Question

The molar enthalpy of fusion of solid ammonia is 5.65 kJ mol^-1, and the molar entropy of fusion is 28.9jK^-1 mol^-1.

(a) Calculate the Gibbs free energy change for the melting of 1.00 molammonia at 170K.

(b) Calculate the Gibbs free energy change for the conversion of3.60 molsolid ammonia to liquid ammonia at 170K.

(c) Will ammonia melt spontaneously at 170K?

(d) At what temperature are solid and liquid ammonia in equilibrium at a pressure of 1 atm?

Short answer

(a) The Gibbs free energy$△G_{1mol}^{°}=740J$

(b) The Gibbs free energy $△G_{3.6mol}^{°}=2.60J$

(c) Ammonia will not melt spontaneously at 170K, because Gibb's free energy for melting is positive

d) The temperature, T=196 K

Step-by-step solution

Given data

The value of standard entropy and enthalpy is given below:

$$\begin{gathered} △H_{fus}=5.65kJmo{l}^{-1} \\ △S_{fus}=28.9J{K}^{-1}mo{l}^{-1} \end{gathered}$$.

Concept of the Gibbs free energy

The Gibbs energy is decreased when a system reaches equilibrium at constant pressure and temperature without being pushed by an applied electrolytic voltage. Its derivative with relevancy the reaction coordinate of the system vanishes at the equilibrium point.

(a) Calculation of Gibbs free energy for one mol ammonia 

Convert unit KJ to J

$\begin{array}{c}\Delta H_{fus}=5.65\frac{kJ}{mol}\times \frac{{10}^{3}J}{1kJ}\\ =5.65\times {10}^{3}Jmo{l}^{-1}\end{array}$

$\Delta H_{fus}and\Delta S_{fus}$1 mol solid ammonia

localid="1663602558259" $\begin{array}{c}△H_{fus}=1mol\times \frac{5.65\times {10}^{3}J}{1mol}\\ =5.65\times {10}^{3}J\end{array}$

$\begin{array}{c}△S_{fus}=1mol\times \frac{28.9J{K}^{-1}}{1mol}\\ =28.9J{K}^{-1}\end{array}$

The Gibb's free energy change for melting of 1 mol solid ammonia at T=170 K is calculated with the formula

$△G_{1mol}^{°}=\Delta H_{fus}-T\Delta S_{fus}$

Where, $△G$is Gibbs free energy, $\Delta H_{fus}$enthalpy of fusion.

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

$\begin{array}{rcl}△G_{1mol}^{°}& =& 5.65x{10}^{3}J-170Kx28.9J{K}^{-1}\\ & =& 5.65x{10}^{3}J-4.91x{10}^{3}J\\ & =& 740J\end{array}$

(b) Calculation Gibbs free energy for 3.60 mol ammonia

The calculation of$△H_{fus}and△S_{fus}$for 3.60 mol ammonia is shown below:

$\begin{array}{c}\Delta H_{fus}=3.60mol\times \frac{5.65\times {10}^{3}J}{1mol}\\ =2.03\times {10}^{4}J\\ \Delta S_{fus}=3.60mol\times \frac{28.9J{K}^{-1}}{1mol}\\ =104J{K}^{-1}\end{array}$

The Gibb's free energy change for melting of 3.60 mol solid ammonia

$△G_{3.6mol}^{°}=△H_{fus}-△S_{fus}$

Where, $△G$is Gibbs free energy, $△H_{fus}$enthalpy of fusion.

Put the value of given data in above equation

$\begin{array}{c}△G°=2.03\times {10}^{4}J-170K\times 104J{K}^{-1}\\ =2.03\times {10}^{4}J-1.77\times {10}^{4}J\\ =2.60\times {10}^{3}J\end{array}$

Convert unit joule into kilo joule

localid="1663602093586" $\begin{array}{l}=2.60\times {10}^{3}J\times \frac{1kJ}{{10}^{3}J}\\ =2.60kJ\end{array}$

(c) explanation on spontaneity 

Ammonia will not melt spontaneously at , because the Gibb's free energy for melting is positive.

(d) Calculation of Gibbs free energy at equilibrium

The calculation of Gibbs free energy is shown below:

$△G^{°}=△H_{fus}-△S_{fus}$

Put $\Delta H_{fus}=5.65\times {10}^{3}Jmo{l}^{-1}$and $\Delta S_{\mathrm{fus}}=28.9{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$in above equation.

It is known that, at equilibrium$△G°=0$

$\begin{array}{c}0=\Delta H_{fus}-T\Delta S_{fus}\\ \Delta H_{fus}=T\Delta S_{fus}\\ T=\frac{\Delta H_{fus}}{\Delta S_{fus}}\\ =\frac{5.65\times {10}^{3}J}{28.9J{K}^{-1}}\end{array}$

By the above calculation, it is found that the value of temperature is 196K