The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (DS) of the reduction process is more on positive side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of DrG V becomes more on negative side and the reduction becomes easier.

K = Ae^-Ea/RT
Ea, absence of catalyst = x KJ/mol
Ea, presence of catalyst = (x-20)KJ/mol
At 500K K = Ae-x/R.500
At 400
K = Ae-(x-20)/R.400
Divide 1 by 2, we get
X = 100KJ/mol
Ea = 100KJ/mol.

(i)

(ii)
 
(iii)

Ag⁺ ion is stronger Lewis acid than Na⁺. Ag⁺ has 18-electron shell configuration (known as pseudo inert gas configuration) and causes greater polarization than Na⁺, an ion with 8-electron shell configuration. This is due to the fact that in the case of 18-electrons shell ion there are 10 d-electrons in addition to 88-and p-electrons. The d-electrons do not shield the nuclear charge effectively and therefore, they have increased the effective nuclear charge. Hence they cause greater polarization and therefore, Ag⁺ is a stronger Lewis acid than Na⁺ ion.

During SNIreaction, the ionization gives 2° carbocation which rearranges by a hydride shift to a more stable 3"-carbocation to form 3°-alcohol as:

For pH = 14, [H⁺]= 10^-14
\([OH^-]={K_w\over [H^+]}={10^{-14}\over 10^{-14}}=1M\)
Cu(OH)₂ ⇌ Cu²⁺+ 2OH^-
K_sp= [Cu²⁺] [OH^-]²
1.0 x 10^-19 = [Cu²⁺](1)²
[Cu²⁺] = 1.0 X 10^-19
For the cell reaction,
Cu²⁺+ 2e^- ⟶ Cu
\(E=E^0{0.059\over 2}log{1\over [Cu^{2+}]}\)
\(=0.34-{0.059\over 2}log{1\over 1.0\times 10^{-19}}\)
\(=0.34-{0.059\over 2}\times19=-0.22V\)

Ethanol and water (95.4% ethanol and 4.6% \({ H }_{ 2 }O\)) from a constant boiling mixture (azeotrope) boiling at 351.5 K. Hence, further water cannot be removed by fractional distillation.