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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.
| 251. |
Find the entropy increment of one mole of a Van der Waals gas due to the isothermal variation of volume from `V_1` to `V_2`. The Van der Walls corrections are assumed to be known. |
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Answer» For a Vander Waal gas `(p + (a)/(V^2))(V - b) = RT` The entropy change along an isotherm can be calculated from `Delta S = int_(V_1)^(V_2) ((del S)/(del V))_T dV` If follows from (2.129) that `((del S)/(del V))_T = ((del p)/(del T))_V = (R)/(V - b)` Assuming `a,b` to be known constants. Thus `Delta S = R 1n (V_2 - b)/(V_1 - b)`. |
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| 252. |
Can water be boiled without heating? |
| Answer» Yes, water can be boiled without heating. This is done by increasing the pressure on the surface of water inside a closed insulated vessel. By doing so, the boiling point of water can be decreased to room temperature. | |
| 253. |
Identify isothermal and adiabatic process in the following diagram |
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Answer» As slope of an adiabatic curve is greater than the slpoe of an isothermal curve, therefore, In (figure)(a): 1 is isothetmal , 2 is adiabatic In (figure) (b), 1 is aidabatic , 2 is isothermal In (figure) (c ), 1 is adiabatic, 2 is isothermal |
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| 254. |
An ideal gas undergoes isothermal process from some initial state `i` to final state `f`. Choose the correct alternatives.A. dU=dQB. dU=-dWC. dU=0D. dU=dW |
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Answer» Correct Answer - C During isothermal process internal energy of systm always remains constant hence change in it will be zero every where fduring the process |
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| 255. |
Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressure `P_(1) and P_(2)`. Then |
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Answer» For a perfect gas, `PV=nRT :. V=(nR)/(P)T` slope of V-T graph with T axis = `(nR)/P` For given amount of gas, slope `prop1/P` Hence `P_(1) gt P_(2)` |
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| 256. |
The specific heat of solids at low temperatures varies with absolute temperature `T` according to the relation `S=AT^(3)`, where `A` is a constant. The heat energy required to raise the temperature of a mass m of such a solid from `T=0` to `T=20K` is:A. `4xx10^(4)mA`B. `2 xx 10^(4)mA`C. `8xx10^(6)mA`D. `2xx10^(6)mA`. |
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Answer» Correct Answer - A `Q=int_(0)^(20)mS" "dT=int_(0)^(20)(AT^3)dT=4xx10^(4)mA`. |
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| 257. |
In the `P-V` diagram shown shown in figure `ABC` is a semicircle.The work done in the process `ABC` is A. `zero`B. `(pi)/(2) atm-L`C. `-(pi)/(2) atm-L`D. `4 atm-L`. |
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Answer» Correct Answer - B `W_(AB)` is negative (volume is decreasing)and `W_(BC)` is positive (volume is increasing) and since, `|W_(BC)|gt|W_(AB)|` `:.` Net work done is positive and area between semicircle which is equal to `(pi)/(2)"atm"-L`. |
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| 258. |
Which one of the following statement is false:A. Work is a state functionB. temperature is a state functionC. change in the state is completely defined when the intial and final states are specifiedD. work appears at the boundary of the system. |
| Answer» Correct Answer - A | |
| 259. |
If x gm of steam at `100^(@)`C is mixed with 5x gm of ice at `0^(@)C`, calculate the final temperature of resultiong mixture. The heat of vapourisation and fusion are 540 cal `gm^(-1)` and 80 cal `gm^(-1)` respectively. |
| Answer» Correct Answer - `40^(@)`C | |
| 260. |
Why is it that different objects kept on a table at room temperature do not exchange heat with the table ? |
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Answer» The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature. |
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| 261. |
When 10 grams of methane is completely burnt in oxygen, the heat evolved is 560 kJ. What is the heat of combustion (in kJ `mol^(-1)`) of methane?A. `-1120`B. `-968`C. `-896`D. `-560` |
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Answer» Correct Answer - C For 10 gm `rarr 560 kJ` `16 gm rarr ?` |
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| 262. |
Calculate the energy required to raise the temperature of 12 g iron from 35°C to 210°C. For iron c = 0.45 J per degree per gram. |
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Answer» Q = m x c x ΔT Q = heat gained m = mass of substance = 12g c = heat capacity of iron = 0.45J /g°C ΔT = change im temperature = ( 210 -35) = 175°C Now put all the given values in the above formula, we get the amount of heat required. Q = 12g x 0.450J /g°C x 175°C Q = 945J Therefore, the amount of heat required is, 945J |
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| 263. |
Which are not correct representation at equilibrium :A. `(V_(1))/(V_(2)) = e^(-Delta S//R)`B. `K = e^(-DeltaG^(@)//RT)`C. `(V_(2))/(V_(1)) = e^(-DeltaS//RT)`D. `(P_(2))/(P_(1)) = e^(Delta H//RT)` |
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Answer» Correct Answer - A::C::D `Delta G = Delta G^(@) +RT ln K_(eq)` at equilibrium `Delta G = 0 rArr Delta G^(@) = -RT ln K_(eq)` `(Delta S) sys = nR ln.(v_(2))/(v_(1)) + nC_(v) ln.(T_(2))/(T_(1))` |
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| 264. |
Select the correct statements:A. State of a system is assumed to be in internal equilibrium and the temperature and pressure are uniform throughout the systemB. Thermal drift in a system with time is more in Dewar flask than in insulated system.C. Thermal drift in a system with time is more in non-insulated system than in insulated system.D. Thermal drift in a system with time is more in insulated system than in non-insulated system |
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Answer» Correct Answer - A::C (a) definition of state (c) thermal drift is more in non-insulated system |
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| 265. |
A heat engine absorbs heat `Q_(1)` at temperature `T_(1)` and `Q_(2)` at temperature `T_(2)` . Work done by the engine is `(Q_(1)+Q_(2))` . This data:A. Violates `1^(st)` law of thermodynamicsB. Violates `1^(st)` law of thermodynamics if `Q_(1)` is -veC. Violates `1^(st)` law of thermodynamics if `Q_(2)` is -veD. Does not violate `1^(st)` law of thermodynamics |
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Answer» Correct Answer - D First law of thermodynamics |
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| 266. |
Find the entropy change when 90 g of `H_(2)O` at `10^(@)`C was converted into steam at `100^(@)`C. [Given `C_(P)(H_(2)O)=75.29 JK^(-1)mol^(-1) and Delta H_("vap")=43.932 JK^(-1)mol^(-1)`] |
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Answer» `H_(2)O(I) rarr H_(2)O(I) hArr H_(2)O("vap")` `10^(@)C rarr 100^(@)C rarr 100^(@)C` `Delta S = underset(283)overset(373)int nC_(P)dIn T+(Delta H_("vap"))/(T_(3))xx` No. of moles `5 xx 2.303 xx 75.29"log"(373)/(283)+(43.932 xx 10^(3))/(373)xx5` = 866.96 log 1.318 + 588.9 = 692.84 `JK^(-1)` |
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| 267. |
Calculate the entropy change of a process possessing ∆Ht = 2090 J mole-1 |
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Answer» Given, Sn(α 13°C) - Sn(β 13°C) ∆Ht = 2090 J mol-1 ∆St = ∆Ht / Tt ∆St = 2090 / 286 ∆St = 7.307 J K-1 mol-1 |
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| 268. |
Which one of the following pairs does not represent example for intensive property ?A. Temperature and densityB. Pressure and molar volumeC. Molar heat capacity and densityD. Heat capacity and enthalpy |
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Answer» Correct Answer - D Heat capacity and enthalpy are extensive. |
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| 269. |
In a measurement of quantum efficiency of photosynthesis in gree plants , it was found that 8 quanta of red light of 6850 `Å` were needed to evolve 1 moleculeof `CO_(2)`. The average energy storage in the photosynthesis process is 112 kcal `//` mol `O_(2)` evolved. Calculatethe percent efficiency of energy conversion in this experiment. |
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Answer» `E = ( hc)/( lambda)= ((6.63 xx 10^(-34)Js) ( 3 xx 10^(8)ms^(-1)))/( ( 6850xx 10^(-10)m)) = 2.90 xx 10^(-19)J` `:. `Energy of 8 quanta `= 8 xx 2.90 xx 10^(-9) J = 2.32 xx10^(-18) J` Energy stored per molecule `=( 112 xx 4.184 xx 10^(3)J mol^(-1))/(6.02 xx 10^(23)mol^(-1)) = 7.78 xx10^(-19)J` `:. %` efficiency of energy conversion`= ("Energy stored per molecule") /( "Energy needed per molecule ") xx 100` `= ( 7.78 xx 10^(-19))/( 2.32 xx 10^(-18)) xx 100 = 33.5%` |
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| 270. |
A room of `10 m xx 15m xx4m` dimenstion having perfectly insulated walls, ceiling and floor has 60 students seated inside. The air inside the room is at `25^(@)C` and 1 atm pressure. If each student loses 200 joules of heat in one second, calculate the rise in temperature noticedin 20 minutes ( Neglect loss of air to the outside as temperature is raised and `C_(p)` for air `= ( 7)/( 2) R)`. |
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Answer» Volume of room `= 10 m xx 15m xx4m = 600m^(3) = 600 xx 10^(3) L =6 xx 10^(5) L` Moles of air in the room at `25^(@)C` at 1 atm pressure `n = ( PV)/( RT) = ( 1 xx 6 xx 10^(5))/( 0.0821 xx 298) = 2.45 xx 10^(4)` Heat producedin1 sec by each student `= 200 J` `:. `Heat produced in 1 sec by 60 students `= 200 xx 60 J = 12000J` Heat produced in 20 minutes `=12000 xx 20 xx60 J = 144 xx 10^(5)J` Change in enthalpy of air, `Delta H = n C_(p) DeltaT` `:. 144 xx 10^(5) = ( 2.45 xx 10^(4)) = ( ( 7)/(2) xx 8.314)xx Delta T ` or ` Delta T = 20.2 K` |
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| 271. |
The value of `log_(10)` K for a reaction `A hArr B` is ( Given `: Delta _(r) H_(298K)^(@)= -54.07kJ mol^(-1),Delta_(r)S_(298K)^(@) = 10J K^(-1)mol^(-1)` and`R = 8.314 JK^(-1) mol^(-1), 2.303 xx 8.314 xx 298 K = 5705) `A. 5B. 10C. 95D. 100 |
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Answer» Correct Answer - b `DeltaG^(@) =DeltaH^(@) - T Delta S^(@)= - 2.303 RT log _(10) K` `:. -2.303 RT log_(10) K= DeltaH ^(@)- T DeltaS^(@)` i.e., `-2.303 xx 8.314 xx 298 xx log_(10)K` `=- 54.07xx 1000 - 298 xx 10` `-5705 log_(10)K =- 54070 - 2980 = - 57050` or `log_(10)K =10` |
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| 272. |
Calculate the magnitude of ring strain energy in `("kJ"//"mol")` of cyclopropane from the following data : `Delta_(f)H[C_(3)H_(6)(g)]=55,Delta_(f)H[C(g)]=715.0`, `Delta_(f)H[H(g)]=220, Be (C-C)=355, " BE "(C-H)=410("all in kJ"//"mole")` |
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Answer» Correct Answer - 115 |
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| 273. |
Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as: `6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)` The energy of one mole of a photon of wave lenght is known as one Einstein. Heats of combustion of graphite and hydrogen at 298K are `-393.5 KJ"mol"^(-1)` and `-285.8 KJ"mol"^(-1)` respectively. If on combustion 1 g of glucose releases 15.58 KJ of energy , Calcualate the heat of formation of glucose at 298 K. |
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Answer» Correct Answer - `C_(6)H_(12)O_(6) + 6O_(2) rarr 6CO_(2) + 6H_(2)O DH =-15.58 xx 2804.4 J` `DeltaH=(H_(f))_(p)-(H_(f))_(R)` `-2804.4 = (6 xx -393.5)+ (6 xx - 285.8) -(H_(f))_("Glucose")` `therefore (H_(f))_("Glucose") =- 1271.4 KJ` Diabetic patients have to monitor their blood glucose level periodically . Home glucometer is an appliance used for this purpose . This device consists fo a disposable test strip with carbon working electrodes and a reference electrode. The working electrode is converted with the enzyme glucose oxidise , which catalyses the reaction the glucose with dissloved oxygen. |
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| 274. |
Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as: `6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)` The energy of one mole of a photon of wave lenght is known as one Einstein. Write balanced equation for the reaction of glucose with `O_(2)` in which `O_(2)` is converted to `H_(2)O_(2)` |
| Answer» Correct Answer - The `H_(2)O_(2)` generated is oxidized at the second working electrode generating a current , which is proportional to the glucose present in the sample. Typically , 5m moles of glucose produces 2.5 mA current. | |
| 275. |
Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as: `6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)` The energy of one mole of a photon of wave lenght is known as one Einstein. Write balanced equations for the half cell reactions corresponding to oxidation of glucose. |
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Answer» Correct Answer - Anode : ` C_(6)H_(12)O_(6) + 2H_(2)O¾ R 6CO_(2) + 24H^(+) + 24e^(-)` Cathode: `O_(2) + 2H_(2)O + 4e^(-) rarr 4oH^(-)` |
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| 276. |
Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as: `6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)` The energy of one mole of a photon of wave lenght is known as one Einstein. 48 Einsteins of 650nm are absorbed by a plant for the production of 1 mole of glucose as given in equation (i). Calculate the % efficiency for the production of glucose by photosynthsis. The energy requried for the formation of 1 mole of glucose is 2870 KJ. |
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Answer» Correct Answer - Energy of 1 mole of photon = `nxx (hC)/(lambda) = 6.023 xx 10^(23) xx (6.625 xx 10^(-34) xx 3 xx 10^(8))/(650 xx 10^(-9) xx 10^(3)) = 184 KJ ` total energy requried `=48 xx 184 = 8823 KJ " "` efficiency `=(2870)/(8832) xx 100 = 32.5%` |
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| 277. |
Carbon monoxide emitted by automobiles is an environmental hazard . A car has an engine of four cylinders with a total cylinder volume of 1600 cc and a fuel consumption fo 7.0 `dm^(3)` per 100 Km , when driving at an average speed of 80 km/hr . In one second , each cylinder goes through 25 burn cycles and consumes 0.400g of fuel . The compression ratio, which is the ratio between the smallest and largest volume within the cylinder as the piston moves forward and backward is 1:8. The gasified fuel and air are compressed to their lowest volume and then ignited . The overall stoichiometric equation for the combustion reaction is `C_(8)H_(18) = 12.10_(2) rarr 0.8 CO + 7.2 CO_(2)+ 9H_(2)O` Calculate the temperature of the i) gases just at the time of maximum compression and ii) exhaust gases leaving the cylinder if the final pressure in the cylinder is 200 Kpa. Relevant data needed for one burn cycle is given below. `{:("compound",DeltaH_(f)(KJ "mol"^(-1)),C_(P)(J "mol"^(-1)K^(-1)),"Composition of gases after combustion" ("mol" xx 10^(-4))),(N_(2)(g),0.0, 29.13 , 101.91),(O_(2)(g),0.0, 29.36,10.1),(CO(g),-110.53, 29.14, 1.12),(CO_(2)(g) ,-395.51,37.11,10.11),(H_(2)(g),-241.82,33.58,12.36),("Isocotane",-187.82,,):}` |
| Answer» Correct Answer - `T_(1) =2060 K " " ; " " T_(2)=708 K` | |
| 278. |
Carbon monoxide emitted by automobiles is an environmental hazard . A car has an engine of four cylinders with a total cylinder volume of 1600 cc and a fuel consumption fo 7.0 `dm^(3)` per 100 Km , when driving at an average speed of 80 km/hr . In one second , each cylinder goes through 25 burn cycles and consumes 0.400g of fuel . The compression ratio, which is the ratio between the smallest and largest volume within the cylinder as the piston moves forward and backward is 1:8. Calculate the air intake of the engine `(m^(3)S^(-1))` , if the gaseous fuel and air are introduced into the cylinder when its volume is largest until the pressure is 101.0 KPa. the temperature of both incoming air and fuel is `100^(@)`C. (Assume the fuel to be isoctane , `C_(8)H_(18))` |
| Answer» Correct Answer - Air constains 21.0% of `O_(2)` and 79.0% of `N_(2)` (by volume) . It is assumed that 10.0% of the carbon of the fuel forms CO upon combustion and that `N_(2)` in air remains insert. | |
| 279. |
For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)`, which one of the following is true :A. `Delta H = Delta E`B. `Delta H = 1//2 Delta E`C. `Delta H lt Delta H`D. `Delta H gt Delta E` |
| Answer» Correct Answer - D | |
| 280. |
The standard free energy change (∆G°) is related to equilibrium constant (K) as ……(a) ∆G° = – 1303 RT in K (b) ∆G° = 2.303 RT log K (c) ∆G° = RT in K (d) ∆G° = -2.303 RT log K |
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Answer» (d) ∆G° = -2.303 RT log K |
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| 281. |
For combustion of 1 mole of benzene at `25^(@)C`, the heat of reaction at constant pressure is `-780.9" kcal."` What will be the heat of reaction at constant volume ? `C_(6)H_(6(l))+7(1)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l))`A. `-781.8" kcal"`B. `-"780.0 kcal"`C. `+"781.8 kcal"`D. `+"780.0 kcal"` |
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Answer» Correct Answer - B `q_(p)=q_(v)+Deltan_(g)RT, q_(v)=q_(p)-Deltan_(g)RT(DeltaH=DeltaU+Deltan_(g)RT)` `q_(v)=-780.9-(-1.5xx2xx298xx10^(-3))=-"780 kcal"` |
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| 282. |
A sample of argon gas at 1 atm pressure and `27^(@)` C expand reversibly and adibatically from `1.25 dm^(3)` to `2.50 dm^(3)`. The enthalpy change in this process will be `C_(v.m.)` for argon is `12.48 JK^(-1)mol^(-1)`.A. 114.52 JB. `-117.14 J`C. `-57.26 J`D. 57.26 J |
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Answer» Correct Answer - D |
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| 283. |
What is the change in enthalapy (kcal) when 1 mole of ideal monoatomic gas is expended reversibly and adibatically from initial volume of 1 L and initial temperature 300 K to final volume of 8 L :A. `-1.125`B. `+1.125`C. 2.25D. `-2.250` |
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Answer» Correct Answer - A |
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| 284. |
The heat of combustion of `CH_(4(g)), C_((g))` and `H_(2(g))` at `25^(@)C` are `-212.4` K cal, -94.0 K cal and -68.4 K cal respectively, the heat of formation of `CH_(4)` will be -A. `+54.4` K calB. `-18.4` K calC. `-375.2` K calD. `+212.8` K cal |
| Answer» Correct Answer - B | |
| 285. |
`Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1)` respectively. The standard enthalpy change `(` in `kJ)` for given reaction is `:` `CO_(2(g))+H_(2(g)) rarr CO_((g))+H_(2)O_((g))`A. `524.1`B. `41.2`C. `-262.5`D. `-41.2` |
| Answer» Correct Answer - B | |
| 286. |
Standard molar enthalpy of formation, `Delta_(f)H^(c-)` is just a special case of enthalpy of reaction, `Delta_(r)H^(c-)` . Is the `Delta_(r)H^(c-)` for the following reaction same as `Delta_(f)H^(c-)` ? Give reaction for your answer. `CaO(s) +CO_(2)(g) rarr CaCO_(3)(s), Delta_(f)H^(c-)= - 178. 3 kJ mol^(-1)` |
| Answer» No, because enthalpy of formation is the enthalpy change when 1 mole of the compound is formed from the elements , i.e., for the reaction `Ca(s) + C(s) + (3)/(2) O_(2)(g) rarr CaCO_(3)(s)`. Thus, this reaction is different from the givien reaction. Hence, `Delta_(r) H^(@) cancel(=) Delta_(f)H^(@)`. | |
| 287. |
The standard enthaply for the reaction `H_(2)(g) + 1//2 O_(2)(g) to H_(2)O(l)` is - 285.76 kJ at 298 K. Calculate the value of `DeltaH` at 373 K .The molar heat capcities at constant pressure `(C_(P))` in the given temperature range of `H_(2)(g),O_(2)(g) and H_(2)O(l)` are respectively 38.83,16 and `75.312 JK^(-1) mol^(-1)`. |
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Answer» Correct Answer - `DeltaH_(373) (H_(2)O(l) )= - 284.11 kJ` `H_(2)- H_(1)= DeltaC_(p)(T_(2)-T_(1))` `H_(2) = - 284.11 kJ` |
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| 288. |
Standard molar enthalpy of formation, `Delta_(f) H^(Θ)` is just a special case of enthalpy of reaction, `Delta_(r) H^(Θ)`. Is the `Delta_(r) H^(Θ)`? Given reason for your answer. `CaO (s) + CO_(2) (g) rarr CaCO_(3) (s) , Delta_(f) H^(Θ) = - 178.3 kJ mol^(-1)` |
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Answer» No, the `Delta_(f)H^(Θ)` for the given reaction is not same as `Delta_(f) H^(Θ)`. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, `Delta(f) H^(Θ)`. `Ca(s) + C(s) + (3)/(2) O_(2) (g) rarr CaCO_(3) (s) , Delta(f) H^(Θ)`. This reaction is different from the given reaction. Hence, `Delta_(f) H^(@) != Delta_(f)H^(@)` |
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| 289. |
Here, in which case can the calculated standard reaction enthaply `(Delta_(1) H^(@))` be identified as the standard molar enthalpy of formation for methanol `(Delta_(g) H_(CH_(3) OH^(2)))`?A. `CH_(4) (g) + (1)/(2) O_(2) (g) rarr CH_(3) OH (g)`B. `C_("diamond") + (1)/(2) O_(2) (g) rarr 2 H_(2) (g) + CH_(3) OH (l)`C. `CO (g) + 2 H_(2) (g) rarr CH_(3) OH (l)`D. `C_("graphite") + (1)/(2) O_(2) rarr 2H_(2) (g) + CH_(3) OH(l)` |
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Answer» Correct Answer - D The standard enthalpy of formation of a substance, denoted by `Delta_(f) H^(@)`, the the enthaply change for the formation of one mole of the substance in its standard state from its constiuent elements in their standard states. Thus, `Delta_(f) H^(@)` refers to specific reaction. |
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| 290. |
`C_(v)` values for monoatomic and diatomic gases respectively areA. `1/2 R, 3/2 R`B. `3/2 R, 5/2 R`C. `5/2 R, 7/2 R`D. `3/2 R, 3/2 R` |
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Answer» Correct Answer - B `C_(V)=3/2 R, C_(P)-C_(V)=R` |
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| 291. |
Find bond enthalpy of C=O (in kJ/mol) using following information : `DeltaH_("atomisation")[C(s)]=700` kJ/mol `DeltaH_(f)[CO_(2)(g)]=-400` kJ/mol `BE_(o=o=500` kJ/mol Resonance energy of `CO_(2)=-150` kJ/mol |
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Answer» Correct Answer - 725 |
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| 292. |
The enthalpy of atomisation for the reaction `CH_(4) (g) rarr C (g) + 4H (g) " is " 1665 kJ mol^(-1)`. What is the bond energy of `C - H` bond ? |
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Answer» In `CH_(4)`, there are four `C - H` bonds. The enthalpy of atomisation of 1 mole of `CH_(4)` means dissociation of four moles of `C - H`bonds. `:. C - H` bond energy per mol `= (1665 kJ)/(4 mol)` `= 416.25 kJ mol^(-1)` |
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| 293. |
The enthalpy of atomisation for the reaction `CH_(4)(g) rarrC(g) + 4H(g)` is 1665kJ `mol^(-1)` . What is the bond energy of C-H bond ? |
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Answer» Enthalpy of atomisation of `CH_(4)` is for dissociation of 4 moles ofC-H bonds. `:. ` C-Hbond energy per mole `= ( 1165)/( 4) kJ mol^(-1)= 416.2 kJ mol^(-1)` |
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| 294. |
The combustion of benzene (l) gives `CO_(2)(g)`and `H_(2)O(l) `. Given that heat of combustion of benzene at constant volumeis `- 3263.9 Kj mol^(-1)` at `25^(@)C`, heatof combustion ( in kJ `mol^(-1))`of benzene at constant pressure will be`( R = 8.314 JK^(-1) mol^(-1))`A. 4152.6B. `-452.46`C. `3260`D. `-3267.6` |
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Answer» Correct Answer - D `C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr6CO_(2)(g)+3H_(2)O(l)` `Deltan_(g)= 6 - ( 15)/( 2) = - (3)/(2)` `DeltaH = DeltaU+ Deltan_(g) RT` `= - 3263.9 + ( - (3)/(2)) xx ( 8.314 xx 10^(-3)) xx ( 298)` `= - 3263.9 + ( -3.71)= - 3267 .6 kJ mol^(-1)` |
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| 295. |
Which of the following has zero value for `D_(f) H^(@)`?A. `O_(3) (g)`B. `O (g)`C. `O_(2) (g)`D. `O_(2) (l)` |
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Answer» Correct Answer - C By convention, the `Delta_(f) H^(@)` value for any elements in its standard state is zero. The standard state of oxygen, i.e., the most stable pure of oxygen at 1 bar pressure and at `25^(@)C` is `O_(2) (g)`. |
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| 296. |
The difference in `DeltaH` and `DeltaE` for the combustion of methane at `27^(@)C` would beA. 2RTB. RTC. -RTD. -2RT |
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Answer» Correct Answer - C `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l))` `Deltan=1-2=-1" "DeltaH=DeltaU+Delta nRT` |
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| 297. |
For the reaction , `2Cl(g)rarrCl_(2)(g)`, what are thesigns of`DeltaH ` and `DeltaS` ? |
| Answer» The given reaction repesents the formation of bonds. Hence, energy is released , ie.,`DeltaH` is `-ve`. Further, 2 moles of atoms have greater randomness than1 mole of molecules. Hence, randomness decreases, i.e.,`DeltaS` is `-ve`. | |
| 298. |
The equilibrium constant for a reaction is 10.What will be thevalue of `DeltaG^(@) ` ? ` R=8.314 JK^(-1)mol^(-1) , T = 300 K `. |
| Answer» `DeltaG^(@) =-2.303 RT log K=-2.303 xx 8.314 JK^(-1) JK^(-1) mol^(-1) xx300 K xxlog10 = -5744.1 J` | |
| 299. |
For the given reaction: `H_(2)(g)+Cl_(2)(g) to2H^(+)(aq)+2Cl^(-)(aq)` `DeltaG^(@)=-262.4kJ` The value of Gibbs free energy of formation `(DeltaG_(r)^(@))` for the ion `Cl^(-)(aq)` is:A. `-131.2 KJ mol^(-1)`B. `+131.2 KJ mol^(-1)`C. `-262.4KJmol^(-1)`D. `+262.4KJmol^(-1)` |
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Answer» Correct Answer - a |
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| 300. |
An ice cube at an unknown temperature is added to `25.0g` of liquid `H_(2)O` at `40.0^(@)C`. The final temperature of the `29.3g` equilibrated mixture of the ice cube? `C_(p)(J//gxx .^(@)C)water=4.184, ice =2.06, deltaH_(fusion)=333J//g^(@)`A. `-6.5^(@)C`B. `-13.1^(@)C`C. `-35.3^(@)C`D. `-56.8^(@)C` |
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Answer» Correct Answer - B |
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