Question

At ${27}^{\circ }\text{C}$ the reaction,
$C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
Proceeds spontaneously because the magnitude of-

• A. $\Delta H>0,T\Delta S<0$
• B. $\Delta H=T\Delta S$
• C. $\Delta H>T\Delta S$
• D. $\Delta H<T\Delta S$

Answer 1

The correct option is C $\Delta H>T\Delta S$

In this reaction, $\Delta S=-ve$ because products have lesser gaseous molecules.
Reaction is exothermic $\therefore \Delta H=–ve$
$\Delta G=\Delta H-T\Delta S$
Since process is spontaneous
$\Delta G=-ve$
This is possible only if magnitude of
$\Delta H>T\Delta S$