Question

CaF2 and MgF2 are mutually insoluble in the solid state and form a simple binary eutectic system. Calculate the composition and temperature of the eutecdc melt assuming that the liquid solutions are Raoultian. The actual eutectic occurs at X CaF2 = 0.45 and T = 1243 K.

Answer 1

Answer:

The temperature of the eutectic melting is 1605 K and the composition is 0.887

Explanation:

The equation for liquids lines are

$X_{CaF_2} = e^{[-\frac{\delta G_M(CaF_2)}{RT}]} \\\\X_{MgF_2} = e^{[-\frac{\delta G_M(MgF_2)}{RT}]}$

At eutectic point liquid and solid line intersect, and

$X_{CaF_2} = 1- X_{MgF_2}$

$X_{CaF_2} = 1- e^{[-\frac{\delta G_M(MgF_2)}{RT}]}$

Also,

ΔGm CaF2 = ΔHm - T(ΔHm/Tm) and ΔGm MgF2 = ΔHm - T(ΔHm/Tm)

$e^{[-\frac{\delta H_m(_{CaF_2}) - T( \frac{\delta H_m}{T_m})(_{CaF_2})}{RT}]} = 1-e^{[-\frac{\delta H_m(_{MgF_2}) - T( \frac{\delta H_m}{T_m})(_{MgF_2})}{RT}]}$

Where;

ΔHm(CaF2) = 31200 J/mol and given Tm(CaF2)= 1691 k and

ΔHm(MgF2) = 58160 J/mol and given Tm(CaF2)= 1563k

$e^{[-\frac{31200- T( \frac{31200}{1691})}{8.314T}]} = 1-e^{[-\frac{58160 - T( \frac{58160}{1563})}{8.314T}]}\\\\e[^{\frac{-3752.71}{T} + 2.219}] = 1 - e[^{\frac{-6995.429}{T} + 4.475}]$

Take natural log of both sides

$[\frac{-3752.71}{T} +2.219] = -[\frac{-6995.429}{T} +4.475]\\\\\frac{-3752.71}{T} +2.219 = \frac{6995.429}{T} - 4.475\\\\\frac{6995.429}{T} +\frac{3752.71}{T} = 2.219 + 4.475\\\\\frac{10748.139}{T} = 6.694\\\\T = \frac{10748.139}{6.694} = 1605 K$

Part (b)

the composition of $X_{CaF_2}$

$X_{CaF_2} = e^{[-\frac{31200- 1605(\frac{31200}{1691})}{8.314*1605}]}\\\\= e[^-2.338 +2.219]\\\\= e^{(-0.119)}\\= 0.887$

XMgF2 = 1-0.887 = 0.133