At $27°C$ the reaction, $C_6{H}_{6\left(l\right)}+15/2O_{2\left(g\right)}\to 6C{O}_{2\left(g\right)}+3H_2{O}_{\left(l\right)}$

proceeds spontaneously because the magnitude of   

For the given reaction,

$\Delta$S = negative  (∵ Lesser number of gaseous products are formed)

$\Rightarrow$ –T$\Delta$S = positive

But $\Delta$H = –ve

Since, it is a combustion reaction and hence exothermic.

∵ $\Delta$G = $\Delta$H – T$\Delta$S ; So for $\Delta$G to be negative, |$\Delta$H>T$\Delta$S|