. Suppose 1mol superheated ice melts to liquid water at 25℃. Assume the specific heats of ice ( 2.1 ) and liquid water ( ) are independent of temperature. kJ kg*K 4.2 kJ kg*K The enthalpy change for the melting of ice at 0℃ is . Calculate ∆H, ∆Ssys 6 , and ∆G kJ mol for this process

Answer / Explanation:

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Question:

Suppose 1.00 mole super heated ice melts to liquid water at 25°C . Assume the specific heats of ice and liquid water have the same value and are independent of temperature. The enthalpy change for the melting of ice at 0°C is 6007 J mol^-1. Calculate ΔH, ΔSsys, and ΔG,. for this process.

If we recall the given parameters :

Enthalpy change of melting ice = 6007 Jole/Mole

Enthalpy change of 1 mole of ice at 25°c = Δ H = 6007 J

Now, if we go ahead to calculate the temperature change at 25°c, we have,

The temperature at 25°c = (275 + 273.15)

                                          = 298.25

Now,

Enthalpy Change =  ∆Ssys = Δ H / T

                             = 6007 / 298.15 = 20.15 J/k

Therefore, Enthalpy Change =  ∆Ssys = 20.15 J/k

Now, recalling the gibbs free energy change:

we have: ∆G ° = Δ H° - T∆S°

                        = 6007 - (298.15) ( 20.15)

                         = 6007 - 6007

                         = 0 Joules

Therefore, ∆G ° = 0 Joules

the correct answer is:

a. formation of precipitation

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