Question

The value $\Delta$S for the process, $H_{2}O\left(s\right)\to H_{2}O\left(l\right)$ at 1 atm pressure and 260K is greater than zero. The value of $\Delta G$ will be :

• A. $>0$
• B. $<0$
• C. $=0$
• D. lies between $-1$ and $0$

Answer 1

Answer: (A)

$>0$

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

$\Delta S$ + ve and $\Delta H$ + ve.

The temperature is low so the process is non-spontaneous.

$\therefore \Delta G>0$.