Delta_(r) H^(@) = [(1)/(2) Delta_(N -= N) H^(@) + (3)/(2) Delta_(F - F) H^(@)] - [3 Delta_(N - F) H^(@)]` `- 114 = [(1)/(2) (946) + (3)/(2) (158)] - [3 Delta_(N - f) H^(@)]` Rearranging to solve for `Delta_(N - F) H^(@)`, we have `Delta_(N - F) H^(@) = ((473 + 237) + (144))/(3) = (824)/(3)` `275 kJ mol^(-1)`

Discussion

4023.	Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid`/ /`base is completely neutralized by base `//`acid in dilute solution . For Strong acid and strong base neutralization net chemical change is`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)` `Delta_(r)H^(@)=-55.84KJ//mol``DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base `DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)` If enthalpy of neutralization of `CH_(3)COOH` by NaOH is -49.86KJ`//`mol then enthalpy of ionization of `CH_(3)COOH` is:A. `5.98 KJ//mol`B. `-5.98 KJ//mol`C. `105.7 KJ//mol`D. None of these
Answer» Correct Answer - A

Discussion

4024.	Assuming a domestic refrigerator as a reversible heat engine working between melting point Of ice and the room temperature at 27°C, calculate the energy in joule that must be supplied to freeze 1Kg of water at 0°C.
Answer» Here T₁ = 27+273 =300K , T₂ =0 +273 = 273 Mass of water to be freezed , M = 1 Kg = 1000g Amount of heat that should be removed to freeze the water Q₂ = ML =1000X 80 cal = 1000X80 X4.2 =3.36 x 10⁵ J Now Q₁ = (T₁/T₂ )X Q₂ = (300/273)X3.36x10⁵ = 3.692 x10⁵ J Therefore energy supplied to freeze the water W =Q₁ – Q₂ = 3.693x10⁵ - 3.36 x10⁵ = 3.32 x10⁵ J

4024.

Assuming a domestic refrigerator as a reversible heat engine working between melting point Of ice and the room temperature at 27°C, calculate the energy in joule that must be supplied to freeze 1Kg of water at 0°C.

Answer»

Here T₁ = 27+273 =300K , T₂ =0 +273 = 273

Mass of water to be freezed , M = 1 Kg = 1000g

Amount of heat that should be removed to freeze the water

Q₂ = ML =1000X 80 cal

= 1000X80 X4.2 =3.36 x 10⁵ J

Now Q₁ = (T₁/T₂ )X Q₂ = (300/273)X3.36x10⁵ = 3.692 x10⁵ J

Therefore energy supplied to freeze the water

W =Q₁ – Q₂ = 3.693x10⁵ - 3.36 x10⁵

= 3.32 x10⁵ J

Discussion

4025.	Assuming the domestic refrigerator as reversible engine working between melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. (Given melting point of ice = 0°C L = 80 cal g-1)
Answer» Given : T₁ = 27 + 273 = 300 K T₂ = 0 + 273 = 273 K m = 1 kg = 1000 g; L = 80 cal g^-1 Heat to be removed, Q₂ = mL = 1000 × 80 cal = 8 × 10⁴ cal From the relation, \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2},\) \(Q_1=\frac{T_1}{T_2}\times Q_2\) \(=\frac{300}{273}\times 8\times10^4\) = 87912.1 cal Energy required to be supplied, W = Q₁ − Q₂ = (87912.1 – 80,000) cal = 7912.1 cal = 7912.1 × 4.2 J = 33230.8 J

4025.

Assuming the domestic refrigerator as reversible engine working between melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. (Given melting point of ice = 0°C L = 80 cal g-1)

Answer»

Given :

T₁ = 27 + 273 = 300 K

T₂ = 0 + 273 = 273 K

m = 1 kg = 1000 g; L = 80 cal g^-1

Heat to be removed,

Q₂ = mL

= 1000 × 80 cal

= 8 × 10⁴ cal

From the relation,

\(\frac{Q_1}{Q_2}=\frac{T_1}{T_2},\)

\(Q_1=\frac{T_1}{T_2}\times Q_2\)

\(=\frac{300}{273}\times 8\times10^4\)

= 87912.1 cal

Energy required to be supplied,

W = Q₁ − Q₂

= (87912.1 – 80,000) cal

= 7912.1 cal

= 7912.1 × 4.2 J

= 33230.8 J

Discussion

4026.	Two identical samples of gases are allowed to expand to the same final volume (i) isothermally (ii) adiabatically. Work done isA. More in the isothermal processB. More in the adiabatic processC. Neither of themD. Equal in both processes
Answer» Correct Answer - A

Discussion

4027.	A refrigerator is driven by `1000 W` electric motor having an efficiency of `60%`. The refrigerator is considerd as a reversible heat engine operating between `273K and 303K`. Calculate the time required by it to freeze `32.5 kg` of water at `0^(@)C`. Heat losses may be rejected. Latent heat of fusion of ice `= 336xx10^(3)J kg^(-1)`.
Answer» Power of motor `= 1000 W` Efficiency of motor `= 60%` `:.` Actual power availiable `= (60)/(100)xx1000W` `=600W` If the refrigerator take `t` seconds to freeze `32.5 kg` of water, then energy supplied `= 600xxt "joule"`. Heat drawn from sink of freeze `32.5 kg` of water `Q_(2)= mL= 32.5xx336xx10^(3)J` `=1.092xx10^(7)J` As `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))` `:. Q_(1)= (T_(1))/(T_(2))xxQ_(2)= (303)/(273)xx1.092xx10^(7)J` `=1.212xx10^(7)J` As `W=Q_(1)-Q_(2)` `:. W= (1.212-1.092) 10^(7)J= 0.12xx10^(7)J` As energy supplied = work done `:. 600 t= 0.12xx10^(7)` `t=(0.12xx10^(7))/(600)= 2000s= 33 min 20 s`

Discussion

4028.	A carnot cycle is performed by `1 mol e of air (gamma=1.4)` initially at `327^(@)C`. Each stage represents a compression or expansion in the ratio `1:6`. Calcultate (a) the lowest temperature (b) net work done during each cycle and (C ) efficiency of the engine. Take `R=8.31J mol e^(-1)K^(-1)`.
Answer» Here, `(V_(1))/(V_(2))= 1/6 , gamma= 1.4`, `T_(1)= 327^(@)C= 327+273= 600K` (a) For an adiabatic change, `T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)` `T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)= 600(1/6)^(1.4-1)` `=600xx0.4884= 293K= 293-273` `=20^(@)C` This is lowest temperature. (b) Net work done during each cycle by 1 mole of air. `W=RT_(1) "log"_(e) (V_(2))/(V_(1))- RT_(2) "log"_(e)(V_(3))/(V_(4))` `=R(T_(1)-T_(2)) "log"_(e) (V_(2))/(V_(1))( :. (V_(3))/(V_(4))=(V_(2))/(V_(1)))` `= 2.303R (T_(1)-T_(2))"log"_(10) (V_(2))/(V_(1))` `=2.303xx8.31(600-293)log_(10)6` `W= 2.303xx8.31xx307xx0.7782` `=4572.2J` (C ) `eta= 1 -(T_(2))/(T_(1))= 1 - (293)/(600)= (307)/600` `=0.512= 51.2%`

Discussion

4029.	`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with `100 cm^(3) 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.
Answer» The initial average temperature of the acid and the base `=(299.95+299.75)/(2) = 299.85K` Rise in temperature `= (302.65-299.85) = 2.80K` Heat evolved during neutralisation `= (100+100+44) xx 4.184 xx 2.8 = 2858.5J` `:.` Ethalpy of neutralisation `=- (2858.5)/(100) xx 1000xx(1)/(0.50)` `=- 57.17 kJ`

Discussion

4030.	The amount of heat exchanged with the surrounding at constant temperature and pressure is called …(a) ΔE (b) ΔH (c) ΔS (d) ΔG
Answer» Answer: (b) ΔH

Discussion

4031.	For the reaction ,`CH_(4)+Cl_(2)toCH_(3)Cl+HCl` which expression given `DeltaH`? `{:("Bond dissociation",Kj.mol^(-1)),(C-H,413),(C-Cl,328),(Cl-Cl,242),(H-Cl,431):}`A. `DeltaH=(413+328)-(242+431)`B. `DeltaH=(413-328)-(242-431)`C. `DeltaH=(413-242)-(328-431)`D. `DeltaH=(413+242)-(328+431)`
Answer» Correct Answer - d

Discussion

4032.	Given chemical equations for these reactions `S(s)+O_(2)(g)toSO_(2)(g)DeltaH=-29608KJxxmol^(-1)` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)Delta=-285.8KJxxmol^(-1)` `H_(2)(g)+S(s)toH_(2)S(g)DeltaH=-20.6Kjxxmol^(-1)` What is the value of `DeltaH` for the reaction below? `2H_(2)s(g)+3O_(2)(g)to 2H_(2)O(l)+2SO_(2)(g)`A. `-603.2KJxxmol^(-1)`B. `-562.0KJxx,mol^(-1)`C. `-1206.4KJxxmol^(-1)`D. `-1124.0 Kjxxmol^(-1)`
Answer» Correct Answer - d

Discussion

4033.	If the bond dissociation energies of `XY`,`X_(2)` and `Y_(2)` are in the ratio of `1:1:0.5` and `DeltaH_(f)` for the formation of `Xy` is `-200 KJ//mol`. The bond dissociation energy of `X_(2)` will be `:-`A. 200 kJ `mol^(-1)`B. 100 kJ `mol^(-1)`C. 800 kJ `mol^(-1)`D. 300 kJ `mol^(-1)`
Answer» Correct Answer - C Given `(1)/(2)X_(2)+(1)/(2)(Y)/(2)rarr XY , Delta H=-200 kJ mol^(_1)` Let the bond dissociation energy of XY = a The ratio of bond dissociation energies of `X_(2), Y_(2)` and XY be `a:(a)/(2):a` (the ratio given) For given equation : `Delta H=(B.E.)_(R )-(B.E.)_(P)` `-200=[(1)/(2)xx a+(1)/(2)xx(a)/(2)]-a` `therefore a=800 kJ mol^(-1)`

Discussion

4034.	Bond disociation enegry of `XY,X_(2)` and `Y_(2)` (all diatomic molecules) are in the ratio of `1: 1:0.5` and `DeltaH_(f) of XY` is `-100 kJ mol^(-1)`. The bond dissociation enegry of `X_(2)` is `100x`. Find the value of `x`.
Answer» `(1)/(2)X_(2) +(1)/(2)Y_(2) rarr XY, DeltaH = - 100 kJ` Let the bond dissociation energy of `X_(2),Y_(2)`, and `XY` be `a: (a)/(2):a` (the given ratio) `kJ mol^(-1)` respectively. `:. (a)/(2)+(a)/(4) -a = - 100` `:. A= 400` `100 x = 400` `x = 4`

Discussion

4035.	Use the thermochemical data given to calculate `Deltah_(f)^(@)` for `N_(2)O_(5)(g)` in KJ/`mol^(-1)` `N_(2)(g)+O_(2)(g)to 2NO(g) DeltaH^(@)=+180.5KJ` `2NO(g)+O_(2)(g)to2NO_(2)(g) DeltaH^(@)=-114.1Kj` `4NO_(2)(g)+O_(2)(g)to 2N_(2)O_(5) DeltaH^(@)=-110.2Kj`A. `-332.8`B. `-43.8`C. `11.3`D. `22.6`
Answer» Correct Answer - c

Discussion

4036.	The heat enegry required to ionise the following molecules is given as follows: `N_(2)(g) overset(DeltaH_(1))rarr N_(2)^(o+(g),O_(2)(g) overset(DeltaH_(2))rarrO_(2)^(o+(g)` `Li_(2)(g) overset(DeltaH_(3))rarrLi_(2)^(o+(g),C_(2)(g) overset(DeltaH_(4))rarrC_(2)^(o+(g)` Arrange the heat terms in decreasing order of enegry:A. `DeltaH_(1) gt DeltaH_(3) gt DeltaH_(2) gt DeltaH_(4)`B. `DeltaH_(2) gt DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4)`C. `DeltaH_(3) gt DeltaH_(4) gt DeltaH_(1) gt DeltaH_(2)`D. `DeltaH_(3) gt DeltaH_(1) gt DeltaH_(4) gt DeltaH_(2)`
Answer» Enthalpy of ionisation depends upon the size of molecules. Larger the size lesser will be ionisation energy.

Discussion

4037.	Which of the following has highest heat of hydrogenation:A. But-I-eneB. `c`is-Bur-2-eneC. trans-Bur-2-eneD. Isobutane
Answer» More stable is an alkene, lesser is its heat of hydrogenation

Discussion

4038.	Calculate the maximum work done in expanding `16g` of oxygen at `300K` occupying volume of `5dm^(3)` and isothermally untill the volume become `25dm^(3)`?
Answer» Since, it is the maximum work, thus, the process is reversible. `nO_(2) = (w)/(m) = (16)/(32),R = 8.314J`, `T = 300K, V_(1) = 5dm^(3), V_(2) = 25dm^(3)` `w =- 2.303nRT log_(10). (V_(2))/(V_(1))` `=- 2.303 xx (16)/(32) xx 8.314 xx 300 log_(10). (25)/(5)` `= - 2.01 xx 10^(3)J`

Discussion

4039.	`16g` oxygen gas expands at `STP` to occupy double of its oxygen volume. The work done during the process is:A. `260 kcal`B. `180 kcal`C. `130 kcal`D. `272.84` kcal
Answer» Correct Answer - D At `STP 16 g O_(2)` or `1/2` mole `O_(2)` will occupy `11.2` litre. Thus if volume is doubled, it means `(V_(2)-V_(1))=22.4-11.2=11.2 litre` `W=Pxx(V_(2)-V_(1))=1xx11.2 litre -atm` `=(1xx11.2xx2)/0.0821 cal =272.84 kcal`

Discussion

4040.	A piece of zinc at a tempreature of `20.0^(@)` C weighing 63.38 g is drooped into 180 g boiling water `(T= 100^(@)C`) . The specific heat of zinc is `0.400 Jg^([email protected]) C` and that of water is `4.20 Jg^([email protected]) C` . What is the finial comman temperature reached by both the zinc and water ?A. `97.3^(@)`CB. `33.4^(@)`CC. `80.1^(@)`CD. `60.0^(@)`C
Answer» Correct Answer - A

Discussion

4041.	For the given reactions `SiO_(2) +4HF rarr SiF_(4) +2H_(2)O, DeltaH =- 10.17 kcal` `SiO_(2) +4HCI rarr SiCI_(4) +2H_(2)O, DeltaH = 36.7 kcal` It may be concluded thatA. `HF` will attack `SiO_(2)` and `HCI` will notB. `HCI` will attack `SiO_(2)` and `HF` will notC. `HF` and `HCI` both attack `SiO_(2)`D. None attacks `SiO_(2)`
Answer» `DeltaG = DeltaH - T DeltaS =- ve` All exothermic reactions are spontaneous, hence `HF` will attack `SiO_(2)`.

Discussion

4042.	A piece of zinc at a temperature of `20.0^(@)` C weighing 63.38 g is dropped into 180 g boiling water `(T= 100^(@)C`) . The specific heat of zinc is `0.400 Jg^([email protected]) C` and that of water is `4.20 Jg^([email protected]) C` . What is the final common temperature reached by both the zinc and water ?A. `97.3 .^(@)C`B. `33.4 .^(@)C`C. `80.1 .^(@)C`D. `60.0 .^(@)C`
Answer» Correct Answer - A `65.38 xx 0.4 (T - 293) = 180 xx 4.2 (373 - T)`

Discussion

4043.	A piece of zinc at a tempreature of `20.0^(@)` C weighing 63.38 g is drooped into 180 g boiling water `(T= 100^(@)C`) . The specific heat of zinc is `0.400 Jg^([email protected]) C` and that of water is `4.20 Jg^([email protected]) C` . What is the finial comman temperature reached by both the zinc and water ?A. `97.3^(@)`CB. `33.4^(@)`CC. `80.1^(@)`CD. `60.0^(@)`C
Answer» Correct Answer - A

Discussion

4044.	Freezing up liquid in a system then:A. `q=0`B. `q gt 0`C. `q lt 0`D. `q gt 0` or `q lt 0` (depending on the nature of liquid)
Answer» Correct Answer - C

Discussion

4045.	A piece of zinc at a temperature of `20^(@)C` weighing `65.38 g` is dropped into `180 g` of boiling water `(T=100^(@)C)` The specific heat of zince is `0.4 J g^(-1)C^(-1)` and that of water is `4.2J g^(-1^(@))C^(-1)`. What is the final common temperature reached by both the zinc and water ?A. `97.3^(@)C`B. `33.4^(@)C`C. `80.1^(@)C`D. `60.0^(@)C`
Answer» Correct Answer - A Let final common temperature is `T_(f)` Heat gained by Zinc piece`=` Heat lost by water `0.4 (T_(f)-20)xx65.38=4.2(100-T_(f))xx180 rArrT_(f)=97.3^(@)C`

Discussion

4046.	A mono-atomic ideal gas of two moles is taken through a cyclic process starting from `A` as shwon in the figure below. The volume ratios are `V_(B)//V_(A) = 2` and `V_(D)//V_(A) = 4`. If the temperature `T_(A)` at `A` is `27^(@)C`. Calculate a. The temperature of gas at `B`. b. Heat absorbed or evolved in each process. c. Total wrok done in cyclic process.
Answer» `A rarrB:` (It is isobaric procee) `(V_(A))/(T_(A)) - (V_(B))/(T_(B))` `:. T_(B) = (V_(B))/(V_(A)) xx T_(A) = 2 xx 300 = 600K` `q_(AB) = nC_(P)Delta_(T) = 2xx (5)/(2)R DeltaT` `= 2 xx (5)/(2) xx2xx300 = 3000cal` `B rarrC`: (Isothermal process) `DeltaU = 0` `:. q_(BC) = w = 2.303nRT log_(10) ((V_(C))/(V_(B)))` `= 2.303 xx2xx2xx600 log.(4)/(2)` `C rarrD:` (Isochric process) `q_(CD) = nC_(v) DeltaT = 2xx (3)/(2) xx 2 (-300) =- 1800cal` `D rarA`: (Isochoric procee) `q_(DA) = 2.303 nRT_(A) log_(10). (V_(A))/(V_(D))` `= 2.303 xx 2xx 2xx200 log.(1)/(4)` Total heat change `= 3000 + (1.663 xx 10^(3)) - 1800 - (1.663 xx 10^(3)) = 1200 cal` Work done `=- 1200 cal`

Discussion

4047.	For ag, Cp (`JK^(-1mol^(-1`) is given be `24+0.006 T//K.` Calculate `DeltaH` (in KJ) if 3 mole of silver are raided from `27^(@)C` to its meltiong point `927^(@)C` under 1 atm pressure.
Answer» Correct Answer - 77

Discussion

4048.	For his 18th birthday in February Peter plants to turn a hut in the garden of his parents into a swimming pool with an artifical beach. In order to estimate the consts for heating the water and the house , peter obtains the data for the natural gas combustion and its price. The desity of natural gas is `0.740 g L^(-1) (1.013 xx 10^(5)Pa, 25^(@)C)` specified by PUC , the public utility company. (a) Calculate the amount of methane and ethane (in moles) in `1.00m^(3)` of natural gas (natural gas, methane and ethane are not ideal gases). (b) Calculate the combustion energy which is released as thermal energy during the buring of `1.00 m^(3)` of natural gas under standard conditions assuming that all products are gaseous .(if you do not have the amount from 1.2 a) assume that `1.00 m^(3)` natural gas corresponds to 40.00 mol natural gas) According to the PUC the combustion energy will be 9.981K Wh per `m^(3)` of natural gas if all products are gaseous . How large is the deviation (in percent) from the value you obtained in b? The swimming pool inside the house is 3.00 m wide ,5.00 m long and 1.50 m deep (below the floor) . The tap water temperature is `8.00^(@)C` and the air temperature in the house (dimensions given in the figure) is `10.0^(@)C.` Assuming a water density of `rho=1.00 Kg L^(-1)` and air behaving like an ideal gas.
Answer» Correct Answer - Amount of methane and ethane in `1 m^(3)` natural gas : `m= gt xx V=0.740 g dm^(-3) xx 1000 dm^(3) =740 g` `M_(av) = sum_(i) xx (i) M(ii)=(0.0024 xx44.01 g "mol"^(-1)) + (0.0134 xx28.02 g "mol"^(-1))+(0.9732 xx 16.05 g "mol"^(-1)) + (0.011 xx 30.08 g "mol"^(-1))=16.43 g "mol"^(-1)` `n_("tol") =m(Mav)^(-1) = 740 gxx (16.43 g//mol)^(-1)=45.83 "mol"` n(i) = x(i) . n_("tot") `n(Ch_(4)) = x(CH_(4)) xx n_("tot") =09732xx 45.04 "mol"=0.495 "mol"` `n(C_(2)H_(6))=x(C_(2)H_(6)xxn_("tot")=0.0110 xx 45.04 "mol" =0.495 "mol"` (b) Energy of combustion , deviation: Ecomb.`(H_(2)O(g)) =sum_(i)n(i) Delta_(c)H-(i)=` `=43.83 "mol" xx (-802.5 "kj mol"^(-1)) + 0.495 "mol" xx 0.5 xx(-2856.8 KJ "mol"^(-1))` `=-35881 KJ` `E_("comb")(H_(2)O(g))=-35881 KJ` Deviation from PUC EPUC `(H_(2)O(g)) =9.981 K Whm^(-3) xx 1 m^(3)xx 3600 KJ (kWh)^(-1) =35932 KJ` Deviation: `DeltaE=(E_("comb")(H_(2)O(g))-E_(PUC)(H_(2)(g)) xx 100%xx [Ecomb.(H2)(g))]^(-1)` `=(35881 KJ -35932 KJ ) xx 100% xx (35881 KJ )^(-1)=.^(-)C0.14%`

Discussion

4049.	For his 18th birthday in February Peter plants to turn a hut in the garden of his parents into a swimming pool with an artifical beach. In order to estimate the consts for heating the water and the house , peter obtains the data for the natural gas combustion and its price. Calculate the energy (in MJ) which is needed to maintain the temperature inside the house at `30.0^(@)C` during the party (12 hours) . `1.00 m^(3)` of natural gas as delivered by PUC costs 0.40 € and `1.00 k` Wh of electricity costs 0.137 € . The rent for the equipment for gas heating will cost him about 150.00 € while the corresponding electrical heaters will only cost 100.00 €.
Answer» Correct Answer - Energy for maintaining the temperatue: Surface area of the house: Ahouse `=3m xx46 m + 8m xx 2m + ((2m)^(2)) + (4m)^(2))^(1//2) xx 2xx 15m=288.16 m^(2)` `Heat conductivity : `lambda_("wall") =1J (s KM)^(-1)` Energy flux along a temperature gradient (wall thickness d=0.2 m) `J=Eloss (Axx Deltat)^(-1) = lambda_("wall") DeltaT d^(-1)` `E_("loss") = 288.16 m^(2) xx (12.60.60s) xx 1 J (s km)^(-1) xx 25 K xx (0.2m) ^(-1) =1556 MJ` `E_("loss") =1556 MJ`

Discussion

4050.	Find `Delta H_(f)^(@)` of `Fe_(2)O_(3)(s)` if the standard heat reaction `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g)` is -6.6 kcal. Given `Delta H_(f)^(@)CO(g)=-26.4 kcal. Delta H_(f)^(@)CO_(2)(g)=-94 kcal`.
Answer» For following thermochemical equation `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g), Delta H=-6.6 kcal` Thus, `Delta H_("of reaction") = [Sigma Delta H_(f)^(@)" of " Fe(s)+3 Delta H_(f)^(@)" of " CO_(2)(g)-(Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)+3xxDelta H_(f)^(@)" of " CO(g))]` `-6.6 = [(2xx0)+(3xx-94)-(Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)+3xx-26.4)]kcal` `-6.6=[0-282-Delta H_(f)^(@) " of " Fe_(2)O_(3)" of " Fe_(2)O_(3)(s)+79.2]` `therefore Delta H_(f)^(@)" of " Fe_(2)O_(3)(s)=-196.2 kcal // mol`

Discussion

Explore topic-wise InterviewSolutions in .

For the hypothetical reaction `A_(2)(g)+B_(2)(g) to 2AB_(g)` `DeltaG_(r)^(@)" and " DeltaS_(r)^(@)" are " 20KJ//moland-20JK^(-1)mol^(-1) "respectively at" 200K` `Delta_(r)C_(r)" is "JK^(-1) " then " DeltaH_(r)^(@) "at" 400K is `A. `20KJ//mol`B. `7.98KJ//mol`C. `28kj//mol`D. None of these

For the hypothetical reaction `A_(2)(g)+B_(2)(g) to 2AB_(g)` `DeltaG_(r)^(@)" and " DeltaS_(r)^(@)" are " 20KJ//moland-20JK^(-1)mol^(-1) "respectively at" 200K` `Delta_(r)C_(p)" is "20 JK^(-1) " then " DeltaH_(r)^(@) "at" 400K is `A. 20 kJ/molB. `7.98` kJ/mosC. 28 kJ/molD. None of these

If one mole of an ideal gas `(C_(p.m.)=(5)/(2)R)` is expanded isothermally at 300K until it’s volume is tripled, then change in entropy of gas is:A. zeroB. infinityC. `(5)/(2)R In 3`D. `R In 3`

If one mole of an ideal gas `(C_(p.m.)=(5)/(2)R)` is expanded isothermally at 300K until it’s volume is tripled, then change in entropy of gas is:A. zeroB. infinityC. `(5)/(2)R ln 3`D. `R ln 3`

One mole of an ideal gas is expanded isothermally at `300` K until its volume is tripled . Find the values of `DeltaS_("sys")` under the condition.

Calculate the entropy change `(J//mol K)` of the given reaction. The molar entropies `(J//K-mol)` are given in brackets after each substance: `2PbS(s)[19.2]+3O_(2)(g)[205.1]` ` to 2PbO(s)[66.5]+2O_(2)(g)[248.2]`A. `-113.5`B. `-168.3`C. `+72.5`D. `-149.2`

An ideal gas `(gamma=1.4)` is used in a carnot cycle as a working substance . Substance. The efficency of the cycle, if as a result of an adiabatic expansion, the gas volume increases `2.75 times` is `[(1.5)^(2.5)=2.75]`A. `(100)/(3)%`B. `(200)/(3)%`C. `50%`D. `25%`

Predict in which of the following, entropy increases or decreases.1. A liquid crystallizes into a solid2. Temperature of a crystallized solid is raised from 0 K to 115 K3. 2NaHCO3(s) → NaCO2CO3(s) + CO2(s) + H2O(s) 4. H2(s) → 2H(s)

Predict in which of the following entropy of the system increases`//`decreases: (P) A liquid crystalizes into a solid (Q) Temperature of a crystalline solid is raised (R ) `2NaHCO_(3)(s)toNa_(2)CO_(3)(s)+H_(2)O(g)+CO_(2)(g)` (S) `H_(2)(g)to2H(g)`

Predict in which of the following , entropy increases `//` decreases . (i) A liquid crystallizes into a solid (ii) Temperaure of a crystalline solid israised from 0Kto 115K (iii) `2 NaHCO_(3)(s) rarr Na_(2)CO_(3)(s) + CO_(2)(g) + H_(2)O(g)` (iv)` H_(2)(g) rarr 2H(g)`

Assuming a domestic refrigerator as a reversible heat engine working between melting point Of ice and the room temperature at 27°C, calculate the energy in joule that must be supplied to freeze 1Kg of water at 0°C.

Assuming the domestic refrigerator as reversible engine working between melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. (Given melting point of ice = 0°C L = 80 cal g-1)

Two identical samples of gases are allowed to expand to the same final volume (i) isothermally (ii) adiabatically. Work done isA. More in the isothermal processB. More in the adiabatic processC. Neither of themD. Equal in both processes

A carnot cycle is performed by `1 mol e of air (gamma=1.4)` initially at `327^(@)C`. Each stage represents a compression or expansion in the ratio `1:6`. Calcultate (a) the lowest temperature (b) net work done during each cycle and (C ) efficiency of the engine. Take `R=8.31J mol e^(-1)K^(-1)`.

`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with `100 cm^(3) 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.

The amount of heat exchanged with the surrounding at constant temperature and pressure is called …(a) ΔE (b) ΔH (c) ΔS (d) ΔG

For the reaction ,`CH_(4)+Cl_(2)toCH_(3)Cl+HCl` which expression given `DeltaH`? `{:("Bond dissociation",Kj.mol^(-1)),(C-H,413),(C-Cl,328),(Cl-Cl,242),(H-Cl,431):}`A. `DeltaH=(413+328)-(242+431)`B. `DeltaH=(413-328)-(242-431)`C. `DeltaH=(413-242)-(328-431)`D. `DeltaH=(413+242)-(328+431)`

If the bond dissociation energies of `XY`,`X_(2)` and `Y_(2)` are in the ratio of `1:1:0.5` and `DeltaH_(f)` for the formation of `Xy` is `-200 KJ//mol`. The bond dissociation energy of `X_(2)` will be `:-`A. 200 kJ `mol^(-1)`B. 100 kJ `mol^(-1)`C. 800 kJ `mol^(-1)`D. 300 kJ `mol^(-1)`

Bond disociation enegry of `XY,X_(2)` and `Y_(2)` (all diatomic molecules) are in the ratio of `1: 1:0.5` and `DeltaH_(f) of XY` is `-100 kJ mol^(-1)`. The bond dissociation enegry of `X_(2)` is `100x`. Find the value of `x`.

Use the thermochemical data given to calculate `Deltah_(f)^(@)` for `N_(2)O_(5)(g)` in KJ/`mol^(-1)` `N_(2)(g)+O_(2)(g)to 2NO(g) DeltaH^(@)=+180.5KJ` `2NO(g)+O_(2)(g)to2NO_(2)(g) DeltaH^(@)=-114.1Kj` `4NO_(2)(g)+O_(2)(g)to 2N_(2)O_(5) DeltaH^(@)=-110.2Kj`A. `-332.8`B. `-43.8`C. `11.3`D. `22.6`

Which of the following has highest heat of hydrogenation:A. But-I-eneB. `c`is-Bur-2-eneC. trans-Bur-2-eneD. Isobutane

Calculate the maximum work done in expanding `16g` of oxygen at `300K` occupying volume of `5dm^(3)` and isothermally untill the volume become `25dm^(3)`?

`16g` oxygen gas expands at `STP` to occupy double of its oxygen volume. The work done during the process is:A. `260 kcal`B. `180 kcal`C. `130 kcal`D. `272.84` kcal

Freezing up liquid in a system then:A. `q=0`B. `q gt 0`C. `q lt 0`D. `q gt 0` or `q lt 0` (depending on the nature of liquid)

For ag, Cp (`JK^(-1mol^(-1`) is given be `24+0.006 T//K.` Calculate `DeltaH` (in KJ) if 3 mole of silver are raided from `27^(@)C` to its meltiong point `927^(@)C` under 1 atm pressure.

Find `Delta H_(f)^(@)` of `Fe_(2)O_(3)(s)` if the standard heat reaction `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g)` is -6.6 kcal. Given `Delta H_(f)^(@)CO(g)=-26.4 kcal. Delta H_(f)^(@)CO_(2)(g)=-94 kcal`.