InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`S_(H_(2(g)))^(0) = 130.6 J K^(-1) mol^(-1)`, `S_(H_(2)O_((l)))^(0)= 69.9 J K^(-1)mol^(-1)` `S_(O_(2(g)))^(0) = 205 J K^(-1)mol^(-1)`, then the absolute entropy change of `H_(2_((g))) + (1)/(2)O_(2_((g))) rarr H_(2)O_((l))` isA. `-163.2 J mol^(-1) K^(-1)`B. `+163.2 J mol^(-1)K^(-1)`C. `-303 J mol^(-1) K^(-1)`D. `+303 J mol^(-1) K^(-1)` |
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Answer» Correct Answer - A `Delta S = S_(P) - S_(R)` |
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| 2. |
An intimate mixture of ferric oxide , `Fe_(2)O_(3)`, and aluminium, Al, is used in solid fuel rockets. Calculate the fuel value per gram and fuel value per cc. of the mixture . Heats of formation and densities are as follows `:-` `DeltaH_(f) (Al_(2)O_(3))= 399kcal//mol`, `DeltaH_(f)(Fe_(2)O_(3))= 199kcal//mol` Density of `Fe_(2)O_(3)= 5.2 g//cc`,Density of Al `= 2.7 g //cc.` |
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Answer» We aim at `Fe_(2)O_(3) +2Al rarr 2 Fe+Al_(2)O_(3), DeltaH = ?` `Delta_(r)H= [2 xx DeltaH _(f)(Fe) + DeltaH_(f)(Al_(2)O_(3))]-[DeltaH_(f)(Fe_(2)O_(3))+ 2 xx DeltaH_(f)(Al) ]` `= ( 0 + 399) - ( 199 + 0 ) = 200 kcal` This is the heat channe that takes place when 1 moleof `Fe_(2)O_(3)` and 2 moles of Al combine i.e., ` 2 xx 56 + 3 xx 16 ) + 2 xx 27 g = 214 g` of the mixture . Hence , heat change per gram `= ( 200)/( 214) = 0.9346kcal` Further, `160g Fe_(2)O_(3)= ( 160g)/( 5.2 g // c c ) = 30.77 c c` and `54g Al= ( 54g)/( 2.7 g //c c)= 20 c c` Total volume of the mixture `= 50.77 c c ` `:. ` Heat change per cc`= ( 200 kcal)/ ( 50.77 c c) ` `= 3.939 kcal` |
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| 3. |
Calculate the entropy change for the conversion of2 moles of liquid water at 373K to vapours , if `Delta_(vap)` H is `37.3kJ mol^(-1)`. |
| Answer» Correct Answer - `200 JK^(-1)` | |
| 4. |
Calculae the entropy change fo n-hexane when one mole of it evaporates at 341.7 K `(Delta_(vap) H = 29.0 kJ mol^(-1))` |
| Answer» Correct Answer - `84.86 JK^(-1) mol^(-1)` | |
| 5. |
An ideal gas has volume`V_(0)` at . `27^(@)C` It is heated at constant pressure so that its volume becomes . `2V_(0).`The final temperature isA. `54^(@)C`B. `32.6^(@)C`C. `327^(@)C`D. 150 k |
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Answer» Correct Answer - D |
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| 6. |
What kind of system does a water filled bucket represents? |
| Answer» Correct Answer - Open system. | |
| 7. |
When a polyatomic gas undergoes an adiabatic process, its tempertaure and volume are related by the equation `TV^(n) =` constant, the value of `n`will beA. `1.33`B. `0.33`C. `2.33`D. `1` |
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Answer» For adiabatic process `TV^(gamma-1) =` constant for poly atomis gas `gamma = (4)/(3) = 1.33 (gamma = (C_(P))/(C_(V)))` `n = 1.33 - 1 = 0.33` |
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| 8. |
A monoatomic gas undergoes adiabatic process. Its volume and temperature are related as `TV^(P)` = constant. The value of p will be `{:((1),1.33,(2),1.67),((3),0.67,(4),0.33):}` |
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Answer» For adiabatic process `TV^(gamma-1)` = constant for monoatomic gas `gamma` = 1.67 p = 1.67 - 1 = 0.67 `therefore "Ans"(3)` |
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| 9. |
For a given reaction `Delta`G obtained was having positive sign convention. State whether the reaction was spontaneous or non-spontaneous. |
| Answer» Correct Answer - Non-spontaneous. | |
| 10. |
A gas is contained in a metallic cylinder fitted with a piston.The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinderA. The pressure decreasesB. The pressure increasesC. The pressure remains the sameD. The pressure may increase or decrease depending upon the nature of the gas |
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Answer» Correct Answer - A |
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| 11. |
When a gas expands adiabaticallyA. No energy is required for expansionB. Energy is required and it comes from the wall of the container of the gasC. Internal energy of the gas is used in doing workD. Law of conservation of energy does not hold |
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Answer» Correct Answer - C |
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| 12. |
For a Carnot engine, the source is at 500K and the sink at 300K.What is efficiency of this engine? A) 0.2 B) 0.4 C) 0.6 D) 0.3 |
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Answer» Answer: B) 0.4 Explanation: Given that, T1 = 500K, T2=300 K \(\eta=\frac{T_1-T_2}{T_1}= \frac{500-300}{500}=\frac{200}{500}=0.4\) |
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| 13. |
A carnot engine has efficiency of 50% when its sink is at `227^(@)C`. What is the temperature of sourceA. `727^(@)C`B. `454^(@)C`C. `113.5^(@)C`D. `500^(@)C`. |
| Answer» Correct Answer - A | |
| 14. |
The entropy change for the conversion of 1 mole of `alpha-" tin"` (at `13^(@)C, 1` atm) to 1 mole of `beta-" tin "(13^(@)C, 1 atm)` if the enthalpy of transition is `2.095 kJ mol^(-1)`A. `7.32 J mol^(-1) K^(-1)`B. `14.62 J K mol^(-1)`C. `56. J mol^(-1) K^(-1)`D. `0` |
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Answer» Correct Answer - A `DeltaS=(DeltaH)/(T)` |
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| 15. |
Explain, Why ?The coolant in a chemical or a nuclear plant (i. e., the liquid used to prevent the different prats of a plant from getting too hot) should have high specific heat. |
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Answer» A Coolant of high specific heat can carry more heat out of the chemical or nuclear plant than a coolant of ordinary specific heat. |
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| 16. |
A gas expands against a constant external pressure so that the work done is 607.8 J. The work done in litre atmosphere is |
| Answer» Correct Answer - 6 | |
| 17. |
Statement-1 : The amount of heat change during the isothermal free expansion of an ideal gas is zero. Statement-2: There are no intermolecular forces of attraction among the gas molecules, in case of real gas at given pressure.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - C | |
| 18. |
Assertion: `C_(P)-C_(V)=R` for an ideal gas. Reason: `((delE)/(delV))_(T)=0` for an ideal gas.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - B | |
| 19. |
In which state, the matter have highest entropyA. SolidB. LiquidC. GasD. Equal in all |
| Answer» Correct Answer - C | |
| 20. |
An ideal gas can be expanded from an initial state to a certain volume through two different processes, (I) `PV^(2)=K` and (II) `P=KV^(2)`, where `K` is a positive constant. Then, choose the correct option from the following.A. Final temperature in `(I)` will be greater than in `(II)`B. Final temperature in `(II)` will be greater then in `(I)`C. Work done by the gas in both the processes would be equalD. Total heat given to the gas in `(I)` is greater then in `(II)` |
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Answer» Correct Answer - B Process(A). `(nRT)/(v)V^(2)=K`,`TV`=constant `Tprop(1)/(2)` Process (B)`(nRT)/(v)=KV^(2)` ,`(T)/(v^(3))`= constant, `Tpropv^(3)` |
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| 21. |
The equilibrium constant for a reaction is one or more if ∆G ° for it is less than zero. Explain. |
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Answer» G⊖ = – RT in K, thus if G⊖ is less than zero Le., it is negative, then In K will be positive and hence K will be greater than one. |
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| 22. |
The spotanity means having the potential to proceed without assistance of external agency. The processess which occur spontaneously areA. flow of heat from colder to warmer bodyB. gas in a container contracting into one cornerC. gas expanding to fill the available volumeD. burining carbon in oxygen to given carbon dioxide |
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Answer» Correct Answer - C::D Gas expands of diffuses in availabel space spontaneously, e.g., leakage of cooling gas gives smell of ethyl mercapatan spontaneously. Moreover burning of carbon to `CO_(2)` is also spontaneous |
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| 23. |
In a constant volume calorimeter, `3.5 g` of a gas with molecular weight `28` was burnt in excess oxygen at `298.0 K`. The temperature of the calorimeter was found to increase from `298.0 K to 298.45 K` due to the combustion process. Given that the heat capacity of the calorimeter is `2.5 kJ K^(-1)`, find the numerical value for the enthalpy of combustion of the gas in `kJ mol^(-1)` |
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Answer» Number of moles of gas `= (3.5)/(28) = 0.125` Heat evolved = Heat capacity `xx DeltaT` `= 2.5 xx (298.45 - 298)` `= 2.5 xx 0.45 = 1.125 kJ` Heat evolved per mole `= (1.125)/(0.125) = 9 kJ` |
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| 24. |
Entropy of a perfectly crystalline solid istaken as zero at the absolute zero. This is a statement of `"…............."` |
| Answer» third law of thermodynamics | |
| 25. |
According to the third law of thermodynamics which one of the following quantities for a perfectly crystalline solid is zero at absolute zero?A. Free energyB. EntropyC. EnthalpyD. Internal energy |
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Answer» Correct Answer - B Entropy is the degree of randomness or disorder of the system. When the temperature of the system is zero kelvin, then all the motion of molecules ceases. According to third law of thermodynamics "At absolute zero the entropy of a perfectly crystalline substance is taken as zero". |
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| 26. |
Why a gas is cooled when expanded ? |
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Answer» Decrease in internal energy. |
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| 27. |
Why does air pressure in car tyre increases during driving ? |
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Answer» During driving, as a result of the friction between the tyre and the road, the temperature of the tyre and the air inside it, increase since volume of the tyre does not change (practically); due to increase in temperature, pressure of the air increases. PV = nRT V = constant, T = increases. So, P also increases P ∝ T |
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| 28. |
The value of coefficient of volume expansion of glycerin is `5 xx 10^(-4) K^(-1)`. The fractional change in the density of glycerin for a rise of `40^(@)C` in its temperature isA. `0.010`B. `0.015`C. `0.020`D. `0.025` |
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Answer» Correct Answer - C Here, `gamma=5xx10^(-4)K^(-1)= 5xx10^(-4).^(@)C^(-1)` Let, `rho_(2), rho_(1)` be the density of glycerine at temperature `t_(2).^(@)C` and `t_(1).^(@)C` respectively. Then `rho_(2)=rho_(1)[1-gamma(t_(2)-t_(1))]=rho_(1)-rho_(1)(t_(2)-t_(1))` or `(rho_(1)-rho_(2))/(rho_(1))=gamma(t_(2)-t_(1))` Frictional change in density of glycerine `=(rho_(1)-rho_(2))/(rho_(1))= gamma(t_(2)-t_(1))= (5xx10^(-4))xx40` `=0.020` |
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| 29. |
Two identical thermally insulated vessels interconnected by a tube with a valve contain one mole of the same ideal gas each. The gas temperature in one vessel is equal to `T_1` and in the other, `T_2`. The molar heat capacity of the gas of constant volume equals `C_v`. the valve having been opened, the gas comes to a new equilibrium state. Find the entropy increment `Delta S` of the gas. Demonstrate that `Delta S gt 0`. |
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Answer» For and ideal gas to internal energy depends on temperature only. We can consider the process in question to be one of simultaneous free expansion. Then the total energy `U = U_1 + U_2` Since `U_1 = C_V T_1, U_2 = C_V T_2, U = 2 C_V (T_1 + T_2)/(2) ` and `(T_1 + T_2)//2` is the final temperature. The entropy change is obtained by considering isochoric processes because in effect, the gas remains confined to its vessel. `Delta S = int _(T_(1))^((T_1 + T_2)//2) (C_V dT)/(T) - int_((T_1 + T_2)//2)^(T_2) C_V (dT)/(T) = C_V 1n ((T_1 + T_2)^2)/(4 T_1 T_2)` Since `(T_1 + T_2)^2 = (T_1 - T_2)^2 + 4 T_1 T_2, Delta S gt 0`. |
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| 30. |
Assertion: Two systems, which are in thermal equilibrium with a third system, are in thermal equilibrium with each other. Reason: The heat flows spontaneously from a system at a higher temp. to a system at a lower temp.A. If both, Assertion and Reason are true and Reason is the correct explanation of the Asserrion.B. If both,Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - A This is as per Zeroth law of thermodynamics. |
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| 31. |
Why is work not a state function? |
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Answer» Work is not a state function, because its value is dependent on the path followed. For example, the work done by a person for reaching the top of a multistoried building by using steps is different from the work done if he goes by lift or ramp. |
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| 32. |
Why is the internal energy of a system or state variable? |
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Answer» Internal energy is a state-function, because it depends only on the state of the system (i.e., temperature, pressure, etc.) and is independent of the method (or number of steps) by which the present state of system has been attained. |
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| 33. |
Assetion :when a bullet is fired from a gun the bullet preces a wooden block and stops changing the temperature of the bullte and the surrounding layers opf ood. Reason : Temperature is related to the energy of motion of the bullet as a whole.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - C When the bullet pierces a wooden block and stops , kinetic energy of the bullet gets converted in to heat. Due to this , the temperature of the wood and the surrounding layers of wood increases .Temperature is related to the energy of the internal motion of the bullet not to the motion of the bullet as a whole. |
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| 34. |
In a gobar gas plant, gobar gas in obtained by bacterial fermentation of animal refuse. The main combustible gas present in the gobar is found to be methane `( 80%` by weight) whose heat of combustion is 809`kJ mol^(-1))` . How much gobar gas would have to produced per day for a village of 100 families, if the average consumption of a family is 20,000kJ per day to meet all its energy requirements . |
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Answer» Total energy requirements for 100 families per day `= 20,000 xx 100kJ = 2 xx 10^(6) kJ` Methane `(CH_(4))` undergoes combustion as follows `:` `CH_(4)(g) rarr 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l), Delta H = - 809 kJ mol^(-1)` For 809 kJ of energy , `CH_(4)`required`=16g` ( Molar mass of `CH_(4) = 16 g mol^(-1))` For ` 2 xx 10^(6)` kJ mol energy, `CH_(4)` required `(16)/(809) xx 2 xx 10^(6)g = 39.56kg` As gobar gas contains `80%` by weight of methane , therefore gobar gas required `=(100)/(80) xx 39.56 kg` `= 49.45kg` |
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| 35. |
Solve the foregoing problem for the case of two concentric spheres of radii `R_1` and `R_2` and temperatures `T_1` and `T_2`. |
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Answer» In equilibrium `4 pi r^2 k (dT)/(dr) = - A = constant` `T = B + (A)/(4 pi k) (1)/( r)` Using `T = T_1` when `r = R_1` and `T = T_2` when `r = R_2`, `T = T_1 + (T_2 - T_1)/((1)/(R_2) -(1)/(R_1)) ((1)/(r) -(1)/(R_1))`. |
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| 36. |
`1` mole of an ideal diatomic gas undergoes a reversible polytropic process `(PV^(2)="constant")`. The gas expand from initial volume of `1` litre and temp `300 K` to final volume `3` lit. Claculate change in internal energy (approx).A. `-40 KJ`B. `-4.2 KJ`C. `-4.4 KJ`D. `-4.6 KJ` |
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Answer» Correct Answer - B For polydropic process `(n=2)` `PV^(2)=C` `TV=C` `T_(1)V_(1)=T_(2)V_(2)` `300xx1=T_(2)xx3 rArrT_(2)=100 K` ` Delta U=nC_(v,m)Delta T=1xx(fR)/(2)(T_(2)-T_(1))=1xx(5xxR)/(2)xx(100-300)=(1xx5xxR)/(2)xx200= -500 R= -4.2 KJ` |
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| 37. |
2 mole of an ideal monoatomic gas undergoes a reversible process for which `PV^(2)=C`. The gas is expanded from initial volume of 1 L to final volume of 3 L starting from initial temperature of 300 K. Find `DeltaH` for the process :A. `-600R`B. `-1000 R`C. `-3000 R`D. `-2000 R` |
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Answer» Correct Answer - B `P_(2)V_(2)^(2) = P_(1)V_(1)^(2)` `(P_(2))/(P_(1)) = ((V_(1))/(V_(2)))^(2)` Now, `(T_(2))/(T_(1)) = (P_(2)V_(2))/(P_(1)V_(1)) = (V_(1))/(V_(2)) = (1)/(3)` `:. T_(2) = (300)/(3) = 100 K , Delta H = nC_(p, m)Delta T` `= 2 xx (5)/(2)R xx (-200 K) = -1000 R` |
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| 38. |
2 mole of an ideal monoatomic gas undergoes a reversible process for which `PV^(2)=C`. The gas is expanded from initial volume of 1 L to final volume of 3 L starting from initial temperature of 300 K. Find `DeltaH` for the process :A. `-600` RB. `-1000` RC. `-3000` RD. None of these |
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Answer» Correct Answer - B `P_(2)V_(2)^(2)=P_(1)V_(1)^(2)` `(P_(2))/(P_(1))=((V_(1))/(V_(2)))^(2)` Now, `(T_(2))/(T_(1))=(P_(2)V_(2))/(P_(1)V_(1))=(V_(1))/(V_(2))=(1)/(3)` `therefore" "T_(2)=(300)/(3)=100K` `DeltaH=nC_(p,m)DeltaT` `=2xx(5)/(2)Rxx(-200K)=1000R` |
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| 39. |
Molecules of a noble gas do not possess vibrational energy because a noble gas_____________ |
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Answer» Molecules of a noble gas do not possess vibrational energy because a noble gas is monoatomic. Explanation: Molecules of noble gas does not possess vibrational energy because noble gases are monoatomic. So, there is no other atom in respect of which it can vibrate. So they do not vibrate. Hence they do not posses vibrational energies. |
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| 40. |
Statement-1 : If same amounts are considered `CO_(2)` requires more heat than `O_(2)`, for the same rise in temperature. And Statement-2 : `CO_(2)` being triatomic has higher heat capacity than `O_(2)`A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 41. |
Statement-1 : Entropy of egg increases after boiling. And Statement-2 : As boiling causes the denaturation of proteins.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 42. |
A sustance undergoes a change represented as show A (l) [1bar,300K]`rarr A` (s) [1bar, 300 K].from the given information, calculate magnitude of change in internal energy (in Joules) when 1 mole of A (l) solidifies Standard melting point of A (s) is 300 K. Latent heat of fusion of A id 0.01 kJ/gm. Specific volume of A (s) is 100 ml/gm. Moar mass of A is 50 gm /mole |
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Answer» Correct Answer - 400 |
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| 43. |
`NO(g)to(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g),DeltaH_(1)^(@)` `2NO(g)toN_(2)O(g)+(1)/(2)O_(2)(g),DeltaH_(2)^(@)` Which relationship is correct ?A. `DeltaH_(1)^(@)=DeltaH_(2)^(@)`B. `DeltaH_(f)^(@)for NO(g)=DeltaH_(1)^(@)`C. `DeltaH_(f)^(@)for N_(2)O(g)=DeltaH_(2)^(@)`D. `DeltaH_(f)^(@) for N_(2)O(g)=DeltaH_(2)^(@)-2DeltaH_(1)^(@)` |
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Answer» Correct Answer - d |
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| 44. |
The combustion reaction occuring in an automobile is `2C_(8)H_(18)+25O_(2)(g) to16CO_(2)(g)+18H_(2)O(g)` This reaction is accompanied with:A. `DeltaH=-ve, DeltaS=+ve,DeltaG=+ve`B. `DeltaH=+ve, DeltaS=-ve,DeltaG=+ve`C. `DeltaH=-ve, DeltaS=-ve,DeltaG=+ve`D. `DeltaH=+ve, DeltaS=+ve,DeltaG=-ve` |
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Answer» Correct Answer - C |
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| 45. |
Enthalpy change for the process, `H_(2)O"(ice")hArr H_(2)O"(water)"` is `"6.01 kJ mol"^(-1)`. The entropy change of 1 mole of ice at its melting point will beA. `"12 J K"^(-1)"mol"^(-1)`B. `"22 J K"^(-1)"mol"^(-1)`C. `"100 J K"^(-1)"mol"^(-1)`D. `"30 J K"^(-1)"mol"^(-1)` |
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Answer» Correct Answer - B `DeltaS_("fusion")=(DeltaH_("fusion"))/(T_("fusion"))=(6.01xx1000)/(273)` `="22 J K"^(-1)"mol"^(-1)` |
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| 46. |
A chemical reaction cannot occur at all isA. `DeltaH` values is positive and `DeltaS` value if negativeB. `DeltaH` value is negative and `DeltaS` value is positiveC. `DeltaH` and `DeltaS` values are negative but `DeltaH lt TDeltaS`D. `DeltaH` and `DeltaS` values are positive but `DeltaH lt T DeltaS` |
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Answer» a. DeltaG is +ve` if `DeltaH is+ve` and `DeltaS is -ve` . Thus, reaction in non-spontaneous. Hence (a) is the correct answer. |
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| 47. |
The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below: `CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))` `DeltaH_(300K)^(@)=-41.16 kJ mol^(-1)` `DeltaS_(300 K)^(@)=-4.24xx10^(-2) kJ mol^(-1)` `DeltaH_(1200 K)^(@)=-32.93 K J mol^(-1)` `DeltaS_(1200 K)^(@)=-2.96xx10^(-2) k J mol^(-1)` Calculate `K_(p)` at each temperature and predict the direction of reaction at `300 K` and `1200 K`, when `P_(CO)=P_(CO_(2))=P_(H_(2))=P_(H_(2)O)=1` atm at initial state. |
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Answer» Correct Answer - At `300 K,DeltaG^(@)=-28.44 kJ,K_(p)=8.94xx10^(4)` `at 1200 K,DeltaG^(@)=+2.59 kJ,K_(p)=0.77` |
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| 48. |
The values of `DeltaH` and `DeltaS` for the reaction, `C_("graphite")+CO_(2)(g)rarr2CO(g)` are `170KJ` and `170JJK^(-)` respectively. This reaction will be spontaneous atA. 710 KB. 910 KC. 1110 KD. 510 K |
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Answer» Correct Answer - C Given , `DeltaH=170 KJ =170xx10^(3)` J `DeltaS=170 JK^(-1), T=?` `DeltaG=DeltaH-TDeltaS` For spontaneous reaction `DeltaG lt 0 implies 0 lt 170 xx 10^(3) -Txx170, T gt 1000` `therefore T=1110` K |
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| 49. |
Given the following reaction : `P : N_(2)(g)+2O_(2)(g)to2NO_(2)(g),DeltaH_(p)=16.18kcal` `Q : N_(2)(g)+2O_(2)(g)toN_(2)O_(4)(g),DeltaH_(Q)=2.31kcal` Based on the above facts :A. `NO_(2)` is more stable than `N_(2)O_(4)` at low temperatureB. `N_(2)O_(4)` is more stable than than `NO_(2)` at low temperatureC. both are equally stable at low temperatureD. none of these |
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Answer» Correct Answer - B |
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| 50. |
Enthalpy of polymerisation of ethylene, as represented by the reaction , `Q : N_(2)(g)+2O_(2)(g)toN_(2)O_(4)(g),DeltaH_(Q)=2.31kcal` is `-100` kJ per mole of ethylene. Given bond enthalpy of `C=C` bond is 600 kJ `mol^(-1)`, enthalpy of `C-C` bond (in kJ mol) will be :A. 116.7B. 350C. 700D. intermediate |
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Answer» Correct Answer - B |
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