What is the entropy change (in J K^$$-$$1 mol^$$-$$1) when one mole of ice is converted into water at 0^oC? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol^$$-$$1 at 0^oC).

Formation of solution from two components can be considered as
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H₁
(ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H₂
(iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H₃
Solution so formed will be ideal if

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^$$-$$3, respectively. If the standard free energy difference $$\left( {\Delta {G^o}} \right)$$ is equal to 1895 J mol^$$-$$1, the pressure at which graphite will be transformed into diamond at 298 K is

For the reaction,
C₃H_8(g) + 5O_2(g) $$ \to $$ 3CO₂(g) + 4H₂O_(l)
at constant temperature, $$\Delta $$H $$-$$ $$\Delta $$E is