Use the References to access important values if needed for this question. The m

Question

Use the References to access important values if needed for this question. The molar enthalpy of fusion of solid tin is 7.07 kJ mor 1, and the molar entropy of fusion is 14.0 J K 1 mol (a) Calculate the Gibbs free energy change for the melting of 1.00 mol of tin at 471 K KJ (b) Calculate the Gibbs free energy change for the conversion of 5.78 mol of solid tin to liquid tin at 471 K. kJ (c) Will tin melt spontaneously at 471 K? (d) At what temperature are solid and liqu Yes uilbrium at a pressure of I atm? uilibrium at a pressure of 1 atm? Submit Answer 5 question attempts remaining Back

Explanation / Answer

Ans. #a.          dG0 = dH - TdS        - equation 1

Putting the values in equation 1-

            dG0 = 7.07 kJ mol-1 - [ 471 K x (14.0 J K-1 mol-1)]

            Or, dG0 = 7.07 kJ mol-1 - 6594 J mol-1

            Or, dG0 = 7.07 kJ mol-1 – 6.594 kJ mol-1

            Hence, dG0 = -0.476 kJ mol-1

Therefore, Gibb’s free energy for melting 1.00 mol of tin = -0.476 kJ

#b. dG0= Molar Gibb’s free energy change x Moles of tin taken

= (-0.476 kJ mol-1) x 5.78 mol

= -2.75128 kJ

#c. YES. Since the dG0 for the reaction is a NEGATIVE value (-0.476 kJ mol-1), it is a SPONTANEOUS reaction under given conditions.

#d. For the solid and liquid phases to co-exist at equilibrium, the value of dG0 must be zero.

Let the required temperature be T.

Now, putting dG0 = 0 in equation 1-

            0 = 7.07 kJ mol-1 - [ T x (14.0 J K-1 mol-1)]

            Or, T x (14.0 J K-1 mol-1) = 7.07 kJ mol-1

            Or, T = 7.07 kJ mol-1 / (0.014 kJ K-1 mol-1)

            Hence, T = 505.0 K

Hence, Required temperature, T = 505.0 K

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