JEE Questions for Chemistry Thermodynamics Quiz 6 - MCQExams.com

At 298 K the entropy of rhombic sulphur is 32.04 J/mol K and that of monoclinic sulphur is 32.68 J/mol K. The heat of their combustion are respectively - 298246 and - 297948 J mol-1. ∆G for the reactions; Srhombic → Smonoclinic will be
  • 107.28 J
  • 10.728 J
  • 107.28 kJ
  • 10728 J
The enthalpy of vaporisation of a certain liquid at its boiling point of 35°C is 24.64 kJ mol-1 . The value of change in entropy for the process is
  • 704 JK-1 mol-1
  • 80 JK-1 mol-1
  • 24.64 KJ-1 mol-1
  • 7.04 JK-1 mol-1
For a spontaneous process, the correct statement (s) is (are)
  • ( ∆Gsystem )T , p > 0
  • ( ∆Ssystem ) + ( ∆Ssurroundings ) > 0
  • ( ∆Gsystem )T , p > 0
  • ( ∆Usystem )T , V > 0
For an ideal gas, consider only p-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure.
Which of the following choice (s) is (are) correct ?
Chemistry-Thermodynamics-8763.png
  • ∆SX→Z = ∆SX→Y + ∆SY→Z
  • WX→Z = WX→Y + WY→Z
  • WX→Y→Z = WX→Y
  • ∆SX→Y→Z = ∆SX→Y

Chemistry-Thermodynamics-8765.png
  • 1165
  • 837
  • 865
  • 815
Which is the best definition of “heat of neutralization”
  • The heat set free when one gram molecule of a base is neutralized by one gram molecule of an acid in dilute solution at a stated temperature
  • The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature
  • The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature
  • The heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in dilute solution at a stated temperature
In view of the signs of ∆r G° for the following reactions
PbO2 + Pb → 2PbO , ∆r G° < 0,
SnO2 + Sn→ 2SnO , ∆r G° > 0,
Which oxidation states are more characteristic for lead and tin ?
  • For lead + 4, for tin + 2
  • For lead + 2, for tin + 2
  • For lead + 4, for tin + 4
  • For lead + 2, for tin + 4
For the reversible reaction,
A(s) + B(g) ⇌ C(g) + D(g); ∆G° =- 350 kJ,
which one of the following statements is true ?
  • The reaction is thermodynamically non-feasible
  • The entropy change is negative
  • Equilibrium constant is greater than one
  • The reaction should be instantaneous
Carbon cannot reduce Fe2 O3 to Fe at a temperature below 983 K because
  • free energy change for the formation of CO is more negative than that of Fe2 O3
  • CO is thermodynamically more stable than Fe2 O3
  • carbon has higher affinity towards oxygen than iron
  • iron has higher affinity towards oxygen than carbon

Chemistry-Thermodynamics-8766.png

  • Chemistry-Thermodynamics-8767.png
  • 2)
    Chemistry-Thermodynamics-8768.png

  • Chemistry-Thermodynamics-8769.png

  • Chemistry-Thermodynamics-8770.png
A container of 1.0 L capacity filled with 1.0 mole of ideal gas is connected to an evacuated vessel of 9.0 L. Calculate change in entropy. (R =1.987 cal)

  • Chemistry-Thermodynamics-8771.png
  • 2)
    Chemistry-Thermodynamics-8772.png

  • Chemistry-Thermodynamics-8773.png

  • Chemistry-Thermodynamics-8774.png

Chemistry-Thermodynamics-8775.png
  • 4.5 V
  • 3.0 V
  • 2.5 V
  • 5.0 V

Chemistry-Thermodynamics-8776.png

  • Chemistry-Thermodynamics-8777.png
  • 2)
    Chemistry-Thermodynamics-8778.png

  • Chemistry-Thermodynamics-8779.png

  • Chemistry-Thermodynamics-8780.png
In which reaction there will be increase in entropy?

  • Chemistry-Thermodynamics-8781.png
  • 2)
    Chemistry-Thermodynamics-8782.png

  • Chemistry-Thermodynamics-8783.png

  • Chemistry-Thermodynamics-8784.png
Which is the correct expression, that relates changes of entropy with the change of pressure for an ideal gas at constant temperature in the following ?

  • Chemistry-Thermodynamics-8785.png
  • 2)
    Chemistry-Thermodynamics-8786.png

  • Chemistry-Thermodynamics-8787.png

  • Chemistry-Thermodynamics-8788.png
For a reversible reaction,
X (g) + 3Y (g) ⇌ 2Z(g); ∆H = - 40 kJ, the standard entropies of X ,Y and Z are 60, 40 and 50 JK-1 mol-1 respectively. The temperature at which the above reaction attains equilibrium is about
  • 400 K
  • 500 K
  • 273 k
  • 373 k
The decomposition of limestone
CaCO3 (g) ⇌ CaO(s) + CO2 (g)
is non-spontaneous at 298 K. The ∆H° and ∆S° values for the reaction are 76.0 kJ and 60 JK-1 respectively. At what temperature the decomposition becomes spontaneous ?
  • At 1000 K
  • Below 500° C
  • Between 500° C and 600° C
  • Above 827°C
Identify the correct statement regarding a spontaneous process.
  • For a spontaneous process in an isolated system, the change in entropy is positive
  • Endothermic processes are never spontaneous
  • Exothermic processes are always spontaneous
  • Lowering of energy in the reaction process is the only criterion for spontaneity
In conversion of limestone to lime,
CaCO3 (s) → CaO(s) + CO2 (g )
the values of ∆H° and ∆S °are + 179.1 kJ mol-1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° and ∆S ° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
  • 1008 K
  • 1200 K
  • 845 K
  • 1118 K
Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1, respectively. Standard Gibbs energy change for the same reaction at 298 K is
  • - 2221.1 kJ mol-1
  • - 339.3 kJ mol-1
  • - 439.3 kJ mol-1
  • - 523.2 kJ mol-1
Which of the following is correct for the reaction ∆H = +ve and ∆S = +ve?
  • Spontaneous at high temperature
  • Spontaneous at low temperature
  • Non-spontaneous at high temperature
  • Non-spontaneous at all temperatures
What is the relation between E° and K ?

  • Chemistry-Thermodynamics-8789.png
  • 2)
    Chemistry-Thermodynamics-8790.png

  • Chemistry-Thermodynamics-8791.png

  • Chemistry-Thermodynamics-8792.png
A schematic plot of In Keq versus inverse of temperature for a reaction is shown below
Chemistry-Thermodynamics-8793.png
  • highly spontaneous at ordinary temperature
  • one with negligible enthalpy change
  • endothermic
  • exothermic
The enthalpy change for the transition of liquid water to steam is 40.8 kJ per mol at 100°C. The entropy change for the process will be
  • 0.408 JK-1 mol-1
  • 408 JK-1 mol-1
  • 109.4 JK-1 mol-1
  • 0.1094 JK-1 mol-1
If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then
  • ∆H is -ve, ∆S is +ve
  • ∆H and ∆S both are +ve
  • ∆H and ∆S both are -ve
  • ∆H is +ve, ∆S is -ve
Consider the following reactions at 1100°C
(I) 2C + O2 → 2CO, ∆G° = - 460 kJ mol-1
(II) 2Zn + O2 → 2 ZnO, ∆G° = - 360 kJ mol-1
Based on these, select correct alternate.
  • Zinc can be oxidised by CO
  • Zinc oxide can be reduced by carbon
  • Both are correct
  • None of the above is correct
For which reaction change of entropy will be positive ?
  • H2 (g) + I2 (g) ⇌ 2HI (g)
  • HCl (g) + NH3 (g) ⇌ NH4Cl (s)
  • NH4 NO3 (s) ⇌ N2O (g) + 2H2O (g)
  • MgO (s) + H2 (g) ⇌ Mg (s) + H2O (l)
The value of log Kc for a reaction
A ⇌ B is (Given, ∆r H° 298 K = - 54.07 kJ mol-1 , ∆r S ° 298 K =10 JK-1 mol-1 and R = 8.314 JK-1 mol-1 2.303 × 8.314 × 298 =
  • 5
  • 10
  • 95
  • 100
Enthalpy is equal to

  • Chemistry-Thermodynamics-8795.png
  • 2)
    Chemistry-Thermodynamics-8796.png

  • Chemistry-Thermodynamics-8797.png

  • Chemistry-Thermodynamics-8798.png
In an irreversible process taking place at constant T and p and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS ), satisfy the criteria.
  • (dS)V ,E < 0, (dG)T ,p < 0
  • (dS)V ,E > 0, (dG)T ,p < 0
  • (dS)V ,E = 0, (dG)T ,p = 0
  • (dS)V ,E = 0, (dG)T ,p > 0
For the reaction at 298 K
A(g) + B(g) ⇌ C(g) + D(g)
∆H° = - 29.8 kcal , ∆S° = - 0.100 kcal K-1
What is the value of ∆G° ?
  • 1
  • 0
  • 2
  • 4
The direct conversion of A to B is difficult, hence it is carried out by the following shown path
Chemistry-Thermodynamics-8799.png
  • + 100 eu
  • + 60 eu
  • - 100 eu
  • - 60 eu
The internal energy of one mole of a monoatomic gas is

  • Chemistry-Thermodynamics-8801.png
  • 2)
    Chemistry-Thermodynamics-8802.png

  • Chemistry-Thermodynamics-8803.png

  • Chemistry-Thermodynamics-8804.png
The correct relationship between ∆H and ∆E for an ideal gas at constant temperature is
  • ∆H = ∆E – P∆V
  • ∆H = ∆E + P∆V
  • ∆H = ∆E + ∆P. ∆V
  • ∆H = E + P∆V
You are given the following two reactions:
(i)CH4(g) + 202(g) → CO2 (g) + 2H2O(g), ∆H = - 890.4 kJ.
(ii)2HgO(s) → 2Hg(l) + O2(g) – 181.6 kJ
Which one of the following statements is correct
  • Both reactions are exothermic
  • Both reactions are endothermic
  • Reaction (i) is endothermic and (ii) is exothermic
  • Reaction (i) is exothermic and (ii) is endothermic
The formation of water from H2 (g) and O2 (g) is an exothermic reaction because
  • H2(g) and O2(g) have a higher chemical energy than water
  • H2(g) and O2(g) have a lower chemical energy than water
  • H2(g) and O2(g) have a higher temperature than water
  • Energy considerations do not arise
Which one of the following expresses Q as the heat of combustion ?
  • 2H2 + O2 → 2H2O + Q
  • C + 1/2 O2 → CO + Q
  • CH4 + 2 O2 → CO2 + 2H2O + Q
  • 2C6H6 + 15 O2 → 12 CO2 + 6H2O + Q
Which one of the following expresses heat of formation of CO2 ?
  • CaCO3 + 2HCl → CaCl2 + H2O + CO2, ∆H = –a kcal
  • CO + 1/2 O2 → CO2 , ∆H = -b kcal
  • C + O2 → CO2, ∆H = -c kcal
  • 2NaHCO3 → Na2CO3 + H2O + CO2, ∆H = +d kcal
The enthalpy of reaction at constant volume (∆E) is related to the enthalpy of reaction at constant pressure (∆H) as
  • ∆H = ∆E − ∆ng RT
  • ∆H = ∆E + ∆ng RT
  • ‒∆H = ‒∆E + ∆ng RT
  • None of these is correct
When 1 mole of H2O2 is decomposed by platinum black, the heat evolved is 96.6 kJ. The heat of formation of 1 mole of H2O2 is
  • 96.6 kJ
  • 193.2 kJ
  • 386.4 kJ
  • 48.3 kJ
The heat of combustion of solid benzoic acid at constant volume is – 321.30 kJ at 27oC. The heat of combustion at constant pressure is
  • –321.30 – 300 R
  • –321.30 + 300 R
  • –321.30 – 150 R
  • –321.30 + 900 R
A hypothetical reaction, A → 2 B, proceeds through following sequence of steps A → C; ∆H = q1 , C → D; ∆H = q2, 1/2 D → B ; ∆H = q3 . The heat of reactions is:
  • q1 ‒ q2 + 2q3
  • q1 + q2 ‒ 2q3
  • q1 + q2 + 2q3
  • q1 + 2q2 ‒ 2q3
The bond energy of H2 is 436.4 kJ. This means that
  • 436.4 kJ of heat is required to break one bond in H2 molecule to give two atoms of hydrogen
  • 436.4 kJ of heat is required to dissociate 6.02 × 1023 molecules of H2 to form H-atoms
  • 436.4 kJ of heat is required to dissociate 3.01 × 1023 molecules of H2 to 6.02 × 1023 atoms of hydrogen
  • 436.4 kJ of electrical energy is required to dissocitate 6.02 × 1023 molecules of H2 to form H+ and H− ions.
If E, R and D represent respectively the tendency for minimum energy, tendency for maximum randomness and overall driving force and arrows represent the direction and magnitude of these tendencies, which one of the following applies correctly to the reaction 2HgO (s) → 2 Hg (l) + O2 (g) ?

  • Chemistry-Thermodynamics-8808.png
  • 2)
    Chemistry-Thermodynamics-8809.png

  • Chemistry-Thermodynamics-8810.png

  • Chemistry-Thermodynamics-8811.png
For a reversible isothermal process in equilibrium, the entropy change is given by the expression

  • Chemistry-Thermodynamics-8812.png
  • 2)
    Chemistry-Thermodynamics-8813.png

  • Chemistry-Thermodynamics-8814.png

  • Chemistry-Thermodynamics-8815.png
The approximate enthalpy of fusion of ice and the enthalpy of vaporisation of water per mole respectively are
  • 18 kJ, 80 kJ
  • 6kJ, 40.6 kJ
  • 80 kJ, 540 kJ
  • 8 kJ, 80kJ
For a reaction R1, ∆G = x kJ mol-1. For a reaction R2, ∆G = y kJ mol-1 Reaction R1 is non-spontaneous but along with R2 it is spontaneous. This means that
  • x is –ve, y is +ve but in magnitude x > y
  • x is +ve, y is – ve but in magnitude y > x
  • both x and y are –ve but not equal
  • both x and y are +ve but not equal
According to latest sign conventions, the correct expression representing the first law of thermodynamic is
  • ∆E = q + w
  • ∆ E = ∆H + PV
  • ∆E = q – w
  • all the expressions are correct
The entropy of a perfectly crystalline solid at absolute zero is
  • negative
  • positive
  • zero
  • not definite
∆G for the reaction Ag2O → 2 Ag + 1/2 O2 at a certain temperature is found to be – 10.0 kJ mol-1. Which one of the following statements is correct at this tempertature?
  • Silver oxide decomposes to give silver and oxygen
  • Silver and oxygen combine to form silver oxide
  • The reaction is in equilibrium
  • The reaction can neither occur in the forward direction nor in the backward direction.
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