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Q1. The free energy change for a reversible reaction at equilibrium is
Solution
d)
d)
Q2.In a flask, colourless N2O4(g) is in equilibrium with brown coloured NO2 (g). At equilibrium when the flask is heated to
100℃, the brown colour deepens and on cooling it becomes coloured.Which statement is incorrect about this observation?
100℃, the brown colour deepens and on cooling it becomes coloured.Which statement is incorrect about this observation?
Solution
(c) ∆H-∆U=∆nRT=1×2×373=746 cal.
(c) ∆H-∆U=∆nRT=1×2×373=746 cal.
Q3. The second law of thermodynamics introduced the concept of:
Solution
c) The second law of thermodynamics has been defined as – the entropy of universe is always increasing in the course of every spontaneous process.
c) The second law of thermodynamics has been defined as – the entropy of universe is always increasing in the course of every spontaneous process.
Q4. ∆G°for the reaction X+Y⇌Z is -4.606 kcal. The value of equilibrium constant of the reaction at 227℃ is (R=2.0 cal mol-1 K-1)
Solution
a) ∆G=-2.303 RT logK -4.606=-2.303×0.002×500 logK logK=2,K=100
a) ∆G=-2.303 RT logK -4.606=-2.303×0.002×500 logK logK=2,K=100
Q5.The standard change is Gibbs energy for the reaction,
H2O⇌H++OH- at 25℃ is:
H2O⇌H++OH- at 25℃ is:
Solution
b) ∆G°=-2.303 RT logK (K for H2 O=10-14/55.6) =-2.303×8.314×298×log10-14/55.6 =-89.84 kJ
b) ∆G°=-2.303 RT logK (K for H2 O=10-14/55.6) =-2.303×8.314×298×log10-14/55.6 =-89.84 kJ
Q6.In an adiabatic process
Solution
d) In the adiabatic process no heat enters or leaves the system i.e., q=0
d) In the adiabatic process no heat enters or leaves the system i.e., q=0
Q7.What is ∆E for system that does 500 cal of work on surrounding and 300 cal of heat is absorbed by the system?
Solution
(a) From first law of thermodynamic. ∆E=q+W Given,q=+300 cal (∵Heat is absorbed) W=-500 cal (∵Work is done on surroundings) ∴ ∆E=q+W=300+(-500) =-200 cal
(a) From first law of thermodynamic. ∆E=q+W Given,q=+300 cal (∵Heat is absorbed) W=-500 cal (∵Work is done on surroundings) ∴ ∆E=q+W=300+(-500) =-200 cal
Q8.An ideal gas undergoing expansion in vacuum shows:
Solution
d) Ideal gas does not show intermolecular forces of attractions
d) Ideal gas does not show intermolecular forces of attractions
Q9.The bond dissociation energy of B-F in BF-3 is 646 kJ mol-1, whereas that of C-F in CF-4 is 515 kJ mol-1.
The correct reason for higher B-F bond dissociation energy as compared to that of C-F is:
The correct reason for higher B-F bond dissociation energy as compared to that of C-F is:
Solution
b) In BF3 pÏ€—pÏ€ interaction leads to back bonding due to vacant p–orbitals of boron and completely filled p-orbitals of F.
b) In BF3 pÏ€—pÏ€ interaction leads to back bonding due to vacant p–orbitals of boron and completely filled p-orbitals of F.
Q10. Boiling point of a liquid is 50 K at 1 atm and ∆Hvap=460.6 cal mol-1. What will be its b.p. at 10 atm?
Solution
(c)
(c)