\(\Delta n_{g}=1-\frac 12=\frac 12\)
As amount of gaseous substance is increasing in product, thus \(\Delta \mathrm{S}\) is positive for this reaction. And we know that \(\Delta G=\Delta \mathrm{H} - T\Delta S\)
As \(\Delta \mathrm{S}\) is positive, thus increase in temperature will make \(T\Delta S\) more negative and \(\Delta \mathrm{G}\) will decrease.
$H_2$$_{(g)} +$ $1/2O_2$ $_{(g)}$ $\rightarrow$ $H_2$$O$$_{(l)}$; $\Delta H= -$ $285.77\, KJ\, mol$$^{-1}$; $H_2$$_{(g)} +$ $1/2O_2$$_{(g)}$ $\rightarrow$ $H_2O$ $_{(g)}$; $\Delta H$ $ = - 241.84\, KJ \,mol$$^{-1}$