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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.
| 351. |
From the following bond energies: `H--H` bond energy: `431.37KJmol^(-1)` `C=C` bond energy: `606.10KJmol^(-1)` `C--C` bond energy: `336.49KJmol^(-1)` `C--H` bond energy: `410.50KJmol^(-1)` Enthalpy for the reaction will be: `overset(H)overset(|)underset(H)underset(|)(C)=overset(H)overset(|)underset(H)underset(|)(C)+H-H to Hoverset(H)overset(|)underset(H)underset(|)(C)-overset(H)overset(|)underset(H)underset(|)(C)-H`A. `553.0 kJ mol^(-1)`B. `1523.6 kJ mol^(-1)`C. `-243. kJ mol^(-1)`D. `-120.0 kJ mol^(-1)` |
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Answer» Correct Answer - D `Delta H= Sigma (B.E.)_(R )- Sigma (B.E.)_(P)` `Delta H=[4xx(B.E.)_(H-H)+1xx(B.E.)_(C=C)+1xx(B.E.)_(H-H)]-[6xx(B.E.)_(H-H)+1xx(B.E.)_(C-C)]` |
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| 352. |
What percentage `T_(1)` is of `T_(2)` for a `10%` efficiency of a heat engine? |
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Answer» Efficiency of heat engine `(eta) = (T_(2)-T_(1))/(T_(2))` or `eta = 1 -(T_(1))/(T_(2)) = 10%` Therefore, `(T_(1))/(T_(2)) = 1- 0.1` `(T_(1))/(T_(2)) xx 100 = (1-0.1) xx 100 = 90` or `T_(1) = 90% of T_(2)` |
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| 353. |
`Xg` of entanal was subjected to combustion in a boub calorimeter and the heat produced is `Y J`. ThenA. `DeltaU` (combustion) `=- XJ`B. `DeltaU`(combustion) `=YJ`C. `DeltaU` (combustion) `=- (44Y)/(X) Jmol^(-1)`D. `DeltaH` (combustion) `= (44Y)/(X) J mol^(-1)` |
| Answer» `:.` Mole `= (W)/("Molecular weight")` | |
| 354. |
Energy required to dissociate 4 gms of `H_(2)(g)` into free atoms is 208 K.cals at `25^(@)C`. The `H-H` bond energy will beA. 104 K.calsB. 1040 K.calsC. 10.4 K. calsD. 208 K. cals |
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Answer» Correct Answer - A `4 gm rarr 2` moles of `H_(2) rarr 208` k cal `2 gm rarr 1` moles of `H_(2) rarr 104` k. cal |
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| 355. |
`Xg` of ethanal was subjected to combustion in a bomb calorimeter and the heat produced is `Y J`. ThenA. `Delta E_(("combustion"))=-XJ`B. `Delta E_(("combustion"))= -YJ`C. `Delta E_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`D. `Delta H_(("combustion"))= -(44Y)/(X)"J mol"^(-1)` |
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Answer» Correct Answer - C Heat of combustion is determined for 1 mole substane. In Bomb calorimeter heat of reaction `= Delta E` `Xg CH_(3)CHO` produces Y Joule heat `therefore` 1 mole `=44 g CH_(3)CHO` will produce `(Y)/(X)xx44` Joule heat `Delta E=(-44Y)/(X)J mol^(-1)` |
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| 356. |
Heat of combustion of benzene is 718 K. cals. When 39 gms of benzene undergoes combustion, the heat liberated isA. 718 K.calsB. 359 K. calsC. 135 K. calsD. 1436 K. cals |
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Answer» Correct Answer - B `78` gm `rarr 718` k.cal `39` gm `rarr ?` |
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| 357. |
The heats of xombustion of carbon, hydrogen and ethane are 94, 68.3 and 373 Kcal/mole respectively. The heat of formation of ethane in K. cals/ mole isA. `+19.9`B. `-19.9`C. `+39.8`D. `-39.8` |
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Answer» Correct Answer - B `2C+3H_(2) rarr C_(2)H_(6), DeltaH=H_(R)-H_(P)` |
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| 358. |
On complete combustion of 2 gm methane -26575 cals heat is generated. The heat of formation of methane will be (given heat of formation of `CO_(2)` and `H_(2)O` are - 97000 and - 68000 cais respectivvely) :A. `+20400` calsB. `+20600` calsC. `-20400` calsD. `-2000` cals |
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Answer» Correct Answer - C `(Delta H_("comb."))_(CH_(4))=(-26575)/(2)xx16=-212600 Cal mol^(-1)` `CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O , Delta H_("comb.")=-212600 Cal mol^(-1)` `Delta H=Sigma (Delta H_(f))_(P)-Sigma (Delta H_(f))_(R )-212600 =[(-97000)+2xx(-68000)]=[Delta H_(f)(CH_(4))+0]` `Delta H_(f)(CH_(4))=-233000+212600` `therefore Delta H_(f)(CH_(4))=-20400 Cal` |
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| 359. |
Thermodynamic efficiency of a cell is given by:A. `-DeltaH//DeltaG`B. `-nFE//DeltaG`C. `-nFE//DeltaH`D. `-nFE^(Theta)` |
| Answer» Correct Answer - `-nFE//DeltaH` | |
| 360. |
Calculate the heat of combustion of benzene form the following data: a. `6C(s) +3H_(2)(g) rarr C_(6)H_(6)(l), DeltaH = 49.0 kJ` b. `H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l), DeltaH =- 285.8 kJ` c. `C(s) +O_(2) (g) rarr CO_(2)(g), DeltaH =- 389.3 kJ` |
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Answer» To calculate `DeltaH` of the following: `C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) -3H_(2)O(l) DeltaH = ?` `DeltaH = 3DeltaH_(2) +6DeltaH_(3) - DeltaH_(2)` `=3 xx 285.8 +6 xx -389.3 - 49.0` `=- 857.4 - 2335.8 - 49` `=- 3242.2 kJ mol^(-1)` |
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| 361. |
Calculate the enthalpy of combustion of benezene from the following data`:-` (i) `6C(s) + 3H_(2)(g) rarr C_(6)H_(6) (l), DeltaH =49.0 kJ mol^(-1) ` (ii) `H_(2) (g) + (1)/(2) O_(2)(g) rarrH_(2)O(l) , DeltaH = - 285.8 kJ mol^(-1)` (iii) `C(s) + O_(2) (g) rarr CO_(2)(g) , DeltaH = - 389. 3 kJ mol^(-1)` |
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Answer» Correct Answer - `-3242.2 kJ mol^(-1)` Aim `: C_(6)H_(6) (l) + (15)/(2) O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(l) , DeltaH = ?` `6 xx `Eqn. (ii) `+ 3 xx `Eqn. (ii) `- `Eqn. (i) gives the required result. |
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| 362. |
When two mole of an ideal gas `(C_(p.m) = (5)/(2)R)` heated from `300 K` to `600 K` at constant pressure. The change in entropy of gas `(DeltaS)` is:A. `(3)/(2)" R " " In " 2`B. `-(3)/(2)" R In " 2`C. `5 "R In" 2`D. `(5)/(2) "R In" 2` |
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Answer» Correct Answer - C `DeltaS = nC_(p.m) "In" (T_(2))/(T_(1)) = 2 xx (5)/(2)"R In" (600)/(300) = 5"R In 2` |
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| 363. |
What is the value of `DeltaH^(@)` (inKJ) for this reaction ?ltbr gt `2CuO(s)toCuO(s)+(1)/(2)O_(2)(g)`A. 141.5B. 14.6C. `-14.6`D. `-141.5` |
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Answer» Correct Answer - a |
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| 364. |
Combustion of fuel in a bomb calorimeter is an example of …(a) adiabatic process (b) isochoric process(c) isobaric process (d) isothemal process |
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Answer» (b) isochoric process |
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| 365. |
Fuel cells are the commercial cells which converts the chemical energy into electrical energy and differ from the normal batteries since fuel cells require constant source of fuel and oxygen to obtain electric current. However, they can be used to produce electricity as long as the supply of fuel and oxygen is maintained. All fuel cells consist of an andoe and a cathode and an electrolyte which allows charges to move between the two sidez of the fuel cell. Electrons are drawn from anode to cathode through an external circuit and hence produce direct current electricity. Under practical cases, efficient of fuel cell is around `40-60%` and in some cases it may be as high as `85-90%` Some common examples of fuel cell are diect methanol fule cell (DMFC), Direct Borohydride fuel cell (DBFC), alkaline fuel cell (AFC) and others.While in DMFC the electrolyte is an ionomer (proton exchange monomer) in case of AFC and DBFC the electrolyte is an aqueous alkaline solution. Also, in case of DBFC the borohydride gets oxidised to metaborate, In all the above three cases, at cathode reduction of `O_(2)`(gas) occurs in a medium controlled by the electrolyte. Based on this information answer the electrolyte. Based on the information answer the question that follow : Data `{:(DeltaH_(f)^(@)CH_(3)OH(l)=-230kJ//"mole",,DeltaH_(f)^(@)CO_(2)`=-390kJ//"mole"),(Delta_(f)^(@)H_(2)O(l) = - 285 kJ//"mole",,S_(mH_(2)O(l))^(@)=130J//K "mole"),(S_(mCO_(2(g))) = 210 J//K "mole",S_(mH_(2)O(l))^(@) = 110 J//K "mol"),(S_(mO_(2(g)))=206J//K "mole",,S_(mH_(2(g))) = 130J//K "mole"),("All data at" 300 K.,1/F = 10^(-5)C^(-1)):}` If in AFC, `H_(2)(g)` is used as anode then identify the options which are not correct for AFC.A. Electricity and water will be obtained due to reactionB. `E_("cell")^(@)` at `300 K = 1241V` approx.C. Magnitude of electrical work obtained will be less than magnitude of heat liberated at constant pressure.D. The cathodic reaction will be `O_(2) + 4e^(-) 4H^(+) rarr 2H_(2)O` |
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Answer» Correct Answer - b,d |
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| 366. |
The heat of a reaction is measured in a bomb calorimeter . This heat is equal to which thermodynamic quantity?A. `DeltaE`B. `DeltaG`C. `DeltaH`D. `DeltaS` |
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Answer» Correct Answer - a |
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| 367. |
State Hess‘s law of constant heat summation ? |
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Answer» The change in enthalpy of reaction remains same, whether the reaction takes place in one step or several steps. |
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| 368. |
Calculate `DeltaG` for (i) `H_(2)O(l,2 atm , 300 K) to H_(2)O(g, 2 atm , 300 K ) ` (ii) `H_(2)O(l,P" "atm , 300 K) to H_(2)O(g, P" "atm , 300 K )` Cacluate p for which `DeltaG = 0` Given: `DeltaH_(373)= 40 kJ` `" "C_(P) (H_(2) O, l) = 75 J//mol //K` `" "C_(P) (H_(2) O, g) = 35 J//mol //K` |
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Answer» Correct Answer - (i) `DeltaG= V (P_(2)- P_(1)) + [DeltaH_(300) - 300 DeltaS_(300)]+ nRT ln .(P_(2))/(P_(1))` (ii) P = 26.28 atm |
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| 369. |
Out of boiling pont (I), entropy (II), pH (III) and density (IV), Intensive properties are :A. I,IIB. I,II,IIIC. I,III,IVD. All of these |
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Answer» Correct Answer - C Boiling point, pH & density are intensive properties . Entrpopy is an extensive property . |
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| 370. |
The cycle in the figure followed by an engine made of an ideal gas in ca cylinder with a piston, the heat exchanged by the engine with the surroundings for adiabatic section AB of cycle is `(C_(V)=(3)/(2)R)` A. `(3)/(2)(P_(B)-P_(A))V_(A)`B. `(5)/(2)P_(A)(V_(A)-V_(B))`C. `(1)/(2)(P_(A)-P_(B))(V_(A)-V_(B))`D. Zero |
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Answer» Correct Answer - D Here from A to B the system is undergoing adiabatic process hence heat exchange of system with surroundings do not take place , hence `Q_(AB)=0` |
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| 371. |
A cycle followed by an engine (made of one mole of an ideal gas in a cyclinder with a piston ) is shown if figure . Find heat exchanged by the enigen, with the surrodings for each section of the cycle . `[C_(V) = (3//2) R]` |
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Answer» (a) For process AB, Volume is constant, hence work done dW = 0 Now, by first law of thermodynamics, `" "dQ =dU+dW=dU+0=dU` `" "=nC_(v)dT=nC_(v)(T_(B)-T_(A))` `" "=(3)/(2)R(T_(B)-T_(A))" "(becausen=1)` `" "=(3)/(2)(RT_(B)-RT_(A))=(3)/(2)(p_(B)V_(B)-p_(A)V_(A))` Heat exchanged = `(3)/(2)(p_(B)V_(B)-p_(A)V_(A))` (b) For process BC, `" "p` = constant `" "dQ=dU+dW=(3)/(2)R(T_(C)-T_(B))+p_(B)(V_(C)-V_(B))` `" "=(3)/(2)(p_(C)V_(C)-p_(B)V_(B))+p_(B)(V_(C)-V_(B))=(5)/(2)p_(B)(V_(C)-V_(B))` Heat exchanged = `=(5)/(2)p_(B)(V_(C)-V_(A))" "(becausep_(B)=p_(C)andp_(B)=V_(A))` ( c ) For process CD, Because CD is adiabatic, dQ = Heat exchanged = 0 (d) DA involes compression of gas from `V_(D)` to `V_(A)` at constant pressure `p_(A)`. `therefore` Heat exchanged can be calculated by similar way as `BC_(1)`, Hence, `" "dQ=(5)/(2)p_(A)(V_(A)-V_(D)).` |
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| 372. |
A satellite in its orbit is open, close or nearly isolated system? |
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Answer» A satellite in its orbit is open system. |
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| 373. |
The earth is open, close or nearly isolated system? |
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Answer» The earth is open system. |
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| 374. |
Ice-cube tray filled with water is open, close or nearly isolated system? |
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Answer» Ice-cube tray filled with water is open system. |
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| 375. |
Which of the following are open, close or nearly isolated systems : (i) Human beings (ii) The Earth (iii) Can of tomato soup (iv) Ice-cube tray filled with water (v) A satellite in a orbit (vi) Coffee in the thermos flask and (vii) Helium filled balloon. |
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Answer» Open : Human beings, Earth, Ice cube tray Isolated : Coffee in thermos flask. |
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| 376. |
Cane of tomato soup is open, close or nearly isolated system? |
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Answer» Cane of tomato soup is open system. |
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| 377. |
During the process `AB` of an ideal gas A. work done on the gas Is zeroB. density of the gas is constantC. slope of line `AB`from the `T`-axis is inversely proportional to the number of moles of the gasD. slope of line `AB` form the `T`-axis is directy proportional to the number of moles of the gas. |
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Answer» Correct Answer - A::B::D `P-T` graph is a straight line passing through origin.Therefore, V = constant `:.` Work done on the gas is zero. Furhter density of the gas `rho=(m)/(V)prop(1)/(V)` Volume of the gas is constant. Therefore, density of gas is also constant. `PV=nRT , or P=((nR)/(V))T` i,e., slope of `P-T` line `prop` n. |
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| 378. |
Two sample `A` and `B` of a gas initially at the same pressure and temperature are compressed from volume `V` to `V//2` (A isothermally and `B` adiabatically). The final pressure of `A` isA. Greater than the final pressure of BB. Equal to the final pressure ofC. Less than the final pressure of BD. wice the final pressure of B |
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Answer» Correct Answer - C |
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| 379. |
Calculate the specific heat of a gas at constant volume from the following data. Desity of the gas at `N.T.P=19xx10^(-2)kg//m^(3)` , `(C_(p)//C_(v))=1.4,J` `=4.2xx10^(3) J//kcal` : atmospheric pressure `=1.013xx10^(5)N//m^(2)` . `("in" kcal//kg k)`A. `2.162`B. `1.612`C. `1.192`D. `2.612` |
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Answer» Correct Answer - C `c_(v)=(R)/(M(gamma-1))impliesc_(v)=(P)/(JrhoT(gamma-1))` |
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| 380. |
If the ratio of specific heats of neon is `1.667` and `R=8312 J//k` mole `K` find the specific heats of neon at constant pressure and constant volume. (Molecular weight of neon `=20.183`)A. `1.029,0.6174`B. `1.831,0.921`C. `1.621,0.421`D. `0.862,0.246` |
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Answer» Correct Answer - A `C_(p)=(gammaR)/((gamma-1)),C_(v)=(R)/((gamma-1))` |
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| 381. |
The first law of thermodynamics for a closed system is dU = dq + dw, where dw = `dw_(pv)+dw_("non-pv")`. The most common type of `w_("non-pv")` is electrical work. As per IUPAC convention work done on the system is positive. A system generates 50 J of electrical energy and delivers 150 J of pressure-volume work against the surroundings while releasing 300 J of heat energy. What is the change in the internal energy of the system?A. `-100`B. `-400`C. `-300`D. `-500` |
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Answer» Correct Answer - D `DeltaU=-300+(-50-150)=-500` |
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| 382. |
The first law of thermodynamics for a closed system is dU = dq + dw, where dw = `dw_(pv)+dw_("non-pv")`. The most common type of `w_("non-pv")` is electrical work. As per IUPAC convention work done on the system is positive. A system generates 50 J electrical energy, has 150 J of pressure-volume work done on it by the surroundings while releasing 300 J of heat energy. What is the change in the internal energy of the sytem?A. `-500`B. `-100`C. `-300`D. `-200` |
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Answer» Correct Answer - D `DeltaU=q+w` `DeltaU=-300+(-50+150)=-200` |
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| 383. |
Suppose `300J` of work done on a system and `70cal`. Of heat heat is extracted from the system. What are the values of `dW , dQ and dU` with proper signs? |
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Answer» As work is done on the system, therefore `dW= -300J` Again as `70 cal.` Of heat is extracted from the system, `:. dQ = -70 cal. = -70xx4.2J` Noe, `dU=dQ-dW` `= -70xx4.2-(-300)J` `dU= -294+300= 6J`. |
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| 384. |
In the relation `dQ=dU+dW`, the quantity which remains the same for all process isA. `dQ`B. `aW`C. `dU`D. all of the above |
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Answer» Correct Answer - C Steeper , an isothermal curve. |
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| 385. |
Which reaction occurs with a decrease in entropy?A. `N_(2)(g) +O_(2)(g) to 2NO(g)`B. `N_(2)O_(4)(g) to 2NO_(2)(g)`C. `2CO(g) to C(s)+CO_(2)(g)`D. `HCl(aq)+Ag_(2)CO_(3)(s)` `to 2AgCl(s)+ CO_(2)(g) +H_(2)O(l)` |
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Answer» Correct Answer - C |
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| 386. |
Which reaction occurs with a decrease in entropy?A. `2H_(2)O(l) to 2H_(2)+O_(2)(g)`B. `2NO(g) to N_(2)(g)+O_(2)(g)`C. `C(s) + O_(2)(g) to CO_(2)(g)`D. `Br_(2)(g) + Cl_(2)(g) to 2BrCl(g)` |
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Answer» Correct Answer - A |
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| 387. |
The reaction `CH_(4)(g)+Cl_(2)(g)to CH_(3)Cl(g)+HCl(g) has DeltaH=-25kcal` `{:(Bond,"Bond Energy ,kcal"),(epsi_(C)-Cl,84),(epsi_(H-Cl),103),(epsi_(C-H),x),(epsi_(Cl-Cl),y):}` x:y=9.5 From the given data , what is the bond energy of `Cl-Cl` bond?A. 70kcalB. 80kcalC. 67.85 kcalD. 57.85 kcal |
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Answer» Correct Answer - d |
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| 388. |
Calculate the ethanly change for the given reaction from data provided (KJ/mole) `{:(HA(g),+,B(g),to, AHB(s)),(("weak acid"),,("Weak base"),,("Salt")):}` `DeltaH_("neutralization "){HA(aq)["at infinte dilution "]//B(aq)` `["at infinite dilution "]=-40kJ//"mole"` `DeltaH_("solution ")[HA(aq)]=-10kJ//"mole"{"at infinite dilution "}` `DeltaH_("soltuion")[B(g)=-5KJ//"mole" {"at infinite dilution" }` `DeltaH_("solution ")[HAB(s)]=+8KJ//"mole"` `{ "at infinite dilution"}`A. `-36`B. `-63`C. `-45`D. `-37` |
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Answer» Correct Answer - b |
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| 389. |
For the reaction: `2NO_(2)(g) to N_(2)O_(4)(g): DeltaH lt0`. What predictions can be made about the sign of `DeltaS` and the temperature conditions under which the reaction would be spontaneous? `{:(,DeltaS_("rxn"),"Temperature Condition"),((a),"negative","low temperature"),((b),"negative","high temperature"),((c),"positive","high temperature"),((d),"postive","low temperature"):}`A. `{:(,DeltaS_("rxn"),"Temperature Condition"),(,"negative","low temperature"):}`B. `{:(,DeltaS_("rxn"),"Temperature Condition"),(,"negative","high temperature"):}`C. `{:(,DeltaS_("rxn"),"Temperature Condition"),(,"positive","high temperature"):}`D. `{:(,DeltaS_("rxn"),"Temperature Condition"),(,"positive","low temperature"):}` |
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Answer» Correct Answer - A |
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| 390. |
Coniseder the following reactions : (P) `2NO_(2)(g)toN_(2)(g)+2O_(2)(g)` (Q) `2IBr(g)toI_(2)(s)+Br_(2)(l)` For which reaction is `DeltaS^(@)lt0`?A. P onlyB. Q onlyC. Both P and QD. Neither P and q |
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Answer» Correct Answer - B |
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| 391. |
In a refrigerator, in one cycle, the external work done on the working substance is 20% of the energy extracted from the cold reservoir. Find the coefficient of performance of the refrigerator. |
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Answer» Data: |W| = 0.2|QC| The coefficient of performance of the refrigerator, K = \(|\frac{Q_C}W|\) = \(\frac{|Q_C|}{0.2|Q_C|}\) = 5 |
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| 392. |
Two different adiabatic curves for the same gas intersect two isothermals at `T_(1), and T_(2)` as shown in `P-V` diagram, (figure). How does the ratio `(V_(a)//V_(d))` compare with the ratio `(V_(b)//V_(c))` ? |
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Answer» Correct Answer - Same For adiabatic curve BC: `T_(1)V_(b)^((gamma-1))= T_(2)V_(c)^((gamma-1))` For adiabatic curve AD: `T_(1)V_(a)^((gamma-1))= T_(2)V_(d)^((gamma-1))` Dividing we get `((V_(a))/(V_(b)))^((gamma-1))= ((V_(d))/(V_(c)))^((gamma-1)) or (V_(a))/(V_(b))=(V_(d))/(V_(c))` `:. (V_(a))/(V_(d))=(V_(b))/(V_(c))` i.e., the ratio remains the same. |
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| 393. |
Energy is neither absorbed nor released during expansion of ideal gas in vacuum. Why? |
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Answer» In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these forces. Like, when a gas expends against vacuum, work done is zero as Pext – Hence, internal energy of the system does not change, there is no absorption nor release of energy during expansion of ideal gas in vacuum. |
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| 394. |
During adiabatic expension of `10 mol es` of a gas , internal energy decrease, by `700 J`. Work done during the process is `xxx10^(2) J`. What is `x`? |
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Answer» From firsrt law of thermodynamics, `dQ= dU+dW` In adiabatic expression, `dQ=0` `:. dW= -dU= -(-700 J)= 700 J= 7xx10^(2)J` `:. X= 7` |
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| 395. |
Heat of formation, `Delta H_(f)^(@)` of an explosive compound like `NCl_(3)` is -A. PositiveB. NagativeC. ZeroD. Positive or negative |
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Answer» Correct Answer - A Formation of `NCl_(3)` is an endothermic process. |
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| 396. |
`S_(("rhombic"))+O_(2(g))rarr SO_(2(g)) , Delta H =-297.5 K J` `S_(("monoclinic"))+O_(2(g))rarr SO_(2) , Delta H = - 300 K J` The data can predict that -A. Rhombic sulphur is yellow in colourB. Monoclinic sulphur has metallic lustureC. Monoclinic sulphur is more stableD. `Delta H` transition of `S_(R )` to `S_(M)` is endothermic |
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Answer» Correct Answer - D `S_(R )+O_(2)rarr SO_(2) , Delta H_(1)=-297.5 kJ` `S_(M)+O_(2)rarr SO_(2), Delta H_(2) = -300 kJ` `S_(R )rarr S_(M) , Delta H_(3) = ?` eq(iii) = eq(i) - eq(ii) `Delta H_(3)=-297.5-(-300)` `Delta H_(3)=+2.5 kJ` so, it endothermic process. |
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| 397. |
When 1g liquid naphthalene `( C_(10) H_(8))` solidifies.149 jouls of heat is evolved. Calculate the enthalpy of fusion of naphthalene. |
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Answer» Molar mass of naphthalene `(C_(10)H_(8))= 128 g mol^(-1)` When 1 g of liquid naphthalene solidified, heat evolved `= 149 xx 128 ` joules `= 19072` Joules Since fuson is reverse of solidification, therefore, heat absorbed for fusion of one mole of naphthalene `=` 19072 joules, i.e., Enthalpy of fusion `(Delta _(fus) H )= + 19072` joules `//` mole |
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| 398. |
Given the following thermochemical equations `:` (i) S( rhombic ) `+ O_(2)(g) rarr SO_(2)(g) , Delta H = - 297.5 kJ mol^(-1)` (ii) S ( monoclinic) `+ O_(2) rarr SO_(2)(g) , Delta H = - 300.0 kJ mol^(-1)` Calculate `Delta H ` for the transformation of one gram atomof rhombic sulphur into monochlinic sulphur. |
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Answer» We aim at `:` S ( rhombic ) `rarr ` S ( monoclinic) , `Delta H = ?` Equation (i) - Equation (ii) gives S ( rhombic ) - S ( monoclinic ) ` rarr 0 , Delta = 297.5 - ( - 300.0) = 2.5 kJ mol^(-1)` or S( rhombic ) `rarr ` S ( monoclinic ) , `Delta H = + 2.5 kJ mol^(-1)` Thus, for the transformation of one gram atom of rhombic sulphur into monoclinic sulphur, `2.5 kJ mol^(-1)` of heat is absorbed. |
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| 399. |
Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining process. If all the capturing of energy from the reaction is done through electrical process (non P-V work), then calculate, maximum available energy which can be captured by combustion of 34.2 g of sucrose : (Given : `DeltaH_("combustion")("sucrose")=-6000kJmol^(-1)` `DeltaS_("combustion")=180j//K-mol` and bodyntemperature is 300 K) |
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Answer» Correct Answer - 6054 |
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| 400. |
Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining process. If all the capturing of energy from the reaction is done through electrical process (non P-V work), then calculate, maximum available energy which can be captured by combustion of 34.2 g of sucrose : (Given : `DeltaH_("combustion")("sucrose")=-6000kJmol^(-1)` `DeltaS_("combustion")=180j//K-mol` and bodyntemperature is 300 K)A. `600` KJB. `5.94.6`KJC. `5.4` KJD. `605.4` KJ |
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Answer» Correct Answer - D (D) No. of moles of sucrose `=(34.2)/(342) = 0.1` `-(DeltaG)_(T.P) `= useful work done by the system `-DeltaG =-DeltaH + T.DeltaS =+ (6000xx0.1) + (180xx 0.1 xx 300)/(1000) = 605.4 `KJ. |
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