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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.
| 51. |
Acetoc acid `CH_(3)COOH` can form a dimer `(CH_(3)COOH)_(2)` in the gas phase. The dimer is held togther by two `H-`bonds with a total strength of `60.0 kJ per` mole of dimer If at `25^(@)C`, the equilibrium constant for the dimerisation is `1.3 xx 10^(3)`, calculate `DeltaS^(Theta)` for the reaction `2CH_(3)COOH (g) hArr (CH_(3)COOH)_(2)(g)` |
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Answer» `DeltaG^(Theta) =- 2.303 RT log_(10)k` `=- 2.303 xx 8.314 xx 298 log (1.3 xx10^(3))` `=- 17767.688 J = - 17.767 kJ` `DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `- 17.767 =- 60.0 - 298 xx DeltaS^(Theta)` `DeltaS^(Theta) = (-60.0+17.767)/(298) =- 0.141 kJ` |
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| 52. |
The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below: `CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)` `DeltaH^(Theta)underset(300K). =- 41.16 kJ mol^(-1)` `DeltaS^(Theta)underset(300K). =- 4.14 xx 10^(-2) kJ mol^(-1)` `DeltaH^(Theta)underset(1200K). =- 31.93 kJ mol^(-1)` `DeltaH^(Theta)underset(1200K). =- 2.96 xx 10^(-2) kJ mol^(-1)` Calculate `K_(p)` at each temperature and predict the direction of reaction at `300K` and `1200k`, when `P_(CO) = P_(CO_(2)) =P_(H_(2)) = P_(H_(2)O) =1` atm at initial state. |
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Answer» At `300K: DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `=- 41.16 - 300 xx (-4.24xx10^(-2))` `=- 28.44 kJ` Since, `DeltaG^(Theta)` is negative, hence recaiton is spontaneous is forward direction. `DeltaG^(Theta) =- 2.3030 RT log K_(p)` `-28.44 = - 2.303 xx 8.314 xx 10^(-3) xx 300 log_(10) K_(p)` `K_(p) = 8.93 xx 10^(4)` At `1200K: DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `=- 32.93 - 1200 (-2.96 xx 10^(-2))` ` =+ 2.59 kJ` Positive value of `DeltaG^(Theta)` shows that the reaction is sponteneous in backward direction `Deltag^(Theta) =- 2.303 RT log_(10) K_(p)` `2.59 =- 2.303 xx 8.314 xx 1200 log K_(p)` `K_(p) = 0.77` |
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| 53. |
For the reaction between `CO_(2)` and graphite: `CO_(2)(g)+C(s)to2CO(g),` `Delta=170kJ and DeltaS=170.JK^(-1). "The reaction will be spontaneous at": `A. 300KB. 500KC. 900KD. 1100K |
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Answer» Correct Answer - D |
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| 54. |
For the reaction, `4C(graphite) +5H_(2)(g) rarr nC_(4)H_(10)(g)`, `DeltaH^(Theta) =- 124.73 kJ mol^(-1), DeltaS^(Theta) =- 365.8 J K^(-1) mol^(-1)` `4C(graphite) +5H_(2)(g) rarr iso-C_(4)H_(10)(g)` `DeltaH^(Theta) =- 131.6 kJ mol^(-1), DeltaS^(Theta) =- 381.079 J K^(-1) mol^(-1)` Indicate whther normal butane can be spontaneously converted to iso-butane or not. |
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Answer» For `nC_(4)H_(10)`, `DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)` `=- 124.73 - 298 (-365.8 xx 10^(-3))` `= - 15.72 kJ` For isobutane (iso-`C_(4)H_(10))`, `DeltaG^(Theta) =- 131.6 - 298 (-381.079 xx 10^(-3)) =- 18.04 kJ` For conversion of `nC_(4)H_(10) rarr iso-C_(4)H_(10)`, `DeltaG^(Theta) =- 18.04 -(-15.72) =- 2.32 kJ` Negative value shows that the process is spontaneous. |
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| 55. |
The value of `DeltaH` and `DeltaS` for the reaction `C_("(graphite)") +CO_92)(g) rarr 2CO(g0` are170 kJ and `170JK^(-1)` respectively. This reaction will be spontaneous atA. 910 KB. 1110KC. 510KD. 710 K |
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Answer» Correct Answer - B `DeltaG =DeltaH - T DeltaS`. For reaction to be spontaneous `DeltaS` should be -ve, i.e.,`DeltaH-TDetlaS lt0 `or `DeltaHlt T DeltaS` or `TDeltaS gt DeltaH ` or`Tgt ( DeltaH )/( DeltaS) , i.e.,T gt (170000J)/( 170 JK^(-1))` or `T gt 1000K`. Hence, options (b) iscorrect. This method can be applied only when `DeltaH` and`DeltaS` both are `+ve`. Alternatively, for equilibrium , `DeltaG =0` `i.e., T DeltaS = DeltaH` or `T= ( DeltaH )/( DeltaS) =( 170000)/( 170) =1000K` For spontaneity , `DeltaG = -ve`which can be so if `T gt 1000 K` ( so that`T DeltaS gt DetalH` in magnitude ) . |
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| 56. |
A certain electric motor produced `16kJ` of energy each second as mechanical work and lost `3kJ` as heat to the surroundings. What is the change in the internal energy of the motor and its power supply each second? |
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Answer» Energy is lost form the system as work, thus `W` is negative. `:. W =- 16kJ` Enegry is lost as heat, so `q =- 3kJ` Therefore, by the first law of thermodnamics, `DeltaU = q +w =- 3 - 16 - 19kJ` |
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| 57. |
A sample consisting of `1mol` of a mono-atomic perfect gas `(C_(V) = (3)/(2)R)` is taken through the cycle as shown. `DeltaU` for the process` (1 rarr2)` isA. `0.00J`B. `+3.40 xx 10^(3)J`C. `-3.40 J`D. `-3.40 xx 10^(3)J` |
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Answer» In expansion from state `1` to state `2`. `P = 1 atm` `V_(1) = 22.44 L` `V_(2) = 44.88 L` `C_(V) = (3)/(2)R` `DeltaT = T_(2) - T_(1) = 546 - 273 = 273 K` `:. DeltaU = nC_(V) DeltaT =1 xx (3)/(2) xx 8.314 xx273 = 3.40 xx 10^(3)J` |
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| 58. |
10 g of argon is compressd isothermally and reversibly at a temperature of`27^(@)C` from 10L to 5L . Calculate q,w,`DeltaE` and `DeltaH ` for this process. `R =2.0 cal K^(-1) mol^(-1) , log= 2 = 0.30`. Atomic wt. of Ar`=40` |
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Answer» `q= 2.303 nRT log. (V_(2))/(V_(1)) = 2.303 xx (10)/( 40) xx 300 xx log. (5)/(10)= -103.635 cal` For isothermal expansion,`DeltaE= 0` `w= DeltaE-q= 0-( - 103.635)= +103.635 cal` Also, when temperature is constant, `P_(1)V_(1) = P_(2)V_(2)` or `PV =` constant `DeltaH = DeltaE +Delta(PV) = 0+0=0` |
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| 59. |
Given: i. `2Fe(s) +(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s), DeltaH^(Theta) =- 193.4 kJ` ii. `Mg(s) +(1)/(2)O_(2)(g) rarr MgO(s), DeltaH^(Theta) =- 140.2kJ` What is `DeltaH^(Theta)`of the reaction? `3Mg +Fe_(2)O_(3)rarr 3MgO +2Fe`A. `-227.2kJ`B. `-272.3kJ`C. `227.2kJ`D. `272.3kJ` |
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Answer» Substracting equation (i) from equation and (ii) multiplied by `3`, we get i. `2Fe(s) +(3)/(2) O_(2)(g) rarr Fe_(2)O_(3)(g), DeltaH^(Theta) =- 193.4kJ` ii. `3Mg(s) +(3)/(2) O_(2)(g)rarr 3MgO(s), DeltaH =- 420.6 kJ` Subtracting (i) from (ii), `3Mg(s) + Fe_(2)O_(3) rarr 3MgO + 2Fe`, `DeltaH^(Theta) =- 420.6 -(-193.4) =- 227.2kJ` |
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| 60. |
A certain electric motor produced 15 kJ of energy each second as numerical work and lost 2kJ as heat to the surroundings. What is the change in the internal energy of the motor and its power supply each second.A. `-17 kJ`B. `17 J`C. `27 kJ`D. `23 kJ` |
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Answer» Correct Answer - A `Q=DeltaU -W` |
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| 61. |
The second law of thermodynamics is a fundamental law of science. In this problem, we consider the thermodynamics of an ideal gas, phase transition, and chemical equilibrium. Three moles of `CO_(2)` gas expands isothermally (in thermal contact with the surroundings, temperature `= 15.0^(@)C`) against a fixed external pressure of `1.00 bar`. The initial and final volumes of the gas are `10.0 L` and `30.0L`, respectively. Select the correct order of the entropy change.A. `Delta_(sys)S gt 0, Delta_(surr)S = 0`B. `Delta_(sys)S lt 0, Delta_(surr)S gt 0`C. `Delta_(sys)S gt 0, Delta_(surr)S lt 0`D. `Delta_(sys)S gt 0, Delta_(surr)S = 0` |
| Answer» In expansion of `CO_(2)` gas molecule, entropy of system `Delta_(sys)S` increases or `Delta_(sys)S gt 0` and entropy of surrounding decreases `Delta_(surr)S lt 0`. | |
| 62. |
Assertion : First law of thermodynamics is a restatement of the principle of conservation. Reason : Energy is fundamental quantity.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - C |
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| 63. |
A sample consisting of `1mol` of a mono-atomic perfect gas `(C_(V) = (3)/(2)R)` is taken through the cycle as shown. `DeltaU` for the process `(2 rarr 3)` isA. `0.00J`B. `+3.40 kJ`C. `-3.40 kJ`D. None of these |
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Answer» In conversion from state `2` to state `3` `DeltaT = T_(2) - T_(1) = 273 - 546 =- 273 K` `C_(V) = (3)/(2)R` `:. DeltaU = nC_(V) DeltaT = 1 xx (3)/(2) xx 8.314 xx (-273)` `=- 3.40 xx 10^(3)J` |
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| 64. |
Water decomposes by absorbing 286.2 kJ of electrical energy per mole. When `H_(2)` and `O_(2)` combine to form one mole of `H_(2)O, 286.2kJ` of heat is produced. Which law is proved ? What statement of the law followsfrom it? |
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Answer» Law of conservation of energy ( or 1st law of thermodynamics) Def.Energy can neither be created or destroyed , although it may be converted from one formto another. |
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| 65. |
The second law of thermodynamics is a fundamental law of science. In this problem, we consider the thermodynamics of an ideal gas, phase transition, and chemical equilibrium. Three moles of `CO_(2)` gas expands isothermally (in thermal contact with the surroundings, temperature `= 15.0^(@)C`) against a fixed external pressure of `1.00 bar`. The initial and final volumes of the gas are `10.0 L` and `30.0L`, respectively. Assuming `CO_(2)` to be an ideal gas, `Delta_(sys)S` isA. `27.4 J K^(-1)`B. `9.1J K^(-1)`C. `-27.4J K^(-1)`D. `-9.1 J K^(-1)` |
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Answer» `-`Work done `=q = nRT In (V_(2))/(V_(1))` `= 3 xx 8.314 xx 288.15 xx In ((30)/(10))` `:. DeltaS = (q)/(T)` or `Delta_(sys)S = (3 xx 8.314 xx 288.15 xx In(30)/(10))/(288.15) = 27.4 J K^(-1)` |
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| 66. |
A sample consisting of `1mol` of a mono-atomic perfect gas `(C_(V) = (3)/(2)R)` is taken through the cycle as shown. `DeltaH` for the overall cycle isA. `+5.67 xx 10^(3)J`B. `-5.67 xx 10^(3)J`C. `-11.34 xx 10^(3)J`D. Zero |
| Answer» In cyclic process, `DeltaH = 0`. | |
| 67. |
Compute the heat of formation of liquid methyl alcohol in `kJ mol^(-1)` , using the following `:` Heat of vaporisation of liquid methyl alcohol `=38 kJ mol^(-1)`. Heat of formation of gaseous atoms from the elements in their standard states `: H, 218 kJ//mol , C,715 kJ // mol, O, 249 kJ //mol`. Averagebond energies `: C-H, 415 kJ //mol ,C-O kJ //mol, O-H, 463 kJ //mol` |
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Answer» We aim at `: C(s) +2H_(2)(g) + (1)/(2)O_(2) rarr CH_(3)OH(l), DeltaH =?` We are given `: (i) CH_(3)OH(l) rarr CH_(3) OH (g), DeltaH = 38 kJ mol^(-1)` (iii) `C(s)rarr C(g), DeltaH =715 kJ mol^(-1)` (iv) `(1)/(2)O_(2)(g) rarrO(g) , DeltaaH -249 kJ mol^(-1)` Also from the given bond energies, we have (v) `CH_(3)OH(g)(H- underset(H) underset(|) overset(H)overset(|) (C) -O-H) rarr C(g) + 4H(g)+O(g), DeltaH = 3 xx 415 +356 +463 = 2064kJ mol^(-1)` Eqn. (iii) `+4 ` Eqn. (ii) `+` Eqn. (iv)- Eqn.(i) - Eqn. (v) gives the requried result i.e., `DeltaH = 715 + 4( 218) + 249-38 -2064 = -266 kJ mol^(-1)` |
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| 68. |
The standard heat of formation values of `SF_(6)(g), S(g)`, and `F(g)` are `-1100, 275`, and `80 kJ mol^(-1)`, respectively. Then the average `S-F` bond enegry in `SF_(6)`A. `310 kJ mol^(-1)`B. `220 kJ mol^(-1)`C. `309 kJ mol^(-1)`D. `280 kJ mol^(-1)` |
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Answer» `S(g) +6F(g) rarr SF_(6)(g), DeltaH =- 1100 kJ mol^(-1)` `S(s) rarr S(g) ,DeltaH =+ 275 kJ mol^(-1)` `(1)/(2) F_(2)(g) rarr F(g),DeltaH = 80 kJ mol^(-1)` Therefore, heta of formation =Bond enegry of reaction -Bond enegry of product `- 1100 = (275 +6 xx 80) - (6 xx S -F)` Thus, bond enegry of `S-F = 309 kJ mol^(-1)` |
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| 69. |
Estimate the average S-F bond energy in `SF_(6)`.The standard heat of formation values of `SF_(6)(g)`, S(g) andF(g) are `: - 1100,275` and`80 Kj mol^(-1)` respectively. |
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Answer» Aim `: (1)/(6) [SF_(6)(g) rarrS(g) +6F(g)],DeltaH=?` For the reaction , `SF_(6)(g) rarrS(g) + 6F(g)` `Delta_(f)H = Delta_(f)h^(@) [S(g)+ 6[Delta_(f)H^(@) F (g)]-[Delta_(f)H^(@) SF_(6)(g)=275 + 6(80)-( -1100) kJ mol^(-1) = 1855kJ mol^(-1)` `:. ` AverageS-F bond energy `= ( 1855)?( 6) =309.16 kJ mol^(-1)` |
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| 70. |
From the following data, calculate the enthalpy change for the combustion of cyclopropane at `298K`. The enthalpy of formation of `CO_(2(g)),H_(2)O_((l))` and `Propen e_((g))` are `-393,-285.8` and ` 20.42 kJ mol^(-1)` respectively. The enthalpy of isomerisation of cyclopropane to propene is `-33.0kJ mol^(-1)` |
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Answer» Correct Answer - `-2091.32 kJ |
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| 71. |
Reactions that have standard free energy change less than zero always have equilibrium constant to beA. unityB. greaterC. lessthan unityD. zero |
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Answer» Correct Answer - B `DeltaG^(@) = - 2.303 RTlog K. `When `Delta G^(@) lt0, `i.e., `-ve, log K `is `+ ve `, i.e., `K gt 1` |
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| 72. |
0.16g of methane was subjected to combustion at `27^(@)C` in a bomb calorimeter system. The temperature of the calorimeter system ( including water ) was found to rise by `0.5^(@)C`. Calculate the heat of combustion of methane at (i) constant volume, and (ii) constant pressure . The thermal capacity of the calorimeter system is `17.7 k J K^(-1) ( R = 8.314 kJ K^(-1) mol^(-1))` |
| Answer» Correct Answer - `q_(v) = - 885 kJ mol^(-1) , q_(p) = -890 kJ mol^(_1)` | |
| 73. |
A sample of `CH_(4)` of `0.08 g` was subjected to combustion at `27^(@)C` in a bomb calorimater. The temperature of the calorimeter system was found to be raised by `0.25^(@)C`. IF heat capacity of calorimeter is 18K J, `Delta H` for combustion of `CH_(4)` at `27^(@)C` isA. `-900` kJ/moleB. `-905` kJ/moleC. `-895` kJ/moleD. `-890` kJ/mole |
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Answer» Correct Answer - B `DeltaU=ZxxthetaxxM/W" , "DeltaH=DeltaU+Delta nRT` |
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| 74. |
A sample of `0.16 g CH_(4)` was subjected to combustion at `27^(@)C` in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by `0.5^(@)C`. Calculate the heat of combustion of methane at (a) constant volume and (b) constant pressure. The thermal capacity of calorimeter system is `17.0 kJ K^(-1)` and `R = 8.314 J K^(-1) mol^(-1)`. |
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Answer» Heat of combustion at constant volume, `DeltaU` `=`Heat capacity of calorimeter system xx Rise in themperature `xx ("Molecular mass of compound")/("Mass of compound")` `Deltan = 1 - 3 = 2,T = 300 K, R = 8.314 xx 10^(-3) kJ K^(-1) mol^(-1)` `DeltaH^(Theta) = DeltaU^(Theta) = DeltanRT` `=- 850 + (-2) xx 8.314 xx 10^(-3) xx 300` `=- 850 - 4.988 =- 854.988 kJ mol^(-1)` |
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| 75. |
The heat of combustion of ethane gas is `-368 kcal mol^(-1)`. Assuming that `60%` of heat is useful, how many `m^(3)` of ethane measured at `NTP` must be burned to supply heat to convert `50 kg` of water at `10^(@)C` to steam at `100^(@)C`? |
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Answer» Heat required per gram of water `= (90 + 540) cal = 630 cal` Total heat needed for `50 kg` of water `= 50 xx 10^(3) xx 630 kcal` As the efficiency is `60%`, the acid amount of heat required `=(50 xx 10^(3) xx 630)/(60) xx 100 = 52500 kcal` Number of mole of ethane required to produce `52500 kcal` `= (52500)/(368) = 142.663 mol` Volume of `142.663` mole at `NTP = 142.663 xx 22.4` `= 3195.65 L = 3.195 m^(3)` |
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| 76. |
Standard entropies of `X_(2), Y_(2)` and`XY_(3)` are 60,40 and `50JK^(-1)mol^(-1)` respectively. For the reaction (1)/(2)X_(2) + (3)/(2)Y_(2) hArr XY_(3) , DeltaH= - 30 kJ ` to be at equilibrium , the temperature should beA. 500KB. 750 KC. 1000KD. 1250 K |
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Answer» Correct Answer - b `(1)/(2) X_(2) + ( 3)/(2)Y_(2) hArr XY_(3)` `DeltaS^(@) = SigmaS_(P)^(@) - Sigma _(R)^(@) = 50-( 30+60) = - 40 JK^(-1) mol^(-1)` `DeltaG^(@) =DeltaH^(@) - T DeltaS^(@) ` . At equilibrium , `DeltaG^(@) = 0`. Hence,`TDeltaS^(@) = DeltaH^(@)` or `T= (DeltaH^(@))/( DeltaS^(@)) = ( -30 xx 10^(3) J mol^(-1))/( -40JK^(-1) mol^(-1))= 750 K` |
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| 77. |
For reversible reaction : `X_((g))+3Y_((g))hArr 2Z_((g)), DeltaH=-"40 kJ"` Standard entropies of X, Y and Z are 60, 40and `"50 J K"^(-1)",ol"^(-1)` respectively. The temperature at which the above reaction is in equilibrium isA. 273 KB. 600 KC. 500 KD. 400 K |
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Answer» Correct Answer - C `X+3Y hArr2Z` `DeltaS=2xx50-(60+3xx40)=-80kJ` `DeltaG=DeltaH-TDeltaS" when "DeltaG=0` `T=(DeltaH)/(DeltaS)=-40xx(1000)/(-80)=500K` |
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| 78. |
Oxygen & ozone are gases at standard temperature. Their molar entropies are in the sequence `O_(2) lt O_(3).` Using molecular properties , explain why ozone is more disordered than oxygen. |
| Answer» Correct Answer - Ozone has three atoms per molecule, whereas `O_(2)` has only two. | |
| 79. |
For the reaction `2H_(2)(g)+ O_(2)(g)to 2H_(2)O(l)`. The standard entropies of `H_(2)(g),O_(2)(g)" and "H_(2)O(l) " at "27^(@)C " are " 30 "cal K"^(-1)mol^(-1),50cal K^(-1) and15cal K^(-1)mol^(-1) "respectivley"`. Assuming `H_(2) " and " O_(2)` to behave as ideal gas (without vibrational degree of freedom ) and `C_((PmH_(2)O(l)))=15.5 calK^(-1)mol^(-1)` . Determine magnitude of standard entropy (in `calK^(-1)`) change for the reaction at `177^(@)C`. ltbrlt [Given : `R =2 calK^(-1)mol^(-1),"In " 1.5 =0.4`] |
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Answer» Correct Answer - 76 |
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| 80. |
Determine the entropy change for the reaction `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(l)` at `300K`. If standard entropies of `H_(2)(g),O_(2)(g)` and `H_(2)O(l)` are `12.6, 201.20` and `68.0 J K^(-1) "mole"^(-1)` respectively. |
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Answer» `DeltaS (Reaction) = sum S(Product) -sumS(Reactants)` `=2 xx S_(H_(2)O) -[2xxS_(H_(2))+S_(O_(2))]` `= 2 xx 68 -[2xx 12.6 +201.20]` `DeltaS =- 318.4 J K^(-1) mol^(-1)`. |
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| 81. |
Change in enthalpy for reaction `2H_(2)O_(2)(l)rarr2H_(2)O(l)+O_(2)(g)` if heat of formation of `H_(2)O_(2)(l)` and `H_(2)O(l)` are `-188` and `-286KJ//mol` respectively isA. `-196` KJ/molB. `+196` KJ/molC. `+948` KJ/molD. `-948` KJ/mol |
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Answer» Correct Answer - A `2H_(2)O_(2)(l) to 2H_(2)O(l) + O_(2)(g) DeltaH=?` `DeltaH=[(2xxDeltaH_(f) " of " H_(2)O(l) )+ (DeltaH_(f) " of " O_(2)(g)]-(2xxDeltaH_(f) " of " H_(2)O_(2)(l))]` `=[(12xx-286)+ (0) -(2xx-188)]` `[-572 + 376 ]=-196` KJ/mol |
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| 82. |
If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The entropy change for the process would beA. `1.0 J "mol"^(-1) K^(-1)`B. `0.1 J "mol"^(-1) K^(-1)`C. `100 J "mol"^(-1) K^(-1)`D. `10 J "mol"^(-1) K^(-1)` |
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Answer» Correct Answer - C `DeltaG^(@) = DeltaH^(@)-TDeltaS^(@)` Given, `DeltaH_("vap")=30 KJ "mol"^(-1)` T=27 + 273 =300 K `DeltaG^(@)=0` at equilibrium `DeltaS_("vap")=(DeltaH_("vap"))/(T)=(30xx10^(3) J "mol"^(-1))/(300 K)` `=100 J "mol"^(-1) K^(-1)` |
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| 83. |
The standard molar enthalpies of formation of `H_(2)O(l) " and " H_(2)O_(2)(l)` are -286 and -188 "kJ"//"mol", respectively. Molar enthalpies of vaporisation of `H_(2)O(l) " and "H_(2)O_(2)(l)` are 44 and 53 kJ respectively. The bond dissociation enthalpy of `O_(2)(g)` is `498 "kJ"//"mol"`. calculate the bond dissociation enthalphy `("in" "kJ"//"mol" )` of `O-O` bond in `H_(2)O_(2)`, assuming that the bond dissociation ethalpy of `O-H` bond is same in both `H_(2) " and " H_(2)O_(2)`. |
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Answer» Correct Answer - 142 |
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| 84. |
Given `C+2S rarr CS_(2), DeltaHf^(@)=+117.0 kJ mol^(-1)` `C+O_(2) rarr CO_(2), DeltaHf^(@)=-393.0 kJ mol^(-1)` `S+O_(2) rarr SO_(2), DeltaHf^(@)=-297.0 kJ mol^(-1)` The heat of combustion of `CS_(2)+3O_(2) rarr CO_(2)+2SO_(2)` isA. `-807 kJ mol^(-1)`B. `-1104 kJ mol^(-1)`C. `+1104 kJ mol^(-1)`D. `+807 kJ mol^(-1)` |
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Answer» Correct Answer - B `DeltaH=H_(P)-H_(R)` |
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| 85. |
Predict the entropy change (positive/negative) in the following:A. A liquid substance crystallises into a solidB. Temperature of a crystal is increasedC. `CaCO_(3)(s) rarr CaO(S) +O_(2)(g)`D. `N_(2)(g)(1atm) rarr N_(2)(g) (0.5 atm)` |
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Answer» a. Entropy decreases, `DeltaS =- ve` b. Entropy increases, `DeltaS = +ve` c. Entropy increases, `DeltaS = +ve` d. Entropy increases, `DeltaS = +ve` |
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| 86. |
Define entropy. Predict whether entropy change in the following processes would be positive or negative. (i) `N_(2)O_(3)(g) rarr N_(2)O(g)+O_(2)(g) ` (ii) Freezing of water (iii) `NH_(3)(g)+HCl(g) rarr NH_(4)Cl(s)` |
| Answer» Correct Answer - (i) `+ve` (ii) -ve (iii) -ve | |
| 87. |
A mixture of 2 moles of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. If `Delta H` is the enthalpy change and `Delta E` is the change in internal energy, then :-A. `Delta H gt Delta E`B. `Delta H lt Delta E`C. `Delta H = Delta E`D. Not definite |
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Answer» Correct Answer - B `2CO+O_(2)rarr CO_(2)` `Delta n_(g)=-1` `therefore Delta H lt Delta E` |
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| 88. |
A thermos flask contains coffee. It is shaken vigorously. (i) Has any heat been added to it. (ii) Has any work been done on it. (iii) Does it internal energy change? (iv) Does its temp. rise? |
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Answer» (a), No, heat is not transferred as the flask is insulated from the surroundings `:. dQ=0` (b) Yes, work is done in shaking (against the viscous), i.e., `dW` is negative. (c ) Yes, internal energy of tea increases. As `dU=dQ-dW=0-dW`, and `dW` is negative, therefore, `dU` is positive. (d) Yes, temperature if tea rises because of increase in internal energy. |
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| 89. |
`1g` of water changes from liquid to vapour phase at constant pressure of `1` atmosphere, the volume increases from `1 mL` to `1671 mL`. The heat of vaporisation at this pressure is `540 cal//g`. Find the: (a) Work done (in `J`)during phase change. (b) Increase in internal energy of water. |
| Answer» Correct Answer - (a)`-168.67 J,(b)2088.53 J`; | |
| 90. |
The given figure shown a change of state `A` to state `C` by two paths `ABC` and `AC` for an ideal gas. Calculate the : (a) Path along which work done is least. (b) Internal energy at `C` if the internal energy of gas at `A` is `10 J` and amount of heat supplied to change its state to `C` through the path `AC` is `200 J`. (c) Amount of heat supplied ot the gas to go from A to B, if internal energy of gas at state `B` is `10 J`. |
| Answer» Correct Answer - `(a) AC` is least, `(b) 170 J, (c) 10 J` | |
| 91. |
A process in which heat can flow from system to surroundings or vice-versa is called `"…......................"` process. |
| Answer» Correct Answer - adiabatic | |
| 92. |
One mole of a monoatomic ideal gas was isobarically heated from 300 K to 600 K at constant pressure of one atm. Subsequently, it is subjected to reversible adiabatic expansion till the volume becomes `4sqrt2` of origingl (starting) value. After that it is subjected to isobaric cooling to original volume. what is the chanege in enthalpy in adiabatic process ?A. `-750`RB. `+500`RC. `+750`RD. `-500`R |
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Answer» Correct Answer - a |
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| 93. |
Statement -1: heat of neutralistion of perchoric acid `,HClO_(4)` with NaOH is same as that of Hcl with NaOH . Statement -2: Both HCl and `HClO_(4)` are strong acidA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) area true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - A | |
| 94. |
Statement -1: heat of neutralistion of perchoric acid `,HClO_(4)` with NaOH is same as that of Hcl with NaOH . Statement -2: Both HCl and `HClO_(4)` are strong acidA. 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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| 95. |
The magnitude of heat of solution ….. On addition of solvent to solutionA. DecreasesB. IncreasesC. Remains constantD. Increases or decreases |
| Answer» Correct Answer - C | |
| 96. |
Determine the heat of transformaiton of `C_(("dimamond"))rarrC_(("graphite"))` form the following data: i. `C_(("diamond"))+O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.5 kcal` ii. `C_(("graphite")) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 94.0 kcal` |
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Answer» Substracting equation (ii) from equation (i), the required equation is obtained. `C_(("diamone")) rarr C(g)` `Delta_("trans")H^(Theta) =- 94.5-(-94.0) =- 94.5 + 94.0` `=- 0.5 kcal` |
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| 97. |
`Cl_(2)(g)rarr2Cl(g)`, In this process value of `Delta H` will be -A. PositiveB. NegativeC. ZeroD. Nothing can be predicted |
| Answer» Correct Answer - A | |
| 98. |
The enthalpy change for the reaction `2C("graphite")+3H_(2)(g)rarrC_(2)H_(6)(g)` is calledA. Enthalpy of formationB. Enthalpy of combustionC. Enthalpy of hydrogenationD. Enthalpy of vaporisation |
| Answer» Correct Answer - A | |
| 99. |
`H_(aq)^(+)+OH_(aq)^(-)toH_(2)O, DeltaH=-105Kj//mol` `2HA(aq) to BaA_(2(aq)+2H_(2)O`, `DeltaH=-105KJ//mol` Iopnistion enthalpy of weak acid Ha (aq)will be :A. 9KJ/molB. 48KJ/molC. 4.5KJ/molD. 24Kj/mol |
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Answer» Correct Answer - c |
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| 100. |
Which of the following is an exothermic reaction?A. `H_(2)(g)+Cl_(2)(g) rarr 2HCl(g), Delta H = -184.6 KJ`B. `N_(2(g))+O_(2(g))` , ? `h = 180.8 kJ`C. `C_(("graphite")) + H_(2)O_((g)) rarr CO_((g))+H_(2(g))-1314 KJ`D. `C_(("graphite"))+2S_((s))+91.9KJ rarr CS_(2(l))` |
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Answer» Correct Answer - A `Delta H = -ve`, exothermic |
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