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Quiz 3
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Q.1
Which fuel has the maximum calorific value? [ MP PET 1990]
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a) Charcoal
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b) Kerosene
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c) Wood
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d) Cow-dung
Explanation
Answer: (b)
Q.2
For the processCO2(s) → CO2(g) .. [ Pb.CET 1988]
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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
Answer: (a)
Q.3
Heat of neutralisation is least when... [ KCET 1988]
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a) NaOH is neutralised by CH3COOH
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b) NaOH is neutralised by HCl
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c) NH4OH is neutralised by CH3COOH
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d) NH4OH is neutralised by HNO3
Explanation
option "c" both acid and base is weak Answer: (c)
Q.4
In which of the following reactions does the heat change represent the heat of formation of water? [ EAMCET 1991]
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a) 2H2(g) + O2(g) → 2H2O(l); ΔH=-116 kcal
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b) H2(g) + ½ O2(g) → H2O(l)
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c) H+(aq) + OH-(aq) → H2O(l) ; ΔH=-13.7 kcal
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d) C2H2(g) + 2.5 O2 → 2CO2(g) + H2O(l); ΔH=-310 kcal
Explanation
Answer: (b)
Q.5
If for H2(g) + ½ O2(g) → H2O(g)ΔH1 is the enthalpy of reaction and for H2(g) + ½ O2(g) → H2O (l)ΔH2 is enthalpy of reaction, then.. [ Pb. CET 1990]
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a) ΔH1 > ΔH2
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b) ΔH1=ΔH2
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c) ΔH1 < ΔH2
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d) ΔH1 + ΔH2=0
Explanation
When H2O (g) condenses to form H2O(l), heat is evolved. Hence ΔH2 > ΔH1 Answer: (c)
Q.6
Heat of neutralisation of strong acid and strong base is nearly equal to ...[ AIIMS 1988]
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a) 10 kJ/mole
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b) 10 cal/mole
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c) -57 kJ/mole
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d) -57 Cal/mole
Explanation
Answer: (c)
Q.7
If ΔH is the change in enthalpy and ΔE, the change in internal energy accompanying a gaseous reaction, then ..[ CBSE PMT 1990]
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a) ΔH is always greater than ΔE
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b) ΔH < ΔE only if the number of moles of products is greater than the number of moles of the reactants
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c) ΔH is always less than ΔE
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d) ΔH < ΔE only if the number of moles of products is less than the number of moles of the reactants
Explanation
From formula ΔH=ΔE + ΔngRTif Δng < 0 then ΔH < ΔEAnswer: (d)
Q.8
In which of the following, the entropy decreases? [ CPMT 1988]
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a) Crystallisation of sucrose from solution
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b) Rusting of iron
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c) Melting of ice
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d) Vaporization of camphor
Explanation
Free sugar molecules of the solution crystallise out. Hence entropy decreases Answer: (a)
Q.9
If H+ + OH- → H2O + 13.7 kcalthen heat of complete neutralisation of one gram mole of H2SO4 with a base will be [ MP 1990]
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a) 13.7 kcal
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b) 27.4 kcal
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c) 6.85 kcal
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d) 3.425 kcal
Explanation
1 mole of H2SO4 give 2 mole of H+ ions Answer: (b)
Q.10
A mixture of 2moles of carbon monoxide and one mole oxygen in a closed vessel in ignited to get carbon dioxide. If ΔH is the enthalpy change and ΔE is the change in internal energy, then ..[ MP PMT 1998]
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a) ΔH > Δ E
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b) Δ H < Δ E
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c) Δ H=Δ H
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d) not definite
Explanation
2CO(g) + O2(g) → 2CO2(g) Δng=2 - 3=-1ΔH=ΔE + ΔngRTΔH=ΔE - RTHence ΔH < ΔEAnswer: (b)
Q.11
The molar neutralisation heat for KOH and HNO3 as compaired to molar neutralisation heat of NaOH and HCl is ..[ MP PMT 1989]
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a) Less
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b) more
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c) equal
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d) depends on pressure
Explanation
In both the cases, acid and base are strong Answer: (c)
Q.12
The enthalpy at 298K of the reactionH2O2 → H2O(l) + ½ O2(g)is -23.5 kcal/mol and enthalpy of formation of H2O2(l) is -44.8 kcal/mol. The enthalpy of formation of H2O(l) is ... [ Pb. CET 1988]
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a) - 68.3 kcal/mol
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b) 68.3 kcal/mol
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c) -91.8 kcal/mol
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d) 91.8 kcal/mol
Explanation
ΔH=Δ(H2O) - Δ(H2O2) -23.5=Δ(H2O) - ( - 44.8 ) Δ(H2O)=-68.3 kcalAnswer: (a)
Q.13
Which of the following is an endothermic reaction? [ IIT 1989]
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a) 2H2 + O2 → 2H2O
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b) N2 + O2 → 2NO
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c) 2NaOH + H2SO4 → Na2SO4 + 2H2O
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d) C2H5OH + 3O2 → 2CO2 + 3H2O
Explanation
Answer: (b)
Q.14
If CH3COOH + OH- → CH3COO- + H2O + q1H+ + OH- → H2O + q2then the enthalpy change for the reactionCH3COOH → CH3COO- + H+ is equal to ..[Pb.CET 1988]
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a) q1 + q2
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b) q1 - q2
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c) q2 - q1
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d) -q1 + q2
Explanation
Equ(2) is for heat of neutralization of strong acid with strong base. Equ(1) is for weak acid with strong base. The difference q2 - q1 is heat of dissociation of weak acid Answer: (c)
Q.15
When 1M H2SO4 is completely neutralised by sodium hydroxide, the heat liberated is 114.64kJ. What is the enthalpy of neutralisation? [ Tamilnadu CET 2000]
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a) + 114.64 kJ
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b) - 114.64 kJ
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c) -57.32 kJ
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d) +57.32 kJ
Explanation
1 M H2SO4=2 N H2SO4 As heat of neutralisation is heat evolved for 1 mole of H+ ions, therefore enthalpy of neutralisation=- 114.64 / 2=-57.32 kJAnswer: (c)
Q.16
H2 + 0.5 O2 → H2O; ΔH°=-68 kcalK + H2O + aq → KOH(aq) + 0.5H2; ΔH°=-48 kcalKOH + aq → KOH(aq) ; ΔH°=-14 kcalFrom the above data, the standard heat of formation of KOH in kcal is ... [ CPMT 1988]
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a) - 68 + 48 - 14
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b) - 68 - 48 + 14
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c) 68 - 48 + 14
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d) 68 + 48 + 14
Explanation
Aim: K(s) + ½H2(g) + ½O2(g) → KOH(s) eqn(2) + eqn(1) - eqn(iii) givesΔH=-48 + (-68) - (-14) ΔH=-68 - 48 + 14Answer: (b)
Q.17
Which of the following statements/relationship is not correct? [ Pb. CET 1989]
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a) In an exothermic reaction, the enthalpy of products is less than that of reactants
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b) ΔHfusion=ΔHsublimation -ΔHvaporisation
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c) A reaction for which ΔH° < 0 and ΔS° > 0 is possible at all temperatures.
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d) ΔH is less than ΔE for the reactionC(s) + ½O2(g) → CO(g)
Explanation
For reaction in option "d" Δng=0.5 From formula ΔH=ΔU + ΔngRTΔH > ΔEAnswer: (d)
Q.18
Which of the following reaction can be used to define the heat of formation of CO2(g)? [ MP PMT 1989]
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a) C(graphite) + O2 → CO2(g)
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b) CH4(g) + O2 → CO2(g) + 2H2O(l)
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c) CO(g) + ½ O2(g) → CO2(g)
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d) C6H6(l) + 7.5 O2(g) → 6CO2(g) + 3H2O
Explanation
Defination of heat of formation Answer: (a)
Q.19
Given C(s) + O2 → CO2(g) ; ΔH=-395 kJS(s) + O2(g) → SO2(g) ; ΔH=-295 kJCS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ; ΔH=-1110 kJThe heat of formation of CS2(l) is.. [ Bihar CEE 1992]
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a) 125 kJ/mol
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b) 31.25 kJ/mol
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c) 32.5 kJ/mol
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d) 250 kJ/mol
Explanation
Aim: C + 2S → CS2 operate eqn(1) + 2×Eqn(2) - eqn(3) We getΔH=-395 + 2(-295) - (-1110)ΔH=125 kJAnswer: (a)
Q.20
If ΔHf(X), ΔHf(Y), ΔHf(R) and ΔHf(S) denote the enthalpies of formation of X, Y,R and S respectively the enthalpy of the reactionX + Y → R + S is given .... [ Pb.CET 1898]
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a) ΔHf(X) + ΔHf(Y)
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b) ΔHf(R) + ΔHf(S)
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c) ΔHf(X) + ΔHf(Y) - ΔHf(R) - ΔHf(S)
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d) ΔHf(S) + ΔHf(R) - ΔHf(X) - ΔHf(Y)
Explanation
ΔHreaction=∑ ΔHproduct - ∑ ΔHreactant Answer: (d)
Q.21
One mole of a non-ideal gas undergoes a change of state (2.0atm, 3.0L, 95K) → (4.0 atm, 5.0L, 245K) with a chnage in internal energy, ΔU=30.0 L atm, The change in enthalpy ( ΔH ) of the process in L atm is ... [ IIT 2002 p
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a) 40.0
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b) 42.3
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c) 44.0
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d) not defined, because pressure is not constant
Explanation
as P is NOT constant ΔH=ΔU - (P2V2 - P1 - V1 ) ΔH=30 + 4 × 5 - 2 ×3=44 L-atmAnswer: (c)
Q.22
(diamond) → C(graphite) ; ΔH=-1.5 kJIt follows that
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a) diamond is exothermic
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b) graphite is endothermic
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c) graphite is stable than diamond
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d) diamond is stable than graphite
Explanation
Answer: (c)
Q.23
For the transition C(diamond) → C(graphite) ; ΔH = -1.5 kJ It follows that ... [ BHU 1981 ]
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a) diamond is exothermic
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b) graphite is endothermic
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c) graphite is stabler than diamond
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d) diamond is stabler than graphite.
Explanation
Answer: (c)
Q.24
Thermodynamic equilibrium involves .. [ Pb CET 1991]
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a) Chemical equilibrium
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b) Thermal equilibrium
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c) Mechanical equilibrium
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d) All the above
Explanation
Answer: (d)
Q.25
An isolated system is that system in which ...[ MP PMT 1993]
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a) there is no exchange of energy with the surroundings
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b) there is exchange of mass and energy with the surroundings
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c) there is no exchange of mass and energy with surroundings
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d) there is exchange of mass with surroundings
Explanation
Answer: (c)
Q.26
In reactionCO2(g) + H2(g) → CO(g) + H2(g); ΔH=2.8 kJΔH represents [ KCET 1993]
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a) heat of reaction
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b) heat of combustion
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c) heat of formation
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d) heat of solution
Explanation
Answer: (a)
Q.27
Evaporation of water is ... [ MP PMT 1993]
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a) An exothermic change
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b) An endothermic change
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c) A process where no heat changes occur
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d) A process accompanied by chemical reaction
Explanation
Answer: (b)
Q.28
At constant T and P which one of the following statements is correct for the reaction? S8(s) + 8O2(g) → 8SO2(g)
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a) ΔH < ΔE
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b) ΔH=ΔE
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c) ΔH > ΔE
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d) ΔH is independent of the physical state of the reactants
Explanation
ΔH=ΔE + ΔngRTSince Δng=0 Answer: (b)
Q.29
Since the enthalpy of elements in their natural state is taken zero, the heat of formation of compounds .... [ KCET 1993]
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a) is always negative
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b) is always positive
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c) may be negative or positive
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d) is zero
Explanation
Answer: (c)
Q.30
The enthalpy change(-ΔH) for the neutralization of 1M HCl by caustic potash in dilute solution at 298K is ... [ BHU 1993]
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a) 68 kJ
0%
b) 65 kJ
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c) 57.3 kJ
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d) 50 kJ
Explanation
Answer: (c)
0 h : 0 m : 1 s
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