Respuesta :
Answer:
The temperature of the eutectic melting is 1605 K and the composition is 0.887
Explanation:
The equation for liquids lines are
[tex]X_{CaF_2} = e^{[-\frac{\delta G_M(CaF_2)}{RT}]} \\\\X_{MgF_2} = e^{[-\frac{\delta G_M(MgF_2)}{RT}]} \\\\[/tex]
At eutectic point liquid and solid line intersect, and
[tex]X_{CaF_2} = 1- X_{MgF_2}[/tex]
[tex]X_{CaF_2} = 1- e^{[-\frac{\delta G_M(MgF_2)}{RT}]} \\[/tex]
Also,
ΔGm CaF2 = ΔHm - T(ΔHm/Tm) and ΔGm MgF2 = ΔHm - T(ΔHm/Tm)
[tex]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}]}[/tex]
Where;
ΔHm(CaF2) = 31200 J/mol and given Tm(CaF2)= 1691 k and
ΔHm(MgF2) = 58160 J/mol and given Tm(CaF2)= 1563k
[tex]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}][/tex]
Take natural log of both sides
[tex][\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[/tex]
Part (b)
the composition of [tex]X_{CaF_2}[/tex]
[tex]X_{CaF_2} = e^{[-\frac{31200- 1605(\frac{31200}{1691})}{8.314*1605}]}\\\\= e[^-2.338 +2.219]\\\\= e^{(-0.119)}\\= 0.887[/tex]
XMgF2 = 1-0.887 = 0.133
Following are the calculation to the given question:
Supposing the liquid and solids behave ideally.
[tex]A = CaF2\\\\B = MgF2[/tex]
These liquidus lines were given by the equation in a binary system that includes two components A and B that have eutectic equilibria with an infinitesimally small solid component.
[tex]Gm = -RT \ ln(aA)\ \text{(in liquidus A melt)}\\\\Gm = -RT \ ln(aB)\ \text{(in liquidus B melt)}[/tex]
Because the liquids are perfect, m denotes melting.
[tex]a_A = X_A \\\\ a_B = X_B[/tex]
So,
[tex]Gm = -RT\ ln(X_{CaF2}) \ \text{(in liquidus A melt)}\\\\Gm = -RT \ ln(X_{MgF2}) \ \text{(in liquidus B melt)}[/tex]
The formulas again for liquidus lines can be obtained by rewriting one entire equation above.
[tex]X_{CaF2} = exp(G_{m(CaF2)}RT)\\\\X_{MgF2} = exp(G_{m(MgF2)}RT)[/tex]
Consider the situation at the eutectic point. These two curves will cross. It means,
[tex]X_{CaF2} = 1- X_{MgF2}\\\\[/tex]
We now understand,
[tex]exp(G_{m(CaF2)}{RT}) = 1- exp(G_{m(MgF2)}}{RT}) \ \ \ \ \ \[/tex]
Now we know that,
[tex]G_{m(CaF2)} = H_{m(CaF2)} -T(\frac{H_{m(CaF2)}}{T_{m(CaF2)}})\\\\[/tex]
And,
[tex]G_{m(MgF2)} = H_{m(MgF2)} -T(\frac{H_{m(MgF2)}}{T_{m(MgF2)}})\\\\[/tex]
Substituting Gm values in (1)
[tex]exp(-[H_{m(CaF2)} - \frac{T(H_{m(CaF2)}}{\frac{T_{m(CaF2)})] } {RT})} \\\\\\= 1- exp(-[H_{m(MgF2)} - \frac{T_{(Hm(MgF2)}}{\frac{T_{m(MgF2)}}{RT)}})]\\\\[/tex]
Considering numbers for Hm while solving for T:
[tex]\to Exp([-31200 + \frac{T(\frac{31200}{1691}])}{8.314T}) = 1-exp([-58160 + \frac{(T(\frac{58160}{1563})]}{8.314T})\\\\\to Exp([-\frac{3752}{T})+2.219) = 1- exp([-\frac{6995}{T}] + 4475)\\\\[/tex]
So,
[tex]\to T = 1329 K\\\\[/tex]
Also, by providing a value for T in the equation for We get, [tex]X_{Caf2}\\\\[/tex]
We get, [tex]X_{CaF2} = 0.546\\\\[/tex]
Therefore, the final answer is "0.546"
Learn more:
brainly.com/question/7180843
