At `25^(@)C` and 1 atm, the heat of formatio of 1 mol of water is -285.8 kJ`*mol^(-1)`. State whether formation reaction will be spontaneous at that temperature and pressure or not. Given: The molar entropies of `H_(2)(s),O_(2)(g)&H_(2)O(l)` at `25^(@)C` and 1 atm are 130.5, 205.0 and 69.9 `J*K^(-1)*mol^(-1)` respectively.

Explanation

Equation for the formation reaction of water:
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.8kJ*mol^(-1)`
The change in entropy for this reaction,
`DeltaS=S_(H_(2)O)-(S_(H_(2)(g))+(1)/(2)S_(O_(2)(g)))`
`=[69.9-130.5-((1)/(2)xx205)]J*K^(-1)=-163.1J*K^(-1)`
Therefore, the change in gibbs free energy at `25^(@)C and 1` atm for the formation of 1 mol of `H_(2)O(l)` from 1 mol of `H_(2)(g) and (1)/(2)` mol of `O_(2)(g),DeltaG=DeltaH-TDeltaS`
`=-285.8xx10^(3)-298xx(-163.1)=-237.196kJ`
As `DeltaG` is negative at `25^(@)C` and 1 atm, so the formation of water at the temperature and pressure will be spontaneous.