Question

For the reaction $2HgO\left(s\right)\to 2Hg\left(l\right)+O_2\left(g\right)$ :

• A. $\Delta H>0$ and $\Delta S<0$
• B. $\Delta H>0$ and $\Delta S>0$
• C. $\Delta H<0$ and $\Delta S<0$
• D. $\Delta H<0$ and $\Delta S>0$

Answer 1

Answer: (C)

$\Delta H<0$ and $\Delta S<0$

The given reaction is

$2HgO\left(s\right)\stackrel{}{\to }2Hg\left(l\right)+O_2\left(l\right)$

If the reaction is sufficiently exothermic.

It can force $\Delta G$ negative only at temperatures below which $|T\Delta S|<|\Delta H|$.

This means that there is a temperature $T=\frac{\Delta H}{\Delta S}$

At which the reaction is at equilibrium.

The reaction will only proceed spontaneously below this temperature.

Hence,

$\Delta H<0and\Delta S<0$

This is the required solution.