Question

Question 23 The enthalpy of the boiling of a compound at its normal boiling point (at 83.0°C) is 111 cal/g. Calculate the entropy (in cal/g. K) of this process. AG=AH-TAS 0.256 0.368 0.100 0.598 0.225 0.456 0.312

Answer 1

Given that ∆H = 111 cal/g

Given boiling point = 83°C + 273 = 356 K

As ∆G is the function of change in temperature and at boiling point there is no change in temperature but only the phase transition occurs from liquid to gas. Hence ∆G = 0 at boiling point.

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

So, $0 = \Delta H - T\Delta S$

Therefore, $\Delta S = \Delta H / T$

$\Delta S = 111 / 356$

$\Delta S = 0.31179 \approx 0.312 cal/g/K$

So, the value of entropy is 0.312 cal/g/K. i.e last option.