Question

Suppose 1.00 mol superheated 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 mol21. Calculate DH, DSsys, and DG for this process.

Answer 1

Answer:

Enthalpy change is 6007 J

Entropy change is $20.15\,JK^{-1}$

Gibb's free energy change is 0 J.

Explanation:

From the given,

The enthalpy change for the melting of ice = $60007\,J\,mol^{-1}$

Temperature = $25^{0}C$

Let's convert the temperature centigrade into Kelvin.

$=25^{0}C \,+273.15\,=298.15K$

Number of moles of ice = 1.00 mol

Enthalpy change of 1.00 mol $25^{0}C$ temperature

$\bigtriangleup H=(1\,mol)(\frac{6007\,J}{mol})$

Therefore, Enthalpy change ($\bigtriangleup H$) for the melting of 1.00 mole of ice at $25^{0}C$ temperature is 6007 J.

Entropy change for the melting of 1.00 mole of ice at $25^{0}C$ :

$\Delta S_{sys}\,=\frac{\Delta H}{T}$

$=\,\frac{6007\,J}{298.15\,K}\,=20.15\,JK^{-1}$

Therefore, Entropy change($\Delta S_{sys}$)  for the melting of 1.00 mole of ice at $25^{0}C$ temperature is $20.15\,JK^{-1}$.

The Gibb's free energy change is expressed by the following formula.

$\Delta G^{o}\,=\Delta H^{o}-T\bigtriangleup S$

$=\,6007\,J-(298.15\,K)(20.15\,J\,K^{-1})$

$=6007\,J-6007\,J=\,0J$

Therefore, Gibb's free enrgy change ($\Delta G^{o}$) for the melting of 1.00 mole of ice at $25^{0}C$ temperature is $0J$