Question

At $27C$ the reaction,
$C_6{H_6}_{\left(l\right)}+\frac{15}{2}{O_2}_{\left(g\right)}\to {{6CO}_2}_{\left(g\right)}+{{3H}_{2}O}_{\left(l\right)}$
proceeds spontaneously because the magnitude of?

• A. $\Delta H=T\Delta S$
• B. $\Delta H>T\Delta S$
• C. $\Delta H$<$T\Delta S$
• D. None of these

Answer 1

Answer: (C)

$\Delta H$<$T\Delta S$

$\Delta H<T\Delta S$

As we know that,

$\Delta G=\Delta H-T\Delta S$

As the reaction is proceeding spontaneously, $\Delta G$ must be negative.

Therefore, for $\Delta G$ to be negative,

$\Delta H<T\Delta S$