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.
| 1351. |
Which of the following is not correct ?A. `Delta G` is zero for a reversible reactionB. `Delta G` is positive for a spontaneous reactionC. `Delta G` is negative for a spontaneous reactionD. `Delta G` is positive for a non-spontaneous reaction |
| Answer» Correct Answer - B | |
| 1352. |
For a cyclic process, which of the following is true?A. `DeltaS =0`B. `DeltaU = 0`C. `DeltaH = 0`D. `DeltaG = 0` |
| Answer» For cyclic process, `DeltaH = DeltaU = 0` | |
| 1353. |
Calculate the `DeltaH^(Theta)` for the reduction of `Fe_(2)O_(3)(s)` by `AI(s)` at `25^(@)C`. The enthalpies of formation of `Fe_(2)O_(3)` and `AI_(2)O_(3)` are `-825.5` and `-1675.7 kJ mol^(-1)` respectively. |
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Answer» `Fe_(2)O_(3)(s) +2AI(s) rarr 2Fe(s) +AI_(2)O_(3), DeltaH = ?` `3Fe(s) +(3)/(2) O_(2)(g) rarr Fe_(2)O_(3), DeltaH_(1) =- 825.5 kJ mol^(-1)` `2AI(s) +(3)/(2)O_(2)(g) rarr AI_(2)O_(3), DeltaH_(2) =- 1675.7 kJ mol^(-1)` Operate: `DeltaH = DeltaH_(1) - DeltaH_(2)` `=- 1675.7 -(-825.5) =- 850.2 kJ mol^(-1)` |
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| 1354. |
Assertion:Spontaneous process is an irreversible process and may be reversed by some external agency. Reason:Decrease in enthalpy is a conrtributory factor for spontaneity.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true and R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 1355. |
In which of the following reactions, heat liberated is known as standard heat of formation of `CO_(2)`?A. `2CO_((g)) + O_(2(g)) rarr 2CO_(2(g))+135.5 K.cal`B. `C("diamond") + O_(2(g)) rarr CO_(2(g))+19.5 k.cal`C. `C("graphite") + O_(2(g)) rarr CO_(2(g))+94.05 k.cal`D. `CH_(4(g))+2O_(2(g)) rarr 2H_(2)O_((l)) +212.8 k.cal` |
| Answer» Correct Answer - C | |
| 1356. |
State and explain first, law of thermodynamics. |
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Answer» The first law of thermodynamics states that the net heat energy supplied to the system is equal to sum of change in internal energy of the system and work done by the system. Let d Q amount of heat be supplied to the system resulting in a change of internal energy by dU. If the work done by the system is dW then, according to first law of thermodynamics, dQ = dU + dW. |
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| 1357. |
What is refrigeration? |
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Answer» The process of transfer of heat from a cold body to a hot body by doing work on the system is called refrigeration. |
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| 1358. |
What is the temperature at which the value in the Celcius scale is equal to that in the Fahrenheit scale? |
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Answer» – 40°C is the temperature at which the value in the Celcius scale is equal to that in the Fahrenheit scale. |
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| 1359. |
A Container having 1 mole of a gas at a temperature `27^(@)` has a movable piston which maintains at constant pressure in container of 1 atm . The gas is compressed until temperature becomes `127^(@)`. The work done is ( C for gas is 7.03 cal / mol – K )A. 703 jB. 814 jC. 121 jD. 2035 j |
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Answer» Correct Answer - D |
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| 1360. |
Which of the following is a slow processA. IsothermalB. AdiabaticC. IsobaricD. None of these |
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Answer» Correct Answer - A |
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| 1361. |
In a reversible isochoric changeA. `DeltaW=0`B. `DeltaQ=0`C. `DeltaT=0`D. `DeltaU=0` |
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Answer» Correct Answer - B |
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| 1362. |
Entropy of a thermodynamic system does not change when this system is used forA. Conduction of heat from a hot reservoir to a cold reservoirB. Conversion of heat into work isobaricallyC. Conversion of heat into internal energy isochoricallyD. Conversion of work into heat isochorically |
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Answer» Correct Answer - A |
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| 1363. |
Select the correct alternatives for an ideal gas :A. The change in internal energy un a constant pressure process from temperature `T_(1) to T_(2)` is equal to `nC_(v)(T_(2)-T_(1)), where C_(v)` is the molar specific heat at constant volume and n the number of moles of the gas.B. The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process.C. The internal energy dose not change in an isothermal process.D. No heat is added or removed in an asiabatic process. |
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Answer» Correct Answer - A::B::C::D (a) `DeltaU=Q-W=Nc_(P)DeltaT-PDeltaV` `=Nc_(P)DeltaT-nRDeltaT=n(C_(P)-R)DeltaT` ltBRgt `=Nc_(V)DeltaT=nC_(V)(T_(2)-T_(1))` (b) ` DeltaQ=DeltaU+DeltaW` But `DeltaQ=0` for adiabatic process, hence `DeltaU=-DeltaW , or,|DeltaU|=|DeltaW|` (c) `Delta U=nC_(V)DeltaT=0 , ( :.DeltaT=0)` (d) `DeltaQ=0` (in adiabatic change) |
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| 1364. |
`P - V` diagram of a cyclic process `ABCA` is as shown in Fig. Choose the correct alternative A. `DeltaQ_(A)rarr_(B)` is negativeB. `DeltaU_(B)rarr_(C)` is positiveC. `DeltaU_(C)rarr_(A)` is negativeD. `DeltaW_(CAB)` is negative |
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Answer» Correct Answer - A::B::D During process `A and B` , pressure and volume both are decreasing. Therefore, temperature and hence internal energy of the gas will decrease `(TalphaPV) or DeltaV_(A)rarr_(B)=` negative. Further, `DeltaW_(A)rarr_(B)` is negative. In process B to C, pressure of the gas is constant while volume is increasing. Hence, temperature should increase or `DeltaU_(B)rarr_(C)` positive. During `C` to `A` volume is constant while pressure is increasing. Therefore, temperature and hence internal energy of the gas should increase off `DeltaU_(C)rarr_(A)` positive. During process `CAB` , volume of the gas in decreasing. Hence, work done by the gas is negative. |
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| 1365. |
If a gas is comprssed adiabatically by doing work of 150 J the change in internal energy of the gas isA. 100 JB. 150 JC. 200 JD. 250 J |
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Answer» Correct Answer - B Since the gas is compressed adiabatically then dQ =dU+dW dU=-dW=-(-150)=150J |
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| 1366. |
An ideal gas having molar specific heat capaicty at constatnt volume is `3/2`R, the molar specific heat capacities at constant pressure isA. `(1)/(2)R`B. `(5)/(2)R`C. `(7)/(2)R`D. `(9)/(2)R` |
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Answer» Correct Answer - B Here ,`C_(V)=(3)/(2)R` Since `C_(P)-C_(V)=R` `therefore C_(P)=C_(V)+R=(3)/(2)R+R=(5)/(2)R` |
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| 1367. |
`DeltaU+DeltaW=0`is valid forA. Adiabatic processB. Isothermal processC. Isobaric processD. Isochoric process |
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Answer» Correct Answer - A |
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| 1368. |
Indicator DiagramA. P-T curveB. P-V curveC. V-T curveD. Q-T curve |
| Answer» Correct Answer - B | |
| 1369. |
During an isochoric processA. V remains constantB. T remains constantC. Q remains constantD. P remains constant |
| Answer» Correct Answer - A | |
| 1370. |
State the practical applications of second law of the thermodynamics. |
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Answer» The second law of thermodynamics decides, whether a given process allowed by first law, will, actually take place or not. For example, 1. Transfer of heat- According to the second law of thermodynamics, dS (entropy) ≥ 0, i.e., the system cannot move towards the direction of decreasing entropy/probability. Hence, heat cannot flow by itself from a hotter body to a colder body. 2. Expansion of a gas- According to the second law of thermodynamics, dS ≥ 0. As entropy cannot decrease, therefore the probability that all the molecules, on their own, go back into one half of the vessel is practically zero. Therefor, this process (of expansion of gas) is irreversible. 3. Diffusion of Gases- According to the second law of thermodynamics, dS ≥ 0. As entropy cannot decrease, therefore process of diffusion of gases is irreversible. 4. A refrigerator or a heat pump is also based on the second law of thermodynamics. Energy in the from of electric energy has to be supplied to enable it to pump heat from inside at lower temperature to the surroundings at higher temperature. |
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| 1371. |
Define cyclic and non-cyclic processes. |
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Answer» Cyclic process - ''A Cyclic process consists of a series of changes which return the system back to its initial state. Non-cyclic process- " In a non-cyclic process, the series of changes involved do not return the system back to its initial state". |
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| 1372. |
Consider a carnot cycle operating between source tempeature 750 K and sink temperature 350 K producing 1.25 J KJ of mechanical work per cycle, the heat transferred to the engine by the reserversA. 1.34 KJB. 2.34 KJC. 3.34 KJD. 4.34 KJ |
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Answer» Correct Answer - B Here `t_(1)=750 K,t_(2)=350 K`, W=1.25 Kj =1250 J Now efficiency `eta=1-(T_(2))/(T_(1))=1-(350)/(750)=(400)/(750)=(8)/(15)` As `eta=(W)/(Q_(1))` `therefore Q_(1)=(W)/(eta)=(1250)/(8//15)=(1250)/(8)xx15=2343.75 J = 2.34 KJ` |
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| 1373. |
A carnot engine takes 900 Kcal of heat from a reservoir at `723^(@)C ` and exhausts it to a sink at `30^(2)C` the work done by the engine isA. `2.73xx10^(6) cal`B. `3.73xx10^(6) cal`C. `6.27xx10^(5) cal`D. `3.73xx10^(5) cal` |
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Answer» Correct Answer - C Here `Q_(1)=900 kcal =900xx10^(3) cal =9xx10^(5) cal` `T_(1)=723^(@)C=723 + 273 =996 K` `t_(2)=30^(@)C = 30 + 273 =303K` `therefore (_(1))/(Q_(2))=(T_(1))/(T_(2))` `therefore Q_(2)=(T_(2))/(T_(1))Q_(1)=(303)/(996)xx9xxx10^(5)=2.73xx10^(5) cal` Now work done by the engine W=`Q_(1)-Q_(2)` `=9x10^(5)-2.73xx10^(5)` `(9-2.73)xx10^(5)=6.27xx10^(5) cal` |
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| 1374. |
When you freeze water in your freezer to make ice cubes,the amount of order in the molecules ofwater increases. However , second law of thermodynamics says that the amount of order in the isolated system can only stay constant or decrease with time. How can thus making of ice violate second law ?A. because water expands during ice formationB. because ice formation takes place at `0^(@)C`C. because ice issolidD. because ice cubes do not constitute isolated system. |
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Answer» Correct Answer - d The statement of sound law is applicable only to isolated system but ice cubes in the freezer do not constitue isolated system. |
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| 1375. |
How many kg of water at `0^(@)C` can a freezer with a coefficient of performance 5 make into ice cubes at `0^(@)C` with a work input of `3.6MJ?` |
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Answer» `m=? beta=5` `W=3.6MJ= 3.6xx10^(6)J` From `beta=(Q_(2))/W, Q_(2)= (beta)W` `mL= 5xx3.6xx10^(6)J` `m=(18xx10^(6))/L= (18xx10^(6))/(80xx1000xx4.2)kg` `m= 54 kg` |
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| 1376. |
The enthalpies of combustion of carbon and carbon monoxide are `-393.5` and `-283` kJ `mol^(-1)` respectively. The enthaly of formation of carbon monoxide per mole is :A. 110.5 kJB. 676.5 kJC. `-676.5 kJ`D. `-110.5 kj` |
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Answer» Correct Answer - B |
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| 1377. |
Suggest a practical way to increase the efficiency of a heat engine. |
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Answer» The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature. |
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| 1378. |
Capacity of an air conditioner is expressed in tonne. Do you know why? |
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Answer» Before refrigerator and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today. (Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.] |
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| 1379. |
Can you explain the thermodynamics involved in cooking food using a pressure cooker ? |
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Answer» Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done. |
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| 1380. |
An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ? |
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Answer» The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = 5 × 107 J. |
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| 1381. |
Explain formation of clouds at high altitude. |
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Answer» As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds. |
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| 1382. |
The unit of entropy isA. `JK^(-1) "mol"^(-1)`B. `J "mol"^(-1)`C. `J^(-1) K^(-1) "mol"^(-1)`D. `JK "mol"^(-1)` |
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Answer» Correct Answer - A Entropy change equal to change in heat per degree. `DeltaS=(q)/(T)` q=required heat per mol T=constant absolute temperature Thus unit of entropy is `KJ^(-1) "mol"^(-1)` |
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| 1383. |
In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?A. `Delta U=W ne 0, q=0`B. `DeltaU=W=0, q ne 0`C. `Delta U=0, W=q ne 0`D. `W=0, Delta U=q ne 0` |
| Answer» Correct Answer - A | |
| 1384. |
In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?A. `DeltaE = W ne ,q=0`B. `DeltaE = W =0 ,qne 0`C. `DeltaE =0, W = qne0`D. `W=0,DeltaE=q ne 0` |
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Answer» Correct Answer - A In closed isulated container a liquid is stirred with a paddle to increase the temperature , therefore it behaves as adiabatic process, so for it q=0 . Hence from first law of thermodynamics `DeltaE=q+W` if, q=0 `therefore DeltaE=W` but not equal to zero. |
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| 1385. |
State the first law of thermodynamics. Express it in mathematical form. |
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Answer» First law of thermodynamics : The change in the internal energy of a system (∆U) is the difference between the heat supplied to the system (Q) and the work done by the system on its surroundings (W). Mathematically, ∆ U = Q – W, which is the same as Q = ∆ U + W. Notes : 1. if Q is positive, it means heat is added to the system. If Q is negative, it means heat is given out by the system or removed from the system, 2. If ∆U is positive, it means there is increase in the internal energy of the system. If ∆ U is negative, it means there is decrease in the internal energy of the system, 3. If W is positive, it means it is the work done by the system on its surroundings. Negative W means work is done on the system by the surroundings, 4. The first law of thermodynamics is largely due to Joule. It is essentially the law of conservation of energy applied to the systems that are not isolated, i.e., the systems that can exchange energy with the surroundings. Thermodynamics was developed in 1850 by Rudolf Clausius (1822-88) German theoretical physicist, His ideas were developed in 1851 by William Thomson [Lord Kelvin] (1824-1907), British physicist and electrical engineer, 5. Q = ∆ U + W. Here, all quantities are expressed in the same units, e.g., cal or joule. If Q and A U are expressed in heat unit (cal, kcal) and W is expressed in mechanical unit (erg, joule) then the above equation takes the form Q = ∆ U + \(\frac{W}{J}\), where J is the mechanical equivalent of heat.] |
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| 1386. |
Which of the following equation gives the values of heat of formation `(DeltaH_(f)^(@))`?A. C(diamond) ` + O_(2)(g) rarr CO_(2)(g)`B. `(1)/(2) H_(2)(g) + (1)/(2)F_(2)(g) rarr HF (g)`C. `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)`D. `H_(2)(g) + F_(2)(g) rarr 3HF(g)` |
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Answer» Correct Answer - B The amount of heat either evoved or absorbed when one gram mole of a substance is formed from its constituent elements, is known as the standard heat of formation `(DeltaH_(f)^(@))`. Standard state temperature os `25C^(@)` or `298K` and pressure of gaseous substance heat of formation of HF. `(1)/(2) H_(2)(g) + (1)/(2)F_(2)(g) rarr HF (g), DeltaH_(f)^(@) = ?` (Standard heat of formation of HF (g)) |
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| 1387. |
The `DeltaH_(f)^(@)` for `CO_(2)(g)`, CO(g) and `H_(2)O(g)` are `-395.5,-110.5` and `-241.8" kJ" mol^(-1)` respectively. The standard enthalpy change in (in kJ) for the reaction `CO_(2)(g)+H_(2)(g)toCO(g)+H_(2)O(g)` isA. 524.1B. 41.2C. `-262.5`D. `-41.2` |
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Answer» Correct Answer - B |
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| 1388. |
`DeltaG` for the conversation of 2 mol of `C_(6)H_(6)(l)` at `80^(@)C` (normal boiling point) to vapour at the same temperature and a pressure of 0.2 atm is:A. `-9.44 kcal//mol`B. `-2.27kcal//mol`C. `-1.135 kcal//mol`D. zero |
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Answer» Correct Answer - B |
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| 1389. |
Calculate the enthalpy change when infinitely dilute solution of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta H_(f)^(0)` for `Ca^(+2)(aq)`. `CO_(3)^(-2)(aq)` and `CaCO_(3)` are `-129.80,-161.7-288.50 kcal mol^(-1)` respectively. |
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Answer» Correct Answer - 3 On mixing `CaCl^(2)(aq)` and `Na_(2)CO_(3)` `CaCl_(2) + Na^(2)CO^(3) rarr CaCO_(3) + 2NaCl` Solutions are very dilute and thus . `100%` dissociation occurs. `Ca^(+2)(aq) + 2Cl^(-)(aq) + 2Na^(+)(aq)+CO_(3)^(-2)(aq)` `" "rarr CaCO_(3) downarrow +Na^(+2)(aq) +2Cl^(-)(aq)` or `" " Ca^(+2)(aq)+ CO_(3)^(-2)(aq) rarr CaCo_(3)(s)` `therefore " "DeltaH = sumH_("products")^(0) - sumH_("reactants")^(0)` or `" " DeltaH =Delta_(f CaCo_(3))^(0) - [DeltaH_(fCa^(+2)) DeltaH_(fCO_(3)^(-2))^(0)]` `therefore " " Delta H^(0)` of a command `=DeltaH_("formation")^(0)` `" " = -288.5 -(-129.80 - 161.7) = 3Kcal` |
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| 1390. |
Diborane is a potential rocket fuel which undergoes combustion according to the reaction, `B_(2)H_(6)(g)+3O_(2)(g)toB_(2)O_(3)(s)+3H_(2)O(g)` from the following data, the enthalpy change for the combustion of diborane will be : `2B(s)+(3)/(2)O_(2)(g)toB_(2)O_(3)(s)," "DeltaH=-1273 " kJ"` `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)," "DeltaH=-286 " kJ"` `H_(2)O(l)toH_(2)O(g)," "DeltaH=44" kJ"` `2B(s)+3H_(2)(g)toB_(2)H_(6)(g)," "DeltaH=46" kJ"`A. `-2079 " kJ" " mol"^(-1)`B. `-1091 " kJ" " mol"^(-1)`C. `-2045 " kJ" " mol"^(-1)`D. `-762 " kJ" " mol"^(-1)` |
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Answer» Correct Answer - C |
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| 1391. |
For the given reaction at 1 atm, `H_(2)(g)+S(s) toH_(2)S(g),` `DeltaH_(r)=100kJ//mol and DeltaS_(r)=400J//mol//K` Temperature at which following reactions occurs reversibly is: ` "Assuming" DeltaH_(r)andDeltaS_(r) "are independent of temperature")`A. 200KB. 250KC. 400KD. None of these |
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Answer» Correct Answer - B |
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| 1392. |
When an ideal monoatomic gas is heated at constant pressure, fraction of heat energy supplied which increases the internal energy of gas, isA. `2/5`B. `3/5`C. `3/7`D. `3/4` |
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Answer» Correct Answer - B |
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| 1393. |
A brick weighing `4.0 kg ` is dropped into a `1.0m` deep river from a height of `2.0m`. Assuming that `80%` of the gravitational potential energy is finally converted into thermal energy, find this thermal energy in calorie.A. `15`B. `17`C. `23`D. `27` |
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Answer» Correct Answer - C `W=JHimplies(80)/(100)(mgh)=JQ` |
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| 1394. |
When an ideal gas `(gamma = 5//3)` is heated under constant pressure, what percentage of given heat energy will be utilized in doing external work ?A. `40%`B. `30%`C. `60%`D. `20%` |
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Answer» Correct Answer - A |
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| 1395. |
A vessel contains a mixture of `7 g` of nitrogen and `11 g` carbon dioxide at temperature ` T = 290 K`. If pressure of the mixure is `1 atm (= 1.01 xx 10^5 N//m^2)`, calculate its density `(R = 8.31 J//mol - K)`. |
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Answer» We know, for the mixture, `N_2` and `CO_2` (being regarded as ideal gases, their mixture too behaves like an ideal gas) `p V = v R T`, so `p_0 V = v R T` Where, `v` is the total number of moles of the gases (mixture) present and `V` is the volume of the vessel. If `v_1` and `v_2` are number of moles of `N_2` and `CO_2` respectively present in the mixture, then `v = v_1 + v_2` Now number of moles of `N_2` and `CO_2` is, by definition, given by `v_1 = (m_1)/(M_1)` and, `v_2 = (m_2)/(M_2)` where, `m_1` is the mass of `N_2` (Moleculer weight `= M_1`) in the mixture and `m_2` is the mass of `CO_2` (Molecular weight `= M_2`) in the mixture. Therefore density of the mixture is given by `rho = (m_1 + m_2)/(V) = (m_1 + m_2)/((v RT//P_0))` =`(p_0)/(RT).(m_1 + m_2)/(v_1 + v_2) = (p_0(m_1 + m_2)M_1 M_2)/(RT(m_1 M_2 +m_2 M_1))` =`1.5 kg//m^3` on substitution. |
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| 1396. |
Thermodynamics true or false questions and please provide the reason:• ________Heat and work are energies that “sit” in the system.• ________There is no heat transfer in a constant-volume process for a system of an ideal gas.• ________Change of enthalpy equals to heat transfer or for all processes.• ________In a reversible adiabatic process, there is no heat transfer into or out of the system.• ________During an isothermal process, the internal energy of an ideal gas depends only on the volume. |
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Answer» (1) Heat and work are energy that "sit" in the system True Reason : ΔU = q + w according to 1st law at thermodynamic change in inertial energy equal to the sum of heat and work. (2) In a reversible adiabatic there is no heat transfer into or out at the system True Reason : For adiabatic process, q = 0 always there is no heat transfer take place (3) During an isothermal process, the internal energy of an ideal gas depends only on the volume False Reason : Ideal gases have no forces between its molecules, no energy is required to maintain a certain volume in ideal gases. so there is no change in internal energy of system. In nut shell internal energy of an ideal gas only depends on Temperature it do not depends on the volume. (4) Change of enthalpy equal to heat transfer for all process False Reason : ΔH = q only at constant pressure not at constant volume temperature (5) There is no heat transfer in a constant volume process for a system of an ideal gas False Reason :According to law of thermodynamies Δ U = q + w Δ U = q - pdv ΔU = q ( at constant volume Δ v = 0) It means at constant volume, change and internal energy equal to heat transfer only on temperature. Hence, the given statement is False |
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| 1397. |
Heat of nrutralization of the acid-base reaction is `57.32` KJ for :A. HCOOH +KOHB. `CH_(3)COOH + NaOH`C. `HNO_(3) + LiOH`D. `HCl + NaOH` |
| Answer» Correct Answer - C::D | |
| 1398. |
For which of the following reaction `DeltaH_("reaction")^(0)` of product.A. `2CO(g) + O_(2)(g) rarr 2CO_(2)`B. `N_(2)(g) + O_(3)(g) rarr N_(2)O_(3)(g)`C. `CH_(4)(g) + 2Cl_(2)(g) rarr CH_(2)Cl_(2)(l) + 2HCl(g)`D. `Xe(g) + 2F_(2)(g) rarr XeF_(4)(g)` |
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Answer» Correct Answer - A::B::C Only in (D) `DeltaH_("reaction") = DeltaH_(r) (XeF_(4)(g)) - 0 = DeltaH_(f)(XeF_(4)(g))` In all other not possible |
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| 1399. |
At 33K. N2H4 is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. |
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Answer» T = 33K N2O4 ⇌ 2NO2 Initial concentration 100% Concentration dissociated 50% Concentration remaining at equilibrium 50% – 100% Keq = 100/50 = 2 ∆G° = -2.303 RT log Keq ∆G° = -2.303 x 8.31 x 33 x log 2 ∆G° = -190.18 J mol-1 |
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| 1400. |
What is a thermodynamic process ? Explain. |
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Answer» A procedure by which the initial state of a system changes to its final state is called a thermodynamic process. During the process/ there may be 1. addition of heat to the system 2. removal of heat from the system 3. change in the temperature of the system 4. change in the volume of the system 5. change in the pressure of the system. |
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