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All questions of Thermodynamics for JEE Exam
Consider the following properties. 
Select intensive and extensive properties.
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'C'. Can you explain this answer?
Consider the following properties.
Select intensive and extensive properties.
a)
a
b)
b
c)
c
d)
d
|
Preeti Khanna answered |
Intensive means that the property is mass independent. We can see that Boiling point, pH, EMF of cell and surface tension are intensive pont .
While Extensive property depends on mass and in given options Volume and entropy are mass dependents.
While Extensive property depends on mass and in given options Volume and entropy are mass dependents.
One mole each of CaC2, AI4C3 and Mg2C3 reacts with H2O in separate open flasks at 25° C. Numerical value of the work done by the system is in order- a)CaC2 < AI4C3 = Mg2C3
- b)CaC2 < AI4C3 < Mg2C3
- c)CaC2 = Mg2C3 < AI4C3
- d)CaC2 = Mg2C3 = AI4C3
Correct answer is option 'C'. Can you explain this answer?
One mole each of CaC2, AI4C3 and Mg2C3 reacts with H2O in separate open flasks at 25° C. Numerical value of the work done by the system is in order
a)
CaC2 < AI4C3 = Mg2C3
b)
CaC2 < AI4C3 < Mg2C3
c)
CaC2 = Mg2C3 < AI4C3
d)
CaC2 = Mg2C3 = AI4C3
|
Ciel Knowledge answered |
i) CaC2 + 2H2O → C2H2(g) + Ca(OH)2
ii) Mg2C3 + 4H2O → CH3-C≡CH(g) + 2Mg(OH)2
iii) Al4C3 + 12H2O → 3CH4(g) + 4Al(OH)3
Work done = -∆ngRT
Or W ∝ ∆ng
Wecan see that in case iii) we have maximum no of moles in the product side(∆ng = 3). So work done will be maximum in case iii). After that in i) and ii), we have the same number of moles on the product side(∆ng = 1), so work done will be the same.
Therefore, option c is correct.
ii) Mg2C3 + 4H2O → CH3-C≡CH(g) + 2Mg(OH)2
iii) Al4C3 + 12H2O → 3CH4(g) + 4Al(OH)3
Work done = -∆ngRT
Or W ∝ ∆ng
Wecan see that in case iii) we have maximum no of moles in the product side(∆ng = 3). So work done will be maximum in case iii). After that in i) and ii), we have the same number of moles on the product side(∆ng = 1), so work done will be the same.
Therefore, option c is correct.
Heat of hydrogenation of ethene is x1 and that of benzene is x2. The resonance energy of benzene is- a)x1 — x2
- b)x1 + x2
- c)3x1 — x2
- d)x1 - 3x2
Correct answer is option 'C'. Can you explain this answer?
Heat of hydrogenation of ethene is x1 and that of benzene is x2. The resonance energy of benzene is
a)
x1 — x2
b)
x1 + x2
c)
3x1 — x2
d)
x1 - 3x2
|
Anaya Patel answered |
Energy of ethene =x1
Number of ethene in benzene=3.
Therefore, Observed energy of hydrogenation of benzene =3x1
Calculated energy of hydrogenation of benzene =x2 (Given)
Therefore,
Resonance energy = observed - calculated =3x1−x2
Number of ethene in benzene=3.
Therefore, Observed energy of hydrogenation of benzene =3x1
Calculated energy of hydrogenation of benzene =x2 (Given)
Therefore,
Resonance energy = observed - calculated =3x1−x2
Specific heat of water is 4.184 JK-1g-1, Rise in temperature when 1 kJ of heat is absorbed by 2 moles of H2O is- a)119.5o
- b)0.37o
- c)0.12o
- d)6.64o
Correct answer is option 'D'. Can you explain this answer?
Specific heat of water is 4.184 JK-1g-1, Rise in temperature when 1 kJ of heat is absorbed by 2 moles of H2O is
a)
119.5o
b)
0.37o
c)
0.12o
d)
6.64o
|
Rajesh Gupta answered |
We have,
Q = ms∆T
1000 = 2*18*4.184*∆T
∆T = 6.63 K
Q = ms∆T
1000 = 2*18*4.184*∆T
∆T = 6.63 K
In which of the following cases, entropy of I is larger than that of II?
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'D'. Can you explain this answer?
In which of the following cases, entropy of I is larger than that of II?
a)
a
b)
b
c)
c
d)
d
|
Infinity Academy answered |
a) More molar mass, more entropy. So ∆SN2O4 > ∆SNO2
b) CO2 has more entropy than dry ice at -78°C
c) Pure alumina iss crystalline solid while ruby is amorphous. And ∆Samorphous > ∆SCrystalline. So alumina has less entropy than ruby.
d) At lower pressure, entropy be higher as gas particles are far from each other. So (∆SN2)1 bar > (∆SN2)5 bar
b) CO2 has more entropy than dry ice at -78°C
c) Pure alumina iss crystalline solid while ruby is amorphous. And ∆Samorphous > ∆SCrystalline. So alumina has less entropy than ruby.
d) At lower pressure, entropy be higher as gas particles are far from each other. So (∆SN2)1 bar > (∆SN2)5 bar
Can you explain the answer of this question below:Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will be
- A:
- 1870.6 J
- B:
- 1870.6J
- C:
-18.47 J
- D:
+18.47 J
The answer is a.
Five moles of an ideal gas at 27° C are allowed to expand isothermally from an initial pressure of 10.0 atm to a final pressure of 4.0 atm against a constant external pressure of 1.0 atm. Thus, work done will be
- 1870.6 J
- 1870.6J
-18.47 J
+18.47 J
|
Suresh Reddy answered |
Work done during isothermal irreversible process is given by the formula w = -pext (nrt/p2-nrt/p1)
The equilibrium constant for a reaction is 10. Calculate the value of ΔG° at 300K , R = R =8.314 JK−1mol−1
- a)-5.527 kJ/mol–
- b)4.744 kJ/mol
- c)-4.744 kJ/mol–
- d)- 5.744 kJ/mol
Correct answer is option 'D'. Can you explain this answer?
The equilibrium constant for a reaction is 10. Calculate the value of ΔG° at 300K , R = R =8.314 JK−1mol−1
a)
-5.527 kJ/mol–
b)
4.744 kJ/mol
c)
-4.744 kJ/mol–
d)
- 5.744 kJ/mol
|
|
Infinity Academy answered |
The value of equilibrium constant(k) = 10
∆G° = -2.303RTlogk
On calculating values, we will get ∆G° = -5.744kJmol-1
∆G° = -2.303RTlogk
On calculating values, we will get ∆G° = -5.744kJmol-1
A gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of 2 x 105 Nm-2. Change in internal energy is - a)- 300 J
- b)- 100 J
- c)+100 J
- d)+300 J
Correct answer is option 'C'. Can you explain this answer?
A gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of 2 x 105 Nm-2. Change in internal energy is
a)
- 300 J
b)
- 100 J
c)
+100 J
d)
+300 J
|
Neha Joshi answered |
The expression for the change in internal energy is given by,
ΔU=q+w
q = heat absorbed = 200 J
w = work done = −P x V
=−2×10⁵ × 500×10−6
= −100 N - m
ΔU=200−100J= 100 J
Hence, the correct option is C.
ΔU=q+w
q = heat absorbed = 200 J
w = work done = −P x V
=−2×10⁵ × 500×10−6
= −100 N - m
ΔU=200−100J= 100 J
Hence, the correct option is C.
When 4.0 mL of 2.0 N solution of weak acid is neutralised by a dilute aqueous solution of sodium hydroxide, 64 cal of heat is liberated. What is the heat of neutralisation in cal/milliequivalent?
Correct answer is '8'. Can you explain this answer?
When 4.0 mL of 2.0 N solution of weak acid is neutralised by a dilute aqueous solution of sodium hydroxide, 64 cal of heat is liberated. What is the heat of neutralisation in cal/milliequivalent?
|
|
Riya Banerjee answered |
Milliequivalents of the acid taken = 4×2 =8.
neutralization of 8 meq liberated 64 cal heat.
So the heat of neutralization is 64/8 = 8cal/meq.
neutralization of 8 meq liberated 64 cal heat.
So the heat of neutralization is 64/8 = 8cal/meq.
What happens to the ions, when an ionic compound dissolves in a solvent?- a)They leave their ordered position in the crystal lattice and gets hydrated.
- b)They leave their ordered position in the crystal lattice and doesn’t hydrate.
- c)They doesn’t leave their ordered position in the crystal lattice and gets hydrated.
- d)None of the above.
Correct answer is option 'A'. Can you explain this answer?
What happens to the ions, when an ionic compound dissolves in a solvent?
a)
They leave their ordered position in the crystal lattice and gets hydrated.
b)
They leave their ordered position in the crystal lattice and doesn’t hydrate.
c)
They doesn’t leave their ordered position in the crystal lattice and gets hydrated.
d)
None of the above.
|
|
Pooja Shah answered |
When an ionic compound is dissolved in the solvent, the compound dissociate into ions and due to external medium i. e. solvent it leave its original position in crystal lattice n gets hydrated..
Eg. NaCl(aq) → Na+(aq) + Cl-(aq)
Eg. NaCl(aq) → Na+(aq) + Cl-(aq)
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0L, 95(K) → (4.0 atm, 5.0 L, 245K) with a change in internal energy, ΔU= 30.0L atm . The change in enthalpy (ΔH) of the process in L atm is (2002S)- a)40.0
- b)42.3
- c)44.0
- d)not defined, because pressure is not constant
Correct answer is option 'C'. Can you explain this answer?
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0L, 95(K) → (4.0 atm, 5.0 L, 245K) with a change in internal energy, ΔU= 30.0L atm . The change in enthalpy (ΔH) of the process in L atm is (2002S)
a)
40.0
b)
42.3
c)
44.0
d)
not defined, because pressure is not constant
|
|
Saptarshi Goyal answered |
In order to fully answer this question, we would need additional information about the gas and the specific change of state it undergoes. Can you please provide more details?
3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)- a)-22.98 JK-1
- b)22.98 JK-1
- c)57.4 JK-1
- d)80.42 JK-1
Correct answer is option 'D'. Can you explain this answer?
3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)
a)
-22.98 JK-1
b)
22.98 JK-1
c)
57.4 JK-1
d)
80.42 JK-1
|
|
Pooja Shah answered |
∆S = nCplnT2/T1 + nRlnP1/P2
Since pressure is constant, so the second term will be zero.
Or ∆S = 3×7/2×8.314×2.303×log(1000/400)
= 80.42 JK-1
Since pressure is constant, so the second term will be zero.
Or ∆S = 3×7/2×8.314×2.303×log(1000/400)
= 80.42 JK-1
Bond energy of (N— H)bond is x kJ mol-1 under standard state. Thus, change in internal energy in the following process is
- a)x kJ mol-1
- b)3x kJ mol-1
- c)- x kJ mol-1
- d)- 3x kJ mol-1
Correct answer is option 'B'. Can you explain this answer?
Bond energy of (N— H)bond is x kJ mol-1 under standard state. Thus, change in internal energy in the following process is
a)
x kJ mol-1
b)
3x kJ mol-1
c)
- x kJ mol-1
d)
- 3x kJ mol-1
|
EduRev Humanities answered |
The correct answer is Option B.
H−N−H (three N-H) bonds
∣
H
Thus, ΔE=3xKJmol-1
H−N−H (three N-H) bonds
∣
H
Thus, ΔE=3xKJmol-1
Direction (Q. Nos. 16 - 18) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q. Temperature of one mole of an ideal gas is increased by 1° at constant pressure. What is mechanical work done (in cals)?
Correct answer is '2'. Can you explain this answer?
Direction (Q. Nos. 16 - 18) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).
Q. Temperature of one mole of an ideal gas is increased by 1° at constant pressure. What is mechanical work done (in cals)?
|
Preeti Iyer answered |
From ideal gas equation
PV=nRT
hence PV=W
W = nR∆T
W = 1×2×1 = 2cal
PV=nRT
hence PV=W
W = nR∆T
W = 1×2×1 = 2cal
In an endothermic reaction, the value of ΔH is always- a)Constant
- b)> 0
- c)< 0
- d)= 0
Correct answer is option 'C'. Can you explain this answer?
In an endothermic reaction, the value of ΔH is always
a)
Constant
b)
> 0
c)
< 0
d)
= 0
|
|
Preeti Iyer answered |
According to me, for an endothermic reaction ,the value of ∆H is always >0 .
Which statement is incorrect ? [AIEEE-2002]- a)All reversible cycles have same efficiency
- b)Reversible cycle has more efficiency than an irreversible one
- c)Carnot cycle is a reversible one
- d)Carnot cycle has the maximum efficiency in all cycles
Correct answer is option 'A'. Can you explain this answer?
Which statement is incorrect ?
[AIEEE-2002]
a)
All reversible cycles have same efficiency
b)
Reversible cycle has more efficiency than an irreversible one
c)
Carnot cycle is a reversible one
d)
Carnot cycle has the maximum efficiency in all cycles
|
|
Preeti Iyer answered |
Out of given statements :
We know that carnot cycle is reversible cycle, and reversible cycle has maximum efficiency of all other cycles and so only statement A is incorrect.
We know that carnot cycle is reversible cycle, and reversible cycle has maximum efficiency of all other cycles and so only statement A is incorrect.
An ideal gas expands in volume from 1 x 10-3 m3 to 1 x 10-2 m2 at 300 K against a constant pressure of 1 x 105 Nm-2. The work done is[AIEEE 2004]- a)- 900 J
- b)- 900 KJ
- c)270 KJ
- d)+900 KJ
Correct answer is option 'A'. Can you explain this answer?
An ideal gas expands in volume from 1 x 10-3 m3 to 1 x 10-2 m2 at 300 K against a constant pressure of 1 x 105 Nm-2. The work done is
[AIEEE 2004]
a)
- 900 J
b)
- 900 KJ
c)
270 KJ
d)
+900 KJ
|
Manas Singh answered |
-900J is a correct option because.10^5[10^-3-10^-2]
For the process,
and 1 atmosphere pressure, the correct choice is[JEE Advanced 2014]- a)ΔS(System) > 0, ΔS(surrounding) > 0
- b)ΔS(System) > 0, ΔS(surrounding) < 0
- c)ΔS(System) < 0, ΔS(surrounding) > 0
- d)ΔS(System) < 0, ΔS(surrounding) < 0
Correct answer is option 'B'. Can you explain this answer?
For the process, and 1 atmosphere pressure, the correct choice is
and 1 atmosphere pressure, the correct choice is[JEE Advanced 2014]
a)
ΔS(System) > 0, ΔS(surrounding) > 0
b)
ΔS(System) > 0, ΔS(surrounding) < 0
c)
ΔS(System) < 0, ΔS(surrounding) > 0
d)
ΔS(System) < 0, ΔS(surrounding) < 0
|
Lavanya Menon answered |
At 100°C and 1 atmosphere pressure H2O (l) ⇋ H2O(g) is at equilibrium. For equilibrium
From the following data, the heat of formation of Ca(OH)2(s) at 18°C is ………..kcal:
- a)-98.69
- b)-235.43
- c)194.91
- d) 98.69
Correct answer is option 'B'. Can you explain this answer?
From the following data, the heat of formation of Ca(OH)2(s) at 18°C is ………..kcal:
a)
-98.69
b)
-235.43
c)
194.91
d)
98.69
|
Knowledge Hub answered |
The correct answer is Option B.
Ca(s) + O2(g) + H2(g) → Ca(OH)2 , ΔHf = ?
Desired equation = eq (iii) + eq(i) - eq (ii)
ΔHf = (−151.80)+(−15.26)−(−68.37)
ΔHf = (-151.80)+(-15.26)-(-68.37)
ΔHf = −235.43KCalmol−1
ΔHf = (-151.80)+(-15.26)-(-68.37)
ΔHf = −235.43KCalmol−1
A cyclic proces ABCD is shown in PV diagram for an ideal gas.
Which of the following diagram represents the same process- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A cyclic proces ABCD is shown in PV diagram for an ideal gas.
Which of the following diagram represents the same process
a)
b)
c)
d)
|
|
Raghav Bansal answered |
In such type of graph conversions, the best way to work is to make a parameter constant and then derive the results between others.
Let us make pressure constant. Then we have, VB>VA
Or
(nrTA/P)<(nrTB/P) or TA<TB
In terms of V-T curve, the thing happens as follow:-
A to B (Volume increase from VA to VB. No change in pressure). B to C (temperature is constant), C to D (Volume remains constant) and D to A (Volume decreases but temperature remains constant. However the constant value is less than from B to C). On these all conditions, only graph d stands.
Let us make pressure constant. Then we have, VB>VA
Or
(nrTA/P)<(nrTB/P) or TA<TB
In terms of V-T curve, the thing happens as follow:-
A to B (Volume increase from VA to VB. No change in pressure). B to C (temperature is constant), C to D (Volume remains constant) and D to A (Volume decreases but temperature remains constant. However the constant value is less than from B to C). On these all conditions, only graph d stands.
Among the following enthalpies, which is always less than zero?- a)ΔcH°
- b)ΔsubH°
- c)ΔmixH°
- d)ΔfH°
Correct answer is option 'A'. Can you explain this answer?
Among the following enthalpies, which is always less than zero?
a)
ΔcH°
b)
ΔsubH°
c)
ΔmixH°
d)
ΔfH°
|
|
Neha Joshi answered |
The symbol represents enthalpy of combustion which is always exothermic. For exothermic rxn delta H is always -ve.
In conversion of lime-stone to lime, CaCO3 (s) → CaO(s)+ CO2(g) the values of ΔH° and ΔS° are + 179.1 kJ mol-1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ΔH° and ΔS° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is [2007] - a)1118 K
- b)1008 K
- c)1200 K
- d)845 K.
Correct answer is option 'A'. Can you explain this answer?
In conversion of lime-stone to lime, CaCO3 (s) → CaO(s)+ CO2(g) the values of ΔH° and ΔS° are + 179.1 kJ mol-1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ΔH° and ΔS° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is [2007]
a)
1118 K
b)
1008 K
c)
1200 K
d)
845 K.
|
|
Vicky Kumar answered |
Standard entropies of X2, Y2 and XY3 are given below the reaction
Q. At what temperature, reaction would be in equilibrium? - a)500 K
- b)750 K
- c)1000 K
- d)1250 K
Correct answer is option 'B'. Can you explain this answer?
Standard entropies of X2, Y2 and XY3 are given below the reaction
Q. At what temperature, reaction would be in equilibrium?
a)
500 K
b)
750 K
c)
1000 K
d)
1250 K
|
|
Suresh Reddy answered |
1/2X2 + 3/2Y2 ⟶XY3,
ΔH= −30 kJ
ΔSreaction = ∑ΔSproduct−∑ΔSreactant
X2 + 3Y2 → 2XY3
ΔH=−60 kJ
ΔSreaction = 2×50−3×40−1×60 =100−120−60=−80 JK−1mol−1
ΔG=ΔH−TΔS=0
ΔH=TΔS
1000×(−60)=−80×T
T=750 K
ΔH= −30 kJ
ΔSreaction = ∑ΔSproduct−∑ΔSreactant
X2 + 3Y2 → 2XY3
ΔH=−60 kJ
ΔSreaction = 2×50−3×40−1×60 =100−120−60=−80 JK−1mol−1
ΔG=ΔH−TΔS=0
ΔH=TΔS
1000×(−60)=−80×T
T=750 K
Direction (Q. No. 10) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.Q.Statement I : Based on the following thermodynamic data
NO2 is more stable than NO.Statement II : NO (g) is an endothermic compound while, NO2(g) is an exothermic compound.- a)Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
- b)Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
- c)Statement I is correct but Statement II is incorrect
- d)Statement II is correct but Statement I is incorrect
Correct answer is option 'C'. Can you explain this answer?
Direction (Q. No. 10) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : Based on the following thermodynamic data
NO2 is more stable than NO.
Statement II : NO (g) is an endothermic compound while, NO2(g) is an exothermic compound.
a)
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
b)
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
c)
Statement I is correct but Statement II is incorrect
d)
Statement II is correct but Statement I is incorrect
|
Krishna Iyer answered |
Reaction number 1 is less stable because heat is not released thus the product side of reaction gains energy which get stored in the product and decreases its stability
While in reaction number 2, the reaction formed is more stable because energy is released and so the product is in a more stable state.
Now there is no such endothermic or exothermic compound , the reaction is exothermic or endothermic and not the product.
While in reaction number 2, the reaction formed is more stable because energy is released and so the product is in a more stable state.
Now there is no such endothermic or exothermic compound , the reaction is exothermic or endothermic and not the product.
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole -
[AIEEE-2004]
- a)110.5 kJ
- b) 676.5 kJ
- c)-676.5 kJ
- d)-110.5 kJ
Correct answer is option 'D'. Can you explain this answer?
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole -
[AIEEE-2004]
a)
110.5 kJ
b)
676.5 kJ
c)
-676.5 kJ
d)
-110.5 kJ
|
|
Riya Banerjee answered |
C + O2 → CO2
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
Consider the following properties.
State functions are : [IITJEE2009]
- a)I, II, III, IV, V
- b)I, II, III, VI
- c)I, II, IV, VI
- d)I, II, III, IV
Correct answer is option `A`. Can you explain this answer?
Consider the following properties.
State functions are : [IITJEE2009]
a)
I, II, III, IV, V
b)
I, II, III, VI
c)
I, II, IV, VI
d)
I, II, III, IV
|
Hansa Sharma answered |
First 4 are fundamental examples of state functions as they are path independent. Since reversible expansion depends on the path followed by process, so it is a path function. However, irreversible expansion work is independent of the path and so, it is a state function.
NH3(g) + 3Cl2(g) 
NCl3(g) + 3HCl(g) ; -ΔH1N2(g) + 3H2(g) 
2NH3(g) ; ΔH2H2(g) + Cl2(g) 
2HCl(g) ; ΔH3The heat of formation of NCl3 (g) in the terms of ΔH1, ΔH2 and ΔH3 is- a)
- b)
- c)
- d) None
Correct answer is option 'A'. Can you explain this answer?
NH3(g) + 3Cl2(g) NCl3(g) + 3HCl(g) ; -ΔH1
NCl3(g) + 3HCl(g) ; -ΔH1N2(g) + 3H2(g) 2NH3(g) ; ΔH2
2NH3(g) ; ΔH2H2(g) + Cl2(g) 2HCl(g) ; ΔH3
2HCl(g) ; ΔH3The heat of formation of NCl3 (g) in the terms of ΔH1, ΔH2 and ΔH3 is
a)
b)
c)
d)
None
|
Naina Sharma answered |
The formation of NCl3 be like
½ N2 + 3/2Cl2 ⇋ NCl3
We can see that for this setup, we need to have eqn (ii) divided by 2, reversing eqn (iii) and multiplying it by 3/2 and then adding all these to equation (i).
So option a is correct.
½ N2 + 3/2Cl2 ⇋ NCl3
We can see that for this setup, we need to have eqn (ii) divided by 2, reversing eqn (iii) and multiplying it by 3/2 and then adding all these to equation (i).
So option a is correct.
Can you explain the answer of this question below:8 g of O2 gas at STP is expanded so that volume is doubled. Thus, work done is
- A:
22.4 L atm
- B:
11.2 L atm
- C:
5.6 L atm
- D:
- 5.6 L atm
The answer is d.
8 g of O2 gas at STP is expanded so that volume is doubled. Thus, work done is
22.4 L atm
11.2 L atm
5.6 L atm
- 5.6 L atm
|
|
Suresh Iyer answered |
Let atomic weight of x = Mx
atomic weight of y = My
atomic weight of y = My
we know,
mole = weight /atomic weight
a/c to question,
mole = weight /atomic weight
a/c to question,
mole of xy2 = 0.1
so,
0.1 = 10g/( Mx +2My)
so,
0.1 = 10g/( Mx +2My)
Mx + 2My = 100g -------(1)
for x3y2 ; mole of x3y2 = 0.05
0.05 = 9/( 3Mx + 2My )
3Mx + 2My = 9/0.05 = 9 × 20 = 180 g ---(2)
solve eqns (1) and (2)
solve eqns (1) and (2)
2Mx = 80
Mx = 40g/mol
Mx = 40g/mol
and My = 30g/mole
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ΔfU0 of formation of CH4 (g) at certain temperature is –393 kJ mol−1. The value of ΔfH0is- a)equal to ΔfU0
- b)< ΔfU0
- c)zero
- d)> ΔfU0
Correct answer is option 'B'. Can you explain this answer?
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ΔfU0 of formation of CH4 (g) at certain temperature is –393 kJ mol−1. The value of ΔfH0is
a)
equal to ΔfU0
b)
< ΔfU0
c)
zero
d)
> ΔfU0
|
Ipsita Sen answered |
For an endothermic reaction when ΔH represents the enthalpy of the reaction in kJ mol-1, the minimum value for the energy of activation will be[IIT JEE 1992]- a)less than ΔH
- b)Zero
- c)More than ΔH
- d)Equal to ΔH
Correct answer is option 'C'. Can you explain this answer?
For an endothermic reaction when ΔH represents the enthalpy of the reaction in kJ mol-1, the minimum value for the energy of activation will be
[IIT JEE 1992]
a)
less than ΔH
b)
Zero
c)
More than ΔH
d)
Equal to ΔH
|
|
Hansa Sharma answered |
,More than ΔH
In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,
Where Ea = activation energy of forwarding reaction
Ea' = activation energy of backwards reaction
ΔH = enthalpy of the reaction
From the above diagram,
Ea = Ea' + ΔH
Thus, Ea > ΔH
The standard enthalpies of formation of CO2(g), H2O (l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is[JEE Advanced 2013]- a)2900 kJ
- b)-2900 kJ
- c)-16.11 kJ
- d)+16.11 kJ
Correct answer is option 'C'. Can you explain this answer?
The standard enthalpies of formation of CO2(g), H2O (l) and glucose (s) at 25°C are - 400 kJ mol-1, - 300 kJ mol-1 and - 1300 kJ mol-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is
[JEE Advanced 2013]
a)
2900 kJ
b)
-2900 kJ
c)
-16.11 kJ
d)
+16.11 kJ
|
|
Anjana Sharma answered |
Direction (Q. Nos. 1-15) This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.Q. Assign the sign of work done (based on SI convention) in the following chemical changes taking place against external atmospheric pressure :
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'A'. Can you explain this answer?
Direction (Q. Nos. 1-15) This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Assign the sign of work done (based on SI convention) in the following chemical changes taking place against external atmospheric pressure :
a)
a
b)
b
c)
c
d)
d
|
|
Lavanya Menon answered |
Work done = - (change in moles) RT For work done positive change in moles negative and vice versa We only cosider the moles that are in gass phase.
For the following process, H2 (g) → 2 H(g), it absorbs 436 kJ mol-1. Thus,
- a)internal energy of the system is 436 kJ mol-1
- b)change in internal energy is 436 kJ mol-1
- c)internal energy of the system is 218 kJ mol-1
- d)change in internal energy of the system is 218 kJ mol-1
Correct answer is option 'B'. Can you explain this answer?
For the following process, H2 (g) → 2 H(g), it absorbs 436 kJ mol-1. Thus,
a)
internal energy of the system is 436 kJ mol-1
b)
change in internal energy is 436 kJ mol-1
c)
internal energy of the system is 218 kJ mol-1
d)
change in internal energy of the system is 218 kJ mol-1
|
Geetika Shah answered |
In this process, one molecule of hydrogen gas dissociated into two gaseous atoms i.e. the number of gaseous particles increase leading to a more disordered state.
.Breaking of bond requires energy
.Breaking of bond requires energy
It is not mentioned that the process is isobaric. If we assume that process is isobaric because volume can not remain constant as number of gaseous particles are increased, then change in internal energy = 436 kJ mol-1. The data of absorption gives us the value of q.
8 g of O2 gas at STP is expanded so that volume is doubled. Thus, work done is- a)22.4 L atm
- b)11.2 L atm
- c)5.6 L atm
- d)- 5.6 L atm
Correct answer is 'D'. Can you explain this answer?
8 g of O2 gas at STP is expanded so that volume is doubled. Thus, work done is
a)
22.4 L atm
b)
11.2 L atm
c)
5.6 L atm
d)
- 5.6 L atm
|
Kamna Science Academy answered |
Let atomic weight of x = Mx
atomic weight of y = My
atomic weight of y = My
we know,
mole = weight /atomic weight
a/c to question,
mole = weight /atomic weight
a/c to question,
mole of xy2 = 0.1
so,
0.1 = 10g/( Mx +2My)
so,
0.1 = 10g/( Mx +2My)
Mx + 2My = 100g -------(1)
for x3y2 ; mole of x3y2 = 0.05
0.05 = 9/( 3Mx + 2My )
3Mx + 2My = 9/0.05 = 9 × 20 = 180 g ---(2)
solve eqns (1) and (2)
solve eqns (1) and (2)
2Mx = 80
Mx = 40g/mol
Mx = 40g/mol
and My = 30g/mole
Consider the following figure representing the increase in entropy of a substance from absolute zero to its gaseous state at some temperature
Q. ΔS° (fusion) and ΔS° (vaporisation) are respectively indicated by - a)AB, BC
- b)BC,CD
- c)BC, DE
- d)CD, DE
Correct answer is option 'C'. Can you explain this answer?
Consider the following figure representing the increase in entropy of a substance from absolute zero to its gaseous state at some temperature
Q. ΔS° (fusion) and ΔS° (vaporisation) are respectively indicated by
a)
AB, BC
b)
BC,CD
c)
BC, DE
d)
CD, DE
|
|
Neha Joshi answered |
You can co-relate both graph and the result will be option c.
The reaction CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) has ΔH = -25 kCal.
From the given data, what is the bond energy of Cl - Cl bond- a)70 kCal
- b)80 kCal
- c) 67.75 kCal
- d) 57.75 kCal
Correct answer is option 'D'. Can you explain this answer?
The reaction CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) has ΔH = -25 kCal.
From the given data, what is the bond energy of Cl - Cl bond
a)
70 kCal
b)
80 kCal
c)
67.75 kCal
d)
57.75 kCal
|
|
Kamna Science Academy answered |
During bond breakage energy is absorbed and during bond formation it is released. From the reaction we can say that 1 C-H bond is broken 1 Cl-Cl bond is broken 1 c-cl bond is formed and 1 h-cl bond is formed. so using the sign conventions the equation becomes
x+y-84-103= -25 (∆H = -25)
5x=9y ..putting x=9/5y we get y = 57.75 kCal
x+y-84-103= -25 (∆H = -25)
5x=9y ..putting x=9/5y we get y = 57.75 kCal
The ΔrH° for CO2(g), CO(g) , and H2O(g) are - 393.5°, - 110.5 and - 241.8 kJ mol-1, respectively. Thus, ΔrH° for the reaction (in kJ)
[IITJEE 2000]- a)+524.1
- b)+41.2
- c)-262.5
- d)-41.2
Correct answer is option 'B'. Can you explain this answer?
The ΔrH° for CO2(g), CO(g) , and H2O(g) are - 393.5°, - 110.5 and - 241.8 kJ mol-1, respectively. Thus, ΔrH° for the reaction (in kJ)
[IITJEE 2000]
a)
+524.1
b)
+41.2
c)
-262.5
d)
-41.2
|
|
Krishna Iyer answered |
∆H= ∆HCO+∆HH2O-∆HCO2 (as ΔH for H2 is 0)
= -110.5-241.8+393.5 = +41.2 kJ mol-1
= -110.5-241.8+393.5 = +41.2 kJ mol-1
Direction (Q. Nos. 19-20) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).The following diagram represents the (p-V) changes of a gas.
Select isochoric changes- a)when pressure changes from p1 to p2
- b)when pressure changes from p2 to p3
- c)when pressure changes from p1 to p3
- d)when pressure changes from p3 to p1
Correct answer is option 'A'. Can you explain this answer?
Direction (Q. Nos. 19-20) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).
The following diagram represents the (p-V) changes of a gas.
Select isochoric changes
a)
when pressure changes from p1 to p2
b)
when pressure changes from p2 to p3
c)
when pressure changes from p1 to p3
d)
when pressure changes from p3 to p1
|
|
Geetika Shah answered |
Isochoric change is one in which volume doesn't change(remains constant) From P1 to P2 , volume is constant(V1) (also work done = 0)
One mole of an ideal gas is put through a series of changes as shown in the figure in which 1,2,3 mark the three stages of the system. Pressure at the stages 1, 2, and 3 respectively will be (in bar)
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'D'. Can you explain this answer?
One mole of an ideal gas is put through a series of changes as shown in the figure in which 1,2,3 mark the three stages of the system. Pressure at the stages 1, 2, and 3 respectively will be (in bar)
a)
a
b)
b
c)
c
d)
d
|
|
Krishna Iyer answered |
Applying ideal gas eqn. at 1
p×22.4 = 1×0.0821×298 (since vol. Is in L. R should haave constaant with lit in it.)
p = 1.09 atm. To convert it into bar, we multiply it by 1.01 or we have 1.03 bar
Similarly at 2 and 3, we get value of pressure at 2 and 3.
The most closest answer is option d
p×22.4 = 1×0.0821×298 (since vol. Is in L. R should haave constaant with lit in it.)
p = 1.09 atm. To convert it into bar, we multiply it by 1.01 or we have 1.03 bar
Similarly at 2 and 3, we get value of pressure at 2 and 3.
The most closest answer is option d
The dissociation energy of CH4 and C2H6 to convert them into gaseous atoms are 360 and 620 kcal mol-1 respectively. Thus, bond energy of (C—C) bond is- a)260 kcal mol-1
- b)180 kcal mol-1
- c)130 kcal mol-1
- d)80 kcal mol-1
Correct answer is option 'D'. Can you explain this answer?
The dissociation energy of CH4 and C2H6 to convert them into gaseous atoms are 360 and 620 kcal mol-1 respectively. Thus, bond energy of (C—C) bond is
a)
260 kcal mol-1
b)
180 kcal mol-1
c)
130 kcal mol-1
d)
80 kcal mol-1
|
|
Krishna Iyer answered |
In CH4 there are 4-C-H bonds.
Let C-H bond energy be X. So 4X = 360. or X = 90.
Now, in C2H6 there are 6 C-H bonds and 1C-C bond.
Let C-C B. E be y.
So 6X + Y = 620. or Y = 620-6×90 = 80.
Let C-H bond energy be X. So 4X = 360. or X = 90.
Now, in C2H6 there are 6 C-H bonds and 1C-C bond.
Let C-C B. E be y.
So 6X + Y = 620. or Y = 620-6×90 = 80.
The heat capacity ratio, r was determined for cyanogen as 1.177. Thus, Cp for this gas is- a)55.3 JK-1 mol-1
- b)33.26 JK-1 mol-1
- c)8.314 JK-1 mol-1
- d)24.94 JK-1 mol-1
Correct answer is option 'A'. Can you explain this answer?
The heat capacity ratio, r was determined for cyanogen as 1.177. Thus, Cp for this gas is
a)
55.3 JK-1 mol-1
b)
33.26 JK-1 mol-1
c)
8.314 JK-1 mol-1
d)
24.94 JK-1 mol-1
|
|
Geetika Shah answered |
γ = 1.177
CV = R/γ-1
= 8.314/0.177 = 46.97
Also γ = CP/CV
1.177 = CP/46.97
Or CP = 55.28
CV = R/γ-1
= 8.314/0.177 = 46.97
Also γ = CP/CV
1.177 = CP/46.97
Or CP = 55.28
The standard enthalpies of formation of SF6 (g), S (g)and F (g) are -1100, + 275 and + 80 kJ mol-1. Thus, average bond energy of (S—F) in SF6 is- a)309.16 kJ mol-1
- b)1855 kJ mol-1
- c)11.130 x 103 kJ mol-1
- d)-309.16 kJ mol-1
Correct answer is option 'A'. Can you explain this answer?
The standard enthalpies of formation of SF6 (g), S (g)and F (g) are -1100, + 275 and + 80 kJ mol-1. Thus, average bond energy of (S—F) in SF6 is
a)
309.16 kJ mol-1
b)
1855 kJ mol-1
c)
11.130 x 103 kJ mol-1
d)
-309.16 kJ mol-1
|
|
Suresh Reddy answered |
S(g) + 6F(g) → SF6(g)
∆rH = ∑∆Hreactant - ∑∆Hproduct
= (275+6 80) - (-1100)
= 1855
So, bond energy of (S—F) = 1855/6 = 309.16 kJ mol-1
∆rH = ∑∆Hreactant - ∑∆Hproduct
= (275+6 80) - (-1100)
= 1855
So, bond energy of (S—F) = 1855/6 = 309.16 kJ mol-1
It has been observed that the ratio of the molar heat of vaporisation of liquid and its normal boiling point is approximately same for most of the liquids (called Triton's rule) Which of the following does not follow this rule- a)C61-16
- b)C6H14
- c)CCI4
- d)NH3
Correct answer is option 'D'. Can you explain this answer?
It has been observed that the ratio of the molar heat of vaporisation of liquid and its normal boiling point is approximately same for most of the liquids (called Triton's rule) Which of the following does not follow this rule
a)
C61-16
b)
C6H14
c)
CCI4
d)
NH3
|
|
Hansa Sharma answered |
The correct answer is Option D.
Due to hydrogen bonding in the liquid state there is less entropy than expected and it shows some negative deviation in the Trouton's rule. In vapor state there is more entropy and it shows positive deviation in the rule. Thus for ammonia the Trouton's rule is not applicable.
= 0.0030 atm. If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of
1 : 1 : 0.5 and ΔfH for the formation of XY is -200 kJ mol-1. The bond dissociation energy of X2 will be[AIEEE-2005]- a)200 kJ mol-1
- b)100 kJ mol-1
- c) 800 kJ mol-1
- d)300 kJ mol-1
Correct answer is option 'C'. Can you explain this answer?
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of
1 : 1 : 0.5 and ΔfH for the formation of XY is -200 kJ mol-1. The bond dissociation energy of X2 will be
1 : 1 : 0.5 and ΔfH for the formation of XY is -200 kJ mol-1. The bond dissociation energy of X2 will be
[AIEEE-2005]
a)
200 kJ mol-1
b)
100 kJ mol-1
c)
800 kJ mol-1
d)
300 kJ mol-1
|
|
Pooja Shah answered |
Sulphur (2.56 g) is burned in a constant volume calorimeter with excess O2(g). The temperature increases from 21.25°C to 26.72°C . The bomb has a heat capacity of 923 JK-1. Calorimeter contains 815 g of water. Thus, change in internal energy per mole of SO2 formed for the reaction is
(specific heat of water is 4.184 JK-1g-1.)- a)- 296.27 kJ
- b)+ 296.27 kJ
- c)- 2370.13 kJ
- d)+ 2370.13 kJ
Correct answer is option 'A'. Can you explain this answer?
Sulphur (2.56 g) is burned in a constant volume calorimeter with excess O2(g). The temperature increases from 21.25°C to 26.72°C . The bomb has a heat capacity of 923 JK-1. Calorimeter contains 815 g of water. Thus, change in internal energy per mole of SO2 formed for the reaction is
(specific heat of water is 4.184 JK-1g-1.)
a)
- 296.27 kJ
b)
+ 296.27 kJ
c)
- 2370.13 kJ
d)
+ 2370.13 kJ
|
Dhanush Kumar answered |
Therila
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