Predict the signs of ! ih, , and 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.)

Explanation:

(a) Given,

Normal melting point of ammonia is -77.7 °C

In this process , ammonia melts at - 60°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.

the answer is option a.