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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.
| 3251. |
Consider the following processes `:` `{:(,DeltaH(kJ //mol) ),(1//2A rarr B ,+ 150 ),(3 B rarr 2C + D,-125),(E+Ararr2D ,+ 350 ):}` For `B +D rarrE +2C , DeltaH ` will beA. `525 kJ //mol`B. ` -175 kJ //mol`C. ` -325kJ // mol`D. ` 325 kJ //mol` |
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Answer» Correct Answer - B `2 xx `Eqn. (i) `+`Eqn. (ii) -Eqn. (iii) gives the required equation. `:. DetlaH =2( 150) + ( - 125) -350 =- 175 kJ //` mol. |
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| 3252. |
Consider the following processes :- `{:(,DeltaH(kJ//mol)),((1)/(2)A rarr B,+150),(3B rarr2C+D,-125),(E+A rarr 2D,+350),("For "B+D rarr E+2C",",Delta H" will be"):}`A. 325 kJ / molB. 525 kJ / molC. `-175` kJ / molD. `-325` kJ / mol |
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Answer» Correct Answer - C eq(iv) = `2xx` eq(i) + eq(ii) - eq(iii) `Delta H_(4)=2xxDelta H_(4)+Delta H_(2)-Delta H_(3)` |
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| 3253. |
The enthalpy of funsion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at `0^(@)C` is :A. 5.260 cal/(mol K)B. 0.526 cal/(mol K)C. 10.52 cal/(mol K)D. 21.04 cal/(mol K) |
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Answer» Correct Answer - A `Delta S=(Delta H)/(T)=(1.435xx1000)/(273)"Cal mol"^(-1)` |
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| 3254. |
Define surroundings. |
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Answer» The rest of the universe which might be in a position to exchange energy and matter with the system is called its surroundings. |
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| 3255. |
Define a system. |
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Answer» A system in thermodynamics refers to that part of the universe in which observations are made. |
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| 3256. |
Silane `(SiH_(4))` burns in air as: `SiH_(4)(g) +2O_(2)(g) rarr SiO_(2)(s) +2H_(2)O(l)` the standard Gibbs energies of formation of `SiH_(4)(g), SiO_(2)(s)`, and `H_(2)O(l)` are `+52.3, -805.0`, and `-228.6kJ mol^(-1)`, respectively. Calculate Gibbs enegry change for the reaction and predict whether the reaction in spontaneous or not. |
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Answer» `SiH_(4)(g) +2O_(2)(g) rarr SiO_(2)(g) +2H_(2)(g)` `Delta_(r)G^(Theta) = Delta_(r)G^(Theta) (SiO_(2)) +2Delta_(f)G^(Theta) (H_(2)O)` `-[Delta_(f)G^(Theta) (SiH_(4))+2Delta_(f)G^(Theta)(O_(2))]` `=- 805.0 +2(-228.6) -[+52.3 +2(0)]` `= - 805.0 -457.2 - 52.3` `=- 1314.5 kJ` Since `Delta_(r)G^(Theta)` is negative, the reaction will be spontaneous. |
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| 3257. |
1000 gm water is heated from `27^(@)` C to `47^(@) C` at a constant pressure of 1 bar . The coefficient of volume expansion of water is `0.002//^(@)`C and the molar volume of water at `0^(@)` is `18.00cm^(3)//"mol"`. The magnitude of work done (in J) by water is |
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Answer» `V_(t^(@)C)= V_(0^(@)C) (1 +gammat)" "gamma="coefficient of volue expansion "` `V_(27^(@)C)= 1000 (1 +0.002 xx27)" "t =" temperature in ".^(@)C ` `= 1054 ml` `V_(47^(@)C) =1000 (1 + 0.002 xx 47)` `=1094 ml` `W =-PDeltaV =(-1 xx(1094 - 1054))/(1000)"bar litre" =(-4)/(100) "bar litre"=(-4)/(100) xx100J = - 4 J` |
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| 3258. |
1 Kg stone at `27^(@)`C falls 100 m into a lake whose temperature is `27^(@)` C. Find the entropy charge of (a) the stone (b) the lake (c) the univers when (i) stone is lowered reversibly (ii) stone is dropped freely Compare the loss of available energy in two cases |
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Answer» (i) If the stone is lowered reversibly no amount of heat will be produced. Also since term of stores remains the same `DeltaS_("stone") = 0` `q_("rev")= 0 rArr DeltaS_("lake") = 0` `and DeltaS_("total") = DeltaS_("universe")= DeltaS_("lake") +DeltaS_("stone")= 0` (ii) If the stone is lowered irreversibly, the potenial energy will be entirely lose in form of heat energy `q_("actual") = q_("irr") =mgh` `DeltaS_("stone") = 0` `DeltaS_("lake") =-(q_("irr"))/(T) = - ((-1 xx9.8 xx1000)/(300)) = (980)/(300)J//K` `=3.26 J//K` `DeltaS_("total") = 3.26 J//K` |
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| 3259. |
The temperture of a definite amount of an ideal monoatomic gas becomes four times in a reversible process for which heat exchange is zero. Which of the following is correct relation between the final and initial parameters of gas ?A. `V_(f)= 8V_(i)`B. `P_(f)=32P_(i)`C. `V_(f)= 16 V_(i)`D. `P_(f)=(1)/(16)P_(i)` |
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Answer» Correct Answer - B `((V_(1))/(V_(2)))= ((T_(2))/(T_(1))) ^((1)/(gamma-1))= (3)^(3//2)=8` `(P_(1))/(P_(2))= ((T_(2))/(T_(1)))^((gamma)/(1-gamma)) = (3)^((5//3)/(-2//3))=(3)^(-5//2) = (1)/(32)` |
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| 3260. |
What is correcto for an ideal gas undergoing reversible adiabatic expansion reagarding temperture?A. Remains sameB. IncreasesC. DecreasesD. May increase or decrease |
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Answer» Correct Answer - C |
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| 3261. |
For the reaction given below the value of standard Gibbs free energy of formation at 298 K are given. What is the nature of the reaction? `I_(2)+H_(2)Srarr2HI+S` `DeltaS_(f)^(@)(HI)="1.8 kJ mol"^(-1), DeltaG_(f)^(@)(H_(2)S)="33.8 kJ mol"^(-1)`A. Non-spontaneous in forward direction.B. Spontaneous in forward direction.C. Spontaneous in backward direction.D. Non-spontaneous in both forward and backward directions. |
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Answer» Correct Answer - B `I_(2)+H_(2)S rarr 2HI +S` `DeltaG^(@)=SigmaG_(f"(Products)")^(@)-SigmaG_(f"(Reactants)")^(@)` `DeltaG^(@)=(2xx1.8+0)-(0+33.8)=-"30.2 kJ"` Hence, reaction is spontaneous in forward direction. |
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| 3262. |
For a reaction to be spontaneous at any temperature, the conditions areA. `DeltaH=+ve, DeltaS=+ve`B. `DeltaH=-ve, DeltaS=-ve`C. `DeltaH=+ve, DeltaS=-ve`D. `DeltaH=-ve, DeltaS=+ve` |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` `DeltaG=-ve` if `DeltaH=-ve ` and `DeltaS` is `+ve.` |
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| 3263. |
An ideal gas taken in an insulated chamber is relased into interstellar space. The statement that is nearly trure for this pupose is :A. `Q=0,Wne0`B. `W=0,Q ne0`C. `delta U=0,Q ne0`D. `Q=W=delta U=0` |
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Answer» Correct Answer - D |
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| 3264. |
Which halogen has the highest standerd entropy, `S^(@)` ?A. `F_(2)(g)`B. `Cl_(2)(g)`C. `Br_(2)(l)`D. `I_(2)(s)` |
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Answer» Correct Answer - B |
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| 3265. |
A chemical reaction has `K_(eq)=1xx10^(-5)` at `25^(@)` C, and the value of `K_(eq)` increases with incresing tempreature. From these statements, what may one conclude?A. `DeltaH^(@)gt0 and DeltaS^(@)gt0`B. `DeltaH^(@)lt0 and DeltaS^(@)lt0`C. `DeltaH^(@)lt0 and DeltaS^(@)gt0`D. `DeltaH^(@)lt0` and no conclusion may be drawn about the sign of `DeltaS^(@)` |
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Answer» Correct Answer - D |
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| 3266. |
Which of the following options correctly represent trure//false nature of statements? Statement-I: `deltaH=deltaU+PdeltaV "for all processes"` Statements-II: For a reaction involving only ideakl gas, `DeltaH_(reaction)` will be independent of the pressure at which reactants and products are taken. Statements-III: Heat taken from a thermal reservoir can be completely converted at work without liberating some heat at lower temperature. Statement-IV: For a chemical reaction, G at equillibrium will be zero.A. All the statement are trueB. Only statement-II are trueC. All the statements are falseD. Only statement-I and III are false |
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Answer» Correct Answer - B |
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| 3267. |
Predict the sign of q, w ,DeltaU and DeltaH for given processes. Hence, find number process for which at least one of `q,omega.DeltaU,DeltaH` are zero |
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Answer» Correct Answer - 4 |
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| 3268. |
Predict the sign of q, w and`Delta U` for given process. Hence , find number of process for which at least one of q,w, `DeltaU,DeltaH` are zero. |
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Answer» Correct Answer - 3 |
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| 3269. |
In how many option (s), enthalpy change is marked incorrectly? `CuSO_(4)(s)+5H_(2)O(g)toCuSO_(4).5H_(2)O(s),Delta_(solution)H^(@)[CuSO_(4)(s)]` `2NH_(3)(g)toN_(2)(g)+3H_(2)(g),H_(atm)^(@)[NH_(3)]` `Mg Cl_(2)(s)to Mg(s)+Cl_(2)(g), Delta_("lattice")H^(@)[MgCl_(2)(s)]` `(1)/(4)P_(4)(g)toP(g), Delta_(P-P) H^(@)` |
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Answer» Correct Answer - 4 |
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| 3270. |
Assertion :- Entropy of system increases for a spontaneous reactions. Reason :- Enthalpy of reaction always decreases for spontaneous reaction.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - D | |
| 3271. |
Assertion :- Catalyst change Gibbs free energy of system. Reason :- Catalyst changes preexponential factor of a chemical reaction.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - D | |
| 3272. |
Assertion :- `Q_("Surr")` is zero for adiabatic process Reason :- Final state of the system remains unchanged in adiabatic process.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - C | |
| 3273. |
Assuming that water vapour in an ideal gas, the internal energy change (`DeltaU)` when 1 mol of water is vaporised at 1 bar of pressure and `100^(@)` C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ `mol^(-1)` and R=8.3 J `mol^(-1)K^(-1)`) will be :A. 37.904 kJ `mol^(-1)`B. 41.00 kJ `mol^(-1)`C. 4.100 kJ `mol^(-1)`D. 3.7904 kJ `mol^(-1)` |
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Answer» Correct Answer - A |
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| 3274. |
Why does absolute zero not correspond to zero energy? |
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Answer» The total energy of a gas is the sum of kinetic and potential energy of its molecules. Since the kinetic energy is a function of the temperature of the gas. Hence at absolute zero, the kinetic energy of the molecules ceases but potential energy is not zero. So, absolute zero temperature is not the temperature of zero energy |
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| 3275. |
State the Second law of thermodynamics and write 2 applications of it? |
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Answer» According to second law of thermodynamics, when a cold body and a hot body are brought into contact with each other, heat always from hot Body to the cold body. Also, that no heat engine that works in cycle completely converts heat into work. Second law of thermodynamics is used in working of heat engine and of refrigerator. |
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| 3276. |
Determine enthalpy change for the following polymerizartion reaction per mole of `N_(2)(g)` consumed `nN_(2)(g)+nH_(2)(g)to(NH-NH)_(n)(g)` Given : bond enthalpy : n-=N=942KJ`//`mole, N-N=163Kj`//`mole H-H=436KJ`//`mole N-H=390Kj`//`moleA. 272Kj/moleB. 140KJ/moleC. `-110KJ//"mole"`D. `-400KJ//"mole"` |
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Answer» Correct Answer - a |
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| 3277. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` final temperature of water is :A. 304.43KB. 296.97KC. 303.93KD. 287K |
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Answer» Correct Answer - b |
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| 3278. |
the combustion of 2- propanol `(M=60.0gxxmol^(-1))` occurs according to the equation , `2CH_(3)CHOHCH_(3)(l)+90_(2)(g)to6CO_(2)(g)+8H_(2) O(l)` What is q for the combustion of 15.0 g of 2- propanol? A. `-5.01xx10^(2)KJ`B. `-1.00xx10^(3)KJ`C. `-2.01xx10^(3)KJ`D. `-4.01xx10^(3)Kj` |
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Answer» Correct Answer - a |
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| 3279. |
At what temperature does the reaction below change from favoring products to favoring A. 162KB. 509KC. 1060KD. 1540K |
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Answer» Correct Answer - c |
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| 3280. |
Standar enthalpy of formation of three combustible isomers compound A,B,C are -20Kcal/mol ,30kcal /mol and 40kcal /mol respectively then what will be order of their enthalpy of combustion ?A. `DeltaH_(c)^(@)[A]ltDeltaH_(c)^(@)[B]lt DeltaH_(c)^(@)[C]`B. `DeltaH_(c)^(@)[B]gtDeltaH_(c)^(@)[A]gt DeltaH_(c)^(@)[C]`C. `DeltaH_(c)^(@)[A]gtDeltaH_(c)^(@)[B]gt DeltaH_(c)^(@)[C]`D. `DeltaH_(c)^(@)[C]gtDeltaH_(c)^(@)[A]gt DeltaH_(c)^(@)[B]` |
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Answer» Correct Answer - c |
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| 3281. |
For an ideal monoatomic gas, an illustration of three different paths A,(B+C) and (D+E) from an initial state `P_(1),V_(1),T_(1)` to a final state `P_(2),V_(2),T_(1)` is shown in the given figure . Path A represents a reversible isothermal from `P_(1)V_(1)" to "P_(2),V_(2)`, path (B+C) represent a reversible adiabatic expansion (B) from `P_(1),V_(1),T_(1)" to " P_(3),V_(2),T_(2)` followed by reversible heating of the gas at constant volume (C) from `P_(3),V_(2),T_(2) " to " P_(2),V_(2),T_(1)`. Path (D+E) represents a reversible expansion at constant pressure `P_(1)(D)` from `P_(1),V_(1),T_(1) " to "P_(1),V_(2),T_(3)` followed by a reversible cooling at constant volume `V_(2)(E)` form `P_(1),V_(2),T_(3) " to "P_(2),V_(2),T_(1)` What is `q_("rev")` for path (A)?A. zeroB. `-nR" In"(V_(2))/(V_(1))`C. `-nRT_(1)" In"(V_(2))/(V_(1))`D. `nRT_(1)" In"(V_(2))/(V_(1))` |
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Answer» Correct Answer - d |
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| 3282. |
For a certain reaction the change in enthalpy and change in entropy are `40.63 kJ mol^(-1)` and `100 JK^(-1)` .What is the value of `Delta G` at `27^(@)C` and indicate whether the reaction is possible or not ? |
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Answer» We know that : `Delta G = Delta H - T Delta S` Given `T=27+273=300 K, Delta H=40.63xx10^(3)J mol^(-1), =40630 J mol^(-1), Delta S = 100 JK^(-1)` `Delta G = 40630-300xx100=40630-30000 =+10630 J` Positive value of `Delta G` indicates that the reaction is not possible. |
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| 3283. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` `DeltaS_(ice)is:`A. `11.04JK^(-1)`B. `3.16JK^(-1)`C. `14.2JK^(-1)`D. `7.84Jk^(-1)` |
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Answer» Correct Answer - c |
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| 3284. |
For an ideal gas, an illustration of the different paths, A `(B + C)` and `(D + E)` from an intial state `P_(1), V_(1), T_(1)` to a final state `P_(2), V_(2), T_(1)` as shown in the given figure. Path (A) represents a reversible isothernal expanison from `P_(1)V_(1)` to `P_(2)V_(2)`. Path `(B + C)` represents a reversible adiabatic expansion (B) from `P_(1), V_(1), T_(1)` to `P_(3), V_(2), T_(2)` followed by reversible heating the gas at constant volume (C) from `P_(3), V_(2), T_(2)` to `P_(2), V_(2) T_(1)` to `P_(1) , V_(2), T_(3)` followed by reversible cooling at a constant volume `V_(2)` (E) from `P_(1), V_(2), T_(3)` to `P_(2), V_(2), T_(1)` What is `Delta S` for path `(D + E)` ?A. zeroB. `underset(T_(1))overset(T_(2))int (C_(V)(T))/(T) dT`C. `-nR ln. (V_(2))/(V_(1))`D. `nR ln. (V_(2))/(V_(1))` |
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Answer» Correct Answer - D path (A) : isothermal `(Delta U = 0)` `q_(rev) = w = nRT_(1) ln. (v_(2))/(v_(1))` path `(B + C) : q_(rev) = 0 + (-nR ln.(v_(2))/(v_(1)))` path `(D + E)`: `(Delta S)_("sys") = nR ln.(v_(2))/(v_(1)) + .^(n)C_(V) ln. (T_(2))/(T_(1)) = nR ln. (v_(2))/(v_(1))` |
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| 3285. |
How much heat energy must be supplied to change36 g of ice at `0^(@)C` to Water at room temperature `25^(@)C`? Data for Water `DeltaH_("fusion")^(@)=9Kj//mol, c_(p("liquid"))=4JK^(-1)g^(-1)`A. 18KJB. 3.6KjC. 22KJD. 21.6Kj |
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Answer» Correct Answer - d |
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| 3286. |
Which of the following statement is incorrect. (i) Entropy is measure of disorder. (ii) Entropy is not a state function. (iii) Entropy is expressed in the units of J/K/mol. (iv) The spontaneous process always accompanies the decrease in free energy. |
| Answer» (ii) Entropy is not a state function | |
| 3287. |
Four moles of an ideal gas expand isothermally from 1 litre to 10 litres are 300 K. Calculated the change in free energy of the gas `(R 8.314 JK^(-1)mol^(-1))`. |
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Answer» For isothermal reversible process `Delta H = 0` and `Delta S = 2.303"nRT log"(V_(2))/(V_(1))` `rArr Delta G=Delta H-T Delta S=-2.303 "nRT log"(V_(2))/(V_(1))=-2.303xx4xx8.314xx300"log"(10)/(1)=-22976.5 J` |
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| 3288. |
Heat capacity `(C_(V))` of an ideal gas is X KJ/mole/K. To rise its temperature from 298 K to 318 K, heat to be supplied per 10g gas will be (in KJ) [MW = 16]A. 16XB. `6.25 X`C. 32XD. `12.5 X` |
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Answer» Correct Answer - D `q_(V) = nc_(V)Delta T` `= (10)/(16) xx x xx 20 = 12.5 x` |
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| 3289. |
For the equilibrium `PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g)` at `298 K, K_(p)=1.8xx10^(-7)`. What is `Delta G^(@)` for the reaction ? |
| Answer» `Delta G = -2.303 "RT log " K_(p)=-2.303xx8.314xx10^(-3)kJ mol^(-1)K^(-1)xx298 K xx log 1.8xx10^(-7)=38.484 kJ mol^(-1)` | |
| 3290. |
On the basic of the following `Delta_(r)G^(Theta)` values at `1073K`: `S_(1)(s) +2O_(2)(g) rarr 2SO_(2)(g) Delta_(r)G^(Theta) =- 544 kJ mol^(-1)` `2Zn(s) +O_(2)(g) rarr 2ZnO(s) Delta_(r)G^(Theta) =- 480 kJ mol^(-1)` `2Zn(s) +S_(2)(s) rarr 2ZnS(s) Delta_(r)G^(Theta) =- 293 KJ mol^(-1)` Show that roasting of zinc sulphide to zince oxide is a spontaneous process. |
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Answer» Roasting of zinc sulphide to zinc oxide may be expressed as: `2ZnS(s) +3O_(2)(g) rarr 2ZnO(s) +2SO_(2)(g)` The given equations are: i. `S_(2)(s) +2O_(2)(g) rarr 2SO_(2)(g), Delta_(r)G^(Theta) =- 544kJ mol^(-1)` ii. `2Zn(s) +O_(2)(g) rarr 2ZnO(s), Delta_(r)G^(Theta) =- 480 kJ mol^(-1)` iii. `2Zn(s) +S_(2)(s) rarr 2ZnS(s), Delta_(r)G^(Theta) =- 293 kJ mol^(-1)` Subtract equation (iii) from equaiton (ii), we get iv. `2ZnS(s) +O_(2)(g) rarr 2ZnO(s) +S_(2)(s), Delta_(r)G^(Theta) =- 187 kJ mol^(-1)` Add equations (i) and (iv) `2ZnS(s) +3O_(2)(g) rarr 2ZnS(s) +2SO_(2)(g), Delta_(r)G^(Theta) =- 731 kJ mol^(-1)` Since `DeltaG^(Theta)` for the process is negative, the process of roasting of zinc sulphide is a spontaneous process. |
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| 3291. |
The relationship between `Delta H` and `Delta E` for the reaction `PCl_(3)(g)+Cl_(2)(g) rarr PCl_(5)(g)` is given asA. `Delta H = Delta E + RT`B. `Delta H = Delta E - RT`C. `Delta H = Delta E-2RT`D. `Delta H=Delta E+2RT` |
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Answer» Correct Answer - B `Delta n = 1-2=-1` |
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| 3292. |
For the system `S (s) + O_(2)(g) rarr SO_(2)(g)` ?A. `Delta H gt Delta E`B. `Delta E gt Delta H`C. `Delta H = 0`D. `Delta H = Delta E` |
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Answer» Correct Answer - D `Delta n = 0` |
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| 3293. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` final temperature of water is :A. 304.43 KB. 296.97 KC. 303.93 KD. 287 K |
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Answer» Correct Answer - B `9 (T - 273) xx 335 = 36 (323 - T) xx 4.18` |
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| 3294. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` What I dthe total entropy change in the process?A. `-1.56 JK^(-1)`B. `-1.60 JK^(-1)`C. `1.5 JK^(-1)`D. `1.60 JK^(-1)` |
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Answer» Correct Answer - C `Delta S = (Delta H)/(T)` |
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| 3295. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` `DeltaS_("water")is:`A. `-12.64JK^(-1)`B. `-0.34JKJ^(-1)`C. `-5.42JK^(-1)`D. `12.64JK^(-1)` |
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Answer» Correct Answer - a |
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| 3296. |
On the basic of the following `Delta_(r)G^(Theta)` values at `1073K`: `S_(1)(s) +2O_(2)(g) rarr 2SO_(2)(g) Delta_(r)G^(Theta) =- 544 kJ mol^(-1)` `2Zn(s) +O_(2)(g) rarr 2ZnO(s) Delta_(r)G^(Theta) =- 480 kJ mol^(-1)` `2Zn(s) +S_(2)(s) rarr 2ZnS(s) Delta_(r)G^(Theta) =- 293 KJ mol^(-1)` Show that roasting of zinc sulphide to zinc oxide is a spontaneous process.A. `-357 KJ`B. `-731 KJ`C. `-773 KJ`D. `-229 KJ` |
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Answer» Correct Answer - B `DeltaG` of formation of different substance are as `2SO_(2) =-544 KJ` `2 ZnS=-293 KJ` `2ZnO=-480 KJ` For the reaction, `2ZnS+ 3O_(2)(g) to 2ZnO(s) + 2SO_(2)(g)` `DeltaG=[(DeltaG_("(products)")-DeltaG_("(reactants)")]` `[(-480) + (-544)-(-293)]` `=-1024 + 293 =- 731 KJ` |
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| 3297. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` `DeltaS_(ice)is:`A. 11.04 `JK^(-1)`B. `3.16 JK^(-1)`C. `14.2 JK^(-1)`D. `7.84 JK^(-1)` |
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Answer» Correct Answer - C `Delta S = (Delta H)/(T)` |
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| 3298. |
9.0 gm ice `0^(@)C` is mixed with 36 gm of water at `50^(@)C` in a thermally insulated container.using the following data , Answer the qusestion that follow : `C_(p)(H_(2)O)=4.18Jg^(-1)k^(-1),DeltaH_("fusion ")(ice)=335Jg^(-1)` `DeltaS_("water")is:`A. `-12.64 JK^(-1)`B. `-0.34 JK^(-1)`C. `-5.42 JK^(-1)`D. `12.64 JK^(-1)` |
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Answer» Correct Answer - A `Delta S = (Delta H)/(T)` |
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| 3299. |
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter `DeltaU`and`w` correspond toA. `Delta U lt 0, w=0`B. `Delta U lt 0, w lt 0`C. `Delta U gt 0, w=0`D. `Delta U gt w gt 0` |
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Answer» Correct Answer - A `Zn+H_(2)SO_(4)rarr ZnSO_(4)+H_(2)`. It is an exothermic reaction so `Delta U lt O`. In Bomb calorimeter the process occurs at constant volume so w = 0. |
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| 3300. |
At `27^@C` latent heat of `I^-` fusion of a compound is `2.7xx10^3 J mol^(-1)`. Calculate the entropy change during fusion.A. `9.77 JK^(-1) "mol"^(-1)`B. `10.73 JK^(-1)"mol"^(-1)`C. `2930 JK^(-1)"mol"^(-1)`D. `108.5 JK^(-1)"mol"^(-1)` |
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Answer» Correct Answer - A Entropy , `DeltaS_(f) =(DeltaH_(f))/(T_(f))=("Fusion enthalpy")/("Temperature")` `DeltaS_(f)=(2930 J "mol"^(-1))/(300 K)=9.77 JK^(-1) "mol"^(-1)` |
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