Predict the signs of !iH, !).S, and !).G of the system for the following processes at 1 atm: (a) ammonia melts at - 60°C, (b) ammonia melts at - 77.7°C, and (c) ammonia melts at - 100°C. (The normal melting point of ammonia is - 77.7°C.)

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is an endothermic process , hence , the change in enthalphy sign will be positive .

Even , the temperature of the ammonia is more than its normal melting point .

Therefore ,

TΔS > ΔH

Hence ,

ΔG < 0

Therefore ,

ΔH = Positive

ΔS = Positive

ΔG = Negative.

(b)Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at -77.7 °C

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is neither endothermic process nor exothermic process, hence , the change in enthalphy sign will be neutral / zero .

Hence ,

ΔG < 0

Therefore ,

ΔH = 0

ΔS = Positive ΔG = Negative.

(c) Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at - 100°C

Now , during the process of melting , ammonia in the solid phase changes to liquid phase, and this increases the number of particles , therefore , the sign of entropy would be positive . And since , the process is an endothermic process , hence , the change in enthalphy sign will be positive .

Even , the temperature of the ammonia is less than its normal melting point .

Hence ,

ΔG > 0

Therefore ,

ΔH = Positive

ΔS = Positive

ΔG = Negative.