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Quiz 2
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Q.1
One gram atom of graphite and one gram atom of diamond were separately burnt to carbon dioxide. The amount of heat liberated were 393.5kJ and 39.4 kJ respectively. It follows that
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a) graphite has greater affinity for oxygen
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b) diamond has greater affinity for oxygen
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c) graphite is stable than diamond
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d) diamond is stable than graphite
Explanation
Heat of combustion or heat liberated on combustion calculated as the difference between the heat of formation of the products and reactants. In chemical reactions energy is liberated if product is more stability than reactant. Now graphite liberate more energy to produce Carbon dioxide it means graphite is less stable than CO2 and diamond liberate less energy to produce CO2, it means diamond is more stable than graphite. Lesser the heat of combustion more the stability Answer: (d)
Q.2
If S + O2 → SO2; ΔH=-298.2 kJSO3 +H2O → H2SO4; ΔH=-130.2 kJH2 + ½O2 → H2O ; ΔH=-287.3 kJ SO2 + ½O2 → SO3 ; ΔH= -98.2 kJ Then enthalpy of formation of H2SO4 at 298K will be ... [ DPMT 1994]
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a) -814.4 kJ
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b) +320.5 kJ
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c) -650.3 kJ
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d) -933.7 kJ
Explanation
Target equation H2 + S + 2O2 → H2SO4add all given equations we getΔH=-287.3 +(-298.2) + (-98.7) + (-130.2)ΔH=-814.4 kJAnswer: (a)
Q.3
Conversion of oxygen to ozone represented by the equation 3O2 → 2O3 is an endothermic reaction. Enthalpy change , ΔH accompanying the reaction.. [ KCET 1988]
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a) is negative
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b) is positive
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c) is zero
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d) depending on temperature
Explanation
Answer: (b)
Q.4
The compound with negative heat of formation is known as ..[ DPMT 1981]
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a) endothermic compound
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b) exothermic compound
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c) endoergonic compound
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d) none of the above
Explanation
Since ΔH is negative Answer: (b)
Q.5
Heat of neutralization of NH4OH and HCl is ... [ Pb. PMT 1985]
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a) 13.7 kcal/mole
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b) < 13.7 kcal/mole
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c) > 13.7 kcal/mole
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d) zero
Explanation
NH4OH is weak base Answer: (b)
Q.6
The bond dissociation energies of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl gas would be.. [ MP PET 1998]
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a) -44 kcal
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b) -88 kcal
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c) -22 kcal
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d) -11 kcal
Explanation
Aim ½H2 + ½C2 → HCLΔH=∑ B.E.(reactant) - ∑ B.E.(product) =B.E.(HCl) - [ ½ B.E.(H2) + ½B.E.(Cl2] =-103 - [ ½(-104) + ½(-58)]=- 22 kcalAnswer: (c)
Q.7
The work done in ergs for reversible expansion of one mole ideal gas from a volume of 1 litres to 20 litres at 25°C is ... [ CMC Vellore 1991]
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a) 2.303 ×298 ×0.082log2
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b) 298 ×107 ×8.31 ×2.303log2
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c) 2.303 ×298 × 0.082log0.5
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d) 2.303×298×2log2
Explanation
Use formula for work isothermal,reversible expansion Answer: (b)
Q.8
N2 + 2O2(g) → 2NO2 + X kJ2NO(g) + O2 (g) → 2NO2(g) + Y kJThe enthalpy of formation of NO is .. [ GUJ CET]
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a) ( 2X - 2Y)
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b) ( X - Y )
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c) ½ ( Y - X)
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d) ½ (X - Y)
Explanation
½ N2 + ½O2 → NO Subtract ½ eq(1) - ½eq(2) . calculation of ΔH=½(X-Y)Answer: (d)
Q.9
A spontaneous change is one in which the system suffers ... [ NCERT 1981]
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a) An increase in internal energy
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b) A lowering of entropy
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c) A lowering of free energy
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d) No energy change
Explanation
Answer: (c)
Q.10
One mole of an ideal gas at 300K is expanded isothermally from an initial volume of 1 litre to 10 lites. The ΔE for this process is ... [ CBSE PMT 1998]( R=2 cal mole-1 K-1)
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a) 163.7 cal
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b) zero
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c) 1381.1 cal
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d) 9 li atom
Explanation
For isothermal process internal energy ΔE=0 Answer: (b)
Q.11
Given that 2C (s) + 2O2(g) → 2CO2(g) ; ΔH=-787 kJH2(g) + ½O2(g) → H2O(l) ; ΔH=-286 kJC2H2(g) + 2.5 O2(g) → 2CO2(g) + H2O (l) ; ΔH=-1301kJHeat of formation of acetylene is ... [ KCET 1989]
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a) -1802 kJ
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b) + 1802 kJ
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c) -800 kJ
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d) +228 kJ
Explanation
Add equation 1 and 2 Then subtract it from equation 3 Answer: (d)
Q.12
All the naturally occurring processes proceed spontaneously in a direction which leads to [ Pb. PMT 1985]
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a) decrease of entropy
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b) increase of enthalpy
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c) increase of free energy
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d) decrease of free energy
Explanation
Answer: (d)
Q.13
A solution of 500 ml of 0.2M KOH and 500 ml of 0.2M HCl is mixed and stirred; the rise in temperature is TThe experiment is repeated using 250ml each solution, the temperature rise is TWhich of the following is true? [ EAMCET 1987]
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a) T1=T2
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b) T1=2T2
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c) T1=4T2
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d) T2=9T1
Explanation
Answer: (a)
Q.14
Which of the following units represents largest amount of energy? [ CPMT 1980]
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a) Calorie
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b) Joule
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c) Erg
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d) Electron volt
Explanation
1eV=1.6 ×10-19J 1 erg=10-7 J1 calorie=4.184 JAnswer: (a)
Q.15
A gas absorbs 250J of heat and expands from 1 litre to 10 litre at constant temperature aganst external pressure of 0.5 atm. The value of q , w and ΔE will be respectively
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a) 250 J, 455 J, 710 J
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b) 250J, -455J, -205 J
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c) -250 J, -455 J, - 205 J
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d) -250 J, 455 J, 205 J
Explanation
q=250J, w=- P(ΔV)=-0.5 × 9=4.5 L atm 4.5 L atm=4.5 ×101J=455 J ΔE=q + w=250 - 455=-205JAnswer: (b)
Q.16
The heat of combustion of CH4(g), C(graphite)., H2(g) are -20kcal, -40 kcal, -10kcal respectively. The heat of ofrmation of methane is... [ EAMCET 1998]
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a) -40 kcal
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b) +40 kcal
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c) -80 kcal
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d) +80 kcal
Explanation
Given CH4 + 2O2 → CO2 + H2O ; ΔH=-20 kcalC(graphite) → CO2 ; Δ=-40kcalH2 + ½ O2 → H2O; ΔH=-10kcalAim: C + 2H2 eq(2) + 2 × eq(3) - eq(1) gives ΔH=-40 + 2 (-10) - (-20)=-40 kcal Answer: (a)
Q.17
Enthalpy of a reaction at 27°C is 15kJ mol-The reaction will be feasible if entropy is ... [ Pb. CET 1988]
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a) 15 J mol-1K-1
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b) -50 J mol-1K-1
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c) greater than 50 J mol-1K-1
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d) less than 50 J mol-1K-1
Explanation
If ΔG < 0, reaction is feasible Thus TΔS > Δ H300 ΔS > 15000 J or ΔS > 50 JAnswer: (c)
Q.18
Which of the following values of heat of formation indicates that the product is least stable? [ MP PMT 1991]
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a) - 94 kcal
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b) -231.6 kcal
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c) +21.4 kcal
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d) +64.8 kcal
Explanation
In formation of compound more the heat absorbed, less stable is the compound Answer: (d)
Q.19
The heat of combustion of yellow P and red P are -9.91 kJ/mol and -8.78 kJ/mol respectively. The heat of transiton of yellow P to red P is .. [ Pb. CET 1992]
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a) - 18.69 kJ
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b) +1.13 kJ
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c) +18.69 kJ
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d) - 1.13 kJ
Explanation
Given 4P(yellow) + 5 O2 → 2P2O5 ΔH=4 × (-9.91)=-39.64 kJ 4P(red) + 5 O2 → 2P2O5 ΔH=4 × (-8.78)=-35.12eq(1) - eq(2) &deltaH=4.52 kJyellow P → red P ΔH -1.13 kJ/molAnswer: (d)
Q.20
Equal volumes of molar hydrocholic acid and sulphuric acid are neutralized by dil.NaOH solution and x kcal and y kcal of heat are liberated respectively. Which of the following is true? [ CBSE PMT 1994]
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a) x=y
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b) x=½ y
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c) x=2y
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d) none of these
Explanation
1 M H2SO4=2g equ. of H2SO4 Hence y=2x or x=½ yAnswer: (b)
Q.21
The occurrence of reaction is impossible if ... [ AIIMS 1994] ΔG = ΔH - TΔS Answer: (c)
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a) Both ΔH and ΔS are positive
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b) ΔH is negative and ΔS is positive
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c) ΔH is positive and ΔS is negative
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d) Both ΔH and ΔS are negative
Explanation
ΔG = ΔH - TΔS If ΔH is positive and ΔS is negative then, ΔG is positive and can not be negative for any value of ΔH and ΔS If Δ G is positive then reaction can not occur on its own Answer: (c)
Q.22
The mutual heat of neutralisation of 40g NaOH and 60g CH3COOH will be.. [ MP PMT 1988]
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a)57.1 kJ
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b) less than 57.1 kJ
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c) more than 57.1 kJ
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d) 13.7 kJ
Explanation
CH3COOH is weak acid Answer: (b)
Q.23
For which one of the following reacts will there be a +ve ΔS? ..[ Hariyana CEET 1991]
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a) H2O(g) → HO(l)
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b) H2 + I2 → 2HI
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c) CaCO3(s) → CaO(s) + CO2(g)
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d) N2(g) + 3H2 → 2NH3(g)
Explanation
In option "c" solid is converted to gas Answer: (c)
Q.24
H2(g) + I2(g) → 2HI(g) ; ΔH=12.4 kcal. According to this reaction heat of formation of HI will be ..[ MP PMT 1990]
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a) 12.4 kcal
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b) -12.4 kcal
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c) -6.20 kcal
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d) + 6.20 kcal
Explanation
two moles of HI formed and 12.4 kcal heat is liberated , thus 6.20 kcal/mol is heat of formation Answer: (d)
Q.25
For a reaction2NH3(g) → N2(g) + 3H2(g)Which of the following statement is correct. .. [ PB.CET1990]
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a) ΔH=ΔU
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b) ΔH < ΔU
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c) ΔH > ΔU
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d) ΔH=0
Explanation
Δng=4-2=2 HEnce ΔH=ΔU + 2RT, so potion "c" is correctAnswer: (c)
Q.26
The amount of energy released when 20 ml of 0.5 M NaOH are mixed with 100ml of 0.1M HCl is x kJ. The heat of neutralisation ( in kJ/mol) is [ manipal PMT 2002]
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a) -100x
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b) -50x
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c) + 100x
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d) + 50x
Explanation
20ml of 0.5M NaOH=20×0.5=10 millimoles 100ml of 0.1M HCl=100 ×0.1=10 millimoles 10 millimoles gives x kJ of heat Thus 1g equivalent=1000 milimole gives heat of 100x kJAnswer: (a)
Q.27
For a reaction to be spontaneous at all temperatures .. [ KCET 1990]
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a) ΔG and ΔH should be negative
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b) ΔH=ΔG=0
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c) ΔG and ΔH shlould be positive
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d) ΔH < ΔG
Explanation
Answer: (a)
Q.28
According to Kirchoff's equation which factor affects the heat of reaction? [ MP PMT 1990]
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a) Pressure
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b) Temperature
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c) Volume
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d) Molecularity
Explanation
Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. In general, enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different. ΔHrf=ΔHri + ΔCp(Tf - Ti)CP is the difference in heat capacity, Heat capaicty final CPfand Heat capaicty inital CPiAnswer: (b)
Q.29
Which of the following reactions is endothermic? [ AFMC 1988]
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a) CaCO3 → CaO + CO2
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b) Fe + S → FeS
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c) NaOH + HCl → NaCl + H2O
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d) CH4 + 2O2 → CO2 + 2H2O
Explanation
CaCO3 absorbs heat for decomposition Answer: (a)
Q.30
For an adiabatic process, which of the following is correct? [ CPMT 1990]
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a) PΔV=0
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b) q=+W
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c) ΔE=q
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d) q=0
Explanation
Answer: (d)
0 h : 0 m : 1 s
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