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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.
| 3801. |
Calculate the enthalpy of vaporisation for water form the following: `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(g),DeltaH =- 57.0 kcal` `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l),DeltaH =- 68.3 kcal` Also calculate the heat required to change `1g H_(2)O(l)` to `H_(2)O(g)` |
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Answer» Given `{:(H_(2)(g)+1//2O_(2)(g)toH_(2)O(g), Delta=-H =- 94.3 kcal ...(i)),(H_(2)(g)+1//2O_(2)(g)toH_(2)O(l),DeltaH =68.3 kcal..(ii)),(" - - - + "),(ulbar{:("Subtracting equation (ii)from equation(i),"),(H_(2)O(l)toH_(2)O(g), DeltaH =+11.3 kcal):}):} ` Therefore, Enthalpy of vaporisation for `H_(2)O = +11.3 kcal` Also `18g H_(2)O` requires enthalpy of vaporisation `=+11.3 kcal` `:. 1 g H_(2)O` required enthalpy of vaporisation `= 11.3//8 kcal = 0.628 kcal` |
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| 3802. |
Use bond energy to estimate `DeltaH` for this rection: `H_(2)(g)+O_(2)(g)to H_(2)O_(2)(g)` `{:(Bond,"BondEnergy"),(H-H,436 KJ mol^(-1)),(O-O,142 KJ mol^(-1)),(O=O,499 KJ mol^(-1)),(H-O,460Kj .mol^(-1)):}`A. `-127KJ`B. `-209KJ `C. `-484KJ `D. `-841 Kj` |
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Answer» Correct Answer - a |
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| 3803. |
An isolated system can exchange_________ with its surrounding.A. energyB. massC. both energy and massD. neither energy nor mass |
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Answer» Correct Answer - D Hot water placed in a flask which is closed as well as insulated constitutes an isolated system as it cannot exchange matter as wellk as energy with its surroundings. |
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| 3804. |
Which of the following equations corresponds to the enthalpy of combustion at `298K`?A. `C_(2)H_(6)(g)+7//2O_(2)9g)rarr 2CO_(2)(g)+3H_(2)O(g)`B. `2C_(2)H_(6)(g)+7O_(2)(g)rarr 4CO_(2)(g)+6H_(2)O(g)`C. `C_(2)H_(6)(g)+7//2O_(2)(g)rarr 2CO_(2)(g)+3H_(2)O(l)`D. `2C_(2)H_(6)(g)+7O_(2)(g)rarr 4CO_(2)(g)+6H_(2)O(l)` |
| Answer» Correct Answer - C | |
| 3805. |
`H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g) DeltaH =- 242 kJ mol^(-1)` Bond energy of `H_(2)` and `O_(2)` is `436` and `500 kJ mol^(-1)`, respectively. What is bond energy of `O-H` bond? |
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Answer» `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g) DeltaH =- 242 kJ mol^(-1)` `DeltaH =` Bond energy of reactants -Bond energy of products `DeltaH = B_(H-H) +(1)/(2) B_(O=O) - 2B_(O-H)`, `- 242 kJ = 436 +(1)/(2) (500) - 2B_(O-H)` `-928 kJ =- 2B_(O-H) rArr B_(O-H) = (928)/(2) = 464 kJ mol^(-1)` |
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| 3806. |
From the following thermochemical equations, calculate the standard enthalpy of formation of HCl(g) . (A) `H_(2)(g) rarr 2H(g) , DeltaH^(0)= + 436.0 kJ mol^(-1) ` (B) ` Cl_(2)(g) rarr 2Cl(g), DeltaH^(0)= + 242.7 kJ mol^(-1)` (C ) `HCl(g) rarr H(g) + Cl(g) , DeltaH^(0) = + 431.8 kJ mol^(-1)` |
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Answer» Correct Answer - `-92.45kJ mol^(-1)` Aim `: (1)/(2) H_(2)(g) + (1)/(2) Cl_(2)(g) rarrHCl(g)` |
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| 3807. |
The amount of heat liberated when one mole of `NH_(4)OH` reacts with one mole of HCl isA. 13.7 KcalB. More than 13.7 KcalC. Less than 13.7 KcalD. Cannot be prodicted |
| Answer» Correct Answer - C | |
| 3808. |
If water is formed from `H^(+)` ions and `OH^(-)` the heat of formation of water is :A. `-13.7` KcalB. 13.7 KcalC. `-63.4` KcalD. More data required |
| Answer» Correct Answer - A | |
| 3809. |
If `H^(+)+OH^(-)rarrH_(2)O+13.7Kcal` , the heat of neutralisation for complete neutralisation of 1 mole of `H_(2)SO_(4)` by base will beA. 13.7 KcalB. 27.4 KcalC. 6.85 KcalD. 3.425 Kcal |
| Answer» Correct Answer - B | |
| 3810. |
Which of the following is a form of potential energy?A. Radiant energyB. Thermal energyC. Chemical energyD. Electrical energy |
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Answer» Correct Answer - C Chemical energy can be regarded as type of potential energy because it is associated with the relative positions and arrangements of atoms within the substance of interest. |
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| 3811. |
Which of the following is not the application of thermodnamics?A. It helps to predict the feasibility of a process.B. It tells the speed of a process.C. It helps in predicting the extent of reversible reaction before equilibrium is attained.D. It help of deduce some important laws. |
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Answer» Correct Answer - B Thermodynamics can predict whether any chemical reaction can occur or not under the given set of conditions but it does not tell anything about the rate at which the reaction takes place, furthemore, law of chamical equilibrium and distribution law are deduced from thermodynamics. |
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| 3812. |
A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butance. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts `(Delta_(C ) H^(@)` of `C_(4) H_(10) = - 2658 kJ mol^(-1))`A. 20B. 50C. 40D. 32 |
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Answer» Correct Answer - D Number of days cylinder will last `("Energy available from 14 kg butane")/("Energy requirement per day")` Energy released out by the combustion of 14 kg of butane is = (Number of moles of butane) `((14 xx 10^(3)g)/(58 g mol^(-1))) (2658 kJ mol^(-1))` `= 641.6 xx 10^(3) kJ` Note that `n_(C_(4)H_(10)) = Mass_(C_(4)H_(10))//"Molar mass"_(C_(4)H_(10))` and molar mass of `C_(4) H_(10` is `58 g mol^(-1)`. Thus, the number of days cylinder will last `= (641.6 xx 10^(3) kJ)/(20 xx 10^(3) kJ day^(-1))` `=32.08 days` |
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| 3813. |
`2CO_((g))+O_(2(g))rarr 2CO_(2(g))+X` KJ In the above equation X KJ refers to :A. Heat of formation of `CO_(2)`B. Heat of vapourisationC. Heat of reactionD. Heat of sublimation |
| Answer» Correct Answer - C | |
| 3814. |
`Delta H` for the reaction, `I_((g))+I_((g))rarr I_(2(g))` will be :-A. ZeroB. `-ve`C. `+ve`D. `oo` |
| Answer» Correct Answer - B | |
| 3815. |
Given that `,` `A(s) rarr A(l)DeltaH=x` `A(l) rarr A(g), DeltaH=y` The heat of sublimation of `A` will be `:`A. `x -y`B. `x +y`C. `x or y`D. `-(x+y)` |
| Answer» `{:(A(s)rarrA(1), DeltaH =x),(A(l)rarrA(g),DeltaH = y),(ulbar("On addition"A(s)rarrA(g),DeltaH =x+y)):}` | |
| 3816. |
Given that `,` `A(s) rarr A(l)DeltaH=x` `A(l) rarr A(g), DeltaH=y` The heat of sublimation of `A` will be `:`A. x + yB. x - yC. x or yD. `(x+y)` |
| Answer» Correct Answer - A | |
| 3817. |
The enthalpy change of a reaction does not depend onA. State of reactants and productsB. Nature of reactants and productsC. Different intermediate reactionsD. Initial and final ethalpy change of reaction |
| Answer» Correct Answer - C | |
| 3818. |
For the precipitation of AgCl by `Ag^(+)` ions and HClA. `Delta H = 0`B. `Delta G = 0`C. `Delta G = -ve`D. `Delta H = Delta G` |
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Answer» Correct Answer - C `Ag^(+)+HCl_((aq))rarr AgCl_((s))darr` It is a spontaneous process so, `Delta G = -ve`. |
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| 3819. |
From the thermochemical reactions, C(graphite)`+1//2O_(2)rarr CO , Delta H = -110.5` KJ `CO+1//2O_(2)rarr CO_(2),Delta H=-283.2` KJ the heat of reaction of C(graphite) `+O_(2)rarr CO_(2)` is :A. 393.7 KJB. `-393.7` KJC. `-172.7` KJD. `+172.7` KJ |
| Answer» Correct Answer - B | |
| 3820. |
From the thermochemical reactions, C(graphite)`+1//2O_(2)rarr CO , Delta H =-110.5`KJ `CO+1//2O_(2)rarr CO_(2),Delta H=-283.2` KJ the heat of reaction of C(graphite) `+O_(2)rarr CO_(2)` is :A. `-393.7 kJ`B. `+393.7 kJ`C. `-172.7 kJ`D. `+172.7 kJ` |
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Answer» Correct Answer - A `DeltaH=H_(P)-H_(R)` |
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| 3821. |
What is the sign of `Delta G` for the process of ice melting at 283 K ?A. `Delta G gt 0`B. `Delta G = 0`C. `Delta G lt 0`D. None of these |
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Answer» Correct Answer - C At `283K(10^(@)C)` ice melts easily so it is a spontaneous process. |
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| 3822. |
Calculate `DeltaH_(f)^(@)` for `Ubr_(4)` from the `DeltaG^(@)` of reaction and the `S^(@)` values at 298 K. `U(s)+2Br_(2)(l)rarrUBr_(4)(s), DeltaG^(@)=-788.6 kJ. " "S^(@)(J//K-"mol")50.3, 152.3, 242.6`A. `-822.1 kJ//mol`B. `-841.2 kJ//mol`C. `-775.6 kJ//mol`D. `-804.3 kJ//mol` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS` `DeltaS=242.6-50.3 -(2xx152.3), DeltaS=-112.3` `DeltaG=-788.6xx1000=DeltaH+298xx112.3` `DeltaH=-788.6 xx 1000 -298 xx 112.3` `DeltaH=-822.1 " kJ mol"^(-1)` |
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| 3823. |
The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `+5.527 KJ mol^(-1)`B. `-5.527 KJ mol^(-1)`C. `+55.27 KJ mol^(-1)`D. `-55.27 KJ mol^(-1)` |
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Answer» Correct Answer - B `Delta G^(@)=-2.303` RT log K `=-2.303xx8 5 10^(-3)xx300xxlog 10` `Delta G^(@)=-5.527 kJ mol^(-1)` |
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| 3824. |
The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`. |
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Answer» From the expression, `DeltaG^(theta)=-2.303RT "log"K_(eq)` `DeltaG^(theta)` for the reaction, `=(2.303)(8.314 JK^(-1)"mol"^(-1)) (300K)` log10 `-5744.14 J"mol"^(-1)` `-5.744 kJ "mol"^(-1)` |
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| 3825. |
Concrete is produced form a mixture of cement, water and small stones. Small amount of gypsum, `CaSO_(4).2H_(2)O` is added in coment production to impove the subsequent hardening of concrete. The elevated temperature during the production of cement may lead to the formation of unwanted hemihydrate `CaSO_(4)(1)/(2)H_(2)O` according to reaction. `CaSO_(4)2H_(2)O(s) rarr CaSO_(4)(1)/(2)H_(2)O(s)+(3)/(2)H_(2)O(g)` The `Delta_(f)H^(Theta) of CaSO_(2).2H_(2)O(s),CaSO_(4)(1)/(2)H_(2)O(s),H_(2)O(g)` are `-2021.0 kJ mol^(-1), -1575.0 kJ mol^(-1)` and `-241.8 kJ mol^(-1)`, respectively. The respective values of their standard entropies are `194.0, 130.0` and `188.0 J K^(-1)mol^(-1)`. The values of `R = 8.314JK^(-1)mol^(-1) = 0.0831L` bar `mol^(-1)K^(-1)`. Answer the following questions on the basis of above information. The equilibrium pressure of water vapour in closed vessel containing `CaSO_(4)2H_(2)O(s),CaSO_(4)(1)/(2)H_(2)O(s)` and `H_(2)O(g)` at `298K` (Antilog `-3.14 = 7.24 xx 10^(-4))` isA. `(17.24 xx 10^(-4))` barB. `(-7.24 xx 10^(-4))^(3)` barC. `(18 xx 10^(-4))^(2//3)`barD. `(7.24 xx 10^(-4))^(2//3)`bar |
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Answer» `DeltaG =- nRT In K_(P)` or `K_(P) =e^(-DeltaG//nRT)` |
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| 3826. |
Concrete is produced form a mixture of cement, water and small stones. Small amount of gypsum, `CaSO_(4).2H_(2)O` is added in coment production to impove the subsequent hardening of concrete. The elevated temperature during the production of cement may lead to the formation of unwanted hemihydrate `CaSO_(4)(1)/(2)H_(2)O` according to reaction. `CaSO_(4)2H_(2)O(s) rarr CaSO_(4)(1)/(2)H_(2)O(s)+(3)/(2)H_(2)O(g)` The `Delta_(f)H^(Theta) of CaSO_(2).2H_(2)O(s),CaSO_(4)(1)/(2)H_(2)O(s),H_(2)O(g)` are `-2021.0 kJ mol^(-1), -1575.0 kJ mol^(-1)` and `-241.8 kJ mol^(-1)`, respectively. The respective values of their standard entropies are `194.0, 130.0` and `188.0 J K^(-1)mol^(-1)`. The values of `R = 8.314JK^(-1)mol^(-1) = 0.0831L` bar`mol^(-1)K^(-1)`. Answer the following questions on the basis of above information. The value of equilibrium for reaction isA. `~~0`B. ` lt 1`C. `gt 1`D. `=1` |
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Answer» `DeltaG =- nRT In K` `:. K = e^(-DeltaG//nRT)` |
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| 3827. |
Calculate the resonance energy of `C_(6)H_(6)` using kekule formula of `C_(6)H_(6)` from the following data. a. `Delta_(f)H^(Theta)` for `C_(6)H_(6) =- 358.5 kJ mol^(-1)` b. Heat of atomisation of `C =716.8 kJ mol^(-1)` c. Bond enegry of `C-H, C-C, C=C`and `H-H` are `490, 620, 436.9 kJ "mole"^(-1)` respectively. |
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Answer» `6C(s) +3H_(2)(g) rarr C_(6)H_(6) Delta_(f)H_((exp)) =- 358.5 kJ` `Delta_(f)H_(cal) = BE(Reactants) -BE(Products)` `BE(Reactants) = 6C_((srarrg)) +3(H-H)` `= (6 xx 716.8 +3 xx 436.9)` `= 4300.8 +1310.7 = 5611.5` `BE(Products) = 3(C-C) +3(C =C) +6(C-H)` `=3 xx 340 +3 xx 620 +6 xx 490 = 5820 kJ` `DeltaH_(cal) = 5611.5 - 5820 =- 208.5 kJ mol^(-1)` `RE = Delta_(f)H_((Exp)) - Delta_(f)H_((Cal))` `=- 358.5 -(-208.5) =- 150.0 kJ mol^(-1)` |
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| 3828. |
Calculate the resonance enegry of toulene (use Kekule structure form the following data `C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l)+ DeltaH, DeltaH^(Theta) =- 3910 kJ mol^(-1)` `C_(7)H_(8)(l) rarr C_(7)H_(8)(g), DeltaH^(Theta) = 38.1 kJ mol^(-1)` `Delta_(f)H^(Theta) (water) =- 285.8 kJ mol^(-1)` `Delta_(f)H^(Theta) [CO_(2)(g)] =- 393.5 kJ mol^(-1)` Heat of atomisaiton of `H_(2)(g) = 436.0 kJ mol^(-1)` Heat of sulimation of `C(g) = 715.0 kJ mol^(-1)` Bond energies of `C-H, C-C`, and `C=C` are `413.0, 345.6`, and `610.0 kJ mol^(-1)`. |
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Answer» From the sata given, we have `C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l) +DeltaH(-3910kJ)` It is known that `DeltaH^(Theta) ("reaction") = (sum_(f)H^(Theta))("products") -(sum_(f)H^(Theta)) ("reactants")` `= (4 xx Delta_(f)H_(H_(2)O)(l)+ 7 xx Delta_(f)H_(CO_(2))^(Theta))-Delta_(f)H[C_(7)H_(8)(l)]` or `-3910 = [4 xx (-285) +7(-393.5)] - Delta_(f)H [C_(7)H_(8)(l)]` or `Delta_(f)H^(Theta) C_(7)H_(8)(l) = (3910 - 2754.5 - 1143.2) kJ` `= 12.3 kJ mol^(-1)` `:. Delta_(f)H^(Theta)` for gaseous toluene `= (12.3 +38.1) kJ = 50.4 kJ` Chemical equation for the formation of gaseous toulene `7C(s) +4H_(2)(g) rarr C_(7)H_(8)(g)` `(Delta_(f)H^(Theta))_(cal) =[ 7 xx ("heat of sublimation of solid" C + 4BE of H_(2)(g))` `=-3 xx C=C +4 xx C-C +8 xx C-H]` `=(7 xx 715 +4 xx 436) -(3 xx 610 +4 xx 345.6 + 8xx 413)` `= (5005 +1744) - (1830 +3304 +1382.4)` `= (6749.0 - 6516.4) kJ = 232.6 kJ` Reasonance enegry `= (232.6 - 50.40) kJ = 182.2 kJ mol^(-1)` |
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| 3829. |
For a given reaction, `C_(7)H_(8_(l))+9I_(2) (g) rarr 7CO_(2)(g) +4H_(2)O(l)` the calculated and observed heats of reaction are respectively `232` and `50.4 kJ mol^(-)`. The resonance energy for it will beA. `160 kJ mol^(-1)`B. `+172.2 kJ mol^(-1)`C. `+182.2 kJ mol^(-1)`D. `-182.2 kJ mol^(-1)` |
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Answer» Correct Answer - D `R.E =-R.E._(cal)-(-R.E._(obs))` |
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| 3830. |
How many calories are required to heat `40 g` of argon from `40^(@)C` to `100^(@)C` at constant volume? `(R = 2 cal mol^(-1) K^(-1))`A. 1200B. 120C. 180D. 2400 |
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Answer» Correct Answer - C Heat required to raise the temperature from `40^(@)C` to `100^(@)C` at constant volume is given is `q = nC_(V) Delta T` where `n` is the number of moles, `C_(V)` is molar heat capacity and `Delta T` is the temperature change. `n_(Ar) = ("Mass"_(Ar))/("Molar mass"_(Ar)) = (40 g)/(40 g mol^(-1)) = 1 mol` For monoatomic gases like argon, `C_(V) = 3 mol^(-1)`. Thus, `q = (1) (3) (100 - 40)` `180 cal` |
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| 3831. |
Heat of reaction `(DeltaH)` is given byA. `DeltaH=(E_(a))_(f)-(E_(a))_(b)`B. `DeltaH=BE_(R)-BE_(P)`C. `DeltaH=H_(P)-H_(R)`D. All the above |
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Answer» Correct Answer - D Standard definition |
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| 3832. |
Which one of the following sets of units represents the smallest and the largest amout of energy, respectively?A. `L` atm and `J`B. `eV` and `L` atmC. cal and eVD. erg and cal |
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Answer» Correct Answer - B `1 L atm = 101.3 J` `1 erg = 10^(-7) J` `1 cal = 4.186 J` `1 eV = 16. xx 10^(-19) J` Thus, the smallest amount of energy is `1 eV` and the largest amount of energy is `L` atm. |
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| 3833. |
Which one of the following sets of units represents the smallest and the largest amount of energy respectively ?A. J and ergB. erg and calC. cal and eVD. eV and lit atm |
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Answer» Correct Answer - d Smallest `=` eV and Largest `=` lit atm `1eV = 1.6 xx 10^(-19) J ,1 cal = 4.186 J` ` 1 erg = 10^(-7) J , 1 `lit atm` =101.3 J` |
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| 3834. |
Calculate `DeltaG` and `DeltaG^(@)` for the reaction `: A + B hArr C + D` at `27^(@)C` .Equilibrium constnat`(K)` for this reaction `= 10^(2)` |
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Answer» `DeltaG = 0` because the reactin is in equilibrium. `DeltaG^(@) = - 2.303 RT logK = -2.303 xx 8.314 JK^(-1) mol^(-1) xx 300 K log 10^(2)` `= - 1148 J mol^(-1) = - 11.488 kJ mol^(-1)` |
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| 3835. |
Work done in expansion of an ideal gas from 4 `dm^(3) ` to `6 dm^(3)`against a constant external pressure of 2.5 atm was usedup to heat1 mole of water at `20^(@)C`. Calculate the final temperature of water ( Given `:` specific heat of water `=4.184 Jg^(-1)K^(-1))`. |
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Answer» As work is being against constant external pressure, the process is irreversible . Hence, ` w= - P _(ext)DeltaV = - 2.5 atm ( 6-4) dm^(3) = - 5L atm ` `( 1 dm^(3) = 1L)` `= - 5 xx 101.3 J = - 506.5 J ` `( 1L atm = 101.3J )` For isothermal expansion of an ideal gas , `DeltaU =0`so that`q= - w=506 .5 J` This heat is used up to heat 1 mole of water. Applying the reaction ` q= m xxc xx DeltaT` `= 506. 5 =18 xx4.184 xxDeltaT` or`DeltaT=6.725^(@)` `:.` Final temperature`=20 +6.725 ^(@) C =26.725^(@) C` |
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| 3836. |
Calculate the value of log `K_(p)` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` at `25^(@)C`. The standard enthalpy of formation of `NH_(3)(g)`is`-46kJ` and standard entropiesof `N_(2)(g), H_(2)(g)` and`NH_(3)(g)` are are 191,130,193`JK^(-1) mol^(-1)` respectively . `( R = 8.3 JK^(-1) mol^(-1))` |
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Answer» `Delta_(f)G^(@) (NH_(3)) = DeltaH^(@) - TDeltaS^(@) = -46000- ( 298 xx192) J mol^(-1) = - 103.2 kJ mol^(-1)` `Delta_(f)G^(@) (N_(2))= DeltaH^(@) - T DeltaS^(@) = 0 -298 xx191 J mol^(-1) = -56.9 kJ mol^(-1)` `Delta_(f)G^(@) ( H_(2)) = DeltaH^(@) - T DeltaS^(@) =0 - 298 xx 130 J mol^(-1) = - 38.74kJ mol^(-1)` `DeltaG^(@)` for reaction `Delta _(r)G^(@) = 2 xx Delta_(f)G^(@) ( NH_(3)) - Delta_(f)G^(@) ( N_(2))-3Delta_(f)(H_(2)) =- 206.4 +56.9 +116.2 = - 33.28 kJ mol^(-1)- DeltaG^(@) =2.303 RT log K` ` 33.28 = 2.303 xx (8.314 xx 10^(-3)) log K` or `log K =1.738 xx 10^(3)` |
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| 3837. |
Consider the following reaction `:` `Ag^(+ ) (aq) +Cl^(-)(aq) rarrAgCl(s)` When 10.0mLof 1.0 M`AgNO_(3)` solution is added to 10 mL of 1.0 mL NaCl solution at `25^(@)C` in a calorimeter , a white ppt. of AgCl is formed and the temperature of the aqueous mixture mixture risesto `32.6^(@)C`. Assuming that the specific heatof the aqueous mixture is `4.18 J //g//^(@)C`, that the density of the mixture is `1.0 g mL^(-1)` and that the calorimeter itself absorbs a negligible amountofheat, calculate the value of enthalpy change accompanying the processin kJ `mol^(-1)` ofAgCl. |
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Answer» Mass of the mixture solution`=`Volume `xx` Density `= 20 mL xx1 g mL^(-1) =20g` Rise is temperature `= 32.6 -25 = 7.6^(@)C` Heat evolved `= m xxC xx Delta t = 20 xx 4.18 xx 7.6 =640J` Moles of`Ag^(+) =( 1)/( 1000) xx 10 =10^(-2)` `:. ` Moles ofAgCl `= 10^(-2)` `:. ` Heat evolved per mole of AgCl ` ( 640 )/( 10^(-2)) = 64000 J mol^(-1) = 64kJ mol^(-1)` |
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| 3838. |
Which one is not correct for a cyclic process as shown in the figure ? A. dU = 0B. q = -WC. W = 314 JD. W = 31.4 J |
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Answer» Correct Answer - A::B::C For a cyclic process `dU = 0` `:. q = Delta U + (-w) " " q = -w` Also, w = area covered by sphere `= pi r^(2)=pi xx[((V_(2)-V_(1)))/(2)]^(2) = (pi xx (20)^(2))/(2^(2)) = 100 xx 3.14 = 3.14 J` |
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| 3839. |
When `100` ml of `0.2 M HCl` is mixed with 100 ml of `0.2 M NaOH`, the rise in temperature is `T_(1)`. When the experiment is repeated using 200 ml each of the same solutions, the rise in temperature is `T_(2)`. ThenA. `T_(1)=T_(2)`B. `2T_(1)=T_(2)`C. `T_(1)=4T_(2)`D. `T_(1)=2T_(2)` |
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Answer» Correct Answer - A The rise in temperature depend on numbers of equivalents of acid and base |
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| 3840. |
When 20 ml of `0.1 M HCl` is mixed with 20 ml of `0.1 M NaOH`, the rise in temperature is `T_(1)`. When the experiment is repeated using 20 ml of `0.2 M HCl` and `0.2 M NaOH` solutions, the rise in temperature is `T_(2)`. ThenA. `T_(1)=T_(2)`B. `T_(1)=2T_(2)`C. `T_(1)=4T_(2)`D. `2T_(1)=T_(2)` |
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Answer» Correct Answer - D 2m mol- `T_(1)` 4m mol - `2T_(1)` |
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| 3841. |
Given, `2Fe_2O_(3(s)) to 4Fe_((s)) + 3 O_(2(g)), Delta_r G_1^@ = +1487kJmol^(-1)` `6CO_((g)) +3O_(2(g)) to 6CO_(2(g)), Delta_r G_2^@ = 1543.2kJmol^(-1)` Select the correct statements:A. `Delta_(r)G^(@)` for reduction of iron oxide by `CO` is `-56.2 k J m^(-1)`B. `Fe_(2)O_(3)` cannot be reduced by `CO` spontaneouslyC. `Fe_(2)O_(3)` can be reduced by `CO` spontaneouslyD. The reduction of `Fe_(2)O_(3)` takes part in higher part of blast furnace |
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Answer» Correct Answer - A::B::D `Delta_(r)G^(@) for 2Fe_(2)O_(3)+6COto4Fe+6CO_(2)` `Delta_(r)G^(@)=Delta_(r)G_(1)^(@)+Delta_(r)G_(2)^(@)=+1487-1543.2=-56.2 kJ mol^(-1)` |
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| 3842. |
Select the incorrect statements for the equillibrium under standard condition: `H_(2)O_((s))hArrH_(2)O_((l)), DeltaS_(1)^(@)` `H_(2)O_((l))hArrH_(2)O_((v)), DeltaS_(2)^(@)` `H_(2)O_((s))hArrH_(2)O_((v)), DeltaS_(3)^(@)`A. `Delta S_(1)^(@) gt Delta S_(2)^(@)`B. `Delta S_(2)^(@) gt gt gt Delta S_(1)^(@)`C. `DeltaS_(3)^(@) gt Delta S_(2)^(@)`D. `Delta S_(3)^(@) gt Delta S_(1)^(@)` |
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Answer» Correct Answer - B::C::D `H_(2)O_((s))` has more ordered arrangement. Also `Delta S_(2)^(@) = S_(H_(2)O_((v)))^(@) - S_(H_(2)O_((l)))^(2), Delta S_(0)^(1) = S_(H_(2)O_((l)))^(@)-S_(H_(2)O_((s)))^(@)` `:. S_(H_(2)O_((v)))^(@)` is maximum and thus `Delta S_(2)^(@) gt gt gt Delta S_(1)^(@)` |
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| 3843. |
Thermodynamics is not concerned about….A. energy changes involved in a chemical reactionB. the externt to which a chemical reaction proceedsC. the rate at which a reaction proceedsD. the feasibility of chemical reaction |
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Answer» Correct Answer - C Thermodynamcs in not concerned with rate at which a reaction proceeds. The rate of reaction is dealt by kinetics. |
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| 3844. |
The amount of heat evolved when `500 cm^(3) 0.1 M HCl` is mixed with `200 cm^(3)` of `0.2 M NaOH` isA. All are correctB. i and ii are correctC. i, ii and iii are correctD. ii and iii are correct |
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Answer» Correct Answer - D `0.25xx57.3 kJ` |
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| 3845. |
The amount of heat evolved when 0.50 mole of HCl is mixed with 0.30 mole of NaOH solution isA. 57.1 kJB. 28.55 kJC. 11.42 kJD. 17.13 kJ |
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Answer» Correct Answer - D `"0.50 mol of HCl "-="0.50 mol of "H^(+)` `"0.30 mol of NaOH "-="0.30 mol of OH"^(-1)` `H^(+)+OH^(-) rarr H_(2)O" (Neutralisation)"` `"0.30 mol of HCl" -="0.30 mol of H"^(+)` `"or 0.30 mol of "H^(+)-=" 0.30 mol of "OH^(-)` `"Heat evolved with 1 mol H"^(+)="57.1kJ"` `"Heat evolved with 0.30 mol H"^(+)=57.1xx0.30="17.13 kJ"` |
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| 3846. |
Which of the following statements is correct ?A. The presence of reacting species in a covered beaker is an example of open system.B. There is an exchange of energy as well as matter between the system and the surounding in a closed systemC. The presence of rectants in aclose vessel made up of copper is an exampleof a closed systemD. The presence of reactants in a thermose flask or any other closed insulated vessel is an example of a closed system |
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Answer» Correct Answer - C |
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| 3847. |
Thermodynamics is not concerned about….A. energy changes involved in a chemical reactionB. the extent to which a chemical reaction proccedesC. the rate at which a reaction proceedesD. the feasibility of a chemical reaction |
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Answer» Correct Answer - C |
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| 3848. |
Which of the following reactions will have the value of enthalpy of neutralisation as `-57.1"kJ mol"^(-1)` ?A. `CH_(3)COOH+NaOH rarr CH_(3)COONa+H_(2)O`B. `HCl+NaOH rarr NaCl +H_(2)O`C. `HCl+NH_(4)OHrarr NH_(4)Cl+H_(2)O`D. `HCOOH+NaOH rarr HCOONa+H_(2)O` |
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Answer» Correct Answer - B Enthalpy of neutralisation is `-"57.1 kJ mol"^(-1)` for a strong acid and a strong base. |
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| 3849. |
Form the given T-S diagram of a reversible carnot engine, find (i) work delivered by engine in one cycle (ii) heat taken from the source in each cycle . (iii) `DeltaS_(sink)` in each cycle . |
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Answer» (i) `W_(AB) = - nRT ln.(V_(2))/(V_(1))` `DeltaS =(q_(rev))/(T) = - (W_(AB))/(T)` `rArr - W_(AB) = TDeltaS = 600 xx100` `" "- W_(BC) = -.^(n)C_(v)(T_(1) - T_(2))` `" "- W_(CD) = TDeltaS = 300 xx(-100)` `" "- W_(DA)= - .^(n)C_(v) (T_(2) - T_(1))` net work delivered during one cycle `= - W_(AB) - W_(BC) - W_(CD) - W_(DA) = 300 xx 100 = 30 kJ` Note: Net work done = area of the rectangle (ii) `(W_("net"))/(q) = eta" "and " " eta= (600-300)/(600) =(1)/(2)` `rArr q=` "heta taken from the the source" `(-W_("net"))/(1//2) = (30kJ)/( 1//2) = + 60 kJ` (iii) `DeltaS_("sink") = - (Q_("sink"))/(T) " "also" "(q_("source") + q_("sink"))= 30` `" "q_("source") = 600 " " rArr " "q_("sink") = - 30 kJ` `rArr " "DeltaS_("sink") = (-q_("sin K"))/(T) = (_(-30000J))/(100) = 100 J//K` |
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| 3850. |
A reaction at 300 K with `DeltaG^(@)` `=- 1743 J//mol` consister of 3 mole of A(g), mole of B(g) and 3 mole of C (g). If A, B and C(g) . If A,B and C are equal are in equilibrium in 1 liter container then the reaction may be `[Given :2=e^(0.7), R=8.3 J//K-mol]` |
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Answer» Correct Answer - C `DeltaG^(@) =- "RT In K"_(eq)` `or,- 1743 = - 8.3 xx300 xx In K_(ep)` `K_(eq) = 2 =K_(eq) " for reaction 2A"hArr B+C " from given data"` |
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