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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 451. |
A particular state of system is arrived at starting from a given state in two different ways `(1)` following reversible path and `(2)` irreversible path. Which of the following relations would be correct if the processes are isothermal ?A. `DeltaS_(rev)ne DeltaS_(irrev)`B. `Deltaq_(rev) = Deltaq_(irrev)`C. `DeltaS_(rev) = DeltaS_(irrev)=(Deltaq_(rev))/(T)`D. `DeltaS_(irrev)=(Deltaq_(irrev))/(T)ne DeltaS_(rev)` |
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Answer» Correct Answer - B Since entropy is a state function, bence, `DeltaS_(rev)=DeltaS_(irrev)=(Deltaq_(rev))/(T)` |
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| 452. |
For a spontaneous reaction, `DeltaG,` equilibrium constant (K) and `E_(cell)^(@)` will be respectively:A. `-ve`, `gt 1`, `+ve`B. `+ve`, `gt 1`, `-ve`C. `-ve`, `lt 1`, `-ve`D. `-ve`, `gt 1`, `-ve` |
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Answer» Correct Answer - A For a spontaneous reaction `DeltaG` is `-ve` `DeltaG=-RT ln K` or `ln K` is `+ve implies K gt1` Now `DeltaG=-nFE_("cell")` As `DeltaG` is `-ve` for a spontaneous reaction. `E_("cell")` for such a reaction is `+ve`. |
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| 453. |
For a spontaneous reaction, `DeltaG` , equilibrium constant `(K)` and `E_(cell)^(0)` will be respectively:A. `-ve,gt1,+ve`B. `+ve,gt1,-ve`C. `-ve,lt1,-ve`D. `-ve,gt1,-ve` |
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Answer» Correct Answer - C For `z` spontaneous reaction `DeltaG=-ve` Also `Kgt1` ans `E_(cell)^(0)=+ve` |
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| 454. |
For the water gas reaction `C(s) + H_(2)O(g) hArr CO(g) + H_(2)(g)` At `1000K` , the standard Gibbs free energy change of the reaction is `-8.314KJ//mol` . Therefore, at `1000K` the equilibrium constant of the above water gas reaction isA. 1B. `10`C. `(1)/(e)`D. `2.718` |
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Answer» Correct Answer - A `Delta_(r)G^(@)=-8.314xx10^(3)` `:. -DeltaG_(r)G^(@)=8.314xx10^(3)=RT` In `K_(aq)` or `8.314xx10^(3)=8.314xx10^(3)` In `K_(aq)` or In `K_(eq)=1="In" e` or `K_(eq)=e=2.718` |
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| 455. |
The heat of formation of the compound in the following reaction is `H_(2)(g)+Cl_(2)(g)to2HCl(g)+44kcal`A. `-44 kcal mol^(-1)`B. `-22 kcal mol^(-1)`C. `11 kcal mol^(-1)`D. `-88 kcal mol^(-1)` |
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Answer» Correct Answer - B `DeltaH_(f)` refer to formation of `1` mole of `HCl(g)` in this case. |
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| 456. |
The calorific value of fat isA. less than that of carbohydrates and proteinB. less than that of protein but more than carbohydratesC. less than that of carbohydrates but more than that of protein.D. more than that of carbohydrate and protein. |
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Answer» Correct Answer - D Calorific value of fat, carbohydrate and proteins are respectively `9.0`, `4.0` and `4.0 kcal//g`. |
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| 457. |
The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `261KJ`B. `+103KJ`C. `+261KJ`D. `-103KJ` |
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Answer» Correct Answer - B `DeltaH^(@)=-2xxe_(H-Br)+e_(H-H)+e_(Br-Br)` `=-2xx364+344+192=-103KJ` |
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| 458. |
At constant temperture and pressure which one of the following statements is correct for the reaction ? `CO(g)+ 1/2O_(2)(g) to CO_(2)(g)`A. `DeltaH = DeltaE`B. `DeltaH lt DeltaE`C. `DeltaH gt DeltaE`D. `DeltaH` is independent of physical state of reactant |
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Answer» Correct Answer - b As we know , `DeltaH=(- 1348.9 xx10^(3))+ 11xx2xx298=-1342.34 kcal` for the reaction , `CO(g)+1/2O_(2)(g) to CO_(2)(g)` ` Deltan=1-(1+1/2) =-1/2` `DeltaH=DeltaE-1/2 RT` `DeltaH lt DeltaE` |
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| 459. |
Enthalpy of an exothermic reaction is alwaysA. positiveB. negativeC. zeroD. may be positive or negative. |
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Answer» Correct Answer - B For exothermic reactions , `sumH_(("Products")) lt sumH_(("Reactants"))` Thus, `DeltaH` which is `(sumH_("Products") - sumH_("Reactants"))` is negative. |
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| 460. |
The reaction is sontaneous if the cell potential isA. PositiveB. NegativeC. ZeroD. Infinite |
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Answer» Correct Answer - A For a reaction to be spontaneous `DeltaG` should be negative. `DeltaG=-nFE^(@)` cell From this reaction, `DeltaG` will be negative when `E^(@)` cell is positive. |
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| 461. |
The equilibrium constant of a reaction is `0.008` at `298 K`. The standard free energy change of the reaction at the same temperture isA. `+11.96 kJ`B. `-11.96 kJ`C. `-5.43 kJ`D. `-8.46 kJ` |
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Answer» Correct Answer - A `DeltaG^(@)=-2.303 RT log K` `=-2.303xx8.314xx298xxlog0.008` `=-2.303xx8.314xx298xx(-2.0969)` `=11964.6 J=11.96 kJ`. |
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| 462. |
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 `(DeltaG^(@)=0)`B. free energy change is less than zero `(DeltaG lt 0)`C. standard free energy change is less than zero `(DeltaG^(@)lt0)`D. standard free energy change is more than zero `(DeltaG^(@)gt0)` |
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Answer» Correct Answer - A For a reversible process at equilibrium. `DeltaG^(@)=0` (at `1` atm and `298 K`) |
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| 463. |
How much energy is released when 6 mole of octane is burnt in air ? Given `DeltaH_(f)^(@)` for `CO_(2)(g),H_(2)O(g)` and `C_(8)H_(18)(l)` respectively are `-490,-240` and `+160KJ//mol`A. `-6.2 MJ`B. `-37.4 MJ`C. `-35.5 MJ`D. `-20.0 MJ` |
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Answer» Correct Answer - B `C_(8)H_(18)+(15)/(2)O_(2)to8CO_(2)9H_(2)O` `DeltaH=8xxDeltaH_(f)CO_(2)+9xxDeltaH_(f)H_(2)O-DeltaH_(f)C_(8)H_(18)` `=8xx(-490)+9xx(-240)-(160)` `=-3920-2160-160=-6240 kJ` Heat produced on burning `6` moles of octane `=-6240 kJxx6=-37440 kJ=-37.440 MJ` |
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| 464. |
Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.A. `-6200 J`B. `-306 J`C. `-3100 J`D. `-1550 J` |
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Answer» Correct Answer - C Volume of `1` mol of water at `373 K`, `1` atm `(V_(1))=18cm^(3)=0.018 L` Volume of `1` mol of water in vapour state at `373 K`, `1 "atm"=(RT)/(P)` `=(0.0821xx373)/(1)=30.62 L` Now `w=-P(V_(2)-V_(1))` `=1(30.62-0.018)` `=-30.6 L-"atm"` `=-30.6xx101.3=-3100 J` |
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| 465. |
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) gtpK_(a)(CH_(3)COOH)`C. `pK_(a)(HCN) ltpK_(a)(CH_(3)COOH)`D. `pK_(a)(HCN) =((45.50)/(2.1))pK_(a)(CH_(3)COOH)` |
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Answer» Correct Answer - B `HCN` has greater heat of ionization, and is a weak acid. Thus it has high `pK_(a)` value. Remember : Smaller the `pK_(a)` value, stronger the acid. |
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| 466. |
Calculate the entropy of ideal mixing when 2 m oles of `N_(2)`, 3 moles of `H_(2) and 2` moles of `NH_(3)` are mixed at constant temperature, assuming no chemical reaction is occurring. |
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Answer» Correct Answer - `62.80JK^(-1)` Use the relation, `DeltaS=-Rsumn_(i)log_(e)x_(i)` when `n_(i)`=no. of moles of component `x_(i)=` mole fraction of the component. |
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| 467. |
Internal energy is sum ofA. parthly potential and partly KineticB. totally KineticC. totally potentialD. none of these |
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Answer» Correct Answer - C The exact value of internal energy is known asa it includes all type of energies of molecules constituting the given mass of matter such as translational, vibrational, rotational. The kenitic and potential energy of the muclei and electron within the individual molecules and the manner in which the `E=E_(translational)+E_(rotational)+E_(vibrational)` Thus, we can say that internal energy is partly potential and partly kinetic. |
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| 468. |
The heat of combustion of graphite and carbon monoxide respectively are `393.5 kJ mol^(-1)` and `283 kJ mol^(-1)`. Thus, heat of formation of carbon monoxide in `kJ mol^(-1)` isA. `+172.5`B. `-110.5`C. `-10760`D. `-676.5` |
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Answer» Correct Answer - B `DeltaH_(f)(CO)=[DeltaH_(comb)"of"C]-[DeltaH_(comb)"of"CO]` `=-393.5-(-283)=-110.5 kJ mol^(-1)` |
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| 469. |
For the transition `C_("(diamond)") rarr C_("(graphite)"), DeltaH=-1.5 kJ`. It follows thatA. graphite is stabler than diamondB. diamond is stabler than graphiteC. graphite is endothermicD. diamond is exothermic |
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Answer» Correct Answer - A Since graphite is formed with the release of energy, it has less energy and is more stable. |
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| 470. |
Given (I) `C`(diamond) `+O_(2)(g)rarrCO_(2)(g),DeltaH^(@)=-92.0Kcal mol^(-1)` (II) `C`(graphite) + `O_(2)(g)rarrCO_(2)(g),DeltaH^(@)=-96.0Kcal mol^(-1)`A. `2.907KcalK^(-1)`B. `2.013KcalK^(-1)`C. `305.4calK^(-1)`D. `-2.013KcalK^(-1)` |
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Answer» Correct Answer - A Eq . `(I)` and Eq . `(II)` give C(diamond)`rarr`(graphite) `DeltaH^(@)=-02-(-96)=+4.0Kcal mol^(-1)` Moles of diamond `=(2.6xx1000g)/(12gmol^(-1))=216.6mol` `DeltaH^(@)` (to convert `200mol` of diamond into graphite) `=216.6xx4.0Kcal=866.4Kcal` `:. DeltaS=(DeltaH^(@))/(T)=(866.4Kcal)/(298K)=2.907KcalK^(-1)` |
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| 471. |
The standared heat of formation listed for gaseous `NH_(3)` is `-11.0kcal mol^(-1)` at `298K`. Given that at `298k`, the constant pressure heat capacities of gaseous `N_(2),H_(2)`, and `NH_(3)` are, respectively, `7.0, 6.0` and `8.0 cal mol^(-1)`. Determine `DeltaH^(Theta)underset(298K)` and `DeltaH_(773K)` for the reactions: `(1)/(2)N_(2)(g) +(3)/(2)H_(2)(g) rarr NH_(3)(g)` |
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Answer» `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)toNH_(3)(g)` `DeltaH_(298K)=sum(H_(f))_(P)-sum(H_(f))_(R)=(-11.02-0)=-11.02kcal" "mol^(-1)` `(DeltaH_(2)-DeltaH_(1))/(T_(2)-T_(1))=DeltaC_(P)` `(DeltaH_(2)-(-11.02))/(773-298)=(8.38-(1)/(2)xx6.96-(3)/(2)xx6.89)xx10^(-3)` `DeltaH_(2)=-13.6" kcal "mol^(-1)`. |
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| 472. |
What is correct about isothermal expansion of the ideal gas?A. `W_(rev)=W_(irr)`B. `W_(rev)+W_(irr)=0`C. `W_(rev)gtW_(irr)`D. `q_(rev)=q_(irr)` |
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Answer» Correct Answer - C Factual Questions |
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| 473. |
Which among the following gives the expression for the reversible isothermal expansion of ideal gas ?A. `-PDeltaV`B. `-nRT ln (V_(2))/(V_(1))`C. Both `A` and `B`D. Neither `A` nor `B` |
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Answer» Correct Answer - B Factual Questions |
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| 474. |
For which of the following cases is `DeltaS=DeltaH//T`A. adiabatic processB. a process for which `DeltaC_(P)=0`C. an isothermal reversible phase changeD. a process at constant pressure. |
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Answer» Correct Answer - C For an isothermal reversible phase transition `DeltaS=(DeltaH)/(T)`. |
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| 475. |
Which of the following expression expressions is true for an ideal gas?A. `((deltaV)/(deltaT))_(P)=0`B. `((deltaP)/(deltaT))_(V)=0`C. `((deltaE)/(deltaV))_(T)=0`D. `((deltaE)/(deltaT))_(V)=0` |
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Answer» Correct Answer - C Factual Questions |
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| 476. |
For a reaction `Br_(2)(l)+Cl_(2)(g)to2BrCl` at `300K` the value of `DeltaH=7kcal` and `DeltaS` is `25 cal K^(-1)`. The free energy change isA. `-500 cal`B. `7493 cal`C. `18 kJ`D. `150 kcal` |
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Answer» Correct Answer - A Apply `DeltaG=DeltaH-T DeltaS` `=7000 cal-300xx25=-500cal` |
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| 477. |
The expression `[deltaR//deltaT]_(V)` representsA. Heat capacity at constant volumeB. Heat capacity at constant pressureC. Enthalpy changeD. Entropy change |
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Answer» Correct Answer - A Factual Questions |
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| 478. |
For the reversible process, the value of `DeltaS` is given by the expression:A. `DeltaH//DeltaT`B. `T-:q_((rev))`C. `q_((rev))xxT`D. `q_((rev))-:T` |
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Answer» Correct Answer - D `DeltaS=(q_(rev))/(T)` |
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| 479. |
Which of the following expression defines the physical significance of free energy change?A. `-DeltaG=w_((non-exp))`B. `-DeltaG=-w_((non-exp))`C. `DeltaG= -w_((exp))`D. `DeltaG= w_((exp))` |
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Answer» Correct Answer - B Decrease in free energy `(-DeltaG)` equals to useful work done by the system `(-W_(non-exp))`. |
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| 480. |
Joule Thomson coefficient is given by the expressionA. `((deltaT)/(deltaP))_(H)`B. `((deltaT)/(deltaV))_(H)`C. `((deltaE)/(deltaP))_(V)`D. `((deltaS)/(deltaT))_(P)` |
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Answer» Correct Answer - A See Joule Thomson effect in Comprehensive Review. |
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| 481. |
The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` isA. `9.92xx106Pa`B. `9.92xx105Pa`C. `9.92xx108Pa`D. `9.92xx107Pa` |
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Answer» Correct Answer - B Volume of graphite `=("mass")/("density")=(112)/(2.25)` Volume of diamond `=(12)/(3.31)` Change in volume, `DeltaV=((12)/(3.31)-(12)/(2.25))xx10^(-3)L=-1.91xx10^(-3)L` `DeltaG^(@)` =work done `=-pDeltaV` `P=(DeltaG^(@))/(DeltaV)` `=(1895Jmol^(-1))/(1.91xx10^(_3)xx 101.3)=9794` `" " [because 1 "atm" =10^(5) xx 1.013 "pa"]` `=9.92 xx 10^(8)` Pa |
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| 482. |
The efficiency of the reversible cycle shown in the given figure is A. 0.3333B. 0.56C. 0.66D. 0.25 |
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Answer» Correct Answer - B Efficiency of cycle `=("Area of closed cycle")/("Area under the curve")xx100` `=((1)/(2)xx(1500-1000)xx(250-150))/((1)/(2)xx(1500-1000)xx(250xx150)+(1500-1000)xx(150-0))xx100` `=((1)/(2)xx500xx100)/((1)/(2)xx500xx100+500xx150)xx100` `=(500xx50xx100)/(500xx50+500xx150)=25`. |
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| 483. |
A sample of argon gas at `1atm` pressure and `27^(@)C` expands reversibly and adiabatically from `1.25 dm^(3)` to `2.50 dm^(3)`. Calculate the enthalpy change in this process. `C_(vm)` for orgon is `12.48J K^(-1) mol^(-1)`. |
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Answer» Number of moles of argon present in the sample `=(PV)/(RT)=(1.25xx1)/(0.0821xx300)=0.05075` For adiabatic expansion, `(T_(1))/(T_(2))=((V_(2))/(V_(1)))^(r-1)` or `(300)/(T_(2))=((2.50)/(1.25))^(1.66-1)` or `T_(2)=188.55K` `C_P=C_(V)+R` `=12.48+8.314` `=20.794JK^(-1)" "mol^(-1)` `DeltaH=nxxC_(P)xxDeltaT` `=0.05075xx20.794xx(300-188.55)` `=117.6J` |
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| 484. |
A system absorbs `600J` of heat and work equivalent to `300J` on its surroundings. The change in internal energy is:A. `300J`B. `400J`C. `500J`D. `600J` |
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Answer» Correct Answer - D We know that `DeltaE=Q+W=600+(-300)=300J` `W=-300` , because the work is done by the system. |
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| 485. |
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. |
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Answer» The required equation is `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g), " "DeltaH=?` `DeltaH=[(1)/(2)DeltaH_(H-H)+(1)/(2)DeltaH_(Cl-Cl)]-[DeltaH_(H-Cl)]` `=(1)/(2)xx104+(1)/(2)xx58-103=-22` kcal `mol^(-1)` |
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| 486. |
When enthalpy and entropy change for a chemical reaction are `-2.5 xx10^(3)` cals and `7.4` cals `deg^(-1)` respectively. Predict that reaction at 298 K isA. spontaneousB. reversibleC. irreversibleD. non-spontaneous |
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Answer» Correct Answer - A `DetlaG=DeltaH-TDeltaS` `=-2500-298xx7.4=-2500-2205` `=-4705 cal` Since `DeltaG` is negative, the reaction is spontaneous |
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| 487. |
when 1.8 g of steam at the normal boiling point of water is converted inot water ,at the same temperature , enthalpy and entropy changes respectively will be ` [given , DeltaH_("vap")for water =40.8 KJ mol^(-1) ]`A. `-8.12kJ,11 .89Jk^(-1)`B. `10.25 kJ,12 .95 Jk ^(-1)`C. `- 4.08 kJ,-10.93 JK^(-1)`D. `10.93 KJ, -4.08 JK^(-1)` |
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Answer» Correct Answer - c ` DeltaH("condensation") for 1.8 g of steam ` ` = (-40.8)kJ xx (1.8/18) mol =- 4.08 KJ` ` DeltaS= (DeltaH)/(T_(b))= (-4.08xx10^(3)J)/(373.15 K ) =- 10.93 JK^(-1)` |
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| 488. |
what the heat of a reaction at constant pressure is `-2.5 xx 10^(3)` cal and entropy change for the reaction is `7.4cal deg ^(-1)` , it is predicted that the reaction at `25^(@)C`isA. reversibleB. spontaneousC. non-spontaneousD. irreversible |
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Answer» Correct Answer - c heat at constant pressure means enthalpy, i.e. we have ` DeltaG= DetlaH- T DeltaS` ` =-2.5 xx 10^(3) - 298 xx 7.4 =- 4705 cal ` ltbr. Hence , the process is spontaneous. |
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| 489. |
the intial state A has the temperature `T_(A)U_(A)`as the internal energy of the system . By appling the mechanical work . New state B is achieved with the temperatutre `T_(B)` and having the interanal energy `U_(B)`. Given that `t_(B) gt T_(A)`. What is the correct expression for the change in internal energy `(DeltaV)` ?A. `U_(B)= U_(A)`B. `U_(B) -U_(A)`C. `U_(A) - U_(B)`D. none of these |
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Answer» Correct Answer - b the change in the internal energy is `DeltaU=U_(B)-U_(A)` |
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| 490. |
An ideal gas expands from an intial volume `V` into vacuume under isothermal conditions. For this process,A. `DeltaUne 0,W=0` and `Qne 0`B. `DeltaU ne 0,Wne 0` and `Q ne O`C. `DeltaUne 0,Wne 0` and `Q= 0`D. `DeltaU=0,W=0` and `Q=0` |
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Answer» Correct Answer - C For an isothermal expansion of an ideal gas `DeltaU=0` Since `P_(ext)` is zero in vacuum. `W=-P_(ext)DeltaV=0` `:. Q=0` |
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| 491. |
What is the value of`DeltaE`(heat change at constant volume)for reversible isothermal evaporation of `90g`water at `100^(@)C`Assuming water vapour behaves as an ideal gas and`(DeltaH_("vap"))_("water")=540calg^(-1)`A. `9xx10^(3)cal`B. `6xx10^(3)cal`C. `4.49cal`D. none of these |
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Answer» Correct Answer - D Heat of system `(DeltaH)_(v)=mxxLv` `=90xx540=486000cal` `DeltaH=DeltaE+P(V_(v)-V_(L))=DeltaE+nRT` or, `48600=DeltaE+(90)/(18)xx2xx373` 0r, `DeltaE48600-3730=44870cal` |
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| 492. |
The standard free energies of formation in the gaseous state of methanol, dimethyl ether and water are -38.7, -27.3 and -54.6 kcal respectively. Is the transformation of methanol to dimehtyl ether and water in gaseous state possible? `2CH_(3)OHtoCH_(3)OCH_(3)+H_(2)O` |
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Answer» `DeltaG^(@)` for the tranformation `=sumDelta_(("product"))^(@)-sumDeltaG_(("reactants"))^(@)` `=-ve` (the transformation is possible). |
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| 493. |
Which of the following has `DeltaS^(@)` greater than zeroA. `CaO+CO_(2)(g)hArrCaCo_(3)`B. `NaCl(aq)hArrNaCl(s)`C. `NaNO_(3)(s)hArrNa^(+)(aq)+NO_(3)^(-)(aq)`D. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` |
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Answer» Correct Answer - D `NaNO_(3)` is a solid, which is converted to liquid ions. |
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| 494. |
The ammount of heat required to raise the temperature of 1 mole of diatomic gas by `1^(@)C` at constant pressure is `60cal` . The amount of heat which goes as internal energy of the gas is nearly.A. `60Cal`B. `30Cal`C. `42.6Cal`D. `49.8Cal` |
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Answer» Correct Answer - B For a diatomic gas `C_(p)=(7)/(2)R` and `C_(v)=(5)/(2)R` Only `(5)/(7)=0.71` of energy supplied increases the internal energy of the gas. The rest is used to do work against external pressure `0.71xx60=42.6cal` |
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| 495. |
`150 mL` of `0.5N` nitric acid solution at `25.35^(@)C` was mixed with `150 mL` of `0.5 N` sodium hydroxide solution at the same temperature. The final temperature was recorded to be `28.77^(@)C`. Calculate the heat of neutralisation of nitric acid with sodium hydroxide. |
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Answer» Total mass of solution `=150+150=300g` Q=total heat produced `=300xx(28.77-25.35)cal` `=300xx3.42=1026cal` Heat of neutralisation `=(Q)/(150)xx1000xx(1)/(0.5)` `=(1026)/(150)xx1000xx(1)/(0.5)=13.68kcal` Since, heat is liberated, heat of neutralisation should be negative. so, heat neutralisation `=-13.68kcal` |
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| 496. |
Ozone is prepared by passing silent electric discharge through oxygen. In this reactionA. energy is given outB. energy is absorbedC. oxygen is loaded with energyD. oxygen is dissociated into atoms |
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Answer» Correct Answer - B `3O_(2) hArr 2O_(3) ` energy is given out. |
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| 497. |
When `4g` of iron is burnt to ferric oxide at constant pressure, `29.28KJ` of heat is evolved. What is the enthalpy of formation of ferric oxide (At wt. of `Fe=56`)A. `-81.98KJ`B. `-819.8KJ`C. `-40.99KJ`D. `+819.8KJ` |
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Answer» Correct Answer - A Given: Weight of iron burnt `=4g` , Heat liberated `-29.28KJ` and atomic weight of iron `(Fe)=56` . We know that in ferric oxide `(Fe_(2)O)` , 2 moles of iron or `2xx56=112g` of iron are burnt. We also know that when `4g` of iron are burnt, then burnt, liberated `=29.28KJ` , therefore when `112g` of the iron are burnt, then heat liberated `=(29.28xx112)/(4)=-819.8KJ` (Minus sign due to liberation of heat). |
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| 498. |
Enthalpy of neutralisation of acetic acid by `NaOH` is `-50.6KJ//mol` and the heat of neutralisation of a storng acid with a strong bases is `-55..9KJ//mol` . What is the value of `DeltaH` for the ionisation of `CH_(3)COOH` ?A. `+5.3KJ//mol`B. `+6.2KJ//mol`C. `+8.2KJ//mol`D. `+9.3KJ//mol` |
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Answer» Correct Answer - B `DeltaH=DeltaH_(ioniz)+DeltaH_("neu")` `=-50.6=DeltaH_(ioniz)+(-55.9)` `DeltaH_(ioniz)=+5.3KJ//mol` |
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| 499. |
given `H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ` `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ` then , calculate the enthalpy of formation of `OH^(-) at 25^(@)C`A. `-228.8kJ`B. `-343.52 kJ`C. `+228.8 kJ`D. `+ 343. 52 kJ` |
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Answer» Correct Answer - a consider the formation of `H_(2)O` `H_(2)(g)+1/2O_(2)(g) to H_(2)O(l),DeltaH=-286.20 kJ` `DeltaH_(r)= DeltaH_(t)[H_(2)O(l)]-DeltaH_(t)[H_(2)(g))]-1/2 DeltaH_(t)[O_(2)(g)] ` ` DeltaH_(t)(H_(2)O(l))=-286.20` now consdier the ionisation of `H_(2)O` `H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ` |
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| 500. |
The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?A. `+229.70KJ`B. `-229.70KJ`C. `+226.70KJ`D. `-22.670KJ` |
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Answer» Correct Answer - A `H^(+)(aq)+OH^(-)(aq)rarrH_(2)O(l),DeltaH=-56.07KJ` `DeltaH=DeltaH_(f)(H_(2)O)-[Delta_(f)(H^(+))+DeltaH_(f)(OH)^(-)]` `[:. DeltaH_(f)(H)^(+)=0]-56.07=-285.77-(0+x)` `:. x=-285.77+56.07=-229.70KJ` |
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