In the reaction, $∆$H and $∆$S both are positive. The condition under which the reaction would not be spontaneous is -

1. $∆$H>T$∆$S                             

2. $∆$S=$∆$H/T

3. $∆$H=T$∆$S                               

4. All of the above

'The free energy change due to a reaction is zero when-

1. The reactants are initially mixed.

2.  A catalyst is added

3. The system is at equilibrium

4. The reactants are completely consumed

18 g of ice is converted into water at 0$°$C and 1 atm. The entropy of H₂O_(s) and H₂O_(l) are 38.2 and 60 J K^-1 mol^-1 respectively. the enthalpy for this conversion will be

(1) 5951.4 J/mol

(2) 595.14 J/mol

(3) -595.14 J/mol

(4) None of these

For a given reaction, $∆$H =35.5 kJ mol^-1 and $∆$S = 83.6JK^-1 mol^-1. The reaction is spontaneous at:
(Assume that $∆$H and $∆$S do not vary with temperature)

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is

$\left(\mathrm{a}\right) ∆\mathrm{H}>0 \mathrm{and} ∆\mathrm{S}<0$
$\left(\mathrm{b}\right) ∆\mathrm{H}<0 \mathrm{and} ∆\mathrm{S}>0$
$\left(\mathrm{c}\right) ∆\mathrm{H}<0 \mathrm{and} ∆\mathrm{S}<0$
$\left(\mathrm{d}\right) ∆\mathrm{H}<0 \mathrm{and} ∆\mathrm{S}=0$