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Thermodynamics Mcq Neet Chemistry
Quiz 6
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Q.1
The enthalpy change at 298K in successive breaking of O - H bond of HOH areH2O(g) → H(g) + OH(g) ; ΔH=498 kJ/molOH(g) → H(g) + O(g) ; °H=428 kJ/molThe bond enthalpy of O-H bond is ... [ CBSE PMT 1994]
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a) 498 kJ/mol
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b) 463 kJ/mol
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c) 428 kJ/mol
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d) 70 kJ/mol
Explanation
Take average of both energy Answer: (b)
Q.2
Mark the correct statement ...[ MP PET 1998]
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a) For a chemical reaction to be feasible, ΔG should be 0
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b) Entropy is a measure of order in a system
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c) For a chemical reaction to be feasible, ΔG should be positive
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d) The total energy of an isolated system is constant
Explanation
Answer: (d)
Q.3
The enthalpy of neutralization of acetic acid and sodium hydroxide is -55.4 kJ. What is the enthalpy of ionization of acetic acid ... [ KCET 1997]
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a) - 1.9 kJ
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b) + 1.9 kJ
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c) + 5.54 kJ
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d) -5.54 kJ
Explanation
Enthalpy of neutralization of strong acid base=-57.3 kJ ∴ enthalpy of ionization of acetic acid=57.3 - 55.4=1.9 kJthis much heat is absorbed for ionization of acetic acidAnswer: (b)
Q.4
The enthalpy of formation of N2O and NO are 82 and 90kJ/mol respectively. The enthalpy of reaction2N2O(g) + O2(g) → 4NO(g) is equal to ... [ CBSE PMT 1991]
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a) 8 kJ
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b) 88 kJ
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c) - 16 kJ
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d) 196 kJ
Explanation
ΔH=∑( ΔH ( product)) - ∑(ΔH (reactant)) ΔH=4×90 - 2×82=196 kJAnswer: (d)
Q.5
The enthalpy change of a reaction does not depend on ... [ AIIMS 1997]
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a) the state of reactants and products
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b) nature of reactants and product
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c) different intermediate reaction
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d) initial and final enthalpy change of reaction
Explanation
By Hess's law Answer: (c)
Q.6
Heat capacity is ... [ MP PMT 1996]
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a) dQ/dT
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b) dQ×dT
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c) ∑Q/dT
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d) none of these
Explanation
Heat capacity is amount of heat required to increase the temperature by one unit Answer: (a)
Q.7
For reaction to occurred spontaneously .. [ CBSE PMT 1995]
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a) ΔS must be negative
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b) (ΔH - TΔS) must be negative
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c) (ΔH + TΔS) must be positive
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d) ΔH must be negative
Explanation
Answer: (b)
Q.8
The calorific value fat is ... [ BHU 1995]
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a) less than that of carbohydrates and protein
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b) less than that of protein but more than carbohydrate
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c) less than carbohydrate but more than that of protein
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d) more than carbohydrate and protein
Explanation
Answer: (d)
Q.9
The heat of formation of the compound in the following reaction is ... [ MP PMT 1995]H2(g) + Cl2(g) → 2HCl(g) + 44 kcal
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a) - 44 kcal/mol
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b) -22 kcal/mol
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c) -11 kcal/mol
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d) -88 kcal/mol
Explanation
ΔH=-22 kcalAnswer: (b)
Q.10
Given the bond energies of N ΞN , H - H and N-H bonds are 945, 436, 391 kJ/mol respectively, the enthalpy of the reaction N2(g) + 3H2(g) → 2NH3(g) is ... [ EAMCET 1992]
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a) -93 kJ
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b) 102 kJ
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c) 90 kJ
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d) 105 kJ
Explanation
Net energy released=∑ (bond energy of reactant) - ∑ ( bond energy of products)ΔH=(945 + 3×436) - 2(3×391)=-93 kJ Answer: (a)
Q.11
The densities of graphite and diamond at 298K are 2.25 and 3.31 g /cc respectively. If the standard free energy difference (ΔG°) is equal to 1895 J/mol, the pressure at which graphite will be transformed into diamond at 298K is ..
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a) ;108Pa
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b) 109Pa
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c) 1010Pa
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d) 107Pa
Explanation
ΔG=P Δ V mass=volume × densityThus Vg=12/2.25=5.3333 cc=5.3333 ×10-6m3 Vg=12/3.31=3.6923 cc=3.6923 ×10-6m3 ΔV=(5.3333 - 3.6923) ×10-6 ΔV=1.641 ×10-6 P=ΔG/ΔV P=1.154 ×109 PaAnswer: (b)
Q.12
If 900 j/g of heat is exchanged at boiling point of water, then what is increase in entropy ... [ BHU 1998]
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a) 43.4 J/mole
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b) 87.2 J/mole
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c) 900 J/mole
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d) Zero
Explanation
Heat exchanged per mole=900×18 ΔS=(900×18)/373=43.4 JK-1mol-1Answer: (a)
Q.13
Under which of the following condition, work is a state function?
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a) Adiabatic process
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b) Isothermal process
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c) Constant pressure
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d) None of these
Explanation
In adiabatic process q=0. Hence ΔE=q+w gives ΔE=w. As ΔE is state function, w become state functionAnswer: (a)
Q.14
If the enthalpy of vaporization of water is 186.5 J/mol, the entropy of its vaporization will be .... [ CPMT 1988]
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a) 0.5 JK-1mol-1
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b) 1.0 JK-1mol-1
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c) 1.5 JK-1mol-1
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d) 2.0 JK-1mol-1
Explanation
ΔSvap=ΔH /T ΔSvap=185.5 /273=0.5 JK-1mol-1 Answer: (c)
Q.15
For precipitation reaction of Ag+ ions with NaCl which of the following statements is correct?
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a) ΔH for the reaction is zero
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b) ΔG for reaction is zero
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c) ΔG for the reaction is negative
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d) ΔG=ΔH
Explanation
Answer: (c)
Q.16
The average molar heat capacity of ice and water are 37.6 and 75.2 J mol-1K-1 respectively and its enthalpy of fusion is 6.02kJ/mol. The amount of heat required to raise the temperature of 10g of water from -10°C to +10°C is equal to...
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a) 2376 J
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b) 4752 J
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c) 1128 J
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d) 3970 J
Explanation
Convert given energy for per gram heat capacity of ice per gram=37.6/18=2.08 heat capacity of water per gram=75.2/18=4.17 enthalpy of fusion per gram=6020/18=334.44 Calculation of energy for -10°C to 0°CE1=mCΔTE1=10×2.08×10=208Calculate enthalpy of fusion for 10g iceE2=334.44×10=3344.4Calculation of energy for 0°C to +10°CE3=mCΔTE3=10×4.17×10=417Total energy=208+3344.4+417=3969.4 JAnswer: (d)
Q.17
The enthalpy of combustion at 25°C of H2, Cyclohexene (C6H10) and cyclohexane (C6H12) are -241, -3800 and -3920 kJ/mol respectively. The heat off hydrogenation of cyclohexane is .. [ IIT 1989]
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a) -121 kJ/mol
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b) +121 kJ/mol
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c) -242 kJ/mol
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d) +242 kJ/mol
Explanation
Given (i) H2 + ½O2 → H2O; ΔH=-2471 kJ(ii) C6H10 + 8.5O2 → 6CO2 + 5H2O ; ΔH=-3800 kJ(iii) C6H12 + 9O2→ 6CO2 +6H2O; ΔH=-3920 kJAim : C6H10 + H2 → C6H12Eqn(i) + Eqn(ii) - Eqn(iii) givesΔH=-241-3800-(-3900)=-121 kJAnswer: (a)
Q.18
Calculate the temperature at which ΔG=-5.2 kJ/mol, ΔH=145.6 kJ/mol and ΔS=216 JK-1mol-1 for a chemical reaction
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a) 698°C
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b) 425°C
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c) 650K
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d) 650°C
Explanation
Use formula ΔG=ΔH - TΔSAnswer: (b)
Q.19
1 mole of ice is converted to liquid at 273K; H2O(s) and H2O(l) have entropies 38.20 and 60.03 J mole-1 deg-Enthalpy change in the conversion will be... [ DPMT 1987]
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a) 59.59 J/mol
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b) 595.95 J/mol
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c) 5959.5 J/mol
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d) 5959.5 J/mol
Explanation
ΔS=Sl - ΔSs ΔS=60.03 - 38.20=21.83 JK-1mol-1ΔH=TΔS=273 ×21.83=5959.5 J mole-1Answer: (c)
Q.20
For which one of the following equations is ΔHreaction equal to ΔH of product? [ CBSE med. 2003]
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a) N2(g) + O3(g) → N2O3(g)
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b) CH4(g) + 2Cl2(g) → CH2Cl2(l) + 2HCl(g)
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c) Xe(g) + 2F2(g) → XeF4(g)
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d) 2CO(g) + O2(g) → 2CO2(g)
Explanation
In option "c" reactants are elementary substances thus enthalpy of reaction=enthalpy of formation Answer: (c)
Q.21
2 mole of an ideal gas at 27°C temperature is expanded reversibly from 2 lit to 20 lit. Find entropy change (R=2cal/mol K) .. [ CBSE PMT 2002]
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a) 92.1
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b) 0
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c) 4
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d) 9.2
Explanation
At constant temperature ΔS=Work Answer: (d)
Q.22
The difference between ΔH and ΔE for the combustion of methane at 27°C will be ( in joule/mol) [ kerala MEE 2001]
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a) 8.314 × 27 × (-3)
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b) 8.314 × 300 × (-3)
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c) 8.314 × 300 × (-2)
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d) 8.314 × 300 × (1)
Explanation
CH4(g) + 2O2 → CO2(g) + 2H2O Δng=1 - 3=0qp - qv=ΔngRT=8.314 ×300 ×(-2)Answer: (c)
Q.23
One mole of an anhydrous salt AB dissolves in water with the evolution of 21.0 J/mol of heat. If the heat of hydration of AB is -29.4 J/mol then the heat of dissociation of the hydrated salt AB is
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a) 50.4 J/mol
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b) 8.4 J/mol
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c) -54.4 J/mol
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d) -8.4 J/mol
Explanation
(i) AB(s) + aq → AB(aq); ΔH=-21 J(ii) AB(s) + xH2O → AB. xH2O; ΔH=-29.4 J Aim: AB x H2O(s) + aq → AB(aq) ; ΔH=?Eqn(i) is equivalent to AB(s) + xH2O → AB.xH2O(s); ΔH=ΔH1AB.xH2O(s) + aq → AB(aq), ΔH=ΔH2ΔH1 + ΔH2=-21-294.4 + ΔH2=-21ΔH2=8.4 J/molAnswer: (b)
Q.24
When Zn dust is added to sufficiently large volume of aqueous solution of copper sulphate 3.175g of copper metal and 20J of heat is evolved. The ΔH for the reactionZn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)(atomic weight of Zn=63.5, Cu=63.5 )
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a) 20 J
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b) 200 J
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c) 400 J
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d) 65.3 J
Explanation
3.175g gives 20 J calculate heat for one mole ( for 63.5 g) Answer: (c)
Q.25
The enthalpy of combusion of cyclohexane, cyclohexene and H2 are respectively -3920, 3800 and -241 kL mol-The heat of hydrogenation of cyclohexene is ...
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a) -121.0 kJ/mol
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b) 121.0 kJ/mol
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c)-242.0 kJ/mol
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d) 242.0 kJ/mol
Explanation
Question is based on Hess's law of heat summation 2C6H12 + 182 → 12CO2 + 12H2O; Δ H = -3920 --- eq(1) 2C6H10 + 172 → 12CO2 + 10H2O; Δ H = -3800 --- eq(2) 2H22 → 2H2O; Δ H = -241 ---eq(3) Now Subtract eq(1) from the addition of eq(2) and eq(3) we get C6H10 + H2 → C6H12 ; Δ -121 kJ/mol Answer: (a)
Q.26
Which one of the following has ΔS greater than zero? [ AIIMS 2003]
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a) CaO(s) + CO2 ⇌ CaCO3
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b) NaCl(aq) ⇌ NaCl(s)
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c) NaNO3(s) ⇌ Na+(aq) + Cl-(aq)
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d) N2(g) + 3H2 ⇌ 2NH3(g)
Explanation
For option (c) randomness of ions in aqueous solution is more thus ΔS > 0 Answer: (c)
Q.27
In which one of the following sets, all the properties belong to same category (all extensive or all intensive)?
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a) mass, volume, pressure
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b) temperature, pressure, volume
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c) heat capacity, density, entropy
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d) enthalpy, internal energy, volume
Explanation
Answer: (d)
Q.28
During isothermal expansion of an ideal gas, its internal energy
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a) decreases
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b) increases
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c) may increases or decrease
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d) remains unchanged
Explanation
Answer: (d)
Q.29
In a change from state A to state B
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a) q depends only on the initial and final state
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b) w depends only on the initial and final state
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c) ΔE depends only on the initial and final state
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d) ΔE depends upon the path adopted by A and change into B
Explanation
Answer: (c)
Q.30
Which of the following statement is not correct
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a) First law of thermodynamics is also called law of thermal equilibrium
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b) Heat of combustion cannot be found at room temperature
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c) On the thermodynamic scale of temperature, one of the reference points is the triple point of water
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d) One degree kelvin is equal to 1/273.16 of the triple point of water
Explanation
Heat of combustion is usually found at room temperature Answer: (b)
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