Question

Suppose \(1.00\;{\rm{mol}}\) superheated ice melts to liquid water at \({25^\circ }{\rm{C}}\). 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 \({0^\circ }{\rm{C}}\) is \(6007\;{\rm{J}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\). Calculate \(\Delta H,\Delta {S_{sys }}\), and \(\Delta G\) for this process.

Short answer

(a)The change in enthalpy,\(\Delta H = 6007\;{\rm{J}}\).

(b)The change of entropy,\(\Delta {S_{{\rm{sys}}}} = 20.16{\rm{J}}{{\rm{K}}^{ - 1}}\).

(c)The Gibbs free energy,\(\Delta G = 0\).

Step-by-step solution

Given data

The heat of vaporization,\(\Delta {H_m} = 6007{\rm{Jmo}}{{\rm{l}}^{ - 1}}\).

The temperature of boiling point, \(T = {25^\circ }C\).

Concept Gibbs free energy

At constant temperature and pressure, the Gibbs free energy is a thermodynamic potential that can be used to compute the maximum work that a thermodynamically closed system can perform.

Calculation of enthalpy

The enthalpy can be determined by the formula:

\(\Delta H = \Delta {H_m} \times n\)

where,\(\Delta H\)is change in enthalpy,\(n\)is number of moles.

Put the value of given data in above equation.

\(\begin{aligned}{}\Delta H = 6007{\rm{Jmo}}{{\rm{l}}^{ - 1}} \times 1\;{\rm{mol}}\\ = 6007\;{\rm{J}}\end{aligned}\)

Calculation of entropy

The entropy is calculated with the help of the formula:

\(\Delta {S_{{\rm{system }}}} = \frac{{\Delta H}}{T}\)

where,\(T\)is work done,\(\Delta H\)is the change in enthalpy.

Put the value of the given data in the above equation.

\(\begin{aligned}{}\Delta {S_{{\rm{system }}}} = \frac{{6007}}{{298}}\\\Delta {S_{{\rm{system }}}} = 20.16{\rm{J}}{{\rm{K}}^{ - 1}}\end{aligned}\)

Calculation of Gibbs free energy

The Gibbs free energy is calculated with the help of the formula:

\(\Delta G = \Delta H - T\Delta S\)

where,\(\Delta G\)is Gibbs free energy,\(\Delta S\)is the entropy of the.

Put the value of the given data in the above equation.

\(\begin{aligned}{}\Delta G = 6007 - 298 \times 20.16\\\Delta G = 0\;{\rm{J}}\end{aligned}\)