At 27°C the reaction, ${\mathrm{C}}_6{\mathrm{H}}_{6\left(\mathrm{l}\right)}+\frac{15}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 6{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$, proceeds spontaneously because the magnitude of

${\mathrm{C}}_6{\mathrm{H}}_{6\left(\mathrm{l}\right)}+\frac{15}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 6{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

It is combustion reaction and ${∆}_{\mathrm{C}}\mathrm{H}=-\mathrm{ve}$;

$∆{\mathrm{n}}_{\mathrm{g}}={\mathrm{n}}_{{\mathrm{CO}}_2}-{\mathrm{n}}_{{\mathrm{O}}_2}=6-\frac{15}{2}=-\frac{3}{2}$

number of moles decreases,volume decreases, entropy also decreases $∆\mathrm{S}=-\mathrm{ve}$

$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

reaction is spontaneous ; $∆\mathrm{G} =-\mathrm{ve}$

This is possible only if magnitude of $∆\mathrm{H}>\mathrm{T}∆\mathrm{S}$