Answer:
[tex]r = 1.46 *10^{-9} m[/tex]
[tex]\Delta G^* = 1.828 *10^{-18} j[/tex]
Explanation:
given data:
latent heat of fusion [tex]\Delta H_f = -1.85*10^9 j/m2[/tex]
surface free energy = 0.204 j/m2
meltinf point = 1538 degree celcius
[tex]\Delta T_c = 273 K[/tex]
critical radius is given as
[tex]r^* = \frac{-2rT_m}{\Delta H_f} *\frac{1}{\DeltaT_c}[/tex]
[tex]= \frac{-2*0.204*(1538+273)}{-1.85*10^9} *\frac{1}{273}[/tex]
[tex]r = 1.46 *10^{-9} m[/tex]
activation free energy is given as
[tex]\Delta G^* = \frac{16\pi r^3 t_m^2}{3\Delta H^2_f} * \frac{1}{\Delta T^2_C}[/tex]
[tex]= \frac{16\pi 0.204^3*(1538+273)^2}{3*(-1.85*10^9)^2} * \frac{1}{273^2}[/tex]
[tex]\Delta G^* = 1.828 *10^{-18} j[/tex]
The critical radius and activation free energy is equal to [tex]1.46 \times 10^{-9}\;m[/tex] and [tex]1.83 \times 10^{-18}\;Joules[/tex] respectively.
Given the following data:
Latent heat of fusion = [tex]-1.85 \times 10^9 \;J/m^3[/tex]
Surface free energy = 0.204 [tex]J/m^2[/tex].
Supercooling value = 273 K.
Melting point of iron = 1538°C to K = [tex]1538+273[/tex] = 1811.
Mathematically, critical radius is given by this formula:
[tex]r=\frac{-2ST_m}{\Delta H_{f}c} \\\\r=\frac{-2 \times 0.0204 \times 1811}{-1.85 \times 10^9 \times 273} \\\\r=\frac{0.0408 \times 1811}{1.85 \times 10^9 \times 273}\\\\r=1.46 \times 10^{-9}\;m[/tex]
Mathematically, activation free energy is given by this formula:
[tex]\Delta G = \frac{16\times 3.142 \times (1.46 \times 10^{-9})^3 \times 1811^2}{3\times 273^2 \times (-1.85 \times 10^9 )^2} \\\\\Delta G = 1.83 \times 10^{-18}\;Joules[/tex]
Read more on critical radius here: https://brainly.com/question/15488835
