InterviewSolution
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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.
| 2001. |
Which of the ions in the table below would have the largest value of enthalpy of hydration? Ionic radius in nm Charge of ionA. 0.0065 + 2B. `0.095 +1C. 0.135 +2D. 0.181 +1 |
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Answer» Correct Answer - A The process of hydration involves an electrostatic interaction between ions and polar water molecules. The enthalpy of hydration is directly proportional to the charge density of the ion. Thus, greater the charge and smaller the size of the ion, greater is the enthalpy of hydration. |
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| 2002. |
Change in internal energy when `4 kJ` of work is done on the system and `1 kJ` of heat is given out of the system isA. `+ 1 kJ`B. `- 5 kJ`C. `+ 5 kJ`D. `+ 3 kJ` |
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Answer» Correct Answer - D According to the first law of thermodynamics, `Delta U = q + w` `= (- 1 kJ) + (4 kJ)` `= + 3 kJ` Thus, the internal energy of the system increases. |
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| 2003. |
Assertion:The change in entropy during melting of ice is negligible in comparison to change in entropy during vaporisation. Reason:The volume occupied by solid and liquids is too less in comparison to volume occupied by gas.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` are correct and `E`C. Both `S` and `E` are correct but `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - c Reson is correct explanation for statement. `DeltaS_(vap.)=S_("vapour)-S_(liq.)` and `DeltaS_(fus.)=S_((L))-S_((S))` `DeltaS_(vap.)gt gt gtDeltaS_("fusion")` Gaseous state has more disordered system. |
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| 2004. |
Determine which of the following reaction at constant pressure represent systems that do work on the surrounding environment (I) `CaO (s) + CO_(2) (g) rarr CaCO_(3) (s)` (II) `NH_(4)Cl (s) rarr NH_(3) (g) + HCl (g)` (III) `2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)`A. IB. IIIC. II & IIID. I and II |
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Answer» Correct Answer - C I. `Delta n - ve`, II `Delta n + ve`, III `Delta n = +ve` |
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| 2005. |
Determine which of the following reactions taking place at constant pressure represents system that do work on the surrounding environment `Ag^(+)(aq)+Cl^(-)(aq)rarrAgCl(s)" " (II) NH_(4)Cl(s)rarrNH_(3)(g)+HCl(g) " " (III) 2NH_(3)(g)rarrN_(2)(g)+3H_(2)(g)`A. IB. IIIC. II and IIID. I and II |
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Answer» Correct Answer - C `W=-Deltan_(g)RT=-ve` `"II. " Deltan_(g)=2, " III. " Deltan_(g)=2` |
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| 2006. |
Fixed mass of an ideal gas collected in a 24.64 litre sealed rigid vessel at 10 atm is heated from `-73^(@)C` to `27^(@)C` calculate change in gibbs free energy if entropy of gas is a function of temperature as `S=2+10^(-2)T(J//K)(1l atm=0.1kJ)`A. `1231.5` JB. `1281.5` JC. `781.5` JD. 0 |
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Answer» Correct Answer - C At constant volume, `(P_(1))/(T_(1))=(P_(2))/(T_(2))` `rArr" "P_(2)=1xx(300)/(200)=(3)/(2)` and `V_(1)=24.63L` for single phase `because" "dG=Vdp-"S dT"` `DeltaG=V.DeltaP-int(2+10^(-2)T).dT` `=1231.5-200-(10^(-2)xx50,000)/(2)` = 781.5 J |
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| 2007. |
From the given graph Which of the following statement is correct ?A. The point B represents the state of equilibriumB. The equilibrium composition strongly favours the reactantC. From the point B formation of product is equally spontaneous as of reactantD. From the point B formation of reactant is more spontaneous than that of product |
| Answer» Correct Answer - A | |
| 2008. |
A substance can exists in two gaseous allotropic forms `A` and `B`. If the equibrium mixture at `2500 K` consists of `80` mole percent `A` then calculate `|DeltaG_("reaction")^(@)|` at `2500 K` for `B(g) to A(g)`. Express answer in `kcal`. Round off your anwer to next higher interger. |
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Answer» Correct Answer - 60 |
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| 2009. |
Which of the following statements is correct?A. Two isothermal curves may intersectB. Two adiabatic curves may intersectC. Two different states of gas can be connected by an isothermal process as well as an adiabatic process on a P-V diagram.D. An isothermal curve may intersect an adiabatic curve |
| Answer» Correct Answer - D | |
| 2010. |
Total kinetic energy of molecules of a gas sample varies `K.E.=(7)/(4)nRT`. This sample containsA. Monatomic gasB. Mixture of amonatomic and diatomic gasC. Mixture of monatomic and polyatomic gasD. Both (2) & (3) are possible |
| Answer» Correct Answer - D | |
| 2011. |
A sample of diatomic gas is heated at constant pressure. If an amount of 280 J of heat is supplied to gas, find ratio of work done by gaa and change in internal energyA. `5:2`B. `2:5`C. `2:3`D. `3:2` |
| Answer» Correct Answer - B | |
| 2012. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will beA. 4 RB. _C. 3RD. `(4R)/(3)` |
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Answer» Correct Answer - C |
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| 2013. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will beA. 4RB. 2.5 RC. 3RD. `(4R)/(3)` |
| Answer» Correct Answer - C | |
| 2014. |
An idel gas undrgoes isothermal process from some initial state `i` to final state `f`. Choose the correct alternatives.A. `dU=0`B. `dQ=0`C. `dQ=dU`D. `dQ=dW` |
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Answer» Correct Answer - A::D For an isothermal process change in temperature of the system `dT=0rArrT=` constant. We know that for an ideal gas `dU` = change in internal energy `=nC_(v)" "dT=0` [where, n is number of moles and `C_(v)` is specific heat capacity at constant volume ] From first law of thermodynamics, `dQ=dU+dW` `=0+dWrArrdQ =dW ` |
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| 2015. |
Consider the given diagram for 1 mole of a gas X and answer the following question. The process `A rarr B` representsA. isobaric changeB. isothermal changeC. adiabatic changeD. isochoric change. |
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Answer» Correct Answer - D There is no change in volume from `A rarr B` |
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| 2016. |
Correct statements about samples of ice and liquid water at `0^(@)` C include which of the following ? (P) Molecules in ice and liquid water have the same kinetic energy. (Q) Liquid water has a greater entropy than ice. (R ) Liquid water has a greater potential energy than iceA. P and Q onlyB. P and R onlyC. Q and R onlyD. P,Q and R |
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Answer» Correct Answer - D |
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| 2017. |
If the heat change at constant volume for decomposition of silver oxide is 80.25 kJ, what will be the heat change at constant pressure ?A. 80.25 kJB. gt 80.25kJC. lt 80.25 kJD. 160.50 kJ |
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Answer» Correct Answer - B `Ag_(2)O_((s)) rarr 2Ag_((s))+(1)/(2)O_(2(g)),q_(v)=DeltaU=80.25kJ` `q_(p)=q_(v)+Deltan_(g)RT` `Deltan_(g) gt 0((1)/(2)-0=(1)/(2)) or q_(p) gt q_(v) or q_(p) gt 80.25kJ` |
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| 2018. |
In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?A. `Delta U = W != 0, q = 0`B. `Delta U = W != 0 = q != 0`C. `Delta U = 0,W = 0, q != 0`D. `U = Delta U = q != 0` |
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Answer» Correct Answer - A In a closed insulated container, no heat can enter or leave the system, thus, `q = 0`. Due increase of temperature, `Delta U != 0`. According to the first law of the thermodynamics, `Delta U = q + W` `= 0 + W` `W != 0` |
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| 2019. |
Change in internal energy of a thermodynatic system is called heat of reaction at contantA. temperatureB. pressureC. volumeD. both (1) and (2) |
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Answer» Correct Answer - C If a process is carried out at constant volume, then according to the first law of thermodynamics, the heat exchanged by the system with its surroundings at constant volume is equal to `Delta U`. As a result of this heat exchange, the temperature of the system also remais constant. |
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| 2020. |
2 mol of an ideal gas at `27^(@)C` temperature is expanded reversibly from `2L` to `20L`. Findk entropy change `(R = 2 cal mol^(-1) K^(-1))`A. 92.1B. 0C. 4D. 9.2 |
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Answer» Correct Answer - D `Delta S` for the isothermal reversible expansion of an ideal gas is given as `Delta S = 2.303 nR log ((V_(2))/(V_(1)))` `= (2.303) (2) (2) "log" (20)/(2)` `= (9.212) log 10` `= 9.2 cal K^(-1) mol^(-1) (log 10 = 1)` |
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| 2021. |
Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` isA. `-17` kcal/molB. `-111` kcal/molC. `-170 ` kcal/molD. `-85 ` kcal/mol |
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Answer» Correct Answer - A For reaction `C(s) + 2H_(2)(g) to CH_(4)(g), DeltaH^(@)C=? ` `C+O_(2) to CO_(2), DeltaH=-94 ` kcal …(i) `2H_(2)+ O_(2) to 2H_(2)O, DeltaH=-68 xx 2` kcal …(ii) `CH_(4)+ 2O_(2) to CO_(2) + 2H_(2)O, DeltaH=-213` Kcal …(iii) On adding Eqs. (i) and (ii) and then subtracting Eq. (ii) `=(-94) + (-2xx68)-(-213)` `-230 + 213 =-17 `K cal/mol |
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| 2022. |
Statement -1: Entropy change in reversible adiabatic expansion of an ideal gas is zero. Statement-2: The increase in entropy due to volume increase just compensates the decrease in entropy due to fall in temperature.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True. |
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Answer» Correct Answer - a |
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| 2023. |
Which of the following ionic solids tend to be the least soluble in water?A. IodidesB. BromidesC. ChloridesD. Fluorides |
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Answer» Correct Answer - D Lattice enthalpy of fluorides is vary high. |
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| 2024. |
Calculate heat of formation of `KOH(s)` using the following equations `K(s) +H_(2)O(l) +aq rarr KOH(aq)+1//2 H_(2)(g),DeltaH =- 48.0 kcal …(i)` `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.4 kcal …(ii)` `KOH(s) +(aq) rarr KOH(aq),DeltaH =- 14.0 kcal ....(iii)`A. `+102.83`B. `+130.85`C. `-102.83`D. `-130.85` |
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Answer» Correct Answer - C `K_((s))+1/2 O_(2) +1/2 H_(2) rarr KOH_((s)) overset(H_(2)O)(rarr) KOH_((aq))` |
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| 2025. |
If `DeltaH` is the enthalpy change and `DeltaU` the change in internal energy accompanying a gaseous reaction, thenA. `DeltaH` is always less than `DeltaU`.B. `DeltaH` is always than `DeltaU`.C. `DeltaH` is less than `DeltaU` if the number of moles of gaseous products is greater than the number of moles of gaseous reactants.D. `DeltaH` is less than `DeltaU` if the number of moles of gaseous products is less than the number of moles of gaseous reatants. |
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Answer» As `DeltaH = DeltaU +Deltan_(g)RT` `DeltaH` value is less than or greater than `DeltaU` depending on the value of `Deltan_(g)` which is the change in number of moles of the gaseous components. As result, (a) and (b) cannot be true. `DeltaH lt DeltaU` when `Deltan_(g) lt 0`, hence (c ) is also false only (d) is correct. Hence, (d) is the correct answer. |
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| 2026. |
Which of the following statements is (are) false?A. The entropy of the universe decreaase and increases at a periodic rateB. The entropy of the universe increases and tends towards the maximum valueC. For endothermic spontaneous process, the total entropy change decreaseD. The entropy of the universe decrease and tends to zero |
| Answer» Correct Answer - A::C::D | |
| 2027. |
What is significance of `T DeltaS` in `DeltaG = DeltaH - T DeltaS`? |
| Answer» It is quantitative measure of the randomness of the system. | |
| 2028. |
What is the relationship between`q_(p)` and `q_(v)` ? |
| Answer» `q_(p)= q_(v)+Deltan_(g)` RTwhere `Deltan_(g)=n_(p)-n_(r)` (gaseous) | |
| 2029. |
The adiabatic compression ratio in a carnot reversible cycle is 9, when the temperature iof the source is `227^(@)C`. Calculate the temperature of the sink. Given `gamma=1.5` |
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Answer» Correct Answer - `-106.33^(@)C` Here,`(V_(2))/(V_(1))=9, T_(1)=227+273= 500K`, `T_(2)=?, gamma=1.5` For an adiabatic change, `T_(2)V_(2)^((gamma-1))=T_(1)V_(1)^((gamma-1))` or `T_(2)=T_(1)((V_(1))/(V_(2)))^((gamma-1))` `=500(1/9)^((1.5-1))=(500)/3= 166.67` `=166.67-273= -106.33^(@)C` |
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| 2030. |
If `DeltaH` is the enthalpy change and `DeltaU` the change in internal energy accompanying a gaseous reaction, thenA. `Delta H` is always greater than `Delta U`B. `Delta H lt Delta U` only if the number of moles of the products is greater than the number of the reactantsC. `Delta H` is always less than `Delta U`D. `Delta H lt Delta U` only if the number of moles of the product is less than the number of moles reactants |
| Answer» Correct Answer - D | |
| 2031. |
If `DeltaH` is the enthalpy change and `DeltaU` the change in internal energy accompanying a gaseous reaction, thenA. `DeltaH` is always greater than `DeltaE`B. `DeltaH lt DeltaE` only if the number of moles of products is greater than the number of moles of the reactantsC. `DeltaH` is always less than `DeltaE`D. `DeltaG lt DeltaE` only if the number of moles of products is less than the number of moles of the reactants |
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Answer» Correct Answer - D Reactions in which there is a decrease in the number of moles of the gaseous components , i.e, `Deltan_(g)` is negative , the enthalpy change `(DeltaH)` is lesser than the internal energy change `(DeltaE)` . Reaction in which there is a increase in the number of moles of gaseous components i.e., `Deltang` is positive , the enthalpy change in greater than the internal energy change. `DeltaH-DeltaE+ Deltan_(g)RT` |
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| 2032. |
For gaseous reactions, if `DeltaH` is the change in enthalpy and `DeltaU` that in internal energy then:A. `DeltaH` is always greater than `DeltaU`B. `DeltaH` is always less than `DeltaU`C. `DeltaH lt DeltaU` only if the number of mole of the gaseous products is less than that of the gaseous reactantsD. `DeltaU lt DeltaH` only if the number of mole of the gaseous reactants is less than that number of the gaseous products |
| Answer» Correct Answer - C::D | |
| 2033. |
Calculate heat of formation of `KOH(s)` using the following equations `K(s) +H_(2)O(l) +aq rarr KOH(aq)+1//2 H_(2)O(g),DeltaH =- 48.0 kcal …(i)` `H_(2)(g) +2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.4 kcal …(ii)` `KOH(s) +(aq) rarr KOH(aq),DeltaH =- 14.0 kcal ....(iii)` |
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Answer» We have to evaluate `DeltaH` for `K(s) +1//2O_(2)(g) +1//2H_(2)(g) rarr KOH(s), DeltaH = ?` Adding equations (i) and (ii), `K(s)+H_(2)O(l) +aq rarrKOH(aq) +1//2H_(2)(g)` `DeltaH =- 48.0 kcal` `H_(2)(g)+1//2O_(2)(g) rarr H_(2)O(l)`, `DeltaH =- 68.4 kcal` `{:ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)+aqunderset(DeltaH=-116.4Kcal...(iv))(rarrKOH(aq)):}` Subtracting equation (iii) from equation (iv) `{:(KOH(s)+aqrarrunderset(DeltaH =- 102.4 kcal ..(v))(KOH(aq),)),("- - - +" ),(ulbar(K(s)+1//2H_(2)(g)+1//2O_(2)(g)rarrunderset(DeltaH =- 102.4 kcal)(KOH(s),))):}` |
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| 2034. |
The lattice enthalpy and hydration enthalpy of four compounds are given below `:` `{:("Compuounds","Lattice enthalpy ( in "kJ mol^(-1)")","Hydration enthalpy ( in kJ " mol^(-1)")"),(P,+780,-920),(Q,+1012,-812),(R,+828,-878),(S,+632,-600):}` The pair of compunds which is soluble in water isA. P and QB. Q and RC. R and SD. P and R |
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Answer» Correct Answer - d A compound is soluble if hydration enthalpy `gt` lattice enthalpy |
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| 2035. |
A substance A has the following variation of vapour presesure with temperature for its solid state and liquild state . Indentify the options which are correct : Date : For solid A: `log_(10)` `P=4-(200)/(T)` For liquind A : `log_(10)` `P=3.48-(1500)/(T)` where P is in mm of Hg and T in K.A. Enthalpy of vapourisation and enthalpy of fusion will be temperature independent.B. `DeltaH_(sub.)` will be approximately `9.212 kcal//mol.`C. `DeltaH_(fusion)` will be approximately `2.303 kcal//mol`.D. `DeltaH_(vap)` will be approximately `6.909 kcal//mol`. |
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Answer» Correct Answer - a,b,c,d |
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| 2036. |
`{:(,"Column-I",,"Column-II"),((a),"Reversible adiabatic compression",(p),"Process in equilibrium"),((b),"Reversible vaporisation of liquid",(q),DeltaS_(system)lt0),((c),2N(g)rarrN_(2)(g),(r),DeltaS_("surrounding")lt0),((d),MgCO_(3)(s)oversetDeltararrMg(s)+CO_(2)(g),(s),DeltaS_("sublimation")=0):}` |
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Answer» Correct Answer - `(a-s);(b-p,r);(c-q);(d-p,r)` |
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| 2037. |
What is the physical significance of free enegry? |
| Answer» Decreases in free energy is the measure of the useful work done by the system. | |
| 2038. |
A substance A has the following variation of vapour presesure with temperature for its solid state and liquild state . Indentify the options which are correct : Date : For solid A: `log_(10)` `P=4-(200)/(T)` For liquind A : `log_(10)` `P=3.48-(1500)/(T)` where P is in mm of Hg and T in K.A. Enthalpy of vapourisation and enthalpy of fusion will be temperature independent.B. `DeltaH_(sub.)` will be approximately `9.212 kcal//mol.`C. `DeltaH_(fusion)` will be approximately `2.303 kcal//mol`.D. `DeltaH_(vap)` will be approximately `6.909 kcal//mol`. |
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Answer» Correct Answer - a,b,c,d |
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| 2039. |
Write the relationship between enthalpy change, `DeltaH `and internal energy change, `DeltaU` , for a process occuring at constant pressure and constant temperature. |
| Answer» `DeltaH = DeltaU + Deltan_(g) RT`. | |
| 2040. |
Which of the following option(s) will show a decrease in Gibbs free energy ?A. Combustion of propane at 1 bar and 500 K.B. Vapourisation of any liquild at 1 atm and above its normal bolling point.C. Fusion of `H_(2)O` at 1 atm and `0^(@)C` if its normal melting point is `0^(@)`C.D. Vapourisation of `H_(2)O` at `100^(@) C`and 1 bar if its normal boling ponts is `100^(@)C`. |
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Answer» Correct Answer - a,b,d |
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| 2041. |
For a reaction at `25^(@)C` enthalpy change `(DeltaH)` and entropy change `(DeltsS)` are `-11.7 K J mol^(-1)` and `-105J mol^(-1) K^(-1)`, respectively. Find out whther this reaction is spontaneous or not? |
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Answer» Given, `DeltaH = - 11.7 xx 10^(3) J mol^(-1)` `DeltaS =- 105 J mol^(-1) K^(-1), T = 298 K` `DeltaG = DeltaH -T DeltaS =- 11700 - 298 xx (-105)` `= 19590J =+ 19.59kJ` Since `DeltaG = +ve` Therefore, reaction is not spontaneous. |
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| 2042. |
`{:(,"Column-I",,"Column-II"),((a),"Heating of an ideal gas at constant pressure",(p),DeltaH=nC_(p.m,)DeltaTne0),((b),"Compression of liquid at constant temperature",(q),DeltaU=0),((c),"Reversible process for anideal gas at constant temperature",(r),DeltaG=V.DeltaP),((d),"Adiabatic free expansion of an ideal gas",(s),DeltaG=nRTln(P_(2)/P_(1))):}` |
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Answer» Correct Answer - `(a-p);(b-r);(c-q,p);(d-q,s)` |
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| 2043. |
`{:(,"Column-I",,"Column-II"),((a),C ("s,graphite")+0_(2)(g)rarrCO_(2)(g),(p),DeltaH^(@)_("Combustion")),((b),C("s,graphite")rarrC(g),(q),DeltaH^(@)_("combustion")),((c),CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),(r),DeltaH^(@)_("atomization")),((d),CH_(4)(g)rarrC(g)+4H(g),(s),DeltaH^(@)("sublimation")):}` |
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Answer» Correct Answer - `(a-p,q);(b-q,r,s);(c-p);(d-r)` |
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| 2044. |
The enthalpy and entropy change of trimerization of gas phase `(3A(g) to A_(3)(g))` at `300 K` are `-100 K//J "mole"` amd `-400J//K"mole"` respectively . If the bolling points of A and `A_(3)` are 300 K and 400 K respectively . Given : `C_("pm"_(A_3(g)))=C_("pm"_(A_3(l))),d_(A(l))=1.2 g//ml`, `d_(A_(3(l)))`= `1.5 g//ml`, molar masss of A =20 garm Assume : Density of `A_(3)(l)` and `A(l)` are independent of pressure.A. `DeltaH` " for trimerisation in liquid phases " at` 300 K` =-` 125 KJ mol^(-1)`B. `DeltaS` "for trimerisation in liquid phase " at `300 K` =-` 275 kJmol^(-1)`C. At standard state of `300 K DeltaG` of the reaction : `3A(l) to A_(3) (l) "is" 7500 J`D. The equilibrium pressure of the reaction `3(A) to A_(3) (l) is 7.501xx10^(8) "Pa at" 300 K` |
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Answer» Correct Answer - c,d |
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| 2045. |
The direct conversion of A to B is difficult, hence is carried out by the following shown path: `Delta(AtoC)=50, DeltaS(CtoD)=30,DeltaS(BtoD)=20` The entropy change for the process `A to B` is :A. `+100e.u`B. `+60e.u`C. `-100e.u`D. `-60e.u` |
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Answer» Correct Answer - B `Delta S = DeltaS_(1)+Delta S_(2)-Delta S_(3)` |
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| 2046. |
The direct conversion of A to B is difficult, hence is carried out by the following shown path: `Delta(AtoC)=50, DeltaS(CtoD)=30,DeltaS(BtoD)=20` The entropy change for the process `A to B` is :A. `100`B. `-60`C. `-100`D. `+60` |
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Answer» Correct Answer - D |
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| 2047. |
Identify the correct statement regarding a spontaneous process `:`A. Exothermic process are always spontaneous.B. Lowering of energy in the reaction process is the only criterion for spontaneity.C. For a spontaneous process in an isolated system, the change in entropy is positive.D. Endothermic process are never spontaneous. |
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Answer» Correct Answer - C |
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| 2048. |
The average `O-H` Bond energy in `H_(2)O` with the help of following data : (P) `H_(2)O(l)toH_(2)O(g)," "DeltaH=+40.6 kJ mol^(-1)` (Q) `2H(g)toH_(2)(g)," "DeltaH=-435.0 kJ mol^(-1)` (R ) `O_(2)(g)to2O(g)," "DeltaH=+489.6 kJ mol^(-1)` (S) `2H_(2)(g)+O_(2)(g)to2H_(2)O(l),` ` " "DeltaH=-571.6 kJ mol^(-1)`A. `584.9 kJ "mole"^(-1)`B. `279.8 kJ"mole"^(-1)`C. `462.5 kJ"mole"^(-1)`D. `925 kJ "mole"^(-1)` |
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Answer» Correct Answer - C |
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| 2049. |
From the following data , the enthalpy change for the sublimation of of ice at 223 K will be [mean heat capacity of ice`=2 JK^(-1)g^(-1)`, mean heat capacity of `H_(2)O(l)=4.2 JK^(-1)g^(-1)`, mean heat capacity of `H_(2)O(v)=1.85 JK^(-1)g^(-1)`, entalpy of fusion of ice at `0^(@) C = 334 Jg^(-1)`.enthalpy of evaporation of water at `100^(@) C = 2255 J g^(-1)`]A. `3000 J g^(-1)`B. `3109 J g^(-1)`C. `3827 J g^(-1)`D. `4000 J g^(-1)` |
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Answer» Correct Answer - B |
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| 2050. |
For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters isA. `DeltaG=0, DeltaS=+ve`B. `DeltaG=0, DeltaS=-ve`C. `DeltaG=+ve, DeltaS=0`D. `DeltaG=-ve, DeltaS=+ve` |
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Answer» Correct Answer - A |
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