Answer:
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.
