For the equilibrium `H_(2)O(l) iff H_(2)O(g)` at 1 atm and 298 K

Options

A. Standard free energy change is equal to zero `(Delta G^(@) = 0)`
B. Free energy change is less than zero `(Delta G lt 0)`
C. Standard free energy change is less than zero `(Delta G^(@) lt 0)`
D. Standard free energy change is greater than zero `(Delta G^(@) gt 0)`

Answer

C

Explanation

At equilibrium `Delta G= 0`
`H_(2)O(l) hArr H_(2)O(g)` at 1 atm and 298K
`Delta H^(@)` is positive because this reaction is endothermic
`Delta S^(@)` is positive because liquid converts to gas
`Delta G^(@) = Delta H^(@)- T Delta S^(@)`
So the value of `Delta G^(@)` will be positive at low temperature and negative at high temperature. At 298K, the value of `Delta G^(@)` is likely to be negative.