Calculate the standard free energy change for the reaction,
`N_(2(g))+3H_(2(g))to2NH_(3(g)) at 298 K`.
Given, `DeltaH^@ =-92.4kJ` and `DeltaS^@ = -198.3 JK^(-1)`. Also, comment on the result.

Explanation

`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`=-92.4-298xx(-198.3)xx10^(-3)`
`( :' DeltaS^(@)=-198.3xx10^(-3)k J K^(-1))`
`=-33.306 k J`
Since `DeltaG^(@)` is negative, ti means that a mixture of `H_(2)` and `N_(2)` at `25^(@)`, each present at a pressure of `1` atm would react spontaneously to form ammonia.